vijayad

Electrostatics

1. Electrostatics Introduction

The branch of physics which deals with the electric effect of static charge is called electrostatics.

2. Electric Charge

The charge of a material body or particle is the property (acquired or natural) due to which it produces and experiences electrical and magnetic effects. Some naturally charged particles are electrons, protons, α- particles, etc.

Charge is a derived physical quantity. Charge is measured in coulomb in S.Ι. unit. In practice, we use mC (10^-3C), μC (10^-6C), nC(10^-9C), etc.

C.G.S. unit of charge = electrostatic unit = esu.

1 coulomb = 3 × 10⁹ esu of charge

Dimensional formula of charge = [MºLºT¹Ι¹]

2.1 Electric Charges

It was observed that if two glass rods rubbed with wool or silk cloth are brought closer they repel each other.

The two strands of wool or two pieces of silk cloth, with which the rods were rubbed, also repel each other. However, the glass rod and wool attracted each other. Similarly, two plastic rods rubbed with a cat’s fur repelled each other but attracted the future. Careful studies by different scientists concluded, that there were only two kinds of an entity called the electric charge.

We say that the bodies like glass or plastic rods, silk, fur, and pith balls are electrified. They acquire an electric charge on rubbing.

The experiments on pith balls suggested that there are two kinds of electrification and we find that (1) like charges repel and (2) unlike charges attract each other

The charges were named as positive and negative by the American scientist Benjamin Franklin. By convention,

The charge on a glass rod or cat’s fur is called positive and that on a plastic rod or silk is termed negative. If an object possesses an electric charge, it is said to be electrified or charged. When it has no charge it is said to be neutral

2.2 Properties of Charge

Charge is a scalar quantity: It adds algebraically and represents excess, or deficiency of electrons.

A charge is of two types : (i) Positive charge and (ii) Negative charge Charging a body implies the transfer of charge (electrons) from one body to another. A positively charged body means loss of electrons, i.e., deficiency of electrons. A negatively charged body means an excess of electrons. This also shows that the mass of a negatively charged body > the mass of a positively charged identical body.

A charge is conserved: In an isolated system, the total charge (sum of positive and negative) remains constant whatever change takes place in that system.

A charge is quantized: A charge on anybody always exists in integral multiples of a fundamental unit of electric charge. This unit is equal to the magnitude of the charge on an electron (1e = 1.6 × 10^-19 coulomb). So charge on anybody Q = ± ne, where n is an integer and e is the charge of the electron. Millikan’s oil drop experiment proved the quantization of charge or atomicity of charge

Note : Recently, the existence of particles of charge ±e\(\frac{1}{3} \text { and } \frac{2}{3}\) ±e has been postulated. These particles are called quarks but st, this is not considered as the quantum of charge because these are unstable (They have ve very short span of life).

Like point charges repel each other while unlike point charges attract each other. (vi) Charge is always associated with mass, i.e., charge can not exist without mass though mass can exist without charge. The Particle as photons neneutrinoshich have no (rest) mass can never have a charge charge is relativistically invariant: This means that charge is independent the of frame of reference, i.e., the charge on a body does not change whatever be its speed. This property is worth mentioning as in contrast to charging, the mass of a body depends on its speed and increases with an increase in speed.

A charge at rest produces only an electric field around itself; a charge having uniform motion produces an electric as well as maa magnetic field around itself while a charge having accelerated motion emits electromagnetic radiation.

2.3 Charging of a body

A body can be charged by using) friction, (b) conduction, (c) induction, (d) thermionic ionization or thermionic emission (e) photoelectric effect, and (f) field emission.

Charging by Friction: When a neutral body is rubbed against another neutral body then some electrons are transferred from one body to another body which can hold electrons tightly, draws some electrons, and the body which can not hold electrons tightly, loses electrons. The body that draws electrons becomes negatively charged and the body that loses electrons becomes positively charged.

For example: Suppose a glass rod is rubbed with a silk cloth. As the silk can hold electrons more tightly and a glass rod can hold electrons less tightly (due to their chemical properties), some electrons will leave the glass rod and get transferred to the silk. So in the glass rod t, heir will be a ficiency of electrons, therefore it will become positively charged. In AnInhe silk there will be some extra electrons, so it will become negatively charged

Charging by conduction (flow): There are three types of material in nature

Conductor: Conductors are the material in which the outermost electrons are very loosely bound, so they are free to move (flow). So in a conductor, there are a large number of free electrons.

Example: Metals like Cu, Ag, Fe, Al………….

Insulator or Dielectric or Nonconductor: Non-conductors are the materials in which outermost electrons are very tightly bound so they cannot move (flow). Hence in a non-conductor, there are no free electrons.

Examples are rubber, wood, etc, etc.

Semiconductor: Semiconductors are theories with free electrons but very less in number.

Now release how the charging is done by conduction. In this method, w, we take a charged conductor ‘A’ and an uncharged conductor ‘B’. When both are connected some charge will flow from the body to the uncharged body. If both the conductors are identical and kept at a large distance if connected then the charge will be divided equally in both the conductors otherwise they will flow till their electric potential becomes sathe. A detailed study will be done in the first section of this chapter.

Charging by Induction: To understand this, let’s have an introduction to induction.

We have studied that there are a lot of free electrons in the conductors. When a cha charge article +Q is brought near a neutral conductor. Due to the reaction of the +Q charge, many electrons (–ve charges) come closer and accumulate on the closer surface. On the other hand a,, positive charge (deficiency of electrons) appears on the other surface. The flow of charge continues till there resultant force on the free electrons of the conductor becomes zero. This phenomenon is called induction, and the charges produced are called induced charges.

A body can be charged by induction in the following two ways :

Method 1 :

Step 1.

• Take an isolated neutral conductor.

Step 2.

• Bring a charged rod near to it. Due to the charged rod, charges will be induced on the conductor.

Step 3.

• Connect another neutral conductor with it.
• Due to the attraction of the rod, some free electrons will move from the right conductor to the left conductor, and due to the deficiency of electrons positive charges
• Will appear on the right conductor, and on the left conductor, there will be an e excess of electrons due to transfer from the right conductor.

Step 4.

• Now disconnect the connecting wire and remove the rod.

The first conductor will be negatively charged and the second conductor will be positively charged.

Method 2:

Step 1.

• Take an isolated neutral conductor.

Step 2.

• Bring a charged rod near to it. Due to the charged rod, charges will be induced on the conductor.

Step 3.

• Connect the conductor to the earth (this process is called grounding or earthing).
• Due to the attraction of the rod, some free electrons will move from the earth to the conductor, so in the conductor.
• There will be an excess of electrons due to transfer from the earth, so the net charge on the conductor will be negative.

Step 4.

• Now disconnect the connecting wire. The conductor becomes negatively charged.

Thermionic emission: When the metal is heated at a high temperature then some electrons of metals are ejected and the metal becomes positively charged.

Photoelectric effect: When light of sufficiently high frequency is incident on the metal surface then some electrons gain energy from light and come out of the metal surface and the remaining metal becomes positively charged.

Field emission: When the electric field of large magnitude is applied near the metal surface then some electrons come out from the metal surface and hence the metal gets positively charged.

2.4 Gold Leaf Electroscope (GLE)

A simple apparatus to detect charge on a body is the gold-leaf electroscope

It consists of a vertical metal rod housed in a box, with two thin gold leaves attached to its bottom end. When a charged object touches the metal knob at the top of the rod, charge flows onto the leaves and they diverge.

Solved Examples

Example 1. If a charged body is placed near a neutral conductor, will it attract the conductor or repel it?
Solution :

If a charged body (+ve) is placed left side near a neutral conductor, (–ve) charge will induce at the left surface, and (+ve) charge will induce at the right surface.

• Due to the positively charged body –ve induced charge will feel attraction and the +ve induced charge will feel repulsion.
• But as the – ve induced charge is nearer, so the attractive force will be greater than the repulsive force.
• So the net force on the conductor due to the positively charged body will be attractive. Similarly, we can prove this for the negatively charged bodies also.
• From the above example, we can conclude that. “A charged body can attract a neutral body.” If there is attraction between two bodies then one of them may be neutral.
• But if there is repulsion between two bodies, both must be charged (similarly charged). So “repulsion is the sure test of electrification”.

Example 2. A positively charged body ‘A’ attracts a body ‘B’ then the charge on body ‘B’ may be:

1. Positive
2. Negative
3. Zero
4. Can’t say

Solution: (2, 3)

Example 3. Five styrofoam balls A, B, C, D, and E are used in an experiment. Several experiments are performed on the balls and the following observations are made :

1. Ball A repels C and attracts B.
2. Ball D attracts B and does not affect E.
3. A negatively charged rod attracts both A and E.

For your information, an electrically neutral Styrofoam ball is very sensitive to charge induction and gets attracted considerably, if placed near a charged body. What are the charges, if any, on each ball?

A   B   C   D  E

(1) +   –   +  0   +

(2) +   –  +  +    0

(3) +   –   +  0   0

(4) –   +   –   0  0

Answer: 3

Solution:

From (1), As A repels C, both A and C must be charged similarly. Either both are +ve or both are –ve. As A also attracts B, so charge on B should be opposite of A or B may be an uncharged conductor.

From (2) As D does not affect E, both D and E should be uncharged, and as B attracts uncharged D, B must be charged and D must be on the uncharged conductor.

From (3) a –ve charged rod attracts the charged ball A, so A must be +ve, and from exp. (i) C must also be +ve and B must be –ve.

Example 4. Charge conservation is always valid. Is it also true for mass?
Solution:

No, mass conservation is not always. In some nuclear reactions, some mass is lost and it is converted into energy.

Example 5. What are the differences between charging by induction and charging by conduction?
Solution:

The major differences between the two methods of charging are as follows :

In induction, two bodies are close to each other but do not touch each other while in conduction they touch each other. (or they are connected by a metallic wire)

In induction, the total charge of a body remains unchanged while in conduction it changes.

In induction, the induced charge is always opposite to that of the source charge while in conduction charge on two bodies finally is of the same nature.

Example 6. If a glass rod is rubbed with silk it acquires a positive charge because :

1. Protons are added to it
2. Protons are removed from it
3. Electrons are added to it
4. Electrons are removed from it.

Answer: 4. Electrons are removed from it.

Example 7. How can you charge a metal sphere positively without touching it?
Solution :

Figure (a) shows an uncharged metallic sphere on an insulating stand. Bring a negatively charged rod close to the metallic sphere, as shown in Fig. (b). As the rod is brought close to the sphere, the free electrons in the sphere move away due to repulsion and start piling up at the farther end. The near end becomes positively charged due to a deficit of electrons. This process of charge distribution stops when the net force on the free electrons inside the metal is zero.

Connect the sphere to the ground by a conducting wire. The electrons will flow to the ground while the positive charges at the near end will remain held there due to the attractive force of the negative charges on the rod, as shown in Fig. (c).

Disconnect the sphere from the ground. The positive charge continues to be held at the near end of Fig.(d) Remove the electrified rod. The positive charge will spread uniformly over the sphere as shown in Fig. (e).

Example 8. If 10⁹ electrons move out of one body to another body every second, how much time is required to get a total charge of 1 C on the other body?
Solution:

In one second 10⁹ electrons move out of the body. Therefore the charge given out in one second is 1.6 × 10^-19 × 10⁹ C = 1.6 × 10^-10 C.

The time required to accumulate a charge of 1 C can then be estimated to be

⇒ \(\frac{1 \mathrm{C}}{1.6 \times 10^{-10} \mathrm{C} / \mathrm{s}}=6.25 \times 10^9 \mathrm{~s}=\frac{6.25 \times 10^9}{365 \times 24 \times 3600} \text { years }=198 \text { years. }\)

Thus to collect a charge of one coulomb, from a body from which 109 electrons move out every second, we will need approximately 200 years. One coulomb is, therefore, a very large unit for many practical purposes.

Example 9. How much positive and negative charge is there in a cup of water?
Solution :

Let us assume that the mass of one cup of water is 250 g.
The molecular mass of water is 18g.

One mole(= 6.02 × 10²³ molecules) of water is 18 g. Therefore the number of molecules in one 9

cup of water is \(\frac{250 \times 10^9}{18} \times 6.02 \times 10^{23}\)

Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e., 10 electrons and 10 protons. Hence the total positive and total negative charge has the same magnitude. It is equal to \(\frac{250 \times 10^9}{18} \times 6.02 \times 10^{23} \times 10 \times 1.6 \times 10^{-19} \mathrm{C}=1.34 \times 10^7 \mathrm{C} \text {. }\)

Example 10. Which is bigger, a coulomb or charge on an electron? How many electronic charges form one coulomb of charge?
Solutions :

A coulomb of charge is bigger than the charge on an electron.

The magnitude of the  charge on one electron, e = 1.6 ×10^-19 coulomb

Number of electronic charge in one coulomb, \(n=\frac{q}{e}=\frac{1}{1.6 \times 10^{-19}}=0.625 \times 10^{19}\)

Example 11. Assume that each atom in a copper wire contributes one free electron. Estimate the number of free electrons in a copper wire having a mass of 6.4 g (take the atomic weight of copper to be 64g/mol).
Solutions :

Number of atoms in 64 g of copper = 6.023 × 10²³

Number of atoms in 6.4 g of copper = \(\frac{6.023 \times 10^{23}}{64} \times 6.4=6.023 \times 10^{22}\)

As each atom contributes one free electron, therefore, number of free electrons in copper wire = 6.023 × 1022.

3. Coulomb’s Law (Inverse Square Law)

Based on experiments Coulomb established the following law known as Coulomb’s law. The magnitude of the electrostatic force between two point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between them.

i.e. F ∝ q1q2 and F ∝\(\frac{1}{r^2} \quad \Rightarrow \quad F \propto \frac{q_1 q_2}{r^2} \quad \Rightarrow \quad F=\frac{K q_1 q_2}{r^2}\)

Important points regarding Coulomb’s law :

1. It is applicable only for point charges.
2. The constant of proportionality K in SI units in a vacuum is expressed as \(\frac{1}{4 \pi \varepsilon_0}\)and in any other medium expressed as \(\frac{1}{4 \pi \varepsilon}\) If charges are dipped in a medium then electrostatic force on one 230
3. Charge is \(\frac{1}{4 \pi \varepsilon_0 \varepsilon_r} \frac{q_1 q_2}{r^2} .\) ε₀ and ε are called the permittivity of the vacuum and absolute permittivity of the medium respectively. The ratio ε/ε₀ = εr is called the relative permittivity of the medium, which is a dimensionless quantity.
4. The value of relative permittivity εris constant for medium and can have values between 1 to ∞. For vacuum, by definition, it is equal to 1. For air, it is nearly equal to 1 and may be taken to be equal to 1 for calculations. For metals, the value of εris ∞ and for water is 81. The material in which more charge can induce εr will be higher.
5. The value of \(\frac{1}{4 \pi \varepsilon_0}\) = 9 × 10⁹ Nm²C–2 ⇒ ε₀= 8.855 × 10^-12C2/Nm². 0
6. Dimensional formula of ε is M^-1 L^-3 T⁴ A²
7. The force acting on one point charge due to the other point charge is always along the line joining these two charges. It is equal in magnitude and opposite in direction on two charges, irrespective of the medium, in which they lie.
8. The force is conservative i.e., work done by electrostatic force in moving a point charge along a close loop of any shape is zero.
9. Since the force is a central force, in the absence of any other external force, the angular momentum of one particle w.r.t. the other particle (in a two-particle system) is conserved,
10. Coulomb Low In Vector Form

Let us consider q₁ and q₂are placed at positions r₁ = x₁iˆ + y₁jˆ + z₁kˆand r = x₂ i + y₂j + z₂ k respectively. If we want to calculate coulomb force on q₂ then q₁ will be considered as a source charge and q₂ will be considered as a test charge.

⇒ \(\vec{F}=\frac{\mathrm{kq}_1 q_2}{|\overrightarrow{\mathrm{r}}|^3} \overrightarrow{\mathrm{r}} \text { where } \overrightarrow{\mathrm{r}}\)= position vector of test charge – position vector source charge.

(r= position vector of test charge w.r.t. source charge)

1. When we use this formula in vector form then we have to put the value of charges with their sign.
2. If the force F₁₂ on charge q₁ due to charge q₂, and F₂₁ is force on charge q2due to charge q₁ then \(\overrightarrow{\mathrm{F}}_{12}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}_{21}^2} \hat{r}_{12}=-\overrightarrow{\mathrm{F}}_{21}\)

Example 12. Find out the electrostatic force between two point charges placed in the air (each of +1 C) if they are separated by 1m.
Solution :

⇒ \(F_e=\frac{k q_1 q_2}{r^2}=\frac{9 \times 10^9 \times 1 \times 1}{1^2}=9 \times 10^9 \mathrm{~N}\)

From the above result, we can say that the 1 C charge is too large to realize. In nature, the charge is usually of the order of μC

Example 13. Two particles having charges q¹ and q² when kept at a certain distance, exert a force F on each other. If the distance between the two particles is reduced to half and the charge on each particle is doubled then what will be the force between the particles :
Solution :

⇒ \(F=\frac{k q_1 q_2}{r^2}\)

If q’₁= 2q₁, q’₂= 2q₂ r’ = \(\frac{r}{2}\)

then F’ = \(F^{\prime}=\frac{k q_1^{\prime} q_2^{\prime}}{r^{\prime 2}}=\frac{k\left(2 q_1\right)\left(2 q_2\right)}{\left(\frac{r}{2}\right)^2} \)

F’ = \(\quad F^{\prime}=\frac{16 k q_1 q_2}{r^2}\)

Example 14. A particle of mass m carrying charge q¹ revolves around a fixed charge –q² in a circular path of radius r. Calculate the period of revolution and its speed.
Solution :

⇒ \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}=m r \omega^2=\frac{4 \pi^2 m r}{T^2}, \quad T^2=\frac{\left(4 \pi \varepsilon_0\right) r^2\left(4 \pi^2 m r\right)}{q_1 q_2} \quad \text { or } \quad T=4 \pi r \sqrt{\frac{\varepsilon_0 m r}{q_1 q_2}}\)

and also we can say that \(\frac{q_1 q_2}{4 \pi \varepsilon_0 r^2}=\frac{m v^2}{r} \quad \Rightarrow \quad v=\sqrt{\frac{q_1 q_2}{4 \pi \varepsilon_0 m r}}\)

Example 15. A point charge q_A= + 100 µc is placed at point A (1, 0, 2) m, and another point charge q_B= +200µc is placed at point B (4, 4, 2) m. Find :

1. The magnitude of the Electrostatic interaction force acting between them
2. F_A(force on A due to B) and F_B(force on B due to A) in vector form

Solution :

Value of F : \(|F|=\frac{k q_A q_B}{r^2}=\frac{\left(9 \times 10^9\right)\left(100 \times 10^{-6}\right)\left(200 \times 10^{-6}\right)}{\sqrt{(4-1)^2+(4-0)^2+(2-2)^2}}=7.2 \mathrm{~N}\)

Force on B \(\vec{F}_B=\frac{k q_A q_B}{|\vec{r}|^3} \vec{r}=\frac{\left(9 \times 10^9\right)\left(100 \times 10^{-6}\right)\left(200 \times 10^{-6}\right)}{\sqrt{(4-1)^2+(4-0)^2+(2-2)^2}}[(4-1) \hat{\mathrm{i}}+(4-0) \hat{\mathrm{j}}+(2-2) \hat{k}]\)

⇒ \(7.2\left(\frac{3}{5} \hat{i}+\frac{4}{5} \hat{j}\right) N\)

Similarly \(\vec{F}_A=7.2\left(-\frac{3}{5} \hat{i}-\frac{4}{5} \hat{j}\right) \mathrm{N}\)

Action(F_A) and Reaction (F_B) are equal but in opposite directions.

Example 16. A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B held by an insulating handle is brought close to A such that the distance between their centers is 10 cm, as shown in Fig. (a). The resulting repulsion of A is noted (for example, by shining a beam of light and measuring the deflection of its shadow on a screen). Spheres A and B are touched by uncharged spheres C and D respectively, as shown in Fig. (b). C and D are then removed and B is brought closer to A distance of 5.0 cm between their centers, as shown in Fig. (c). What is the expected repulsion of A based on Coulomb’s law? Spheres A and C and spheres B and D have identical sizes. Ignore the sizes of A and B in comparison to the separation between their centers.

Solution :

Let the original charge on sphere A be q and that on B be q’. At a distance r between their centers, the magnitude of the electrostatic force on each is given by

⇒ \(F=\frac{1}{4 \pi \varepsilon_0} \frac{q^{\prime}}{\mathrm{r}^2}\)

neglecting the sizes of spheres A and B in comparison to r. When an identical but uncharged sphere C touches A, the charges redistribute on A and C, and, by symmetry, each sphere carries a charge q/2. Similarly, after D touches B, the redistributed charge on each is q’/2. Now, if the separation between A and B is halved, the magnitude of the electrostatic force on each is

⇒ \(\mathrm{F}^{\prime}=\frac{1}{4 \pi \varepsilon_0} \frac{(\mathrm{q} / 2)\left(\mathrm{q}^{\prime} / 2\right)}{(\mathrm{r} / 2)^2}=\frac{1}{4 \pi \varepsilon_0} \frac{\left(\mathrm{qq}^{\prime}\right)}{\mathrm{r}^2}=F\)

Thus the electrostatic force on A, due to B, remains unaltered.

4. Principle Of Superposition

The electrostatic force is a two-body interaction i.e. electrical force

between two point charges are independent of the presence or absence of other charges and so the principle of superposition is valid i.e. force on the charged particle due to several point charges is the resultant of forces due to individual point charges.

Consider that n point charges q1, q2, q3, …. are discretely distributed in space. The charges are interacting with each other. Let us find the total force on the charge, say due to all other remaining charges. If the charge q2, q3, …. qnexert forces\(\overrightarrow{\mathrm{F}}_{12}, \overrightarrow{\mathrm{F}}_{13}, \ldots . \overrightarrow{\mathrm{F}}_{1 \mathrm{n}}\)on the charge q1, then according to principle of superposition, the total force on charge q1is given by \(\overrightarrow{\mathrm{F}}_1=\overrightarrow{\mathrm{F}}_{12}+\overrightarrow{\mathrm{F}}_{13}+\ldots \ldots \overrightarrow{\mathrm{F}}_{1 \mathrm{n}}\)

Solved Examples

Example 17. Two point charges of charge value Q and q are placed at a distance of x and x/2 respectively from a third charge of charge value 4q, all charges being in the same straight line. Calculate the magnitude and nature of charge Q, such that the net force experienced by the charge q is zero.
Solution :

Suppose that the charge 4q is located at point A. The charges Q and q are placed at points B and C, such that AB = x and AC = x/2. Also, all the charges lie in the same straight line. We assume that the charges of 4q and q are of the same nature, a say positive.

Then, force on the charge q due to 4q,

⇒ \(\mathrm{F}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{4 \mathrm{q} \cdot \mathrm{q}}{(\mathrm{x} / 2)^2}\)

The net force experienced by charge q will be zero only if the charge Q exerts force on the charge q equal and opposite to that exerted by the charge 4q. Thus, the charge Q should exert force FBon charge q equal to FA(in magnitude) and along CA. For this, charge Q has to be positive (i.e. of the nature same as that of 4q or q).

Now, force on the charge q due to charge Q

FB= \(\mathrm{F}_{\mathrm{B}}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{Q} . \mathrm{q}}{(\mathrm{BC})^2}\)

FB=\(\mathrm{F}_{\mathrm{B}}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{Q} . \mathrm{q}}{(\mathrm{x} / 2)^2}\) (along CA)

For net force on the charge q to be zero, FB= FA

⇒ \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q . q}{(x / 2)^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{4 q \cdot q}{(x / 2)^2}=Q=4 q\)

Example 18. Consider three point charges each having charge q at the vertices of an equilateral triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in the figure?

Solution :

In the given equilateral triangle ABC of sides of length l, if we draw a perpendicular AD to the side BC, AD = AC cos 30º = ( 3 / 2) l and the distance AO of the centroid O from A is (2/3) AD = Thus, (1/ 3)l. By symmetry AO = BO = CO.

Force \(\overrightarrow{\mathrm{F}}_1\) on Q due to charge q at A = \(\frac{3}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\ell^2}\) along AO

Force \(\overrightarrow{\mathrm{F}}_2\)on Q due to charge q at B = \(\frac{3}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\ell^2}\) along BO

Force \(\overrightarrow{\mathrm{F}}_3\)on Q due to charge q at C = \(\frac{3}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\ell^2}\)along CO

The resultant of forces \(\overrightarrow{\mathrm{F}}_2 and \overrightarrow{\mathrm{F}}_3\) is \(\frac{3}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\ell^2}\) along OA, by the parallelogram law. Therefore, the total force on Q = \(\frac{3}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\ell^2}\) = 0, where rˆ is the unit vector along OA.

It is clear also by symmetry that the three forces will sum to zero. Suppose that the resultant force was non-zero but in some direction. Consider what would happen if the system was rotated through 60º about O.

Example 19. Consider the charges q, q, and –q placed at the vertices of an equilateral triangle of side l. Calculate the force on each charge.?
Solution :

The forces acting on charge q at A due to charges q at B and –q at C are \(\overrightarrow{\mathrm{F}}_{\mathrm{AB}}\) along BA and \(\overrightarrow{\mathrm{F}}_{\mathrm{Ac}}\) along AC respectively as shown in Fig.

The force of attraction or repulsion for each pair of charges has the same magnitude 2

⇒ \(\mathrm{F}=\frac{\mathrm{q}^2}{4 \pi \varepsilon_0 \ell^2}\)

⇒ \(\left|F_A\right|=\left|F_B\right|=F\)

⇒ \(\left|F_C\right|=\sqrt{3} F\)

⇒ \(\overrightarrow{\mathrm{F}}_{\mathrm{A}}+\overrightarrow{\mathrm{F}}_{\mathrm{B}}+\overrightarrow{\mathrm{F}}_{\mathrm{C}}=0 .\)

It is interesting to see that the sum of the forces on three charges is zero.

Example 20. Two pith-balls each of mass weighing 10^-4kg are suspended from the same point using silk threads 0.5 m long. On charging the pith balls equally, they are found to repel each other to a distance of 0.6 m. Calculate the charge on each ball. (g = 10m/s²)
Solution :

Consider two pith balls A and B each having charge q and mass 10^-13 kg. When the pith balls are suspended from point S by two threads each 0.5 m long, they repel each other to the distance AB = 0.2 m as shown in Fig.

Each of the two pith-balls is in equilibrium under the action of the following three forces :

1. The electrostatic repulsive force F.
2. The weight acts vertically downwards.
3. The tension T in the string is directed towards point S. The three forces mg, F, and T can be represented by the therefore, according to the  triangle law of forces,

⇒ \(\frac{F}{O A}=\frac{m g}{S O}=\frac{T}{A S}\)…….(1)

Here, \(\mathrm{F}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}^2}{(\mathrm{AB})^2}=9 \times 10^9 \times \frac{\mathrm{q}^2}{(0.6)^2} \mathrm{~N}\)

mg = 10^-4 × 10 = 10^-3 N

From the equation (i), we have

F = mg × \(\frac{O A}{S O} \quad \text { or } \quad 9 \times 10^9 \times \frac{q^2}{(0.6)^2}=10^{-3} \times \frac{0.3}{\sqrt{(0.5)^2-(0.3)^2}} \text { or } \quad q=\sqrt{3} \times 10^{-6} \mathrm{C}\)

Example 21 Three equal point charges of charge +qise moving along a circle of radius R and a point charge –2q is also placed at the center of the circle as (shown in the figure), if charges are revolving with constant and same speed then calculate speed.

Solution :

F₂– 2F₁cos 30 = \(\frac{m v^2}{R} \Rightarrow \frac{K(q)(2 q)}{R^2}-\frac{2\left(K q^2\right)}{(\sqrt{3} R)^2} \cos 30=\frac{m v^2}{R} \Rightarrow v=\sqrt{\frac{k q^2}{R m}}\left[2-\frac{1}{\sqrt{3}}\right]\)

Example 22 Two equally charged identical small metallic spheres A and B repel each other with a force 2 × 10^-5N when placed in air (neglect gravitation attraction). Another identical uncharged sphere C is touched to B and then placed at the midpoint of the line joining A and B. What is the net electrostatic force on C?
Solution :

Let initially the charge on each sphere be q and separation between their centers be r; then according to a given problem.

⇒ \(\mathrm{F}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q} \times \mathrm{q}}{\mathrm{r}^2}=2 \times 10^{-5} \mathrm{~N}\)

When sphere C touches B, the charge of B, q will distribute equally on B and C as spheres are identical conductors, i.e., now charges on spheres;

q_B= q_C= (q/2)

So sphere C will experience a force

⇒ \(\mathrm{F}_{\mathrm{CA}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}(\mathrm{q} / 2)}{(\mathrm{r} / 2)^2}=2 \mathrm{~F} \text { along } \overrightarrow{\mathrm{AB}}\) due to charge on A

and,\(\mathrm{F}_{\mathrm{CB}}=\frac{1}{4 \pi \varepsilon_0} \frac{(\mathrm{q} / 2)(\mathrm{q} / 2)}{(\mathrm{r} / 2)^2}=\mathrm{F} \text { along } \overrightarrow{\mathrm{BA}}\) due to charge on B

So the net force FCon C due to charges on A and B,

F_C= F_CA – F_CB = 2F – F = 2 × 10–5 N along AB.

Example 23 Five point charges, each of value q are placed on five vertices of a regular hexagon of side L. What is the magnitude of the force on a point charge of value – q coulomb placed at the center of the hexagon?
Solution :

Method: 1

If there had been a sixth charge +q at the remaining vertex of hexagon force due to all the six charges on –q at O would be zero (as the forces due to individual charges will balance each other), i.e.,

⇒ \(\overrightarrow{F_R}=0\)

Now if \(\vec{f}\) is the force due to the sixth charge and \(\vec{f}\) due to the remaining five charges.

⇒ \(\vec{F}+\vec{f}=0 \quad \text { i.e. } \quad \vec{F}=-\vec{f} \text { or, } \quad|F|=|f|=\frac{1}{4 \pi \varepsilon_0} \frac{q \times q}{L^2}=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{L^2}\)

⇒ \(\vec{F}_{\text {Net }}=\vec{F}_{\mathrm{CO}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}^2}{\mathrm{~L}^2} \text { along } \mathrm{CO}\)

Method: 2

In the diagram we can see that force due to charges A and D are opposite to each other

Similarly

⇒ \(\vec{F}_{D O}+\vec{F}_{A O}=0\)

⇒ \(\vec{F}_{B O}+\vec{F}_{E O}=0\)

Using (1) and (2)\(\vec{F}_{\text {Net }}=\overrightarrow{\mathrm{F}}_{\mathrm{CO}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}^2}{\mathrm{~L}^2} \text { along } \mathrm{CO} \text {. }\) along CO. 2

Note: The total charge of A rod cannot be considered to be placed at the center of the rod as we do in mechanics for mass in many problems.

Note: If a >> l then F =\(\frac{K Q q}{a^2}\) behavior of the rod is just like a point charge.

5. Electrostatic Equilibrium

The point where the resultant force on a charged particle becomes zero is called the equilibrium position.

5.1 Stable Equilibrium: A charge is initially in an equilibrium position and is displaced by a small distance. If the charge tries to return to the same equilibrium position then this equilibrium is called the position of stable equilibrium.

5.2 Unstable Equilibrium: If the charge is displaced by a small distance from its equilibrium position and the charge does not tend to return to the same equilibrium position. Instead, it goes away from the equilibrium position.

5.3 Neutral Equilibrium: If the charge is displaced by a small distance and it is still in equilibrium condition then it is called neutral equilibrium.

Solved Examples

Example 24 Two equal positive point charges ‘Q’ are fixed at points B(a, 0) and A(–a, 0). Another test charge q0 is also placed at O(0, 0). Show that the equilibrium at ‘O’ is

1. Stable for displacement along the X-axis.
2. Unstable for displacement along the Y-axis.

Solution :

Initially\(\overrightarrow{\mathrm{F}}_{\mathrm{AO}}+\overrightarrow{\mathrm{F}}_{\mathrm{BO}}=0 \Rightarrow\left|\overrightarrow{\mathrm{F}}_{\mathrm{AO}}\right|=\left|\overrightarrow{\mathrm{F}}_{\mathrm{BO}}\right|=\frac{K Q q_0}{\mathrm{a}^2}\)

When the charge is slightly shifted towards the + x-axis by a small distance Δx, then.

⇒ \(\left|\vec{F}_{\mathrm{AO}}\right|<\left|\dot{\vec{F}}_{\mathrm{BO}}\right|\)

Therefore the particle will move toward the origin (its original position) hence the equilibrium is stable.

When the charge is shifted along the axis

After resolving components net force will be along the y-axis so the particle will not return to its original position so it is an unstable equilibrium. Finally, the charge will move to infinity.

Example 25. Two point charges of charge q₁ and q₂(both of the same sign) and each of mass m are placed such that gravitation attraction between them balances the electrostatic repulsion. Are they in stable equilibrium? If not then what is the nature of equilibrium?
Solution :

In given example :

⇒ \(\frac{\mathrm{Kq}_1 \mathrm{q}_2}{\mathrm{r}^2}=\frac{\mathrm{Gm}^2}{\mathrm{r}^2}\)

We can see that irrespective of the distance between them charges will remain in equilibrium. If now distance is increased or decreased then there is no effect on their equilibrium. Therefore it is a neutral equilibrium.

Example 26. A particle of mass m and charge q is located midway between two fixed charged particles each having a charge q and a distance 2l apart. Prove that the motion of the particle will be SHM if it is displaced slightly along the line connecting them and released. Also, find its time.
Solution :

Let the charge q at the mid-point the displaced slightly to the left.

The force on the displaced charge q due to charge q at A,

F1=\(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}^2}{(\ell+\mathrm{x})^2}\)

The force on the displaced charge q due to charge at B,

F₂= \(\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{(\ell-x)^2}\)

Net restoring force on the displaced charge q.

F = F₂– F₁or F = F₂= \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}^2}{(\ell-\mathrm{x})^2}-\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}^2}{(\ell+\mathrm{x})^2}\)

or F = \(\mathrm{F}=\frac{\mathrm{q}^2}{4 \pi \varepsilon_0}\left[\frac{1}{(\ell-\mathrm{x})^2}-\frac{1}{(\ell+\mathrm{x})^2}\right]=\frac{\mathrm{q}^2}{4 \pi \varepsilon_0} \frac{4 \ell \mathrm{x}}{\left(\ell^2-\mathrm{x}^2\right)^2}\)

Since l >> x, ∴ F = \(\frac{q^2 \ell x}{\pi \varepsilon_0 \ell^4} \text { or } F=\frac{q^2 x}{\pi \varepsilon_0 \ell^3}\)

We see that F ∝ x and it is opposite to the direction of displacement. Therefore, the motion is

SHM. T =\(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \text {, here } \mathrm{k}=\frac{\mathrm{q}^2}{\pi \epsilon_0 \ell^3}=2 \pi \sqrt{\frac{\mathrm{m} \pi \epsilon_0 \ell^3}{\mathrm{q}^2}}\)

Example 27. Two identical charged spheres are suspended by strings of equal length. Each string makes an angle θ with the vertical. When suspended in a liquid of density σ = 0.8 gm/cc, the angle remains the same. What is the dielectric constant of the liquid? (The density of the material of the sphere is ρ = 1.6 gm/cc.)
Solution :

Initially, as the forces acting on each ball are tension T, weight mg, and electric force F, for its equilibrium along vertically,

T cos θ = mg …(1)

and along horizontal

T sin θ = F …(2)

Dividing Eqn. (2) by (1), we have

tan θ = Fmg… (3)

When the balls are suspended in a liquid of density σ and dielectric constant K, the electric force will become (1/K) times, i.e., F’ = (F/K) while weight

mg’ = mg – FB= mg – Vσg [as FB= Vσg, where σ is the density of the material of the sphere]

i.e., mg’ = mg \(\) So for equilibrium of ball,

tan θ’ = \(\frac{F^{\prime}}{m g^{\prime}}=\frac{F}{K m g[1-(\sigma / \rho)]}\)… (4)

According to the given information θ’ = θ; so from equations (4) and (3), we have

K = \(\frac{\rho}{(\rho-\sigma)}=\frac{1.6}{(1.6-0.8)}=2\)

6. Electric Field

The electric field is the region around a charged particle or charged body in which if another charge is placed, it experiences electrostatic force.

6.1 Electric field intensity

E: Electric field intensity at a point is equal to the electrostatic force experienced by a unit positive point charge both in magnitude and direction.

If a test charge q0is placed at a point in an electric field and experiences a force \(\overrightarrow{\mathrm{F}}\) due to some charges (called source charges), the electric field intensity at that point due to source charges is given

⇒ \(\overrightarrow{\mathrm{E}}=\frac{\overrightarrow{\mathrm{F}}}{\mathrm{q}_0} \text {; }\)

If the \(\overrightarrow{\mathrm{E}}\)is to be determined practically then the test charge q0 should be small otherwise it will affect the charge distribution on the source which is producing the electric field and hence modify the quantity which is measured.

Example 28. A positively charged ball hangs from a long silk thread. We wish to measure E at point P in the same horizontal plane as that of the hanging charge. To do so, we put a positive test charge q0 at the point and measure F/q₀. Will F/q₀ be less than, equal to, or greater than E at the point in question?
Solution :

When we try to measure the electric field at point P then after placing the test charge at P it repels the source charge (suspended charge) and the measured value of the electric field

Emeasured =\(\frac{F}{q_0}\) will be less than the actual value Eact that we wanted to measure.

6.2 Properties of electric field intensity E:

It is a vector quantity. Its direction is the same as the force experienced by a positive charge.

The direction of the electric field due to the positive charge is always away from it while due to the negative charge is always towards it.

Its S.Ι. unit is Newton/Coulomb.

Its dimensional formula is [MLT^-3A^-1]

The electric force on a charge q placed in a region of the electric field at a point where the electric field intensity is Eis given by \(\overrightarrow{\mathrm{F}}=\mathrm{q} \overrightarrow{\mathrm{E}} \text {. }\).

The electric force on the point charge is in the same direction as the electric field on a positive charge and in the opposite direction on a negative charge.

It obeys the superposition principle, that is, the field intensity at a point due to a system of charges is a vector sum of the field intensities due to individual point charges.

E = E₁ + E₂ + E₃ + …….

It is produced by source charges. The electric field will be a fixed value at a point unless we change the distribution of source charges.

6.3 Electric field due to a point charge :

Consider that a point charge +q is placed at the origin O of the coordinate frame. Let P be the point,

where the electric field due to the point charge +q is to be determined. Let \(\overline{\mathrm{OP}}=\overrightarrow{\mathrm{r}}\) the position vector of the point P.

To find the electric field at point P, place a vanishingly small positive test charge q0at point P. According to Coulomb’s law, force on the test charge due to charge q is given by :

⇒ \(\overrightarrow{\mathrm{F}}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{qq}}{\mathrm{r}_0} \hat{\mathrm{r}}\)

where rˆis unit vector along OP. If \(\) is the electric field at point P, then

⇒ \(\vec{E}=\frac{\vec{F}}{q_0}=\left(\frac{1}{q_0} \cdot \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q q_0}{r^2} \hat{r}\right) \quad \text { or } \quad \vec{E}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2} \hat{r}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^3} \vec{r}\)

The magnitude of the electric field at point P is given by

⇒ \(E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}\)

Solved Examples

Example 29. The electrostatic force experienced by –3μC charge placed at point ‘P’ due to a system ‘S’ of fixed point charges as shown in the figure is \(\overrightarrow{\mathrm{F}}=(2 \hat{1} \hat{i}+9 \hat{\mathbf{j}}) \mu \mathrm{N}.\)

1. Find out electric field intensity at point P due to S.
2. If now 2μC charge is placed and –3 μC is removed at point P then the force experienced by it will be.

Solution :

⇒ \(\vec{F}=q \vec{E} \quad \Rightarrow \quad(21 \hat{i}+9 \hat{j}) \mu N=-3 \mu C(\vec{E})\)

⇒ \(\vec{E}=-7 \hat{i}-3 \hat{j} \frac{\mu \mathrm{N}}{C}\)

Since the source charges are not disturbed the electric field intensity at ‘P’ will remain the same.

⇒ \(\vec{F}_{2 \mu C}=+2(\vec{E})=2(-7 \hat{i}-3 \hat{j})=-14 \hat{i}-6 \hat{j} \mu N\)

Example 30. Calculate the electric field intensity which would be just sufficient to balance the weight of a particle of charge –10 μc and mass 10 mg. (take g = 10 ms²)
Solution :

As the force on a charge q in an electric field \( \overrightarrow{\mathrm{E}}\) is

⇒ \(\overrightarrow{\mathrm{F}}_{\mathrm{q}}=\mathrm{q} \overrightarrow{\mathrm{E}}\)

So according to the given problem Fe

⇒ \(\left|\vec{F}_{\mathrm{q}}\right|=|\vec{W}| \quad \text { i.e., } \quad|\mathrm{q}| E=m g\)

⇒ \(E=\frac{\mathrm{mg}}{|\mathrm{q}|}=10 \mathrm{~N} / \mathrm{C} .\) in the downward direction.

List of formulas for Electric Field Intensity due to various types of charge distribution :

Electric Field Due To Point Charge

⇒ \(E=\frac{k q}{r^2} \Rightarrow \quad \vec{E}=\frac{k q}{r^2} \hat{r} \quad \Rightarrow \quad \vec{E}=\frac{k q}{r^3} \vec{r}\)

⇒ \(\overrightarrow{\mathrm{r}}\)= position vector of test point with respect to source charge

⇒ \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{r}}_{\text {testpoint }}-\overrightarrow{\mathrm{r}}_{\text {sourcecharge }}\)

Solved Examples

Example 31. Find out the electric field intensity at point A (0, 1m, 2m) due to a point charge –20μC situated at point B( 2m, 0, 1m).
Solution :

⇒ E = \(E=\frac{K Q}{|\vec{r}|^3} \vec{r}=\frac{K Q}{|\vec{r}|^2} \hat{r}\) ⇒r= P.V. of A – P.V. of B (P.V. = Position vector)

⇒ \(=(-\sqrt{2} \hat{i}+\hat{j}+\hat{k}) \quad|\vec{r}|==2\) = 2

⇒ E = \(E=\frac{9 \times 10^9 \times\left(-20 \times 10^{-6}\right)}{8}(-\sqrt{2} \hat{i}+\hat{j}+\hat{k})=-22.5 \times 10^3(-\sqrt{2} \hat{i}+\hat{j}+\hat{k}) N / C .\)

Example 32. Two point charges 2μc and – 2μc are placed at points A and B as shown in the figure. Find out electric field intensity at points C and D. [All the distances are measured in meters].

Solution :

The electric field at point C (E_A, are magnitudes only, and arrows represent directions)

The electric field due to positive charge is away from it while due to negative charge, it is towards the charge. It is clear that E_B> E_A.

∴ E_net = (E_B– E_A) towards negative X-axis

⇒ \(\frac{\mathrm{K}(2 \mu \mathrm{c})}{(\sqrt{2})^2}-\frac{\mathrm{K}(2 \mu \mathrm{c})}{(3 \sqrt{2})^2}\) towards negative X-axis = 8000 (–ˆi) N/C

The electric field at point D :

Since the magnitude of charges are same and also AD = BD

So E_A= E_B

Vertical components of\(\overrightarrow{\mathrm{E}}_{\mathrm{A}} \text { and } \overrightarrow{\mathrm{E}}_{\mathrm{B}}\) cancel each other while horizontal components are in the same direction.

So, E_net = 2E_Acosθ = \(\frac{2 . K(2 \mu c)}{2^2} \cos 45^{\circ}\)

⇒ \(=\frac{\mathrm{K} \times 10^{-6}}{\sqrt{2}}=\frac{9000}{\sqrt{2}} \hat{\mathrm{i}} \mathrm{N} / \mathrm{C} .\)

Example 33. Six equal point charges are placed at the corners of a regular hexagon of side ‘a’. Calculate the electric field intensity at the center of the hexagon.

Solution: Zero

Similarly electric field due to a uniformly charged ring at the center of the ring :

Note :

• The net charge on a conductor remains only on the outer surface of a conductor. This property will be discussed in the article on the conductor. (article no.17)
• On the surface of the isolated spherical conductor, the charge is uniformly distributed.

6.4 Electric Field Due To A Uniformly Charged Ring And Arc.

Example 34. Find out the electric field intensity at the center of a uniformly charged semicircular ring of radius R and linear charge density λ.
Solution :

λ = linear charge density.

The arc is the collection of large no. of point charges. Consider a part of the ring as an element of length Rdθ which subtends an angle dθ at the center of the ring and it lies between θ and θ + dθ

⇒ \(\overrightarrow{d E}=d E_x \hat{i}+d E_y \hat{j} \quad E_x=\int d E_x=0\)

⇒ \(E_y=\int d E_y=\int_0^\pi d E \sin \theta=\frac{K \lambda}{R} \int_0^\pi \sin \theta \cdot d \theta=\frac{2 K \lambda}{R}\)

Example 35. Find out the electric field intensity at the center of a uniformly charged quarter ring of radius R and linear charge density λ.
Solution :

Refer to the previous question \(\overrightarrow{d E}=d E_x \hat{i}+d E_y \hat{j}\) on solving E_net= \(E_{\text {net }}=\frac{\mathrm{K} \lambda}{\mathrm{R}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}}) \text {, }\)

Electric field due to ring on its axis :

⇒ \(E_{\text {net }}=\frac{K Q x}{\left[R^2+x^2\right]^{3 / 2}}\)

E will be max when \(\frac{\mathrm{dE}}{\mathrm{dx}}\) = 0, that is at x= \(\frac{R}{\sqrt{2}} \text { and } E_{\max }=\frac{2 K Q}{3 \sqrt{3} \mathrm{R}^2}\)

Case (1) : if x>>R, E = \(\frac{K Q}{x^2}\)

Case (2) : if x<<R, E = \(\frac{K Q x}{R^3}\) Hence the ring will act like a point charge

Solved Examples

Example 36. Positive charge Q is distributed uniformly over a circular ring of radius R. A point particle having a mass m and a negative charge –q is placed on its axis at a distance x from the center. Find the force on the particle. Assuming x << R, find the period of oscillation of the particle if it is released from there. (Neglect gravity)
Solution :

When the negative charge is shifted at a distance x from the center of the ring along its axis then force acting on the point charge due to the ring:

FE= qE (towards centre) = q \(\left[\frac{K Q x}{\left(R^2+x^2\right)^{3 / 2}}\right]\)

if R >>x then

R² + x² ~ R²

⇒ \(F_E=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qqx}}{\mathrm{R}^3}\)(Towards centre)

Since restoring force FE ∝ x, therefore the motion of charge the particle will be S.H.M. period of SHM.

T = 2π \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \quad=2 \pi \sqrt{\frac{\mathrm{m}}{\left(\frac{\mathrm{Qq}}{4 \pi \varepsilon_0 \mathrm{R}^3}\right)}}=\left[\frac{16 \pi^3 \varepsilon_0 \mathrm{mR}^3}{\mathrm{Qq}}\right]^{1 / 2}\)

6.5 Electric Field Due To Uniformly Charged Wire

1. Line charge of finite length: Derivation of expression for intensity of the electric field at a point due to line charge of finite size of uniform linear charge density λ. The perpendicular distance of the point from the line charge is r and lines joining ends of line charge distribution make angle θ₁and θ₂ with the perpendicular line.

Ex = \(\frac{K \lambda}{\mathbf{K}}\)[sinθ₁+ sinθ₂] ………..(1) Kλ

EY= \(\frac{K \lambda}{\mathbf{K}}\)[cosθ₂– cosθ₁]

Net electric field at the point

E_net = \(\sqrt{E_x^2+E_y^2}\)

2. We can derive a result for infinitely long line charge: In the above eq. (1) and (2) if we put θ₁= θ₂ = 90º we can get the required result. 2K

E_net = Ex=\(\frac{2 \mathrm{~K} \lambda}{\mathrm{r}}\)

3. For Semi-infinite wire

θ₁= 90º and θ₂ = 0º so Ex=\(E_x=\frac{K \lambda}{r}, E_y=\frac{K \lambda}{r}\)

Solved Examples

Example 37. A point charge q is placed at a distance r from a very long charge thread of uniform linear charge density λ. Find out the total electric force experienced by the line charge due to the point charge. (Neglect gravity).
Solution :

Force on charge q due to the thread,

F = \(\left(\frac{2 K \lambda}{r}\right) \cdot q\)

By Newton’s 3 law, every action has equal and

opposite reaction so force on the thread = \(\frac{2 K \lambda}{r} \cdot q\) (away from point charge)

6.6 Electric Field Due To Uniformly Charged Infinite Sheet

E_net = \(\frac{\sigma}{2 \varepsilon_0}\)toward normal direction.

Note:

• The direction of the electric field is always perpendicular to the sheet.
• The magnitude of the electric field is independent of the distance from the sheet.

Solved Examples

Example 38. An infinitely large plate of surface charge density +σ is lying in a horizontal xy plane. A particle having charge –qo and mass m is projected from the plate with velocity u making an angle θ with a sheet. Find :

1. The time is taken by the particle to return to the plate.
2. Maximum height achieved by the particle.
3. At what distance will it strike the plate (Neglecting gravitational force on the particle)

Solution :

Electric force acting on the particle Fe= qoE : Fe= (qo)\(\) downward

So acceleration of the particle : a = \(\left(\frac{\sigma}{2 \varepsilon_0}\right)\) = uniform

this acceleration will act like ‘g’ (acceleration due to gravity)

So the particle will perform projectile motion.

T = \(\frac{2 u \sin \theta}{g}=\frac{2 u \sin \theta}{\left(\frac{q_0 \sigma}{2 \varepsilon_0 m}\right)}\)

H = \(H=\frac{u^2 \sin ^2 \theta}{2 g}=\frac{2 u^2 \sin ^2 \theta}{2\left(\frac{q_0 \sigma}{2 \varepsilon_0 m}\right)}\)

R = \(R=\frac{u^2 \sin 2 \theta}{g}=\frac{u^2 \sin 2 \theta}{\left(\frac{q_0 \sigma}{2 \varepsilon_0 m}\right)}\)

Example 39. A block having mass m and charge –q is resting on a frictionless plane at a distance L from a fixed large non-conducting infinite sheet of uniform charge density σ as shown in Figure. Discuss the motion of the block assuming that the collision of the block with the sheet is perfectly elastic. Is it SHM?
Solution :

The situation is shown in Figure. The electric force produced by the sheet will accelerate the block towards the sheet producing an acceleration. Acceleration will be uniform because electric field E due to the sheet is uniform.

⇒ \(a=\frac{F}{m}=\frac{q E}{m} \text {, where } E=\sigma / 2 \varepsilon_0\)

As initially, the block is at rest and acceleration is constant, from the second equation of motion, the time taken by the block to reach the wall

⇒ \(\mathrm{L}=\frac{1}{2} \mathrm{at}^2 \quad \text { i.e., } \quad \mathrm{t}=\sqrt{\frac{2 \mathrm{~L}}{\mathrm{a}}}=\sqrt{\frac{2 \mathrm{~mL}}{\mathrm{aE}}}=\sqrt{\frac{4 \mathrm{~mL} \varepsilon_0}{\mathrm{a} \sigma}}\)

As collision with the wall is perfectly elastic, the block will rebound with the same speed and as now its motion is opposite to the acceleration, it will come to rest after traveling the same distance L in the same time t. After stopping it will be again accelerated towards the wall and so the block will execute oscillatory motion with ‘span’ L and period.

⇒ \(\mathrm{T}=2 \mathrm{t}=2 \sqrt{\frac{2 \mathrm{~mL}}{\mathrm{aE}}}=2 \sqrt{\frac{4 \mathrm{~mL} \varepsilon_0}{\mathrm{a \sigma}}}\)

However, as the restoring force F = qE is constant and not proportional to displacement x, the motion is not simply harmonic.

Example 40. If an isolated infinite sheet contains charge Q₁ on its one surface and charge Q2on its other surface then prove that electric field intensity at a point in front of the sheet will be \(\frac{Q}{2 \mathrm{~A} \varepsilon_{\mathrm{O}}},\), where Q = Q₁+ Q₂

Solution: Electric field at point P :

⇒ \(\overrightarrow{\mathrm{E}}=\overrightarrow{\mathrm{E}}_{\mathrm{Q}_1}+\overrightarrow{\mathrm{E}}_{\mathrm{Q}_2}\)

⇒ \(=\frac{Q_1}{2 A \varepsilon_0} \hat{n}+\frac{Q_2}{2 A \varepsilon_0} \hat{n}=\frac{Q_1+Q_2}{2 A \varepsilon_0} \hat{n}=\frac{Q}{2 A \varepsilon_0} \hat{n}\)

[This shows that the resultant field due to a sheet depends only on the total charge of the sheet and not on the distribution of charge on individual surfaces].

Example 41. Three large conducting parallel sheets are placed at a finite distance from each other as shown in the figure. Find out the electric field intensity at points A, B, C, and D.

Solution: For point A

⇒ \(\vec{E}_{\text {net }}=\vec{E}_Q+\vec{E}_{3 Q}+\vec{E}_{-2 Q}=-\frac{Q}{2 A \varepsilon_0} \hat{i}-\frac{3 Q}{2 A \varepsilon_0} \hat{i}+\frac{2 Q}{2 A \varepsilon_0} \hat{i}=-\frac{Q}{A \varepsilon_0} \hat{i}\)

for point B

⇒ \(\vec{E}_{n e t}=\vec{E}_{3 Q}+\vec{E}_{-2 Q}+\vec{E}_Q=-\frac{3 Q}{2 A \varepsilon_0} \hat{i}+\frac{2 Q}{2 A \varepsilon_0} \hat{i}+\frac{Q}{2 A \varepsilon_0} \hat{i}=0\)

for point C

⇒ \(\vec{E}_{\text {net }}=\vec{E}_Q+\vec{E}_{3 Q}+\vec{E}_{-2 Q}=+\frac{Q}{2 A \varepsilon_0} \hat{i}-\frac{3 Q}{2 A \varepsilon_0} \hat{i}-\frac{2 Q}{2 A \varepsilon_0} \hat{i}=-\frac{2 Q}{A \varepsilon_0} \hat{i}\)

for point D

⇒ \(\overrightarrow{\mathrm{E}}_{\text {net }}=\overrightarrow{\mathrm{E}}_{\mathrm{Q}}+\overrightarrow{\mathrm{E}}_{3 Q}+\overrightarrow{\mathrm{E}}_{-2 Q}=+\frac{\mathrm{Q}}{2 \mathrm{~A} \varepsilon_0} \hat{\mathrm{i}}+\frac{3 Q}{2 \mathrm{~A} \varepsilon_0} \hat{\mathrm{i}}-\frac{2 Q}{2 \mathrm{~A} \varepsilon_0} \hat{\mathrm{i}}=\frac{\mathrm{Q}}{\mathrm{A} \varepsilon_0} \hat{\mathrm{i}}\)

6.7 Electric field due to uniformly charged spherical shell

⇒ \(E=\frac{K Q}{r^2} \quad r \geq R\) ⇒ For the outside points and points on the surface the uniformly

charged spherical shell behaves as a point charge placed at the center

E = 0 r < R

The electric field due to the spherical shell outside is always along the radial direction.

Solved Examples

Example 42. The figure shows a uniformly charged sphere of radius R and total charge Q. A point charge q is situated outside the sphere at a distance r from the center of the sphere. Find out the following :

1. Force acting on the point charge q due to the sphere.

2. Force acting on the sphere due to the point charge.

Solution :

The electric field at the position of a point charge

⇒ \(\vec{E}=\frac{K Q}{r^2} \hat{r} \quad \text { so, } \quad \vec{F}=\frac{K q Q}{r^2} \hat{r} \quad|\vec{F}|=\frac{K q Q}{r^2}\)

Since we know that every action has equal and opposite reactions so

⇒ \(\overrightarrow{\mathrm{F}}_{\text {sphere }}=-\frac{\mathrm{KqQ}}{\mathrm{r}^2} \hat{\mathrm{r}}\)

⇒ \(\left|\overrightarrow{\mathrm{F}}_{\text {sphere }}\right|=\frac{\mathrm{KqQ}}{\mathrm{r}^2}\)

Example 43. The figure shows a uniformly charged thin sphere of total charge Q and radius R. A point charge q is also situated at the center of the sphere. Find out the following :

1. Force on charge q
2. Electric field intensity at A.
3. Electric field intensity at B.

Solution :

The electric field at the center of the uniformly charged hollow sphere = 0 So force on charge q = 0

The electric field at A

⇒ \(\overrightarrow{\mathrm{E}}_{\mathrm{A}}=\overrightarrow{\mathrm{E}}_{\text {Sphere }}+\overrightarrow{\mathrm{E}}_{\mathrm{q}}=0+\frac{\mathrm{Kq}}{\mathrm{r}^2} ; \mathrm{r}=\mathrm{CA}\)

E due to sphere = 0, because the point lies inside the charged hollow sphere.

Electric field \(\)

⇒ \(=\frac{K Q}{r^2} \hat{r}+\frac{K q}{r^2} \hat{r}=\frac{K(Q+q)}{r^2} \hat{r} ; r=C B\)

Note :

Here we can also assume that the total charge of the sphere is concentrated at the center, for the calculation of the electric field at B.

Example 44. Two concentric uniformly charged spherical shells of radius R₁ and R₂(R₂> R₁) have total charges Q₁ and respectively. Derive an expression of the electric field as a function of r for the following positions.

r < R₁

R₁≤ r < R₂

r ≥ R₂

Solution :

For r < R₁,

Therefore point lies inside both spheres

E_net = E_Inner + E_Outer = 0 + 0

For R₁≤ r < R₂

Therefore point lies outside the inner sphere but inside the outer sphere:

E_net= E_Inner + E_Outer

⇒ \(=\frac{K Q_1}{r^2}+0 \quad=\frac{K Q_1}{r^2} \hat{r}\)

For r ≥ R₂

The point lies outside the inner as well as outer sphere, therefore.

E_net = Einner+ Eouter

⇒ \(\frac{K Q_1}{r^2} \hat{r}+\frac{K Q_2}{r^2} \hat{r}=\frac{K\left(Q_1+Q_2\right)}{r^2} \hat{r}\)

Example 45. A spherical shell having charge +Q (uniformly distributed) and a point charge + q₀ is placed as shown. Find the force between the shell and the point charge(r>>R).

Force on the point charge + q₀ due to the shell

⇒ \(=q_0 \vec{E}_{\text {shell }}=\left(q_0\right)\left(\frac{K Q}{r^2}\right) \hat{r}=\frac{K Q q_0}{r^2} \hat{r}\) where rˆis unit vector along OP.

From the action-reaction principle, force on the shell due to the point charge will also be Fshell = \(F_{\text {shell }}=\frac{K Q q_0}{r^2}(-\hat{r})\)

Conclusion – To find the force on a hollow sphere due to outside charges, we can replace the sphere with a point charge kept at the center.

Example 46. Find the force acting between two shells of radius R₁ and R₂ which have uniformly distributed charges Q₁ and Q₂ respectively and the distance between their centre is r.

Solution :

The shells can be replaced by point charges kept at the center so force between them \(F=\frac{K Q_1 Q_2}{r^2}\)

6.8 Electric Field Due To Uniformly Charged Solid Sphere

Derive an expression for electric field due to solid sphere of radius R and total charge Q which is uniformly distributed in the volume, at a point which is at a distance r from center for the given two cases.

r ≥ R

r ≤ R

Assume an elementary concentric shell of charge dq. Due to this shell, the electric field at the point (r > R) will be

⇒ \(\mathrm{dE}=\frac{\mathrm{Kdq}}{\mathrm{r}^2}\)[from above result of hollow sphere]

E_net = \(E_{\text {net }}=\int d E=\frac{K Q}{r^2}\)

For r < R, there will be no electric field due to the shell of radius greater than r, so an electric field at the point will be present only due to shells having a radius less than r.

E´_net = \(E_{\text {net }}^{\prime}=\frac{K Q^{\prime}}{r^2}\)

here Q’ = \(\frac{Q}{\frac{4}{3} \pi R^3} \times \frac{4}{3} \pi r^3=\frac{Q r^3}{R^3}\)

⇒ \(E_{\text {net }}^{\prime}=\frac{K Q^{\prime}}{r^2}=\frac{K Q r}{R^3}\)

Note: The electric field inside and outside the sphere is always in a radial direction.

7. Electric Potential

In an electrostatic field, the electric potential (due to some source charges) at point P is defined as the work done by an external agent in taking a point unit positive charge from a reference point (generally taken at infinity) to that point P without changing its kinetic energy.

7.1 Mathematical Representation :

If (W ∞ → P)ext is the work required in moving a point charge q from infinity to a point P, the electric potential of the point P is

Note :

(W∝ → P)ext can also be called the work done by an external agent against the electric force on a unit positive charge due to the source charge.

Write both W and q with the proper sign.

7.2 Properties :

Potential is a scalar quantity, its value may be positive, negative, or zero.

S.Ι. Unit of potential is volt = \(\frac{\text { joule }}{\text { coulmb }}\) and its dimensional formula is [M¹L2T–3Ι^-1].

The electric potential at a point is also equal to the negative of the work done by the electric field in taking the point charge from the reference point (i.e. infinity) to that point.

Electric potential due to a positive charge is always positive and due to a negative charge, it is always negative except at infinite. (taking V∞= 0).

Potential decreases in the direction of the electric field.

V = V₁+ V₂+ V₃+ …….

7.3 Use of potential :

If we know the potential at some point (in terms of numerical value or terms of formula) then we can find out the work done by electric force when charge moves from point ‘P’ to ∞ by the formula Wel )p ∞ = qVp

Solved Examples

Example 47. A charge 2μC is taken from infinity to a point in an electric field, without changing its velocity. If work done against electrostatic forces is –40μJ then find the potential at that point.
Solution :

⇒ \(V=\frac{W_{\text {ext }}}{q}=\frac{-40 \mu \mathrm{J}}{2 \mu \mathrm{C}}=-20 \mathrm{~V}\)

Example 48. When charge 10 μC is shifted from infinity to a point in an electric field, it is found that work done by electrostatic forces is –10 μJ. If the charge is doubled and taken again from infinity to the same point without accelerating it, then find the amount of work done by an electric field and against an electric field.
Solution :

W_est)∞ p = –well)∞ p= well)p ∞= 10 μJ

because ΔKE = 0

Vp= \(\frac{\left.W_{\text {ext }}\right)_{\infty x p}}{q}=\frac{10 \mu \mathrm{J}}{10 \mu \mathrm{C}}=1 \mathrm{~V}\) = 1V 10 C

So if now the charge is doubled and taken from infinity then

1 = \(\left.\left.\frac{\left.w_{e x t}\right)_{\infty p}}{20 \mu \mathrm{C}} \quad \Rightarrow \quad \mathrm{W}_{\text {exx }}\right)_{\infty \mathrm{P}}=20 \mu \mathrm{J} \quad \Rightarrow \quad \mathrm{W}_{\mathrm{el}}\right)_{\infty \mathrm{P}}=-20 \mu \mathrm{J}\)

Example 49. A charge 3μC is released at rest from a point P where the electric potential is 20 V then its kinetic energy when it reaches infinite is :
Solution :

Wel = ΔK = Kf– 0

Wel)P → ∞= qVP= 60 μJ

so, Kf= 60 μJ

Electric Potential due to various charge distributions are given in the table.

7.4 Potential Due To A Point Charge :

The electrostatic potential at a point in an electric field due to the point charge may be defined as the amount of work done per unit positive test charge in moving it from infinity to that point (without acceleration) against the electrostatic force due to the electric field of a point charge. It is a scalar quantity.

Consider a point charge +q placed at point O. Suppose that V_Ais electric potential at point A, whose distance from the source charge +q is rA

If W∞A is work done in moving a vanishingly small positive test charge from infinity to point A, then

⇒ \(V_A=\frac{W_{\infty A}}{q_0}\)

Derivation :

Consider a positive point charge Q at the origin. We wish to determine the potential at any point A with position vector r from the origin.

Work done in bringing a unit positive test charge from infinity to point A. For Q > 0. The work done against the repulsive force on the test charge is positive.

Since work done is independent of the path, we choose a convenient path – along the radial direction from infinity to point A.

At some intermediate point A’ on the path, the electrostatic force on a unit positive charge is \(\) where
rˆis the unit vector along OP’. Work done by electric field on test charge for small \(\overrightarrow{\mathrm{F}}_{\mathrm{E}}=\frac{\mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{r}^2} \hat{\mathrm{r}}\)displacement as shown in figure.

⇒ \(\overrightarrow{\mathrm{F}}_{\mathrm{ext}}=-\overrightarrow{\mathrm{F}}_{\mathrm{E}}\)

⇒ \(\mathrm{dw}=\vec{F}_{\text {ext }} \cdot d \overrightarrow{\mathrm{r}} \quad \Rightarrow \quad\left(-\vec{F}_E\right) \cdot(\mathrm{d} \overrightarrow{\mathrm{r}})\)

⇒ \(d w=F_E(-d r)\) =( Here r is decreasing so we will take dr negative)

⇒ \(w_{\infty A}=-\int_{\infty}^{r_A} d w=-\int_{\infty}^{r_A} \frac{Q}{4 \pi \varepsilon_0 r^2} d r\)

⇒ \(V_A-V_{\infty}=\frac{Q}{4 \pi \varepsilon_0 r_A}\left(V_{\infty}=0\right)\) (reference point is taken at infinity)

⇒ \(V_{\mathrm{A}}=\frac{\mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{r}_{\mathrm{A}}}\)

In this case, the distance of point A from the charge +Q is denoted by

⇒ \(V=\frac{Q}{4 \pi \varepsilon_0 r}\)

Electric Potential Due To A System Of Charges

Let us now find the electrostatic potential at point P due to a group of point charges q1, q2, q3… flying at distances r1, r2, r3…. from point P (fig.). The electrostatic potential at point P due to these charges is found by calculating the electrostatic potential P due to each charge, considering the other charges to be absent, and then adding up these electrostatic potentials algebraically.

The electrostatic potential at point P due to charge q1, when other charges are considered absent,

⇒ \(V_1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{r_1}\)

Similarly, electrostatic potentials at point P due to the individual charges q2, q3,…..qn(when other charges are absent) are given by

⇒ \(V_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2}{r_2} ; V_3=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_3}{r_3} ; \ldots \ldots . . ; V_n=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_n}{r_n}\)

Hence, the electrostatic potential at point P due to the group of n point charges, V = V1+ V2+ V3+ ….. + Vn

⇒ \(=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{r_1}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2}{r_2}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_3}{r_3}+\ldots . .+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_n}{r_n}\)

⇒ \(=\frac{1}{4 \pi \varepsilon_0}\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}+\frac{q_3}{r_3}+\ldots . .+\frac{q_n}{r_n}\right) \Rightarrow \quad V=\frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{r_i}\)

Solved Examples

Example 50. Four-point charges are placed at the corners of a square of side l to calculate the potential at the center of the square.

Solution: V = 0 at ‘C’.

Example 51. Two point charges 2μC and – 4μC are situated at points (–2m, 0m) and (2 m, 0 m) respectively. Find out potential at point C. (4 m, 0 m) and. D (0 v5m)

Solution: Potential at point C

VC= \(\mathrm{V}_{\mathrm{q}_1}+\mathrm{V}_{\mathrm{q}_2}=\frac{\mathrm{K}(2 \mu \mathrm{C})}{6}+\frac{\mathrm{K}(-4 \mu \mathrm{C})}{2}=\frac{9 \times 10^9 \times 2 \times 10^{-6}}{6}-\frac{9 \times 10^9 \times 4 \times 10^{-6}}{2}=-15000 \mathrm{~V}\)

Similarly, VD= \(V_D=V_{Q_1}+V_{Q_2}=\frac{K(2 \mu C)}{\sqrt{(\sqrt{5})^2+2^2}}+\frac{K(-4 \mu C)}{\sqrt{(\sqrt{5})^2+2^2}}=\frac{K(2 \mu C)}{3}+\frac{K(-4 \mu C)}{3}=-6000 \mathrm{~V} .\)

7.5 Potential due to a ring :

Potential at the center of uniformly charged ring: Potential due to the small element dq

⇒ \(\mathrm{dV}=\frac{\mathrm{Kdq}}{\mathrm{R}}\)

Net Potential \(V=\int \frac{K d q}{R} \quad \Rightarrow \quad V=\frac{K}{R} \int d q=\frac{K q}{R}\)

For non-uniformly charged ring potential at the center is

⇒ \(\mathrm{V}=\frac{\mathrm{Kq} \text { total }}{\mathrm{R}}\)

Potential due to half ring at the center is: Kq V

⇒ \(V=\frac{K q}{R}\)

Potential at the axis of a ring:

V = \(\frac{K Q}{\sqrt{R^2+x^2}}\)

Solved Examples

Example 52. The figure shows two rings having charges Q and – 5Q. Find the Potential difference between A and B (V_A– V_B).

Solution : \(V_A=\frac{K Q}{R}+\frac{K(-\sqrt{5} Q)}{\sqrt{(2 R)^2+(R)^2}} \quad V_B=\frac{K(-\sqrt{5} Q)}{2 R}+\frac{K(Q)}{\sqrt{(R)^2+(R)^2}}\)

From above we can easily find V_A– V_B

Example 53. A point charge q0 is placed at the center of a uniformly charged ring of total charge Q and radius R. If the point charge is slightly displaced with negligible force along the axis of the ring then find out its speed when it reaches a large distance.
Solution :

Only electric force is acting on q0

∴ Wel = ΔK = \(=\frac{1}{2} m v^2\) – 0 ⇒ Now Wel)c→∞= q0 Vc= q0.\(\frac{\mathrm{KQ}}{\mathrm{R}}\)

∴ \(\frac{K q_0 \mathrm{Q}}{\mathrm{R}}=\frac{1}{2} m v^2\)mv2 ⇒ v = \(\sqrt{\frac{2 \mathrm{Kq}_0 \mathrm{Q}}{\mathrm{mR}}}\)

7.6 Potential Due To Uniformly Charged Disc :

∴ \(\mathrm{V}=\frac{\sigma}{2 \varepsilon_0}\left(\sqrt{\mathrm{R}^2+\mathrm{x}^2}-\mathrm{x}\right)\), where σ is the charged density and x is the distance of the point on the axis from the center of the disc, R is the radius of disc.

7.7 Potential Due To Uniformly Charged Spherical Shell :

Derivation of expression for potential due to the uniformly charged hollow sphere of radius R and total charge Q, at a point which is at a distance r from the center for the following situation

r > R

r < R

As the formula of E is easy , we use V = \(V=-\int_{r \rightarrow \infty}^{r=r} \vec{E} \cdot d \vec{r}\)

At outside point (r > R):

⇒ \(V_{\text {out }}=-\int_{r \rightarrow \infty}^{r=r}\left(\frac{K Q}{r^2}\right) d r \Rightarrow V_{\text {out }}=\frac{K Q}{r}=\frac{K Q}{\text { (Distance from centre) }}\)

For the outside point, the hollow sphere acts like a point charge.

Potential at the center of the sphere (r=0) :

As all the charges Are at a distance R from the center,

So V_Center = \(\frac{\mathrm{KQ}}{\mathrm{R}}=\frac{\mathrm{KQ}}{\text { (Radius of the sphere) }}\)

Potential at inside point ( r<R ) :

Suppose we want to find potential at point P, inside the sphere. +Q,

The potential difference between Point P and O :

⇒ \(-\int_0^P \vec{E}_{\text {in }} \cdot d \vec{r} \text { Where } E_{\text {in }}=0\)

Where Ein = 0

So V_p – V_o = 0

⇒ V_p = V_o = \(\frac{\mathrm{KQ}}{\mathrm{R}}\)

⇒ V_IN = \(\frac{\mathrm{KQ}}{\mathrm{R}}=\frac{\mathrm{KQ}}{\text { (Radius of the sphere) }}\)

7.8 Potential Due To Uniformly Charged Solid Sphere :

For R ≥ R (Outside)

⇒ \(V=\frac{K Q}{r}\)

For r ≤ R (inside)

⇒ \(V=\frac{K Q}{2 R^3}\left(3 R^2-r^2\right) \quad \Rightarrow \text { Here } \rho=\frac{Q}{\frac{4}{3} \pi R^3}\)

Example 54. Two concentric spherical shells of radius R₁ and R₂(R₂> R₁) have uniformly distributed charges Q₁ and Q₂ respectively. Find out potential

1. At point A
2. At the surface of the smaller shell (i.e. at point B)
3. At the surface of the larger shell (i.e. at point C)
4. At r ≤ R₁
5. At R₁≤ r ≤ R₂
6. At r ≥ R₂image

Solution :

Using the results of the hollow sphere as given in table 7.4.

⇒ \(V_A=\frac{K Q_1}{R_1}+\frac{K Q_2}{R_2}\)

⇒ \(V_B=\frac{K Q_1}{R_1}+\frac{K Q_2}{R_2}\)

⇒ \(V_c=\frac{K Q_1}{R_2}+\frac{K Q_2}{R_2}\)

For \(r \leq R_1 \quad V=\frac{K Q_1}{R_1}+\frac{K {R_2}\)

For \(R_1 \leq r \leq R_2\)

⇒ \(V=\frac{K Q_1}{r}+\frac{K Q_2}{R_2}\)

For \(r \geq R_2\)

⇒ \(V=\frac{K Q_1}{r}+\frac{K Q_2}{r}\)

Example 55. Two hollow concentric nonconducting spheres of radius a and b (a > b) contain charges Qa and Qbrespectively. Prove that the potential difference between two spheres is independent of the charge on the outer sphere. If the outer sphere is given an extra charge, is there any change in potential difference?
Solution :

Vinner sphere = \(\frac{K Q_b}{b}+\frac{K Q_a}{a}\)

Vouter sphere = \(\frac{K Q_b}{a}+\frac{K Q_a}{a}\)

Vinner sphere – Vouter sphere = \(\frac{K Q_b}{b}-\frac{K Q_b}{a}\)

⇒ \(\Delta V=K Q_b\left[\frac{1}{b}-\frac{1}{a}\right]\)

Which is independent of the charge on the outer sphere.

If the outer sphere is given any extra charge then there will be no change in potential difference.

8. Potential Difference

The potential difference between two points A and B is work done by an external agent against the electric field in taking a unit positive charge from A to B without acceleration (or keeping Kinetic Energy constant or Ki= Kf))

Mathematical Representation :

If (WA → B)ext = work done by an external agent against the electric field in taking the unit charge from A to B

V_B– V_A= \(\left.V_B-V_A=\frac{\left(W_{A \rightarrow B}\right)_{\text {ext }}}{q}\right)_{\Delta K=0}=\frac{-\left(W_{A \rightarrow B}\right)_{\text {electric }}}{q}=\frac{U_B-U_A}{q}=\frac{-\int_A^B \vec{F}_e \cdot \overrightarrow{d r}}{q}=-\int_A^B \vec{E} \cdot d r\)

Note: Take W and q both with a sign

Properties :

The difference in potential between two points is called the potential difference. It is also called voltage.

The potential difference is a scalar quantity. Its S.I. unit is also volt.

If V_A and V_B are the potential of two points A and B, then work done by an external agent in taking the charge q from A to B is (W_est)AB= q (V_B– V_A) or (W_el) AB = q (V_A– V_B).

The potential difference between the two points is independent of the reference point.

8.1 Potential difference in a uniform electric field :

V_B– V_A= – E_AB ⋅

⇒ V_B– V_A= – |E| |AB| cos θ

= – |E| d

= – Ed

d = effective distance between A and B along the electric field.

or we can also say that E = \(\frac{\Delta \mathrm{V}}{\Delta \mathrm{d}}\)

Special Cases :

Case 1. Line AB is parallel to the electric field.

∴ V_A– V_B= Ed

Case 2. Line AB is perpendicular to the electric field.

∴ V_A– V_B= 0 ⇒ V_A= V_D

Note: In the direction of the electric field potential always decreases.

Example 56. 1μC charge is shifted from A to B and it is found that work done by an external force is 40μJ in doing so against electrostatic forces then, find the potential difference V_A– V_B
Solution :

(WAB)ext = q(V_B– V_A) ⇒ 40 μJ = 1μC (V_B– V_B) ⇒ V_A– V_B= – 40

Example 57. A uniform electric field is present in the positive x-direction. If the intensity of the field is 5N/C then find the potential difference (V_B–V_A) between two points A (0m, 2 m) and B (5 m,3 m)
Solution :

V_B– V_A= – E.AB = –25V. Δ

The electric field intensity in uniform electric field, E = \(\frac{\Delta \mathrm{V}}{\Delta \mathrm{d}}\)

Where ΔV = potential difference between two points.

Δd = effective distance between the two points.

(projection of the displacement along the direction of the electric field.)

Example 58. Find out following

1. V_A– V_B
2. V_B– V_C
3. V_C – V_A
4. V_D – V_C
5. V_A– V_D

Arrange the order of potential for points A, B, C, and D.

Solution :

⇒ \(\left|\Delta V_{A B}\right|=E d=20 \times 2 \times 10^{-2}=0.4\)

So, V_A– V_B= 0.4 V because In the direction of the electric field potential always decreases.

⇒ \(\left|\Delta V_{B C}\right|=E d=20 \times 2 \times 10^{-2}=0.4 \quad \text { so, } \quad V_B-V_C=0.4 \mathrm{~V}\)

⇒ \(\left|\Delta \mathrm{V}_{\mathrm{CA}}\right|=\mathrm{Ed}=20 \times 4 \times 10^{-2}=0.8 \quad \text { so, } \quad \mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{A}}=-0.8 \mathrm{~V}\)

because In the direction of the electric field potential always decreases.

⇒ \(\left|\Delta V_{D C}\right|=E d=20 \times 0=0so, \quad V_D-V_C=0\)

because the effective distance between D and C is zero.

⇒ \(\left|\Delta V_{A D}\right|=E d=20 \times 4 \times 10^{-2}=0.8 \quad \text { so, } \quad V_A-V_D=0.8 \mathrm{~V}\)

because In the direction of the electric field potential always decreases.

The order of potential

V_A> V_B> V_C= V_D.

8.2 Potential difference due to infinitely long wire :

Derivation of expression for the potential difference between two points, which have perpendicular distance rAand from infinitely long line charge of uniform linear charge density λ.

From the definition of potential difference

⇒ \(V_{A B}=V_B-V_A=-\int_{r_A}^{r_B} \vec{E} \cdot \overrightarrow{d r}=-\int_{r_A}^{r_E} \frac{2 K \lambda}{r} \hat{r} \cdot \overrightarrow{d r}\)

⇒ \(V_{\mathrm{AB}}=-2 \mathrm{~K} \lambda \ell \mathrm{n}\left(\frac{\mathrm{r}_{\mathrm{B}}}{\mathrm{r}_{\mathrm{A}}}\right)\)

8.3 Potential Difference Due To Infinitely Long Thin Sheet:

Derivation of expression for the potential difference between two points, having separation d in the direction perpendicularly to a very large uniformly charged thin sheet of uniform surface charge density σ.

Let the points A and B have perpendicular distances A and B respectively then from the definition of potential difference.

⇒ \(V_{A B}=V_B-V_A=-\int_{r_A}^{T_B} \vec{E} \cdot \overrightarrow{d r}=-\int_{r_A}^{r_B} \frac{\sigma}{2 \varepsilon_0} \hat{r} \cdot \overrightarrow{d r} \Rightarrow V_{A B}=-\frac{\sigma}{2 \varepsilon_0}\left(r_B-r_A\right) \quad=-\frac{\sigma d}{2 \varepsilon_0}\)

9. Equipotential Surface :

9.1 Equipotential Surface Definition: If the potential of a surface (imaginary or physically existing) is the same throughout then such surface is known as an equipotential surface.

9.2 Properties of Equipotential Surfaces :

The following properties are associated with the equipotential surfaces :

No work was done in moving a test charge over an equipotential surface. Let A and B be two points on an equipotential surface (Fig. ). If a positive test charge q0is moved from point A to B, then work done in moving the test charge is related to the electrostatic potential difference between the two points as

⇒ \(V_B-V_A=\frac{W_{A B}}{q_0}\)

Since the two points A and B are on the same equipotential surface, V_B– V_A= 0

∴ \(\frac{W_{A B}}{q_0}=0\)

Hence, no work is done in moving a test charge between two points on an equipotential surface.

The electric field is always at right angles to the equipotential surface Since the work done in moving a test charge between two points on an equipotential surface is zero, the displacement of the test charge and the force applied to it must be perpendicular to each other. Since displacement is along the equipotential surface, and force on test charge is \(\mathrm{q}_0 \overrightarrow{\mathrm{E}}\), then the electric field \((\vec{E})\) must be at right angles to the equipotential surface.

The equipotential surfaces help to distinguish regions of strong fields from those of weak fields. We know that

E = \(-\frac{\mathrm{dV}}{\mathrm{dr}}\) or dr = \(-\frac{\mathrm{dV}}{\mathrm{E}}\)

For the same change in the value of dV i.e. dV = constant, we have

⇒ \(\mathrm{dr} \propto \frac{1}{\mathrm{E}}\)

i.e. the spacing between the equipotential surfaces will be denser in the regions, where the electric field is stronger and vice-versa. Therefore, the equipotential surfaces are closer together, where the electric field is stronger, and farther apart, where the field is weaker.

The equipotential surfaces tell the direction of the electric field.

Again E = \(-\frac{\mathrm{dV}}{\mathrm{dr}}\)

The negative sign tells that the electric field is directed in the direction of electric potential with distance. Therefore, the direction of the electric field is from the equipotential surfaces that are close to each other to those that are more and farther away from each other, provided such surfaces have been drawn for the same change in the value of DV.

No two equipotential surfaces can intersect each other.

In case, two equipotential surfaces intersect each other, then at their point of intersection, there will be two values of electric potential. As it is not possible, the two equipotential surfaces can not intersect each other.

9.3 Examples of equipotential surfaces :

For a uniform electric field: In a uniform electric field, the strength and direction of the field are the same at every point inside it.

In a uniform electric field, equipotential surfaces differing by the same amount of potential difference will be equidistant from each other.

For an isolated point charge: The electric field due to an isolated point charge is radial and varies inversely as the square of the distance from the charge. The potential at all the points equidistant from the charge is the same. All such points lie on the surface of a spherical shell, such that the charge lies at its center. Therefore, for a point charge, equipotential surfaces will be a series of concentric spherical shells.

For a system of two-point charges: the dotted lines represent electric lines of force. The thick circles around the charges represent equipotential surfaces due to individual charges.

Line charge: Equipotential surfaces have curved surfaces as that of coaxial cylinders of different radii.

Solved Examples

Example 59. Some equipotential surfaces are shown in the figure. What can you say about the magnitude and the direction of the electric field?

Solution: Here we can say that the electric will be perpendicular to equipotential surfaces.

Also \(|\vec{E}|=\frac{\Delta V}{\Delta d}\)

where ΔV = potential difference between two equipotential surfaces. Δd = perpendicular distance between two equipotential surfaces.

So \(|\vec{E}|=\frac{10}{\left(10 \sin 30^{\circ}\right) \times 10^{-2}}=200 \mathrm{~V} / \mathrm{m}\)

Now there are two perpendicular directions either direction 1 or direction 2 as shown in the figure, but since we know that in the direction of the electric field electric potential decreases so the correct direction is direction 2.

Hence E = 200 V/m, making an angle 120° with the x-axis

Example 60. The figure shows some equipotential surfaces produced by some charges. At which point the value of the electric field is greatest?

Solution :

E is larger where equipotential surfaces are closer. ELOF are ⊥ to equipotential surfaces. In the figure we can see that for point B they are closer so E at point B is the maximum

11. Electrostatic Potential Energy

11.1 Electrostatic potential energy of a point charge due to many charges: The electrostatic potential energy of a point charge at a point in an electric field is the work done in taking the charge from a reference point (generally at infinity) to that point without acceleration (or keeping KE const. or Ki= Kf)).

Its Mathematical formula is

U = W∝P)_ext acc = 0 = qV = – WP∝)el

Here q is the charge whose potential energy is being calculated and V is the potential at its position due to the source charges.

Note: Always put q and V with a sign.

11.2 Properties :

Electric potential energy is a scalar quantity but may be positive, negative, or zero.

Its unit is the same as the unit of work or energy that is the joule (in the S.Ι. system).

Some times energy is also given in electron-volts 1eV = 1.6 × 10^-19 J

Electric potential energy depends on the reference point. (Generally, Potential Energy at r= ∞ is taken zero)

Solved Examples

Example 61 The four identical charges q each are placed at the corners of a square of side a. Find the potential energy of one of the charges due to the remaining charges.

Solution :

The electric potential of point A due to the charges placed at B, C, and D is

⇒ \(\mathrm{V}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{a}}+\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\sqrt{2 \mathrm{a}}}+\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{a}}=\frac{1}{4 \pi \varepsilon_0}\left(2+\frac{1}{\sqrt{2}}\right) \frac{\mathrm{q}}{\mathrm{a}}\)

∴ Potential energy of the charge at A is = qV =\(\frac{1}{4 \pi \varepsilon_0}\left(2+\frac{1}{\sqrt{2}}\right) \frac{q^2}{\mathrm{a}}\)

Example 62. A particle of mass 40 mg and carrying a charge 5 × 10^-9 C is moving directly towards a fixed positive point charge of magnitude 10^-8 C. When it is at a distance of 10 cm from the fixed point charge it has a speed of 50 cm/s. At what distance from the fixed point charge will the particle come momentarily to rest? Is the acceleration constant during the motion?
Solution :

If the particle comes to rest momentarily at a distance r from the fixed charge, then from the conservation of energy we have

⇒ \(\frac{1}{2} m u^2+\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{a}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}}\)

Substituting the given data, we get

⇒ \(\frac{1}{2} \times 40 \times 10^{-6} \times \frac{1}{2} \times \frac{1}{2}=9 \times 10^9 \times 5 \times 10^{-8} \times 10^{-9}\left[\frac{1}{r}-10\right]\)

or, \(\frac{1}{r}-10=\frac{5 \times 10^{-6}}{9 \times 5 \times 10^{-8}}=\frac{100}{9} \quad \Rightarrow \frac{1}{r}=\frac{190}{9} \quad \Rightarrow r=\frac{9}{190} \mathrm{~m}\)

or, i.e., r = 4.7 × 10^-2 m

As here, F = \(\mathrm{F}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}^2} \quad \text { so } \quad \text { acc. }=\frac{\mathrm{F}}{\mathrm{m}} \propto \frac{1}{\mathrm{r}^2}\)

i.e., acceleration is not constant during the motion.

Example 63. A proton moves from a large distance with a speed u m/s directly towards a free proton originally at rest. Find the distance of the closest approach for the two protons in terms of the mass of proton m and its charge e.
Solution :

As here the particle at rest is free to move, when one particle approaches the other, due to electrostatic repulsion other will also start moving so the velocity of the first particle will decrease while of other will increase and at the closest approach both will move with same velocity.

So if v is the common velocity of each particle at closest approach, then by ‘conservation of momentum’ of the two protons system.

mu = mv + mv i.e., v = \(\frac{1}{2} u\)

And by conservation of energy

⇒ \(\frac{1}{2} m u^2=\frac{1}{2} m v^2+\frac{1}{2} m v^2+\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r}\)

r = \(\frac{\mathrm{e}^2}{\pi \mathrm{m} \varepsilon_0 \mathrm{u}^2}\)

12. Electrostatic Potential Energy Of A System Of Charges

(This concept is useful when more than one charges move.) It is the work done by an external agent against the internal electric field required to make a system of charges in a particular configuration from infinite separation without accelerating it.

12.1 Types of system of charge

• Point charge system
• Continuous charge system.

12.2 Derivation For A System Of Point Charges:

Keep all the charges at infinity. Now bring the charges one by one to their corresponding position and find the work required. PE of the system is an algebric sum of all the works.

Let W1= work done in bringing the first charge

W2= work done in bringing the second charge against force due to 1st charge.

W3 work done in bringing the third charge against force due to the 1st and 2nd charges. n(n 1)

PE = W₁+ W₂ + W₃ + …… . (This will contain\(\frac{n(n-1)}{2}={ }^n C_2\)terms)

Method of calculation (to be used in problems)

U = sum of the interaction energies of the charges.

= (U₁₂ + U₁₃ + …….. + U_1n) + (U₂₃ + U₂₄ + …….. + U_2n) + (U₂₄ + U₃₅ + …….. + U_3n)

The method of calculation is useful for symmetrical point charge systems. Find the PE of each charge due to the rest of the charges.

If U₁= PE of the first charge due to all other charges.

= (U₁₂ + U₁₃ + …….. + U_1n)

U₂= PE of the second charge due to all other charges.

= (U₂₁ + U₂₃ + …….. + U_2n) then U = PE of the system

⇒ \(=\frac{U_1+U_2+\ldots . . U_n}{2}\)

Solved Examples

Example 64 Find out the potential energy of the two-point charge system having q1 and q2 charges separated by distance r.
Solution :

Let both the charges be placed at a very large separation initially.

Let W₁= work done in bringing charge q₁ in absence of q₂= q(V_f– V_i) = 0

W₂= work done in bringing charge q₂ in presence of q₁= q(V_f– V_i) = q1(Kq2/r – 0)

PE = W₁+ W₂ = 0 + Kq₁q₂/ r = Kq₁q₂/ r

Example 65 Figure shows an arrangement of three-point charges. The total potential energy of this arrangement is zero. Calculate the ratio \(\frac{\mathrm{q}}{\mathrm{Q}} .\)

Solution : Usys= \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{-q Q}{r}+\frac{(+q)(+q)}{2 r}+\frac{Q(-q)}{r}\right]=0\)

Example 66. Two point charges each of mass m and charge q are released when they are at a distance r from each other. What is the speed of each charged particle when they are at a distance of 2r?
Solution :

According to momentum conservation both the charge particles will move with the same speed now applying energy conservation.

⇒ \(0+0+\frac{K q^2}{r}=2 \frac{1}{2} m v^2+\frac{K q^2}{2 r} \Rightarrow \quad v=\sqrt{\frac{K q^2}{2 r m}}\)

Example 67. Two charged particles each having equal charges 2 × 10^-5 C are brought from infinity to within a separation of 10 cm. Calculate the increase in potential energy during the process and the work required for this purpose.
Solution :

ΔU = U_f– U_I = U_f– 0 = U_f

We have to simply calculate the electrostatic potential energy of the given system of charges

ΔU = Uf= \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r}=\frac{9 \times 10^9 \times 2 \times 10^{-5} \times 2 \times 10^{-5} \times 100}{10} \mathrm{~J}=36 \mathrm{~J}\)

work required = 36 J.

Example 68. Three equal charges q are placed at the corners of an equilateral triangle of side a.

1. Find out the potential energy of the charge system.
2. Calculate the work required to decrease the side of the triangle to a/2.
3. If the charges are released from the shown position and each of them has the same mass m then find the speed of each particle when they lie on the triangle of side 2a.

Solution :

Method I (Derivation)

Assume all the charges are at infinity initially.

work done in putting charge q at corner A

W₁= q (v_f– v_i) = q (0 – 0)

Since potential at A is zero in the absence of charges, work done in putting q at corner B in the presence of charge at A :

⇒ \(\mathrm{W}_2=\left(\frac{\mathrm{Kq}}{\mathrm{a}}-0\right)=\frac{\mathrm{Kq}^2}{\mathrm{a}}\)

Similarly, work done in putting charge q at corner C in the presence of charge at A and B.

⇒ \(\mathrm{W}_3=\mathrm{q}\left(\mathrm{v}_{\mathrm{f}}-v_{\mathrm{i}}\right) \quad=\mathrm{q}\left[\left(\frac{\mathrm{Kq}}{\mathrm{a}}+\frac{\mathrm{Kq}}{\mathrm{a}}\right)-0\right]\)

So-net potential energy PE = \(=W_1+W_2+W_3=0+\frac{K q^2}{a}+\frac{2 K q^2}{a}=\frac{3 K q^2}{a}\)

Method 2 (using direct formula)

U = U₁₂ + U₁₃ + U₂₃ = \(\frac{K^2}{a}+\frac{K q^2}{a}+\frac{K q^2}{a}=\frac{3 K q^2}{a}\)

Work required to decrease the sides

W = U_f– U_i = \(\frac{3 K q^2}{a / 2}-\frac{3 K q^2}{a}=\frac{3 K q^2}{a}\)

Work done by electrostatic forces = change is the kinetic energy of particles.

U_i– U_f= K_f– K_i ⇒ \(\frac{3 K q^2}{a}-\frac{3 K q^2}{2 a}=3\left(\frac{1}{2} m v^2\right)-0 \Rightarrow \quad v=\sqrt{\frac{K q^2}{a m}}\)

Example 69 Four identical point charges q are placed at four corners of a square of side a. Find out the potential energy of the charge system

Solution :

Method 1 (using direct formula) :

U = U₁₂ + U₁₃ + U₁₄ + U₂₃ + U₂₄ + U₃₄

⇒ \(=\frac{K^2}{a}+\frac{K^2}{a \sqrt{2}}+\frac{K^2}{a}+\frac{K^2}{a}+\frac{K^2}{a \sqrt{2}}+\frac{K^2}{a}=\left[\frac{4 K^2}{a}+\frac{2 K q^2}{a \sqrt{2}}\right]=\frac{2 K q^2}{a}\left[2+\frac{1}{\sqrt{2}}\right]\)

Method 2 [using U = 2(U₁+ U₂+ ……)] :

U₁= total P.E. of charge at corner 1 due to all other charges

U₂= total P.E. of charge at corner 2 due to all other charges

U₃= total P.E. of charge at corner 3 due to all other charges

U₄= total P.E. of charge at corner 4 due to all other charges

Due to symmetry U₁= U₂= U₃= U₄

U_net = \(\frac{\mathrm{U}_1+\mathrm{U}_2+\mathrm{U}_3+\mathrm{U}_4}{2}=2 \mathrm{U}_1=2\left[\frac{\mathrm{Kq}^2}{\mathrm{a}}+\frac{\mathrm{Kq}^2}{\mathrm{a}}+\frac{\mathrm{Kq}^2}{\sqrt{2} \mathrm{a}}\right]=\frac{2 \mathrm{Kq}^2}{\mathrm{a}}\left[2+\frac{1}{\sqrt{2}}\right]\)

Example 70. Six equal point charges q are placed at six corners of a hexagon of side a. Find out the potential energy of the charge system

Solution :

Unit = \(\frac{\mathrm{U}_1+\mathrm{U}_2+\mathrm{U}_3+\mathrm{U}_4+\mathrm{U}_5+\mathrm{U}_6}{2}\)

Due to symmetry U₁= U₂= U₃= U₄= U₅= U₆ so U_net = 3U₁= \(\frac{3 K q^2}{a}\left[2+\frac{2}{\sqrt{3}}+\frac{1}{2}\right]\)

12.3 Electric Potential Energy For Continues Charge System :

This energy is also known as self-energy.

P.E. (Self Energy) of a uniformly charged spherical shell:-

Uself = \(\frac{K Q^2}{2 R}\)

Self-energy of the uniformly charged solid sphere :

For a solid sphere P.E. is Uself = \(\frac{3}{5} \frac{K Q^2}{R}\)

Example 71 A spherical shell of radius R with uniform charge q is expanded to a radius 2R. Find the work performed by the electric forces and external agents against electric forces in this process.
Solution :

W_ext = U_f– U_i= \(\frac{q^2}{16 \pi \varepsilon_0 R}-\frac{q^2}{8 \pi \varepsilon_0 R}=-\frac{q^2}{16 \pi \varepsilon_0 R}\)

W_ext = U_f– U_i= \(\frac{q^2}{8 \pi \varepsilon_0 R}-\frac{q^2}{16 \pi \varepsilon_0 R}=\frac{q^2}{16 \pi \varepsilon_0 R}\)

Example 72 Two nonconducting hollow uniformly charged spheres of radii and R2 with charge and Q2 respectively are placed at a distance r. Find out the total energy of the system.

Solution : Utotal = Uself + UInteraction = \(\frac{\mathrm{Q}_1^2}{8 \pi \varepsilon_0 \mathrm{R}_1}+\frac{\mathrm{Q}_2^2}{8 \pi \varepsilon_0 \mathrm{R}_2}+\frac{\mathrm{Q}_1 \mathrm{Q}_2}{4 \pi \varepsilon_0 \mathrm{r}}\)

Example 73. Two concentric spherical shells of radius R1 and R2(R2> R1) have uniformly distributed charges Q1 and Q2respectively. Find out the total energy of the system.

Solution : Utotal = Uself 1 + Uself 2 + UInteraction = \(\frac{\mathrm{Q}_1^2}{8 \pi \varepsilon_0 \mathrm{R}_1}+\frac{\mathrm{Q}_2^2}{8 \pi \varepsilon_0 \mathrm{R}_2}+\frac{\mathrm{Q}_1 \mathrm{Q}_2}{4 \pi \varepsilon_0 \mathrm{R}_2}\)

12.4 Energy Density :

Energy Density Definition: Energy density is defined as energy stored in unit volume in any electric field. Its mathematical formula is given as following Energy density =\(\frac{1}{2}\) where E = electric field intensity at that point ε = ε₀εrelectric permittivity of medium

Example 74 Find out energy stored in an imaginary cubical volume of side a in front of an infinitely large nonconducting sheet of uniform charge density σ.
Solution :

Energy stored

⇒ \(\mathrm{U}=\int \frac{1}{2} \varepsilon_0 \mathrm{E}^2 \mathrm{dV}\)where dV is small volume = \(\frac{1}{2} \varepsilon_0 \mathrm{E}^2 \int \mathrm{dV}\)

⇒ \(\frac{1}{2} \varepsilon_0 \frac{\sigma^2}{4 \varepsilon_0^2} \cdot a^3 \cdot=\frac{\sigma^2 a^3}{8 \varepsilon_0}\)

13. Relation Between Electric Field Intensity And Electric Potential

13.1 For uniform electric field :

The potential difference between two points A and B

VB – VA = – E . AB

13.2 Nonuniform electric field

⇒ \(E_x=-\frac{\partial V}{\partial x}, E_y=-\frac{\partial V}{\partial y}, E_z=-\frac{\partial V}{\partial z} \quad \Rightarrow \quad \vec{E}=E_x \hat{i}+E_y \hat{j}+E_z \hat{k}\)

⇒ \(-\left[\hat{\mathrm{i}} \frac{\partial}{\partial \mathbf{x}} \mathrm{V}+\hat{\mathrm{j}} \frac{\partial}{\partial \mathbf{y}} \mathrm{V}+\hat{\mathrm{k}} \frac{\partial}{\partial \mathbf{z}} \mathrm{V}\right]=-\left[\hat{\mathrm{i}} \frac{\partial}{\partial \mathbf{x}}+\hat{\mathrm{j}} \frac{\partial}{\partial \mathbf{y}}+\hat{\mathrm{k}} \frac{\partial}{\partial \mathbf{z}}\right] \mathrm{V}=-\nabla \mathrm{V}=- \text { grad } V\)

Where\(\frac{\partial V}{\partial X}\)= derivative of V with respect to x (keeping y and z constant)

⇒ \(\frac{\partial V}{\partial y}\)= derivative of V with respect to y (keeping z and x constant)

⇒ \(\frac{\partial \mathrm{V}}{\partial \mathrm{z}}\)= derivative of V with respect to z (keeping x and y constant)

13.3 If Electric Potential And Electric Field Depends Only On One Coordinate, Say R :

⇒ \(\overrightarrow{\mathrm{E}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{r}} \hat{\mathrm{r}}\)

where rˆis a unit vector along increasing r.

⇒ \(\int d V=-\int \vec{E} \cdot \overrightarrow{d r} \quad \Rightarrow \quad V_B-V_A=-\int_{r_A}^{r_B} \vec{E} \cdot \overrightarrow{d r}\)

⇒ \(\overrightarrow{\mathrm{dr}}\)is along the increasing direction of r.

The potential of a point V = \(V=-\int_{\infty}^r \vec{E} \cdot \overrightarrow{d r}\)

Example 75 A uniform electric field is along x – the x-axis. The potential difference VA– VB = 10 V between two points A (2m, 3m) and B (4m, 8m). Find the electric field intensity.

Solution : E = \(\frac{\Delta \mathrm{V}}{\Delta \mathrm{d}}=\frac{10}{2}\) = 5 V / m. It is along the + ve x-axis.

Example 76 V = x2 + y , Find \(\vec{E}\)
Solution :

⇒ \(\frac{\partial V}{\partial x}=2 x, \frac{\partial V}{\partial y}=1 \quad \text { and } \frac{\partial V}{\partial z}=0\)

⇒ \(\vec{E}=-\left(\hat{i} \frac{\partial V}{\partial x}+\hat{j} \frac{\partial V}{\partial y}+\hat{k} \frac{\partial V}{\partial z}\right)=-(2 x \hat{i}+\hat{j})\) Electric field is nonuniform.

Example 77 For given \(\vec{E}=2 x \hat{i}+3 y \hat{j}\) find the potential at (x, y) if V at origin is 5 volts.
Solution :

⇒ \(\int_5^V d V=-\int \vec{E} \cdot \overline{d r}=-\int_0^x E_x d x-\int_0^y E_y d y\)

⇒ \(V-5=-\frac{2 x^2}{2}-\frac{3 y^2}{2} \Rightarrow V=-\frac{2 x^2}{2}-\frac{3 y^2}{2}+5\)

14. Electric Dipole

14.1 Electric Dipole

If two point charges are equal in magnitude q and opposite in sign separated by a distance a such that the distance of field point r>>a, the system is called a dipole. The electric dipole moment is defined as a vector quantity having magnitude p = (q × a) and direction from negative charge to positive charge.

Note: [In chemistry, the direction of dipole moment is assumed to be from positive to negative charge.] The C.G.S unit of electric dipole moment is debye which is defined as the dipole moment of two equal and opposite point charges each having charge 10–10 frankline and separation of 1 Å, i.e.,

1 debye (D) = 10–10 × 10–8 = 10–18 Fr × cm

1D = 10–18 ×\(\frac{C}{3 \times 10^9}\) × 10–2 m = 3.3 × 10–30 C × m.

S.I. Unit is coulomb × metre = C . m

Solved Examples

Example 78 A system has two charges qA= 2.5 × 10–7 C and qB= – 2.5 × 10–7 C located at points A : (0, 0, – 0.15 m) and B ; (0, 0, + 0.15 m) respectively. What is the net charge and electric dipole moment of the system?
Solution :

Net charge = 2.5 × 10–7 – 2.5 × 10–7 = 0

Electric dipole moment,

P = (Magnitude of charge) × (Separation between charges)

= 2.5 × 10–7[0.15 + 0.15] C m = 7.5 × 10–8 C m

The direction of the dipole moment is from B to A.

14.2 Electric Field Intensity Due to Dipole :

At the axial point:-

⇒ \(\vec{E}=\frac{K q}{\left(r-\frac{a}{2}\right)}-\frac{K q}{\left(r+\frac{a}{2}\right)^2} \text { along the } \hat{P}=\frac{K q(2 r a)}{\left(r^2-\frac{a^2}{4}\right)^2} \hat{P}\)

If r >> a then

⇒ \(\vec{E}=\frac{K q 2 r a}{r^4} \hat{P}=\frac{2 K \vec{P}}{r^3},\)

As the direction of the electric field at the axial position is along the dipole moment \((\vec{P})\)

⇒ \(\vec{E}_{\text {axial }}=\frac{2 K \vec{P}}{r^3}\)

The electric field at the perpendicular Bisector (Equitorial Position)

E_net= \(2 \mathrm{E} \cos \theta \text { (along }-\hat{P} \text { ) }\)

⇒ \(\overrightarrow{\mathrm{E}}_{\text {net }}=2\left(\frac{\mathrm{Kq}}{\left(\sqrt{r^2+\left(\frac{a}{2}\right)^2}\right)^2}\right) \frac{\frac{a}{2}}{\sqrt{r^2+\left(\frac{a}{2}\right)^2}}(-\hat{P})=2 \frac{K q a}{\left(r^2+\left(\frac{a}{2}\right)^2\right)^{3 / 2}}(-\hat{P})\)

If r >> a then

⇒ \(\vec{E}_{\text {net }}=\frac{K P}{r^3}(-\hat{P})\)

As the direction of \(\vec{E}\) at equitorial position is opposite of \(\vec{P}\) so we can write in vector form:

⇒ \(\vec{E}_{\text {eqt }}=-\frac{K \vec{P}}{r^3}\)

Electric field at general point (r, θ) :

⇒ \(\mathrm{E}_{\text {net }}=\frac{\mathrm{KP}}{\mathrm{r}^3} \sqrt{1+3 \cos ^2 \theta} ; \quad \tan \phi=\frac{\tan \theta}{2}\)

Solved Examples

Example 79 The electric field due to a short dipole at a distance r, on the axial line, from its midpoint is the same as that of the electric field at a distance r’, on the equatorial line, from its mid-point. Determine the ratio \(\)
Solution :

⇒ \(\frac{1}{4 \pi \varepsilon_0} \frac{2 \mathrm{p}}{\mathrm{r}^3}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{p}}{\mathrm{r}^{\prime 3}} \text { or } \quad \frac{2}{\mathrm{r}^3}=\frac{1}{\mathrm{r}^3} \quad \text { or } \quad \frac{\mathrm{r}^3}{\mathrm{r}^3}=2 \quad \text { or } \quad \frac{\mathrm{r}}{\mathrm{r}^{\prime}}=2^{1 / 3}\)

Example 80 Two charges, each of 5 μC but opposite in sign, are placed 4 cm apart. Calculate the electric field intensity of a point that is at a distance of 4 cm from the midpoint on the axial line of the dipole.
Solution :

We can not use the formula of short dipole here because the distance of the point is comparable to the distance between the two point charges.

q = 5 × 10–6 C, a = 4 ×10–2 m, r = 4 × 10–2 m

image

E_ess= E++ E– = \(\frac{\mathrm{K}(5 \mu \mathrm{C})}{(2 \mathrm{~cm})^2}-\frac{\mathrm{K}(5 \mu \mathrm{C})}{(6 \mathrm{~cm})^2}=\frac{144}{144 \times 10^{-8}} N \mathrm{NC}^{-1}=10^8 \mathrm{~N} \mathrm{C}^{-1}\)

Example 81 Two charges ± 10 μC are placed 5 × 10–3 m apart. Determine the electric field at a point Q which is 0.15 m away from O, on the equatorial line.
Solution: In the given problem, r >> an

∴ E = \(\frac{1}{4 \pi \varepsilon_0} \frac{p}{r^3}=\frac{1}{4 \pi \varepsilon_0} \frac{q(a)}{r^3}\)

or E = 9 × 109 \(\frac{10 \times 10^{-6} \times 5 \times 10^{-3}}{0.15 \times 0.15 \times 0.15} \mathrm{NC}^{-1}\)

= 1.33 ×105 NC–1

14.3 Electric Potential due to a small dipole :

Potential at axial position :

V = \(\frac{K q}{\left(r-\frac{a}{2}\right)}+\frac{K(-q)}{\left(r+\frac{a}{2}\right)}\)

V = \(\frac{K q a}{\left(r^2-\left(\frac{a}{2}\right)^2\right)}\)

If r >> a than

V = \(\frac{\text { Kqa }}{r^2}\)

where qa = p

V_axial= \(\frac{K P}{r^2}\)

Potential at equatorial position :

V = \(\frac{K q}{\sqrt{r^2+\left(\frac{a}{2}\right)^2}}+\frac{K(-q)}{\sqrt{r^2+\left(\frac{a}{2}\right)^2}}=0\)

Veqt = 0

Potential at general point (r,θ) :

V = \(\frac{K(\vec{P} \cdot \vec{r})}{r^3}\)

Example 82

1. Find the potential at points A and B due to the small charge – system fixed near the origin. (the distance between the charges is negligible).
2. Find work done to bring a test charge from point A to point B, slowly. All parameters are in S.I. units.

Solution :

The dipole moment of the system is

⇒ \(\vec{P}=(q a) \hat{i}+(q a) \hat{j}\)

Potential at point A due to the dipole

VA = \(K \frac{(\vec{P} \cdot \vec{r})}{r^B}=\frac{K[(q a) \hat{i}+(q a) \hat{j}] \cdot(4 \hat{i}+3 \hat{j})}{5^3}=\frac{k(q a)}{125}(7)\)

⇒ V_B= \(\frac{K[(q a) \hat{i}+(q a) \hat{j}] \cdot(3 \hat{i}-4 \hat{j})}{(5)^3}=\frac{K(q a)}{125}\)

WA → B = UB– UA= q0(V_B– V_A) =\(\left[-\frac{\mathrm{K}(\mathrm{qa})}{125}-\left(\frac{\mathrm{K}(\mathrm{qa})(7)}{125}\right)\right] \Rightarrow \mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}=\frac{\mathrm{Kqq}_0 \mathrm{a}}{125} \text { (8) }\)

14.4 Dipole in a uniform electric field

Dipole is placed along the electric field :

In this case, F_net = 0, τ_net = 0 so it is an equilibrium state. And it is a stable equilibrium position.

If the dipole is placed at θ angle from \(\vec{E}\):

In this case F_net = 0 but

Net torque τ = (qEsinθ) (a)

Here qa = P ⇒ τ = PE sinθ in vector form τ= \(\vec{P} \cdot \vec{E}\)

Example 83 A dipole is formed by two point charges –q and +q, each of mass m, and both the point charges are connected by a rod of length l and mass m1. This dipole is placed in a uniform electric field E. If the dipole is disturbed by a small angle θ from a stable equilibrium position, prove that its motion will be almost SHM. Also, find its period.
Solution :

If the dipole is disturbed by θ angle,

τ_net = –PE sinθ (here – ve sign indicates that the direction of the torque is opposite of θ). If θ is very small, sinθ = θ

τ_net = –(PE)θ

τ_net∝ (–θ) so motion will be almost SHM.

T = 2π \(\sqrt{\frac{\mathrm{I}}{\mathrm{K}}}\)

The potential energy of a dipole placed in a uniform electric field :

UB– UA= \(-\int_A^B \vec{F} \cdot \overrightarrow{d r} \quad \text { Here } \quad U_B-U_A=-\int_A^B \vec{\tau} \cdot \overrightarrow{d \theta}\)

In the case of dipole, at θ = 90°, P.E. is assumed to be zero.

Uθ– U90° =\(-\int_{\theta=90^{\circ}}^{\theta=\theta}(-P E \sin \theta)(d \theta)\)

(As the direction of torque is opposite of θ)

Uθ– 0 = – PE cos θ

θ = 90° is chosen as a reference so that the lower limit comes out to be zero.

Uθ= \(-\vec{P} \cdot \vec{E}\)

From the potential energy curve, we can conclude :

at θ = 0, there is a minimum of P.E. so it is a stable equilibrium position.

at θ = 180°, there is a maxima of P.E. so it is a position of unstable equilibrium.

Example 84 Two point masses of mass m and equal and opposite charge of magnitude q are attached on the corners of a non-conducting uniform rod of mass m and the system is released from rest in uniform electric field E as shown in figure from θ = 53°

1. Find the angular acceleration of the rod just after releasing
2. What will be the angular velocity of the rod when it passes through stable equilibrium?
3. Find the work required to rotate the system by 180°.

Solution :

t_net= PE sin53° = I α

⇒ \(\alpha=\frac{(\mathrm{q} \ell) \mathrm{E}\left(\frac{4}{5}\right)}{\frac{\mathrm{m} \ell^2}{12}+\mathrm{m}\left(\frac{\ell}{2}\right)^2+\mathrm{m}\left(\frac{\ell}{2}\right)^2}=\frac{48 \mathrm{qE}}{35 \mathrm{~m} \ell}\)

From energy conservation :

Ki+ Ui= Kf+ Uf

0 + (– PE cos 53°) = \(\frac{1}{2}\)Ιω2 + (–PE cos 0°)

where I = \(\mathrm{I}=\frac{\mathrm{m} \ell^2}{12}+\mathrm{m}\left(\frac{\ell}{2}\right)^2+\mathrm{m}\left(\frac{\ell}{2}\right)^2 \quad \Rightarrow \quad \omega=\sqrt{\frac{48 \mathrm{qE}}{35 \mathrm{~m} \ell}}\)

W_ext= Uf– Ui

W_ext = (–PE cos(180° + 53°)) – (–PEcos 53°)

W_ext = (ql)E\(\left(\frac{4}{5}\right)+(q \ell) E\left(\frac{4}{5}\right) \quad \Rightarrow \quad W_{\text {ext }}=\left(\frac{8}{5}\right) q \ell E\)

15. Electric Lines Of Force (ELOF)

The line of force in an electric field is an imaginary line, the tangent to which at any point on it represents the direction of the electric field at the given point.

15.1 Properties :

A line of force originates from a positive charge and terminates on a negative charge. If there is only one positive charge then lines start from a positive charge and terminate at ∞. If there is only one negative charge then lines start from ∞ and terminate at a negative charge.

Two lines of force never intersect each other because there cannot be two directions of \(\vec{E}\)

Electric lines of force produced by static charges do not form a closed loop.

If lines of force make a closed loop, then work done to move a +q charge along the loop will be non-zero. So it will not be a conservative field. So these types of lines of force are not possible in electrostatics.

The Number of lines per unit area (line density) represents the magnitude of the electric field.

If lines are dense, ⇒ E will be more

If Lines are rare, ⇒ E will be less, and if E = O, no line of force will be found here

The number of lines originating (terminating) is proportional to the charge.

Example 85 If several electric lines of force from charge q is 10 then find out several electric lines of force from 2q charge.
Solution :

No. of ELOF ∝ charge

10 ∝ q ⇒ 20 ∝ 2q

• So the number of ELOF will be 20.
• Electric lines of force end or start perpendicularly on the surface of a conductor.
• Electric lines of force never enter into conductors.

Example 86 Some electric lines of force are shown in the figure, for points A and B

1. E_A> E_B
2. EB> E_A
3. V_A> V_B
4. V_B> B_A

Solution: lines are denser at B so E_A> EBIn the direction of the Electric field, potential decreases so V_A> V_B

Example 87 If a charge is released in an electric field, will it follow lines of force?
Solution :

Case I :

If lines of force are parallel (in a uniform electric field):-

In this type of field, if a charge is released, the force on it will be QoE and its direction will be along \(\vec{E}\). So the charge will move in a straight line, along the lines of force.

Case 2: –

If lines of force are curved (in a non-uniform electric field):-

The charge will not follow lines of force

Example 88 A charge + Q is fixed at a distance of d in front of an infinite metal plate. Draw the lines of force indicating the directions.
Solution :

There will be induced charge on two surfaces of the conducting plate, so ELOF will start from the +Q charge and terminate at the conductor and then will again start from the other surface of the conductor.

15.2 Solid Angle :

A solid angle is a measure of a cone. Consider the intersection of the given cone with a sphere of radius R. The solid angle ΔΩ of the cone is defined to be equal to ΔS/R2, where ΔS is the area on the sphere cut out by the cone.

16. Electric Flux

Consider some surface in an electric field The electric flux of the field over the \(\overrightarrow{\mathrm{E}}\). Let us select a small area element \(\overrightarrow{\mathrm{dS}}\) on this surface.

area element is given by dφE= \(\overrightarrow{\mathrm{E}} \overrightarrow{\mathrm{dS}}\)

The direction of \(\overrightarrow{\mathrm{dS}}\) is normal to the surface. It is along ˆn

or dφE= EdS cos θ or dφE= (E cos θ) dS or dφE= EndS

where En is the component of the electric field in the direction of \(\overrightarrow{\mathrm{dS}}\).

The electric flux over the whole area is given by φE= \(\)

If the electric field is uniform over that area then φE=

Special Cases :

Case I: If the electric field is normal to the surface, then the angle of the electric field \(\vec{E}\) with normal will be zero

So φ = ES cos 0

φ = ES

Case II: If the electric field is parallel to the surface (glazing), then the angle made by \(\vec{E}\) with normal = 90º

So φ = ES cos 90º = 0

16.1 Physical Meaning :

The electric flux through a surface inside an electric field represents the total number of electric lines of force crossing the surface. It is a property of the electric field

16.2 Unit

• The SI unit of electric flux is Nm2 C–1(gauss) or J m C–1.
• Electric flux is a scalar quantity. (It can be positive, negative, or zero)

Example 89. If the electric field is given by \((6 \hat{i}+3 \hat{j}+4 \hat{k}) N / C\) calculate the electric flux through a surface of area 20 m2 lying in YZ plane.
Solution :

Here,\(\overrightarrow{\mathrm{E}}=6 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\)

They are vectors representing the surface of area 20 units in the YZ-plane given by

⇒ \(\overrightarrow{\mathrm{S}}=20 \hat{\mathrm{i}}\)

Therefore, electric flux through the surface,

⇒ \(\phi=\vec{F} \cdot \vec{S}=(6 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot 20 \hat{i}=120 \mathrm{~N}-\mathrm{m}^2 / \mathrm{C}\)

Example 90. A rectangular surface of sides 10 cm and 15 cm is placed inside a uniform electric field of 25 Vm–1, such that normal to the surface makes an angle of 60º with the direction of the electric field. Find the flux of the electric field through the rectangular surface.
Solution :

The flux through the rectangular surface given by

φ =\(\overrightarrow{\mathrm{E}} . \Delta \overrightarrow{\mathrm{S}}=\) EΔ S cos

Here, E = 25 V m–1;

ΔS = 10 × 15 = 150 cm2 = 150 × 10–4 m2

and θ = 60º

∴ φ = 25 × 150 × 10–4 cos 60º = \(\frac{3 \sqrt{3}}{16} \mathrm{Nm}^2 \mathrm{C}^{-1}\)

Example 91 The electric field in a region is given by \(\vec{E}=\frac{3}{5} E_0 \vec{i}+\frac{4}{5} E_0 \vec{j}\) with E₀= 2.0 × 103 N/C. Find the flux of this field through a rectangular surface of area 0.2m2 parallel to the Y–Z plane.
Solution :

⇒ \(\phi_E=\vec{E} \cdot \vec{S}=\left(\frac{3}{5} E_0 \vec{i}+\frac{4}{5} E_0 \vec{j}\right) \cdot(0.2 \hat{i})=240 \frac{N-m^2}{C}\)

Example 92 A point charge Q is placed at the corner of a square of side a, then find the flux through the square.

Solution :

The electric field due to Q at any point of the square will be along the plane of the square and the electric field line is perpendicular to the square; so φ = 0.

In other words, we can say that no line is crossing the square so flux = 0.

Case-III: Curved surface in uniform electric field Suppose a circular surface of radius R is placed in a uniform electric field as shown.

Flux passing through the surface φ = E (πR²)

Now suppose, a hemispherical surface is placed in the electric field flux through the hemispherical surface

φ = ∫ Eds cos θ φ = E ∫ ds cos θ

where ∫ ds cos θ is a projection of the spherical surface Area on the base.

∫ds cosθ = πR²

so φ = E(πR²) = same Ans. as in previous case

so we can conclude that

If the number of electric field lines passing through two surfaces are same, then flux passing through these surfaces will also be the same, irrespective of the shape of the surface

φ1= φ2= φ3= E(πR²)

Case 4:

Flux through a closed surface :

Suppose there is a spherical surface and a charge ‘placed at the center. flux through the spherical surface

φ = \(\int \vec{E} \cdot \overrightarrow{d s}=\int E d s \quad \text { as } \vec{E} \text { is along } \overrightarrow{d s} \text { (normal) }\)

φ= \(\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{Q}}{\mathrm{R}^2} \int \quad \text { ds } \quad \text { where } \int \quad \mathrm{ds}=4 \pi \mathrm{R}^2\)

φ = \(\left(\frac{1}{4 \pi \mathrm{R}^2} \frac{\mathrm{Q}}{\mathrm{R}^2}\right)\left(4 \pi \mathrm{R}^2\right) \Rightarrow \phi=\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}\)

Now if the charge Q is enclosed by any other closed surface, still same lines of force will still pass through the surface.

So here also flux will be φ = \(\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}\), that’s what Gauss Theorem is. ο

17. Gauss’s Law In Electrostatics Or Gauss’s Theorem

This law was stated by a mathematician Karl F Gauss. This law gives the relation between the electric field at a point on a closed surface and the net charge enclosed by that surface. This surface is called the Gaussian surface. It is a closed hypothetical surface. Its validity is shown by experiments. It is used to determine the electric field due to some symmetric charge distributions.

17.1 Statement and Details :

Gauss’s law is stated as given below.

The surface integral of the electric field intensity over any closed hypothetical surface (called Gaussian surface) in free space is equal to \(\frac{1}{\varepsilon_0}\) times the total charge enclosed within the surface. Here, ε₀is the permittivity of free space.

If S is the Gaussian surface and \(\sum_{i=1}^n q_i\)is the total charge enclosed by the Gaussian surface, then according to Gauss’s law,

⇒ \(\phi_E=\int \vec{E} \cdot \overrightarrow{d S}=\frac{1}{\varepsilon_0} \sum_{i=1}^n q_i\)

The circle on the sign of integration indicates that the integration is to be carried out over the closed surface.

Note :

Flux through the Gaussian surface is independent of its shape.

Flux through the Gaussian surface depends only on the total charge present inside the Gaussian surface.

Flux through the Gaussian surface is independent of the position of charges in the ide Gaussian surface.

Electric field intensity at the Gaussian surface is due to all the charges present inside as well as outside the Gaussian surface.

In a close surface, incoming flux is taken negatively while outgoing flux is taken positively because ˆis taken positively in an outward direction.

In a Gaussian surface φ = 0 does not imply E = 0 at every point of the surface but E = 0 at every point implies φ = 0.

Example 93 Find out flux through the given Gaussian surface.

Solution : φ = \(\frac{\mathrm{Q}_{\mathrm{in}}}{\varepsilon_0}=\frac{2 \mu \mathrm{C}-3 \mu \mathrm{C}+4 \mu \mathrm{C}}{\varepsilon_0}=\frac{3 \times 10^{-6}}{\varepsilon_0} \mathrm{Nm}^2 / \mathrm{C}\)

Example 94 If a point charge q is placed at the centre of a cube then find out flux through any one surface of the cube.
Solution :

Flux through 6 surfaces = \(\frac{\mathrm{q}}{\varepsilon_0}\). Since all the surfaces are symmetrical 0

so, flux through one surfaces = \(\frac{1}{6} \frac{\mathrm{q}}{\varepsilon_0}\)

17.2 Flux through open surfaces using Gauss’s Theorem :

Example 95 A point charge +q is placed at the centre of curvature of a hemisphere. Find flux through the hemispherical surface.

Solution :

Let’s put an upper-half hemisphere. Now flux passing through the entire sphere = \(=\frac{\mathrm{q}}{\varepsilon_0}\)

As the charge q is symmetrical to the upper half and lower half hemispheres, so half-half flux will emit from both surfaces.

Example 96 A charge Q is placed at a distance a/2 above the center of a horizontal, square surface of the edge as shown in the figure. Find the flux of the electric field through the square surface.

Solution: We can consider imaginary faces of the cube such that the charge lies at the center of the cube. Due to symmetry, we can say that flux through the given area (which is one face of the cube)

φ = \(\phi=\frac{\mathrm{Q}}{6 \varepsilon_0}\)

Example 97 Find flux through the hemispherical surface

Solution :

Flux through the hemispherical surface due to +q = \(\frac{\mathrm{q}}{2 \varepsilon_0}\) (we have seen in previous examples)

Flux through the hemispherical surface due to +q0charge = 0, because due to +q0charge field lines entering the surface = field lines coming out of the surface.

17.3 Finding qin from flux :

Solved Examples

Example 98 Flux (in S.I. units) coming out and entering a closed surface is shown in the figure. Find the charge enclosed by the closed surface.

Solution :

Net flux through the closed surface = + 20 + 30 + 10 -15 = 45 N.m2/c from Gauss`s theorem

φ_net = \(\frac{q_{\text {in }}}{\varepsilon_0} \quad \Rightarrow \quad 45=\frac{q_{\text {in }}}{\varepsilon_0} \quad \Rightarrow \quad q_{\text {in }}=(45) \varepsilon_0\)

17.4 Finding electric field from Gauss`s Theorem :

From Gauss`s theorem, we can say

⇒ \(\int \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{ds}}=\phi_{\text {net }}=\frac{\mathrm{q}_{\mathrm{in}}}{\varepsilon_0}\)

17.4.1 Finding E due to a spherical shell:-

The electric field outside the Sphere :

Since the electric field due to a shell will be radially outwards. So let’s choose a spherical Gaussian surface. Gauss`s theorem for this spherical Gauss`s surface,

⇒ \(\int \overrightarrow{\mathrm{E}} \overrightarrow{\mathrm{ds}}=\phi_{\text {net }}=\frac{\mathrm{q}_{\text {in }}}{\varepsilon_{\mathrm{o}}}=\frac{\mathrm{q}}{\varepsilon_{\mathrm{o}}}\)

⇒ \(\int|\vec{E}||\overrightarrow{d s}| \cos 0\) (because the E→is normal to the surface)

⇒ \(E \int \mathrm{ds}\) (because the value of E is constant at the surface) E

E (4πr²) (ds ∫total area of the spherical surface = 4 πr²)

⇒ E (4πr²) = \(\frac{q_{\text {in }}}{\varepsilon_0} \quad \Rightarrow \quad E_{\text {out }}=\frac{\mathrm{q}}{4 \pi \varepsilon_0 r^2}\)

The electric field inside a spherical shell :

Let’s choose a spherical Gaussian surface inside the shell. Applying Gauss`s theorem for this surface q, R

⇒ \(\int \vec{E} \overrightarrow{\mathrm{ds}}=\phi_{\text {net }}=\frac{\mathrm{q}_{\text {in }}}{\varepsilon_0}=0\)

⇒ \(\int|\vec{E} \| \overrightarrow{d s}| \cos 0\)

⇒ \(E \int d s\)

↓
E (4πr²) ⇒ E (4πr²) = 0 ⇒ Ein = 0

17.4.2 Electric field due to solid sphere (having uniformly distributed charge Q and radius R) :

The electric field outside the sphere :

The direction of the electric field is radially outwards, so we will choose a spherical Gaussian surface Applying Gauss`s theorem

⇒ \(\int \overrightarrow{\mathrm{E}} \overrightarrow{\mathrm{ds}}=\phi_{\text {net }}=\frac{\mathrm{q}_{\text {in }}}{\varepsilon_{\mathrm{o}}}=\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}\)

⇒ \(\int|\vec{E} \| \overrightarrow{d s}| \cos 0\)

⇒ \(\mathrm{E} \int \mathrm{ds}\)

⇒ \(E\left(4 \pi r^2\right)\)

⇒ E (4πr²) = \(\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}} \Rightarrow \mathrm{E}_{\text {out }}=\frac{\mathrm{Q}}{4 \pi \varepsilon_{\mathrm{o}} \mathrm{r}^2}\)

The electric field inside a solid sphere :

For this choose a spherical Gaussian surface inside the solid sphere Applying Gauss`s theorem to this surface

⇒ \(\int^{\downarrow}Ed s\)

⇒ \(E\left(4 \pi r^2\right) \quad \Rightarrow \quad E\left(4 \pi r^2\right)=\frac{q r^3}{\varepsilon_0 R^3}\)

⇒ \(E=\frac{q r}{4 \pi \varepsilon_0 R^3} \Rightarrow E_{\text {in }}=\frac{k Q}{R^3} r\)

17.4.3 Electric field due to infinite line charge (having uniformly distributed charged of charge density λ ) :

The electric field due to infinite wire is radial so we will choose a cylindrical Gaussian surface as shown in the figure.

⇒ \(\frac{\mathrm{q}_{\mathrm{in}}}{\varepsilon_{\mathrm{o}}}=\frac{\lambda \ell}{\varepsilon_{\mathrm{o}}}\)

φ₃ = \(\int \vec{E} \cdot \overrightarrow{d s} \quad=\int E d s=E \int d s=E(2 \pi r \ell)\)

E (2πrl) = \(\frac{\lambda \ell}{\varepsilon_o}\)

E = \(\frac{\lambda}{2 \pi \varepsilon_0 r}=\frac{2 \mathrm{k} \lambda}{\mathrm{r}}\)

17.4.4 Electric field due to infinity long charged tube (having uniform surface charge density σ and radius R)):

E outside the tube:- let’s choose a cylindrical Gaussian surface

φ_net = \(\frac{\mathrm{q}_{\text {in }}}{\varepsilon_{\mathrm{o}}}=\frac{\sigma 2 \pi \mathrm{R} \ell}{\varepsilon_{\mathrm{o}}}\)

About × 2πrl = \(\frac{\sigma 2 \pi \mathrm{R} \ell}{\varepsilon_{\mathrm{o}}}\)

E = \(\frac{\sigma \mathrm{R}}{\mathrm{r} \varepsilon_0}\)

E inside the tube :

let’s choose a cylindrical Gaussian surface inside the tube.

φ_net = \(\phi_{\text {net }}=\frac{\mathrm{q}_{\text {in }}}{\varepsilon_{\mathrm{o}}}=0 \quad \text { So } \quad \mathrm{E}_{\text {in }}=0\)

17.4.5 E due to infinitely long solid cylinder of radius R (having uniformly distributed charge in volume (charge density ρ)) :

E at outside point:-

Let’s choose a cylindrical Gaussian surface. Applying Gauss`s theorem

E × 2πrl = \(\mathrm{E} \times 2 \pi \mathrm{r} \ell=\frac{\mathrm{q}_{\text {in }}}{\varepsilon_{\mathrm{o}}}=\frac{\rho \times \pi \mathrm{R}^2 \ell}{\varepsilon_{\mathrm{o}}} \quad \Rightarrow \quad \mathrm{E}_{\text {out }}=\frac{\rho \mathrm{R}^2}{2 \mathrm{r} \varepsilon_0}\)

E at inside point :

let’s choose a cylindrical Gaussian surface inside the solid cylinder. Applying Gauss`s theorem

E × 2πrl= \(\frac{q_{\text {in }}}{\varepsilon_{\mathrm{o}}}=\frac{\rho \times \pi \mathrm{r}^2 \ell}{\varepsilon_{\mathrm{o}}} \quad \Rightarrow \quad \mathrm{E}_{\text {in }}=\frac{\rho \mathrm{r}}{2 \varepsilon_0}\)

18. Conductor

18.1 Conductor and its properties [For electrostatic condition]

• Conductors are materials that contain a large number of free electrons that can move freely inside the conductor.
• Ιn electrostatics conductors are always equipotential surfaces.
• Charge always resides on the outer surface of the conductor.

If there is a cavity inside the conductor having no charge then a charge will always reside only on the outer surface of the conductor.

• The electric field is always perpendicular to the conducting surface.
• Electric lines of force never enter into conductors.
• Electric field intensity near the conducting surface is given by the formula

⇒ \(\overrightarrow{\mathrm{E}}=\frac{\sigma}{\varepsilon_0} \hat{n}\)

⇒ \(\overrightarrow{E_A}=\frac{\sigma_{\mathrm{A}}}{\varepsilon_0} \hat{n} ; \overrightarrow{E_{\mathrm{B}}}=\frac{\sigma_{\mathrm{B}}}{\varepsilon_0} \hat{n} \text { and } \overrightarrow{\mathrm{E}_{\mathrm{C}}}=\frac{\sigma_{\mathrm{C}}}{\varepsilon_0} \hat{n}\)

When a conductor is grounded its potential becomes zero.

When an isolated conductor is grounded then its charge becomes zero.

When two conductors are connected there will be charge flow till their potential becomes equal

Electric pressure: Electric pressure at the surface of a conductor is given by formula P = \(\frac{\sigma^2}{2 \varepsilon_0}\)where σ is the local surface charge density.

18.2 Finding field due to a conductor

Suppose we have a conductor and at any ‘A’, local surface charge density = σ. We have to find an electric field just outside the conductor surface.

For this let’s consider a small cylindrical Gaussian surface, which is partly inside and partly outside the conductor surface, as shown in the figure. It has a small cross-section area ds and negligible height.

Applying Gauss’s theorem to this surface

So, \(\mathrm{Eds}=\frac{\sigma \mathrm{ds}}{\varepsilon_0} \quad \mathrm{E}=\frac{\sigma}{\varepsilon_0}\)

Electric field just outside the surface of conductor E = \(\mathrm{E}=\frac{\sigma}{\varepsilon_0}\) direction will be normal to the surface in vector form \(\overrightarrow{\mathrm{E}}=\frac{\sigma}{\varepsilon_0} \hat{\mathbf{n}}\)(here n = unit vector normal to the conductor surface)

18.3 Electrostatic pressure at the surface of the conductor

Electrostatic pressure at the surface of the conductor in \(\mathrm{P}=\frac{\sigma^2}{2 \varepsilon_0}\)

where σ = local surface charge density.

Electrostatic shielding

Consider a conductor with a cavity of any shape and size, with no charges inside the cavity. The electric field inside the cavity is zero, whatever the charge on the conductor and the external

Any cavity in a conductor remains shielded from outside electric influence: the field inside the cavity is always zero (If cava it has no charge). This is known as electrostatic shielding.

This effect can be made use of in protecting sensitive instruments from outside electrical influence.

18.4 Electric field due to a conducting and nonconducting uniformly charge infinite sheets

Example 99 Prove that if an isolated (isolated means no charges are near the sheet) large conducting sheet is given a charge then the charge distributes equally on its two surfaces.
Solution :

Let there be x charge on the left side of the sheet and Q–x charge on the right side of the sheet. Since point P lies inside the conductor so

EP= O

⇒ \(\frac{x}{2 A \varepsilon_O}-\frac{Q-x}{2 A \varepsilon_0}=0 \quad \Rightarrow \frac{2 x}{2 A \varepsilon_0}=\frac{Q}{2 A \varepsilon_0} \quad \Rightarrow x=\frac{Q}{2} \quad Q-x=\frac{Q}{2}\)

So charges are equally distributed on both sides

Example 100 If an isolated infinite sheet contains charge Q₁on its one surface and charge Q₂on its other surface then prove that electric field intensity at a point in front of the sheet will be\(\frac{Q}{2 A \varepsilon_0}\) where Q = Q₁+ Q₂

Solution:

The electric field at point P :

⇒ \(\vec{E}=\vec{E}_{Q_1}+\vec{E}_{Q_2}\)

⇒ \(\frac{Q_1}{2 A \varepsilon_0} \hat{n}+\frac{Q_2}{2 A \varepsilon_0} \hat{n}=\frac{Q_1+Q_2}{2 A \varepsilon_0} \hat{n}=\frac{Q}{2 A \varepsilon_0} \hat{n}\)

[This shows that the resultant field due to a sheet depends only on the total charge of the sheet and not on the distribution of charge on individual surfaces].

Example 101 Two large parallel conducting sheets (placed at a finite distance ) are given charges Q and 2Q respectively. Find out charges appearing on all the surfaces.

Solution :

Let there be x amount of charge on the left side of the first plate, so on its right side charge will be Q–x, similarly for the second plate there is y charge on the left side, and 2Q – y charge is on the right side of the second plate Ep= 0 ( By the property of conductor)

⇒ \(\frac{x}{2 A \varepsilon_{\mathrm{o}}}-\left\{\frac{\mathrm{Q}-\mathrm{x}}{2 \mathrm{~A} \varepsilon_{\mathrm{o}}}+\frac{\mathrm{y}}{2 \mathrm{~A} \varepsilon_{\mathrm{o}}}+\frac{2 \mathrm{Q}-\mathrm{y}}{2 \mathrm{~A} \varepsilon_{\mathrm{o}}}\right\}=0\)

we can also say that the charge on the left side of P = charge on the right side of P

x = Q – x + y + 2Q – y ⇒ x = \(x=\frac{3 Q}{2}, Q-x=\frac{-Q}{2}\)

Similarly for point Q:

x + Q – x + y = 2Q – y ⇒ y = Q/2 , 2Q – y = 3Q/2

So final charge distribution of plates is: –

Example 102 An isolated conducting sheet of area A and carrying a charge Q is placed in a uniform electric field E, such that the electric field is perpendicular to the sheet and covers all the sheets. Find out charges appearing on its two surfaces.

Solution: Let there be x charge on the left side of the plate and Q – x charge on the right side of the plate EP= 0

⇒ \(\frac{X}{2 \mathrm{~A} \varepsilon_0}+E=\frac{Q-X}{2 \mathrm{~A} \varepsilon_0}\)

⇒\(\frac{\mathrm{x}}{\mathrm{A} \varepsilon_0}=\frac{\mathrm{Q}}{2 \mathrm{~A} \varepsilon_0}-\mathrm{E} \quad \Rightarrow \quad \mathrm{x}=\frac{\mathrm{Q}}{2}-E A \varepsilon_0 \text { and } \mathrm{Q}-\mathrm{x}=\frac{\mathrm{Q}}{2}+E A \varepsilon_0\)

So charge on one side is \(\frac{Q}{2}\)– EAε_o and other side \(\frac{\mathrm{Q}}{2}+\mathrm{EA} \varepsilon_0\)

Note: Solve this question for Q = 0 without using the above answer and match that answer with the answers that you will get by putting Q = 0 in the above answer.

18.5 Some other important results for a closed conductor.

If a charge q is kept in the cavity then –q will be induced on the inner surface and +q will be induced on the outer surface of the conductor (it can be proved using Gauss theorem)

If a charge q is kept inside the cavity of a conductor and the conductor is given a charge Q then –q charge will be induced on the inner surface and the total charge on the outer surface will be q + Q. (it can be proved using Gauss theorem)

The resultant field, due to q (which is inside the cavity) and induced charge on S₁, at any point outside S₁(like B, C) is zero. The resultant field due to q + Q on and any other charge outside S₂, at any point inside of surface S₂(like A, B) is zero

The resulting field in a charge-free cavity in a closed conductor is zero. There can be charges outside the conductor and on the surface also. Then also this result is true. No charge will be induced on the innermost surface of the conductor.

Charge distribution for different types of cavities in conductors

Using the result that \(\)charges are not at the geometrical center Eresin the conducting material should be zero and using result (iii) We can show

Note: In all cases charge on inner surface S₁= –q and on outer surface S₂= q. The distribution of charge on ‘S₁’ will not change even if some charges are kept outside the conductor (i.e. outside the surface S₂). However the charge distribution on ‘S₂’ may change if some charges(s) are/are kept outside the conductor.

Sharing of charges: Two conducting hollow spherical shells of radii R₁ and R₂ having charges Q₁ and Q₂respectively and separated by a large distance, are joined by a conducting wire Let the final charges on spheres be q₁ and q₂ respectively.

Potential on both spherical shells becomes equal after joining, therefore

⇒ \(\frac{K q_1}{R_1}=\frac{K q_2}{R_2} \Rightarrow \frac{q_1}{q_2}=\frac{R_1}{R_2}\)……(i) 2

and, q₁+ q₂= Q₁+ Q₂……(ii)

from (i) and (ii) q₁= \(q_1=\frac{\left(Q_1+Q_2\right) R_1}{R_1+R_2} \quad q_2=\frac{\left(Q_1+Q_2\right) R_2}{R_1+R_2}\)

ratio of charges \(\frac{q_1}{q_2}=\frac{R_1}{R_2} \quad \Rightarrow \quad \frac{\sigma_1 4 \pi R_1^2}{\sigma_2 4 \pi R_2^2}=\frac{R_1}{R_2}\)

ratio of surface charge densities \(\frac{\sigma_1}{\sigma_2}=\frac{R_2}{R_1}}\)

Ratio of final charges \(\frac{q_1}{q_2}=\frac{R_1}{R_2}\)

Ratio of final surface charge densities.\(\frac{\sigma_1}{\sigma_2}=\frac{R_2}{R_1}\)

Example 103 The two conducting spherical shells are joined by a conducting wire and cut after some time when the charge stops flowing. Find out the charge on each sphere after that.

Solution :

After cutting the wire, the potential of both the shells is equal

Thus, potential of inner shell Vin = \(\frac{K x}{R}+\frac{K(-2 Q-x)}{2 R}=\frac{k(x-2 Q)}{2 R}\) and potential of outer shell 2R

⇒ \(V_{\text {out }}=\frac{K x}{2 R}+\frac{K(-2 Q-x)}{2 R}=\frac{-K Q}{R}\)

As Vout = Vin ⇒\(\frac{-K R}{R}=\frac{K(x-2 Q)}{2 R}\) ⇒ –2Q = x – 2Q ⇒ x = 0

So charge on inner spherical shell = 0 and outer spherical shell = – 2Q.

Example 104 Two conducting hollow spherical shells of radii R and 2R carry charges – Q and 3Q respectively. How much charge will flow into the earth if the inner shell is grounded?

Solution: When the inner shell is grounded to the Earth then the potential of the inner shell will become zero because the potential of the Earth is taken to be zero.

⇒ \(\frac{K x}{R}+\frac{K 3 Q}{2 R}=0 \quad \Rightarrow \quad x=\frac{-3 Q}{2},\), the charge that has increased

⇒ \(=\frac{-3 Q}{2}-(-Q)=\frac{-Q}{2}\)hence charge flows into the Earth = \(\frac{Q}{2}\)

Example 105 An isolated conducting sphere of charge Q and radius R is connected to a similar uncharged sphere (kept at a large distance) by using a high-resistance wire. After a long time, what is the amount of heat loss?
Solution :

When two conducting spheres of equal radius are connected charge is equally distributed on them (Result VI). So we can say that heat loss in the system

ΔH = Ui– Uf

⇒ \(=\left(\frac{\mathrm{Q}^2}{8 \pi \varepsilon_0 \mathrm{R}}-0\right)-\left(\frac{\mathrm{Q}^2 / 4}{8 \pi \varepsilon_0 \mathrm{R}}+\frac{\mathrm{Q}^2 / 4}{8 \pi \varepsilon_0 \mathrm{R}}\right)=\frac{\mathrm{Q}^2}{16 \pi \varepsilon_0 \mathrm{R}}\)

10. Van De Graff Generator

This is a machine that can build up high voltages of the order of a few million volts. The resulting large electric fields are used to accelerate charged particles (electrons, protons, ions) to high energies needed for experiments to probe the small-scale structure of matter.

Designed by R.J. Van de Graaff in 1931.

It is an electrostatic generator capable of generating a very high potential of the order of 5 × 106 V.

This high potential is used in accelerating the charged particles.

Principle :

• It is based on the following two electrostatic phenomena :
• The electric discharge takes place in air or gases readily at pointed conductors.
• If a hollow conductor is in contact with another conductor, which lies inside the hollow conductor. Then, the charge is supplied to the the inner conductor. The charge immediately shifts to the outer surface of the hollow conductor.
• Consider a large spherical conducting shell A having radius R and charge +Q, potential inside the shell is constant and it is equal to that at its surface.
• Therefore, the  potential inside the charged conducting shell A,
• \(\mathrm{V}_1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{Q}}{\mathrm{R}}\)
• Suppose that a small conducting sphere B having radius r and charge +q is placed at the center of shell A.
• Then, potential due to the sphere B at the surface of shell A,
• \(\mathrm{V}_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}}{\mathrm{R}}\)
• and potential due to the sphere B at its surface, 1 q
• \(V_3=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}\)

Thus, the total potential at the surface of shell A due to the charges Q and q,

• \(V_A=V_1+V_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{R}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{R} \quad \text { or } \quad V_A=\frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{R}+\frac{q}{R}\right)\)
• and the total potential at the surface of sphere B due to the charges Q and q,
• \(V_B=V_1+V_3=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{R}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r} \quad \text { or } \quad V_B=\frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{R}+\frac{q}{r}\right)\)
• It follows that VB> VA. Hence, the potential difference between the sphere and the shell,
• \(V=V_B-V_A=\frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{R}+\frac{q}{r}\right)-\frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{R}+\frac{\mathrm{q}}{\mathrm{R}}\right) \Rightarrow \quad V=\frac{1}{4 \pi \varepsilon_0} \cdot \mathrm{q}\left(\frac{1}{r}-\frac{1}{R}\right)\)
• It follows that the potential difference between the sphere and the shell is independent of the charge Q on the shell. Therefore, if the sphere is connected to the shell by a wire, the charge supplied to the sphere will immediately flow to the shell.
• It is because the potential of the sphere is higher than that of the shell and the charge always flows from higher to lower potential.
• It forms the basic principle of the Van de Graaff generator.

Construction :

Working :

• An endless belt of insulating material is made to run on two pulleys P₁ and P₂ with the help of an electric motor.
• The metal comb C₁, called spray comb is held potential with the help of E.H.T. source (≈ 104 V), it produces ions in its vicinity. The positive ions get sprayed on the belt due to the repulsive action of comb C₁.
• These positive ions are carried upward by the moving belt. A comb C₂, called a collecting comb is positioned near the upper end of the belt, such that the pointed ends touch the belt and the other end is in contact with the inner surface of the metallic sphere S. The comb C₂ collects the positive ions and transfers them to the metallic sphere.
• The charge transferred by the comb immediately moves onto the outer surface of the hollow sphere. As the belt goes on moving, the accumulation of positive charge on the sphere also keeps on taking place continuously and its potential rises considerably.
• With the increase of charge on the sphere, its leakage due to ionization of surrounding air also becomes faster.
• The maximum potential to which the sphere can be raised is reached when the rate of loss of charge due to leakage becomes equal to the rate at which the charge is transferred to the sphere.
• To prevent the leakage of charge from the sphere, the generator is completely enclosed inside an earth-connected steel tank, which is filled with air under pressure.
• If the charged particles, such as protons, neutrons, etc. are now generated in the discharge tube D with the lower end earthed and the upper end inside the hollow sphere, they get accelerated in a downward direction along the length of the tube. At the other end, they come to hit the target with large kinetic energy.
• Van de Graaff generator of this type was installed at the Carnegie Institute in Washington in 1937. One such generator was installed at the Indian Institute of Technology, Kanpur in 1970 and it accelerates particles to 2 MeV energy.

Problem 1. Two charges of Q each are placed at two opposite corners of a square. A charge q is placed at each of the other two corners.

1. If the resultant force on Q is zero, how are Q and q related?
2. Could q be chosen to make the resultant force on each charge zero?

Solution : (a) Let at a square ABCD charge be placed as shown

Now forces on charge Q (at point A) due to other charges are \(\) respectively shown in the figure.

Fnet on Q = \(\)(at point A)

But Fnet = 0 So, ΣFx= 0

ΣFx= – FQQ cos45° – FQq

⇒ \( \frac{K Q^2}{(\sqrt{2} a)^2} \cdot \frac{1}{\sqrt{2}}+\frac{K Q q}{a^2}=0 \quad \Rightarrow\)

For the resultant force on each charge to be zero :

From previous data, force on charge Q is zero when q = \(-\frac{Q}{2 \sqrt{2}}\) if for this value of charge q, force on q is zero then and only then the value of q exists for which the resultant force on each charge is zero.

Force on q:-

Forces on charge q (at point D) due to other three charges are \(\overrightarrow{\mathrm{F}}_{\mathrm{qQ}}, \overrightarrow{\mathrm{F}}_{\mathrm{qq}} \text { and } \overrightarrow{\mathrm{F}}_{\mathrm{qQ}}\) respectively shown in figure.

The net force on charge q:-

⇒ \(\vec{F}_{\text {net }}=\vec{F}_{\mathrm{qq}}+\overrightarrow{\mathrm{F}}_{\mathrm{qQ}}+\overrightarrow{\mathrm{F}}_{\mathrm{qQ}} \quad \text { But } \overrightarrow{\mathrm{F}}_{\text {net }}=0\)

But So, ΣFx= 0

ΣFx= \(-\frac{K q^2}{(\sqrt{2} a)^2} \cdot \frac{1}{\sqrt{2}}-\frac{K Q q}{(a)^2} \Rightarrow q=-2 \sqrt{2} Q\)

But from previous condition, q = \(-\frac{Q}{2 \sqrt{2}}\)

So, no value of q makes the resultant force on each charge zero.

Problem 2. The figure shows a uniformly charged thin non-conducting sphere of total charge Q and radius R. If point charge q is situated at point ‘A’ which is at a distance r < R from the center of the sphere then find out the following

1. Force acting on charge q.
2. The electric field at the center of the sphere.
3. The electric field at point B.

Solution :

The electric field inside a hollow sphere = 0

∴ Force on charge q.

F = qE = q × 0 = 0

The net electric field at the center of the sphere

E_net = \(\overrightarrow{\mathrm{E}}_1+\overrightarrow{\mathrm{E}}_2\)

E₁= Field due to sphere = 0

E₂= field due to this charge = \(=\frac{\mathrm{Kq}}{\mathrm{r}^2}\)

E_net = \(\frac{\mathrm{Kq}}{\mathrm{r}^2}\)

Electric field at B due to charge on sphere\(\overrightarrow{\mathrm{E}}_1=\frac{\mathrm{KQ}}{\mathrm{r}_1^2} \hat{\mathrm{r}}_1\)

and due to charge q at A ,\(\vec{E}_2=\frac{K q}{r_2{ }^2} \hat{r}_2 \quad \text { So, } \vec{E}_{\text {net }}=\vec{E}_1+\vec{E}_2 \quad=\frac{K Q}{r_1{ }^2} \hat{r}_1+\frac{K q}{r_2{ }^2} \hat{r}_2\)

where r₁= CB and r₂= AB

Problem 3. The figure shows two concentric spheres of radius and R₂(R₂> R₁) which contain uniformly distributed charges Q and –Q respectively. Find out electric field intensities at the following positions :

1. r < R₁
2. R₁≤ r < R₂
3. r ≥ R₂

Solution :

Net electric field = E₁+ E₂

E₁= field due to the sphere of radius R₁

E₂= field due to the sphere of radius R₂

E₁= 0, E₂= 0

E_net = 0

E₁= \(\frac{\mathrm{KQ}}{\mathrm{r}^2}, \mathrm{E}_2=0 \quad \Rightarrow \quad \overrightarrow{\mathrm{E}}=\frac{\mathrm{Kq}}{\mathrm{r}^2} \hat{r}\)

⇒ \(\vec{E}_1=\frac{K q}{r^2} \hat{r} \vec{E}_2=\frac{K q}{r^2}(-\hat{r}) \quad \Rightarrow \quad \vec{E}_{\text {net }}=\vec{E}_1+\vec{E}_2=0\)

Problem 4. Three identical spheres each having a charge q (uniformly distributed) and radius R, are kept in such a way that each touches the other two. Find the magnitude of the electric force on any sphere due to the other two.
Solution :

Given three identical spheres each having a charge q and radius R are kept as shown:-

For any external point; the sphere behaves like a point charge. So it becomes a triangle having point charges on its corner.

⇒ \(\left|\vec{F}_{\mathrm{qq}}\right|=\frac{\mathrm{kq}^2}{4 \mathrm{R}^2}\)

So net force (F) = 2.\(\frac{k q^2}{4 R^2} \cdot \cos \frac{60}{2}=2 \cdot \frac{k q^2}{4 R^2} \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4} \frac{k q^2}{R^2}\)

Problem 5. A uniform electric field of 10 N/C exists in the vertically downward direction. Find the increase in the electric potential as one goes up through a height of 50cm.
Solution :

E =\(-\frac{\mathrm{dv}}{\mathrm{dr}} \Rightarrow \mathrm{dv}=-\overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{r}}\)

⇒ \(\overrightarrow{\mathrm{E}}=\text { constant } \quad \Rightarrow \quad \Delta v=-\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\Delta r}\)

⇒ \(\Delta v=-10(-\hat{j}) \cdot\left(50 \times 10^{-2}\right) \hat{j}\) = 5 volts.

Problem 6. An electric field of 20 N/C exists along the x-axis in space. Calculate the potential difference V_B – V_A where the points A and B are given by –

1. A = (0,0) ; B = (4m , 2m)
2. A = (4m,2m) ; B = (6m , 5m)
3. A = (0,0) ; B = (6m , 5m)

Solution: Electric fields-axis axis mean \(\vec{E}=20 \hat{i}\)

1. \(\left|\Delta V_{A B}\right|=\vec{E} \cdot \vec{d}=20 \hat{i} \cdot 4 \hat{i}=80 \mathrm{~V} \Rightarrow \quad V_B-V_A=-80 \mathrm{~V}\)
2. \(\left|\Delta V_{B C}\right|=\vec{E} \cdot \vec{d}=20 \hat{i} \cdot 2 \hat{i}=40 \text { volt } \quad \Rightarrow \quad V_C-V_B=-40 V\)
3. \(\left|\Delta V_{A C}\right|=\vec{E} \cdot \vec{d}=20 \hat{i} \cdot 6 \hat{i}=120 \text { volt } \quad \Rightarrow \quad V_c-V_A=-120 \mathrm{~V}\)

Problem 7. A point charge of charge –q and mass m is released with negligible speed from a distance 3Ron the axis of a fixed uniformly charged ring of charge Q and radius R. Find out its velocity when it reaches the center of the ring.

Solution :

As potential due to uniform charged ring at its axis (at x distance)

⇒ \(V=\frac{k Q}{\sqrt{R^2+x^2}}\)

So potential at point A due to ring

⇒ \(V_1=\frac{k Q}{\sqrt{R^2+3 R^2}}=\frac{k Q}{2 R}\)

So potential energy of charge –q at point A

⇒ \(\text { P.E.E. }=\frac{-k Q q}{2 R}\)

and potential at point B V2= \(V_2=\frac{k Q}{R}\)

So potential energy of charge –q at point B

⇒ \(P. E_{.2}=\frac{-k Q q}{R}\)

Now by energy conservation P.E.1+ K.E.1= P.E2+ K.E2

⇒ \(\frac{-k Q q}{2 R}+0=\frac{-k Q q}{R}+\frac{1}{2} m v^2 \quad \Rightarrow \quad v^2=\frac{k Q q}{m R}\)

So the velocity of charge – q at point B v \(v=\sqrt{\frac{k Q q}{m R}}\)

Problem 8. Four small point charges each of equal magnitude q are placed at four corners of a regular tetrahedron of side a. Find out the potential energy of the charge system

Solution: Potential energy of the system :

U = U₁₂ + U₁₃ + U₁₄ + U₂₃ + U24+ U₃₄

⇒ \(U=\frac{-k q^2}{a}+\frac{k q^2}{a}+\frac{-k q^2}{a}+\frac{-k q^2}{a}+\frac{k q^2}{a}+\frac{-k q^2}{a}\)

Total potential energy of this charge system U = \(\frac{-2 k q^2}{a}\)

Problem 9. If V = x₂y + y₂z then find \(\) at (x, y, z)
Solution :

Given V = x2y + y2z and \(\vec{E}=-\frac{\partial v}{\partial r}\)

⇒ \(\overrightarrow{\mathrm{E}}=-\left[\frac{\partial \mathrm{V}}{\partial \mathbf{x}} \hat{\mathbf{i}}+\frac{\partial \mathrm{V}}{\partial \mathbf{y}} \hat{\mathbf{j}}+\frac{\partial \mathrm{V}}{\partial \mathbf{z}} \hat{\mathbf{k}}\right]\)

⇒ \(\vec{E}=-\left[2 x y \hat{i}+\left(x^2+2 y z\right) \hat{j}+y^2 \hat{k}\right]\)

Problem 10. If E = 2r2then find V(r)
Solution :

Given E = 2r2

we know that \(\int d v=-\int \vec{E} \cdot d r=-\int 2 r^2 d r\)

⇒ \(V(r)=\frac{-2 r^3}{3}+c\)

Problem 11. A charge Q is uniformly distributed over a rod of length l. Consider a hypothetical cube of edge l with the center of the cube at one end of the rod. Find the minimum possible flux of the electric field through the entire surface of the cube.
Solution :

According to Gauss law: flux depends upon charge inside the closed hypothetical surface so for minimum possible flux through the entire surface of the cube = the charge inside should be minimal.

Linear charge density of rod = \(\frac{Q}{\ell}\) and minimum length of rod inside the cube = \(\frac{\ell}{2}\)

So charge inside the cube = \(\frac{\ell}{2} \cdot \frac{Q}{\ell}=\frac{Q}{2}\)

So flux through the entire surface of the cube = \(\frac{\Sigma \mathrm{q}}{\epsilon_{\mathrm{o}}}=\frac{\mathrm{Q}}{2 \epsilon_0}\)

Problem 12. A charge Q is placed at a corner of a cube. Find the flux of the electric field through the six surfaces of the cube.

Solution :

By Gauss law, φ = \(\frac{\mathrm{q}_{\text {in }}}{\varepsilon_0}\)

Here, since Q is kept at the corner so only \(\frac{q}{8}\) charge is inside the cube. (since complete

charge can be enclosed by 8 such cubes) ∴ qin = \(\frac{Q}{8}\)

So, φ = \(\phi=\frac{q_{\text {in }}}{\varepsilon_0}=\frac{Q}{8 \varepsilon_0}\)

Problem 13. An isolated conducting sphere of charge Q and radius R is grounded by using a high-resistance wire. What is the amount of heat loss?

Solution :

When the sphere is ground, its potential becomes zero which means all charge goes to earth (due to the sphere being conducting and isolated) so all energy in the sphere is converted into heat, a total of 2

Heat loss = \(\frac{k Q^2}{2 R}\)

Problem 14. Two uncharged and parallel conducting sheets each of area A are placed in a uniform electric field E at a finite distance from each other. Such that the electric field is perpendicular to the sheets and covers all the sheets. Find out charges appearing on its two surfaces.

Solution :

Plates are conducting so the net electric field inside these plates should be zero. So, the electric field due to induced charges (on the surface of the plate) balances the outside electric field.

Here \(\overrightarrow{\mathrm{E}}_{\mathrm{i}}\) = induced electrid field

For both plates \(\overrightarrow{\mathrm{E}}_{\mathrm{i}}+\overrightarrow{\mathrm{E}}=0 \quad \Rightarrow \quad \overrightarrow{\mathrm{E}}_{\mathrm{i}}=-\overrightarrow{\mathrm{E}}\) …………….. (1)

Let charge induced on surfaces are +Q and – Q then

⇒ \(\left|\vec{E}_i\right|=\frac{Q}{A \varepsilon_0}\)

by equation (1)

⇒ \(\frac{Q}{A \varepsilon_0}=E\) ⇒ Q = AEε₀

Problem 15. A positive charge q is placed in front of a conducting solid cube at a distance d from its center. Find the electric field at the center of the cube due to the charges appearing on its surface.
Solution :

Here Ei= electric field due to induced charges

and Eq= electric field due to charge q

We know that the net electric field in a conducting cavity is equal to zero.

This means E= 0 at the center of the cube

⇒ \(\overrightarrow{\mathrm{E}}_{\mathrm{i}}+\overrightarrow{\mathrm{E}}_{\mathrm{q}}=\)

⇒ \(\vec{E}_i=-\vec{E}_i\)

⇒ \(\vec{E}_i=-\frac{k q}{d^2} \overrightarrow{\mathrm{PO}}\)

Electrostatics Multiple Choice Question And Answers

Question 1. A body can be negatively charged by

1. Giving excess electrons to it
2. Removing some electrons from it
3. Giving some protons from it
4. Removing some neutrons from it

Solution: 1. Giving excess of electrons to it

Question 2. The minimum charge on an object is

1. 1 coulomb
2. 1 stat coulomb
3. 1.6×10^-19 coulomb
4. 3.2×10^-19 coulomb

Solution: 3. 1.6×10^-19 coulomb

Question 3. A total charge Q is broken into two parts Q1and and they are placed at a distance R from each other. the maximum force of repulsion between them will occur, when

1. \(Q_2=\frac{Q}{R}, Q_1=Q-\frac{Q}{R}\)
2. \(Q_2=\frac{Q}{4}, Q=Q-\frac{2 Q}{3}\)
3. \(Q_2=\frac{Q}{4}, Q_1=\frac{3 Q}{4}\)
4. \(Q_1=\frac{Q}{2}, Q_2=\frac{Q}{2}\)

Solution: 4. \(Q_1=\frac{Q}{2}, Q_2=\frac{Q}{2}\)

Question 4. +2C and +6C two charges are repelling each other with a force of 12N. if each charge is given –2C of charge, the value of the force will be

1. 4N(Attractive)
2. 4N (Repulsive)
3. 8N (Repulsive)
4. Zero

Solution: 4. Zero

Question 5. The magnitude of elective field intensity E is such that an electron placed in it would experience an electrical force equal to its weight given by

1. age
2. \(\frac{\mathrm{mg}}{\mathrm{e}}\)
3. \(\frac{\mathrm{e}}{\mathrm{mg}}\)
4. \(\frac{e^2}{m^2} g\)

Solution: 2. \(\frac{\mathrm{mg}}{\mathrm{e}}\)

Question 6. The distance between the two charges 25μC and 36μC is 11cm At what point on the line joining the two, the intensity will be zero

1. At a distance of 5cm from 25μC
2. At a distance of 5 cm from 36μC
3. At a distance of 10cm from 25μC
4. At a distance of 11cm from 36μC

Solution: 1. At a distance of 5cm from 25μC

Question 7. A charge produces an electric field of 1 N\C at a point distant 0.1 m from it. The magnitude of the charge is

1. 1.11×10^-12C
2. 9.11×10¹²C
3. 7.11×10^-6C
4. None of these

Solution: 1. 1.11×10-12C

Question 8. The angle between the equipotential surface and lines of force is

1. Zero
2. 1800
3. 900
4. 450

Solution: 3. 900

Question 9. A charge of 5C experiences a force of 5000N when it is kept in a uniform electric field. What is the potential difference between two points separated by a distance of 1cm

1. 10 V
2. 250 V
3. 1000 V
4. 2500 V

Solution: 1. 10 V

Question 10. Two equal charges q are placed at a distance of 2a and a third charge –2q is placed at the midpoint, The potential energy of the system is

1. \(\frac{\mathrm{q}^2}{8 \pi \varepsilon_0 \mathrm{a}}\)
2. \(\frac{6 \mathrm{q}^2}{8 \pi \varepsilon_0 \mathrm{a}}\)
3. \(-\frac{7 q^2}{8 \pi \varepsilon_0 a}\)
4. \(\frac{9 q^2}{8 \pi \varepsilon_0 a}\)

Solution: 3. \(-\frac{7 q^2}{8 \pi \varepsilon_0 a}\)

Question 11. A particle of mass ‘m’ and charge ‘q’ is accelerated through a potential difference of V unit, its energy will be

1. qV
2. mqV
3. \(\left(\frac{q}{m}\right) V\)
4. \(\frac{\mathrm{q}}{\mathrm{mV}}\)

Solution: 1. qV

Question 12. When one electron is taken towards the other electron, then the electric potential energy of the system

1. Decreases
2. Increase
3. Remains unchanged
4. Becomes zero

Solution: 2. Increase

Question 13. The electric potential at a point on the axis of an electric dipole depends on the distance r of the point from the dipole as

1. \(\propto \frac{1}{r}\)
2. \(\propto \frac{1}{r^2}\)
3. ∝r
4. \(\propto \frac{1}{r^3}\)

Solution: 2. \(\propto \frac{1}{r^2}\)

Question 14. An electric dipole when placed in a uniform electric field E will have minimum potential energy if the positive direction of dipole moment makes the following angle with E

1. π
2. π/2
3. Zero
4. 3π/2

Solution: 3. Zero

Question 15. An electric dipole of moment P is placed in the position of stable equilibrium in the uniform electric field of intensity E. It is rotated through an angle θ from the initial position. the potential energy of the electric dipole in the final position is :

1. PE cos θ
2. PE sin θ
3. PE (1-cos θ)
4. – PE cos θ

Solution: 4. – PE cos θ

Question 16. The figure shows the electric lines of force emerging from a charged body. If the electric field at A and B are EA and EBrespectively and if the displacement between A and B is r then

1. E_A> E_B
2. E_A < E_B
3. E_A= E_B
4. E_A = E_B

Solution: 1. E_A> E_B

Question 17. The figure shows some of the elective field lines corresponding to an elective field. The figure suggests

1. E_A> E_B> EC
2. E_A= E_B= EC
3. E_A= E_C> E_B
4. E_A= E_C< E_B

Solution: 3. E_A= E_C> E_B

Question 18. A cylinder of radius R and length L is placed in a uniform electric field E parallel to the cylinder axis. The total flux for the surface of the cylinder is given by

1. 2πR²E
2. πR²/E
3. (πR² – πR)E
4. zero

Solution: 4. zero

Question 19. The electric field at a point varies as r0for

1. An electric dipole
2. A point charge
3. A plane infinite sheet of charge
4. A line charge of infinite length

Solution: 3. A plane infinite sheet of charge

Question 20. An electric charge q is placed at the center of a cube of side α. The electric flux on one of its faces will be

1. \(\frac{q}{6 \varepsilon_0}\)
2. \(\frac{\mathrm{q}}{\varepsilon_0 \mathrm{a}^2}\)
3. \(\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{a}^2}\)
4. \(\frac{\mathrm{q}}{\varepsilon_0}\)

Solution: 1. \(\frac{q}{6 \varepsilon_0}\)

Question 21. The total electric flux coming out of a unit positive charge put in air is

1. ε₀
2. ε₀1
3. (4πε₀)–1
4. 4πε₀ 0

Solution: 2. ε₀₁

Question 22. According to Gauss Theorem electric field of an infinitely long straight wire is proportional to r(3) 31r(4) 1r

1. r
2. \(\frac{1}{r^2}\)
3. \(\frac{1}{r^3}\)
4. \(\frac{1}{r}\)

Solution: 4. \(\frac{1}{r}\)

Question 23. The S.I unit of electric flux is

1. Weber
2. Newton per coulomb
3. Volt ×meter
4. Joule per coulomb

Solution: 3. Volt ×meter

Question 24. A metallic solid sphere is placed in a uniform elective field. The lines of force follow the path (s) shown in the  figure as

1. 1
2. 2
3. 3
4. 4

Solution: 4. 4

Question 25. Inside a hollow charged spherical conductor, the potential

1. Is Constant
2. Varies directly as the distance from the center
3. Varies inversely as the distance from the center
4. Varies inversely as the space of the distance from the center

Solution: 1. Is Constant

Question 26. If q is the charge per unit area on the surface of a conductor, then the electric field intensity at a point on the surface is

1. \(\left(\frac{\mathrm{q}}{\varepsilon_0}\right) \text { normal to surface }\)
2. \(\left(\frac{\mathrm{q}}{2 \varepsilon_0}\right) \text { normal to surface }\)
3. \(\left(\frac{\mathrm{q}}{\varepsilon_0}\right) \text { tangential to surface }\)
4. \(\left(\frac{\mathrm{q}}{2 \varepsilon_0}\right) \text { tangential to surface }\)

Solution: 1. \(\left(\frac{\mathrm{q}}{\varepsilon_0}\right) \text { normal to surface }\)

Question 27. A hollow conducting sphere of radius R has a charge (+Q) on its surface. What is the electric potential within the sphere at a distance r = \(\)From its center

1. Zero (2)
2. \(\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}\)
3. \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{r}^2}\)
4. \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{r}}\)

Solution: 3. \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{r}^2}\)

Electrostatics Exercise – 1

Section (1)  Properties Of Charge And Coulomb’s Law

Question 1. The relative permittivity of mica is :

1. One
2. Less than one
3. More then one
4. Infinite

Solution: 3. More than one

Question 2. Two identical metallic spheres are charged with 10 and -20 units of charge. If both the spheres are first brought into contact with each other and then are placed in their previous positions, then the ratio of the force in the two situations will be:-

1. -8  1
2. 1: 8
3. -2  1
4. 1: 2

Solution: 1. -8  1

Question 3. Two equal and like charges when placed 5 cm apart experience a repulsive force of 0.144 newtons. The magnitude of the charge in the microcolumn will be :

1. 0.2
2. 2
3. 20
4. 12

Solution: 1. 0.2

Question 4. Two charges of +1 μC and + 5 μC are placed 4 cm apart, the ratio of the force exerted by both charges on each other will be –

1. 1  1
2. 1: 5
3. 5: 1
4. 25: 1

Solution: 1. 1: 1

Question 5. A negative charge is placed at some point on the line joining the two +Q charges at rest. The direction of motion of the negative charge will depend upon the :

1. Position of negative charge alone
2. The magnitude of the negative charge alone
3. Both on the magnitude and position of the negative charge
4. Magnitude of positive charge.

Solution: 1. Position of negative charge alone

Question 6. A body has –80 microcoulomb of charge. Several additional electrons on it will be :

1. 8 x 10^-5
2. 80 x 10¹⁵
3. 5 x 10¹⁴
4. 1.28 x 10^-17

Solution: 3. 5 x 10¹⁴

Question 7. Coulomb’s law for the force between electric charges most closely resembles the following:

1. Law of conservation of energy
2. Newton’s law of gravitation
3. Newton’s 2nd law of motion
4. The law of conservation of charge

Solution: 2. Newton’s law of gravitation

Question 8. A charge Q₁ exerts force on a second charge Q₂. If a 3rd charge Q₃ is brought near, the force of Q₁ is exerted on Q₂.

1. Will increase
2. Will decrease
3. Will remain unchanged
4. Will increase if Q₃ is of the same sign as Q₁ and will decrease if Q₃ is of the opposite sign

Solution: 3. Will remain unchanged

Question 9. A charge particle q₁ is at position (2, – 1, 3). The electrostatic force on another charged particle q₂ at (0, 0, 0) is :

1. \(\\frac{q_1 q_2}{56 \pi \epsilon_0}(2 \hat{i}-\hat{j}+3 \hat{k})\)
2. \(\frac{q_1 q_2}{56 \sqrt{14} \pi \epsilon_0}(2 \hat{i}-\hat{j}+3 \hat{k})\)
3. \(\frac{q_1 q_2}{56 \pi \epsilon_0}(\hat{j}-2 \hat{\mathbf{i}}-3 \hat{k})\)
4. \(\frac{q_1 q_2}{56 \sqrt{14} \pi \epsilon_0}(\hat{j}-2 \hat{i}-3 \hat{k})\)

Solution: 4. \(\frac{q_1 q_2}{56 \sqrt{14} \pi \epsilon_0}(\hat{j}-2 \hat{i}-3 \hat{k})\)

Question 10. Three charges +4q, Q, and q are placed in a straight line of length l at points distance x = 0, x = l/2, and x = respectively. What should the value of Q be to make the net force on q zero?

1. –q
2. –2q
3. –q/2
4. 4q

Solution: 1. –q

Question 11. Two point charges placed at a distance r in the air exert a force F on each other. The value of distance R at which they experience force 4F when placed in a medium of dielectric constant K = 16 is :

1. r
2. r/4
3. r/8
4. 2r

Solution: 3. r/8

Question 12. Two point charges of the same magnitude and opposite sign are fixed at points A and B. A third small point charge is to be balanced at point P by the electrostatic force due to these two charges. The point P:

1. lies on the perpendicular bisector of line AB
2. At the midpoint of line AB
3. Lies to the left of A
4. None of these.

Solution: 4. None of these.

Question 13. A total charge of 20 μC is divided into two parts and placed some distance apart. If the charges experience maximum Colombian repulsion, the charges should be :

1. 5μC, 15 μC
2. 10 μC, 10 μC
3. 12 μC, 8 μC
4. \(\frac{40}{3} \mu C, \frac{20}{3} \mu C\)

Solution: 2. 10 μC, 10 μC

Question 14. Two small spherical balls each carrying a charge Q = 10 μC (10 micro-coulomb) are suspended by two insulating threads of equal lengths 1 each, from a point fixed in the ceiling. It is found that equilibrium threads are separated by an angle of 60º between them, as shown in Fig. What is the tension in the threads (Given \(\frac{1}{\left(4 \pi \varepsilon_0\right)}\)= 9 × 10⁹ Nm/C²) – 0

1. 18 N
2. 1.8 N
3. 0.18 N
4. None of the above

Solution: 2. 1.8 N

Question 15. The separation between the two charges +q and – q becomes double. The value of force will be

1. Twofoldld
2. HalFour
3. folded
4. One fourth

Solution: 4. One fourth

Question 16. The dielectric constant K of an insulator can be :

1. 5
2. 0.5
3. –1
4. zero

Solution: 1. 5

Question 17. Two spherical conductors B and C having equal radii and carrying equal charges in them repel each other with a force F when kept apart at some distance. A third spherical conductor having the same radius as that of B but uncharged is brought in contact with B, then brought in contact with C, and finally removed away from both. The new force of repulsion between B and C is

1. \(\frac{F}{4}\)
2. \(\frac{3 F}{4}\)
3. \(\frac{F}{8}\)
4. \(\frac{3 F}{8}\)

Solution: 4. \(\frac{3 F}{8}\)

Question 18. Three charges –q₁, + q_2, and –are placed as shown in the figure. The x-component of the force on –q₁ is proportional to :

1. \(\frac{q_2}{b^2}-\frac{q_3}{a^2} \cos \theta\)
2. \(\frac{q_2}{b^2}+\frac{q_3}{a^2} \sin \theta\)
3. \(\frac{q_2}{b^2}+\frac{q_3}{a^2} \cos \theta\)
4. \(\frac{q_2}{b^2}-\frac{q_3}{a^2} \sin \theta\)

Solution: 2. \(\frac{q_2}{b^2}+\frac{q_3}{a^2} \sin \theta\)

Question 19. Two spherical conductors B and C having equal radii and carrying equal charges repel each other with a force F when kept apart at some distance. A third spherical conductor having the same radius as that of B but uncharged is brought in contact with B, then brought in contact with C, and finally removed away from both. The new force of repulsion between B and C is :

1. \(\frac{F}{4}\)
2. \(\frac{3 F}{4}\)
3. \(\frac{F}{8}\)
4. \(\frac{3 F}{8}\)

Solution: 4. \(\frac{3 F}{8}\)

Question 20. Under the influence of the Coulomb field of charge +Q, a charge –q is moving around it in an elliptical orbit. Find out the correct statement(s).

1. The angular momentum of the charge –q is constant
2. The linear momentum of the charge –q is constant
3. The angular velocity of the charge – q is constant
4. The linear speed of the charge –q is constant

Solution: 1. The angular momentum of the charge –q is constant

Question 21. When the charge is given to a soap bubble, it shows :

1. An increase in size `
2. Sometimes an increase and sometimes a decrease in size
3. No change in size
4. None of these

Solution: 1. An increase in size `

Section (2): Electric Field

Question 1. If an electron is placed in a uniform electric field, then the electron will :

1. Experience no force.
2. Moving with constant velocity in the direction of the field.
3. Move with constant velocity in the direction opposite to the field.
4. Accelerate in a direction opposite to the field.

Solution: 4. Accelerate in a direction opposite to the field.

Question 2. If Q = 2 column and force on it is F = 100 newton, then the value of field intensity will be :

1. 100 N/C
2. 50 N/C
3. 200 N/C
4. 10 N/C

Solution: 2. 200 N/C

Question 3. Two infinite linear charges are placed parallel at 0.1 m apart. If each has a charge density of 5μ C/m, then the force per unit length of one of the linear charges in N/m is :

1. 2.5
2. 3.25
3. 4.5
4. 7.5

Solution: 3. 4.5

Question 4. The electric field intensity due to a uniformly charged sphere is zero :

1. At the center
2. At infinity
3. At the center and an infinite distance
4. On the surface

Solution: 3. At the center and an infinite distance

Question 5. Two spheres of radii 2 cm and 4 cm are charged equally, then the ratio of charge density on the surfaces of the spheres will be –

1. 1: 2
2. 4: 1
3. 8: 1
4. 1: 4

Solution: 2. 4: 1

Question 6. The total charge on a sphere of radii 10 cm is 1 μC. The maximum electric field due to the sphere in N/C will be –

1. 9 x 10^-5
2. 9 x 10³
3. 9 x 10⁵
4. 9 x 10¹⁵

Solution: 3. 9 x 10⁵

Question 7. A charged water drop of radius 0.1 μm is under equilibrium in some electric field. The charge on the drop is equivalent to an electronic charge. The intensity of the electric field is (g = 10 m/s²)-

1. 1.61 NC^-1
2. 26.2 NC^-1
3. 262 NC^-1
4. 1610 NC^-1

Solution: 3. 262 NC^-1

Question 8. Two large-sized charged plates have a charge density of +σ and -σ. The resultant force on the proton located midway between them will be –

1. σ ∈ e/ 0
2. σ e / 2 ∈ 0
3. 2 e/ σ ∈ 0
4. zero

Solution: 1. σ ∈ e/ 0

Question 9. Two parallel charged plates have a charge density of +σ and -σ. The resultant force on the proton located outside the plates at some distance will be –

1. 2e/ σ∈0
2. σe/∈0
3. σe / 2 ∈ 0
4. zero

Solution: 4. zero

Question 10. The charge density of an insulating infinite surface is (e/π) C/m² then the field intensity at a nearby point in volt/meter will be –

1. 2.88 x 10^-12
2. 2.88 x 10^-10
3. 2.88 x 10^-9
4. 2.88 x 10^-19

Solution: 3. 2.88 x 10^-9

Question 11. There is a uniform electric field in the x-direction. If the work done by an external agent in moving a charge of 0.2 C through a distance of 2 meters slowly along the line making an angle of 60º with x-direction is 4 joule, then the magnitude of E is :

1. 3 N / C
2. 4 N/C
3. 5 N/C
4. 20 N/C

Solution: 4. 20 N/C

Question 12. A simple pendulum has a length of L and a mass of bob m. The bob is given a charge q coulomb. The pendulum is suspended in a uniform horizontal electric field of strength E as shown in the figure, then calculate the period of oscillation when the bob is slightly displaced from its mean position is: E

1. \(2 \pi \sqrt{\frac{\ell}{g}}\)
2. \(2 \pi \sqrt{\left\{\frac{\ell}{g+\frac{q E}{m}}\right\}}\)
3. \(2 \pi \sqrt{\left\{\frac{\ell}{g-\frac{q E}{m}}\right\}}\)
4. \(2 \pi \sqrt{\frac{\ell}{\sqrt{g^2+\left(\frac{q E}{m}\right)^2}}}\)

Solution: 4. \(2 \pi \sqrt{\frac{\ell}{\sqrt{g^2+\left(\frac{q E}{m}\right)^2}}}\)

Question 13. Charge 2Q and –Q are placed as shown in the figure. The point at which electric field intensity is zero will be:

1. Somewhere between –Q and 2Q
2. Somewhere on the left of –Q
3. Somewhere on the right of 2Q
4. Somewhere on the right bisector of line joining –Q and 2Q

Solution: 2. Somewhere on the left of –Q

Question 14. The maximum electric field intensity on the axis of a uniformly charged ring of charge q and radius R will be :

1. \(\frac{1}{4 \pi \varepsilon_0} \frac{q}{3 \sqrt{3} R^2}\)
2. \(\frac{1}{4 \pi \varepsilon_0} \frac{2 q}{3 R^2}\)
3. \(\frac{1}{4 \pi \varepsilon_0} \frac{2 q}{3 \sqrt{3} R^2}\)
4. \(\frac{1}{4 \pi \varepsilon_0} \quad \frac{3 q}{2 \sqrt{3} R^2}\)

Solution: 3. \(\frac{1}{4 \pi \varepsilon_0} \frac{2 q}{3 \sqrt{3} R^2}\)

Question 15. A charged particle of charge q and mass m is released from rest in a uniform electric field E. Neglecting the effect of gravity, the kinetic energy of the charged particle after time ‘t’ seconds is

1. \(\frac{\text { Eqm }}{\mathrm{t}}\)
2. \(\frac{E^2 q^2 t^2}{2 m}\)
3. \(\frac{2 E^2 t^2}{m q}\)
4. \(\frac{E q^2 m}{2 t^2}\)

Solution: 2. \(\frac{E^2 q^2 t^2}{2 m}\)

Question 16. The electric field above a uniformly charged nonconducting sheet is E. If the nonconducting sheet is now replaced by a conducting sheet, with the charge same as before, the new electric field at the same point is :

1. 2E
2. E
3. E/2
4. None of these

Solution: 2. E

Question 17. The linear charge density on the upper half of a segment of a ring is λ and at the lower half, it is – λ. The direction of the electric field at the center O of the ring is :

1. along OA
2. along OB
3. along OC
4. along OD

Solution: 3. along OC

Question 18. The given figure gives electric lines of force due to two charges q₁ and q₂. What are the signs of the two charges?

1. imageBoth are negative
2. Both are positive
3. q₁is positive but q₂is negative
4. q₁is negative but q₂is positive

Solution: 1. Both are negative

Question 19. A positively charged pendulum is oscillating in a uniform electric field as shown in Figure. Its period of SHM is compared to that when it was uncharged. (mg > qE)

1. Will increase
2. Will decrease
3. Will not change
4. Will first increase then decrease

Solution: 1. Will increase

Question 20. A +q₁charge is at the center of an imaginary spherical Gaussian surface ‘S’, and a – q₁ charge is placed near this +q₁charge inside ‘S’. A charge +q₂is located outside this Gaussian surface. Then electric field on the Gaussian surface will be

1. Due to – q₁and q₂
2. Uniform
3. Due to all charges
4. Zero

Solution: 3. Due to all charges

Question 21. Three large parallel plates have uniform surface charge densities as shown in the figure. Find out the electric field intensity at point P.

1. \(-\frac{4 \sigma}{\epsilon_0} \hat{\mathrm{k}}\)
2. \(\frac{4 \sigma}{\epsilon_0} \hat{k}\)
3. \(-\frac{2 \sigma}{\epsilon_0} \hat{\mathrm{k}}\)
4. \(\frac{2 \sigma}{\epsilon_0} \hat{k}\)

Solution: 3. \(-\frac{2 \sigma}{\epsilon_0} \hat{\mathrm{k}}\)

Question 22. The wrong statement about electric lines of force is –

1. These originate from positive charge and end on negative charge
2. They do not intersect each other at a point
3. They have the same form for a point charge and a sphere(outside the sphere)
4. They have physical existences

Solution: 4. They have physical existences

Question 23. The insulation property of air breaks down at intensity as 3 × 10⁶ V/m. The maximum charge that can be given to a sphere of diameter 5 m is :

1. 2 × 10^-2 C
2. 2 × 10^-3 C
3. 2 × 10^-4 C
4. 0

Solution: 2. 2 × 10^-3 C

Question 24. Choose the correct statement regarding electric lines of force :

1. Emerges from (–υe) charge and meet from (+υe) charge
2. Where the electric lines of force are close electric field in that region is strong
3. Just as it is shown for a point system in the same way it represents for a solid sphere
4. Has a physical nature

Solution: 2. Where the electric lines of force are close electric field in that region is strong

Question 25. The electric field required to keep a water drop of mass m and charge e just to remain suspended is :

1. mg
2. EMG
3. mg/e
4. em/g

Solution: 3. mg/e

Question 26. Two parallel large thin metal sheets have equal surface charge densities (σ = 26.4 × 10^-12 C/m2) of opposite signs. The electric field between these sheets is

1. 1.5 N/C
2. 1.5 × 10^-10 N/C
3. 3 N/C
4. 3 × 10^-10 N/C

Solution: 3. 3 N/C

Question 27. A charged ball B hangs from a silk thread S, which makes an angle θ with a large charged conducting sheet P, as shown in the figure. The surface charge density σ of the sheet is proportional to

1. cot θ
2. cos θ
3. tan θ
4. sin θ

Solution: 3. tan θ

Question 28. The electric potential at a point in free space due to a charge Q coulomb is Q × 10¹¹ V. The electric field at that point is

1. 4π ε₀ Q × 10²² V/m
2. 12π ε₀ Q × 10²⁰ V/m
3. 4π ε₀ Q × 10²⁰ V/m
4. 12π ε₀ Q × 10²² V/m

Solution: 1. 4π ε₀ Q × 10²² V/m

Question 29. A thin conducting ring of radius R is given a charge +Q. The electric field at the center O of the ring due to the charge on the part AKB of the ring is E. The electric field at the center due to the charge on the part ACDB of the ring is

1. 3E along KO
2. E along OK
3. E along KO
4. 3 E along OK

Solution: 2. E along OK

Question 30. A charged ball B hangs from a silk thread S, which makes an angle θ with a large charged conducting sheet P, as shown in the figure. The surface charge density σ of the sheet is proportional to:

1. sin θ
2. tanθ
3. cosθ
4. cot θ

Solution: 2. tanθ

Question 31. Two point charges + 8 q and – 2q are located at x = 0 and x = L respectively. The location of a point on the x-axis at which the net electric field due to these two point charges is zero is:

1. 8L
2. 4L
3. 2L
4. L/4

Solution: 3. 2L

Question 32. Two spherical conductors A and B of radii 1 mm and 2mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is :

1. 2: 1
2. 1: 4
3. 4: 1
4. 1: 2

Solution: 1. 2: 1

Question 33. A thin spherical shell of radius R has charge Q spread uniformly over its surface. Which of the following graphs most closely represents the electric field E (r) produced by the shell in the range 0 < r < ∞, where r is the distance from the center of the shell?

Solution: 4.

Question 34. The figure shows the electric lines of force emerging from a charged body. If the electric fields at A and B are and respectively and if the distance between A and B is r, then

1. E_A< E_B
2. E_A> E_B
3. \(E_A=\frac{E_B}{r}\)
4. \(E_A=\frac{E_B}{r^2}\)

Solution: 2. E_A> E_B

Question 35. Two point charges a and b, whose magnitudes are the same are positioned at a certain distance from each other with a at the origin. The graph is drawn between electric field strength at points between a and b and distance x from a. E is taken positive if it is along the line joining from a to b. From the graph, it can be decided that

1. A is positive, B is negative
2. A and B both are positive
3. A and B both are negative
4. A is negative, B is positive

Solution: 1. A is positive, B is negative

Question 36. A wooden block performs SHM on a frictionless surface with frequency, ν0. The block carries a charge +Q on its surface. If now a uniform electric field E is switched on as shown, then the SHM of the block will be

1. Of the same frequency and with a shifted mean position.
2. Of the same frequency and with the same mean position.
3. Of changed frequency and with shifted mean position.
4. Of changed frequency and with the same mean position.

Solution: 1. Of the same frequency and with shifted mean position.

Question 37. A charged oil drop is suspended in a uniform field of 3 × 10⁴ V/m so that it neither falls nor rises. The charge on the drop will be (Take the mass of the drop = 9.9 × 10^-15 kg and g = 10 m/s2)

1. 3.3 × 10^-18C
2. 3.2 × 10^-18 C
3. 1.6 × 10^-18 C
4. 4.8 × 10^-18 C

Solution: 1. 3.3 × 10^-18 C

Question 38. Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in :

Solution: 2.

Section (3): Electric Potential And Potential Difference

Question 1. If we move in a direction opposite to the electric lines of force :

1. Electrical potential decreases.
2. Electrical potential increases.
3. Electrical potential remains uncharged
4. Nothing can be said.

Solution: 2. Electrical potential increases.

Question 2. The distance between two plates is 2 cm when an electric potential of 10 volts is applied to both plates, then the value of the electric field will be –

1. 20 N/C
2. 500 N/C
3. 5 N/C
4. 250 N/C

Solution: 2. 500 N/C

Question 3. Two objects A and B are charged with equal charge Q. The potential of A relative to B will be –

1. More
2. Equal
3. Less
4. Indefinite

Solution: 4. Indefinite

Question 4. In electrostatics the potential is equivalent to –

1. Temperature in heat
2. Height of levels in liquids
3. Pressure in gases
4. All of the above

Solution: 4. All of the above

Question 5. The potential due to a point charge at distance r is –

1. Proportional to r.
2. Inversely proportional to r.
3. Proportional to r₂.
4. Inversely proportional to r₂

Solution: 2. Inversely proportional to r.

Question 6. The dimensions of potential difference are –

1. ML²T–2Q^-1
2. MLT–2Q^-1
3. MT^-2Q^-2
4. ML2T^-2Q^-1

Solution: 1. ML2T^-2Q–1

Question 7. An object is charged with a positive charge. The potential at that object will be –

1. Positive only
2. Negative only
3. Zero always
4. May be positive, negative, or zero.

Solution: 4. May be positive, negative or zero.

Question 8. Two points (0, a) and (0, -a) have charges q and -q respectively then the electrical potential at the origin will be

1. Zero
2. Q/a
3. Q/2a
4. Q/4a₂

Solution: 1. Zero

Question 9. The charges of the same magnitude q are placed at four corners of a square of side a. The value of the potential at the center of the square will be –

1. 4kq/a
2. 4 √2kq /a
3. 4kq √2a
4. kg / a √2

Solution: 2. 4 V2kq /a

Question 10. Three equal charges are placed at the three corners of an isosceles triangle as shown in the figure. The statement which is true for electric potential V and the field intensity E at the center of the triangle –

1. V = 0, E = 0
2. V = 0, E ≠ 0
3. V ≠ 0, E = 0
4. V ≠ 0, E ≠ 0

Solution: 3. V ≠ 0, E = 0

Question 11. The potential at 0.5 Å from a proton is –

1. 0.5 volt
2. 8μ volt
3. 28.8 volt
4. 2 volt

Solution: 3. 28.8 volt

Question 12. A wire of 5 m in length carries a steady current. If it has an electric field of 0.2 V/m, the potential difference across the wire in volts will be –

1. 25
2. 0.04
3. 1.0
4. None of the above

Solution: 3. 1.0

Question 13. An infinite number of charges of equal magnitude q, but alternate charges of opposite sign are placed along the x-axis at x = 1, x = 2, x = 4, x =8,… and so on. The electric potential at the point x = 0 due to all these charges will be –

1. kq/2
2. kq/3
3. 2kq/3
4. 3kq/2

Solution: 3. 2kq/3

Question 14. The electric potential inside a uniformly positively charged non-conducting solid sphere has the value which –

1. Increase with increases in distance from the center.
2. Decreases with increases in distance from the center.
3. Is equal at all the points.
4. Is zero at all the points.

Solution: 2. Decreases with increases in distance from the center.

Question 15. Two metallic spheres which have equal charges, but their radii are different, are made to touch each other and then separated apart. The potential spheres will be –

1. Same as before
2. More for bigger
3. More for smaller
4. Equal

Solution: 4. Equal

Question 16. Two spheres of radii R and 2R are given a source equally positively charged and then connected by a long conducting wire, then the positive charge will

1. Flow from the smaller sphere to the bigger sphere
2. Flow from the bigger sphere to the smaller sphere
3. Not flow.
4. Oscillate between the spheres.

Solution: 1. Flow from the smaller sphere to the bigger sphere

Question 17. The potential difference between two isolated spheres of radii r₁ and r₂ is zero. The ratio of their charges Q₁/Q₂ will be

1. r₁/r₂
2. r₂/r₁
3. r₁₂/r₂₂
4. r₁₃/r₂₃

Solution: 1. r1/r₂

Question 18. The potential on the conducting spheres of radii r₁ and r₂ is the same, the ratio of their charge densities will be

1. r₁/r₂
2. r₂/r₁
3. r₁₂/r₂₂
4. r₂₂/r₁₂

Solution: 2. r₂/r1

Question 19. 64 charged drops coalesce to form a bigger charged drop. The potential of the bigger drop will be times that of a smaller drop –

1. 4
2. 16
3. 64
4. 8

Solution: 2. 16

Question 20. The electric potential outside a uniformly charged sphere at a distance ‘r’ is (‘a’ being the radius of the sphere)-

1. Directly proportional to a3
2. Directly proportional to r.
3. Inversely proportional to r.
4. Inversely proportional to a3.

Solution: 3. Inversely proportional to r.

Question 21. A conducting shell of radius 10 cm is charged with 3.2 x 10^-19 C. The electric potential at a distance of 4cm from its center in volt be –

1. 9 x 10^-9
2. 288
3. 2.88 x 10^-8
4. Zero

Solution: 3. 2.88 x 10^-8

Question 22. At a certain distance from a point charge the electric field is 500 V/m and the potential is 3000 V. What is the distance?

1. 6 m
2. 12 m
3. 36 m
4. 144 m

Solution: 1. 6 m

Question 23. The figure represents a square carrying charges +q, +q, –q, –q at its four corners as shown. Then the potential will be zero at points

1. A, B, C, P, and Q
2. A, B, and C
3. A, P, C, and Q
4. P, B, and Q

Solution: 2. A, B, and C

Question 24. Two equal positive charges are kept at points A and B. The electric potential at the points between A and B (excluding these points) is studied while moving from A to B. The potential

1. Continuously increases
2. Continuously decreases
3. Increases then decreases
4. Decreases than increases

Solution: 4. Decreases than increases

Question 25. A semicircular ring of radius 0.5 m is uniformly charged with a total charge of 1.5 × 10–9 coul. The electric potential at the center of this ring is :

1. 27 V
2. 13.5 V
3. 54 V
4. 45.5 V

Solution: 1. 27 V

Question 26. The kinetic energy that an electron acquires when accelerated (from rest) through a potential difference of 1 volt is called :

1. 1 joule
2. 1 electron volt
3. 1 erg
4. 1 watt

Solution: 2. 1 electron volt

Question 27. The potential difference between points A and B in the given uniform electric field is :

1. Ea
2. \(E \sqrt{\left(a^2+b^2\right)}\)
3. Eb
4. (Eb/√2)

Solution: 3. Eb

Question 28. A particle of charge Q and mass m travels through a potential difference V from rest. The final momentum of the particle is :

1. \(\frac{\mathrm{mV}}{\mathrm{Q}}\)
2. \(2 Q \sqrt{m V}\)
3. \(\sqrt{2 m Q V}\)
4. \(\sqrt{\frac{2 Q V}{m}}\)

Solution: 3. \(\sqrt{2 m Q V}\)

Question 29. If a uniformly charged spherical shell of radius 10 cm has a potential V at a point distant 5 cm from its center, then the potential at a point distant 15 cm from the center will be :

1. \(\frac{V}{3}\)
2. \(\frac{2 V}{3}\)
3. \(\frac{3}{2} \mathrm{~V}\)
4. 3V

Solution: 2. \(\frac{2 V}{3}\)

Question 30. A hollow conducting sphere of radius R has a charge (+Q) on its surface. What is the electric potential within the sphere at a distance r =\(\frac{R}{3}\)from its center

1. zero
2. \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{r}}\)
3. \(\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}\)
4. \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{r}^2}\)

Solution: 3. \(\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}\)

Question 31. Consider a thin spherical shell of radius R with its center at the origin, carrying uniform positive surface charge density. The variation of the magnitude of the electric field \(|\vec{E}(r)|\)and the electric potential V(r) with the distance r from the center, is best represented by which graph?

Solution: 4.

Question 32. The electric field at a point 20 cm away from the center of the dielectric sphere is 100 V/m, the radius of the sphere is 10 cm, then the value of the electric field at a distance 3 cm from the center is :

1. 100 V/m
2. 125 V/m
3. 120 V/m
4. 0

Solution: 4. 0

Question 33. If n drops of potential V merge, find new potential on the big drop :

1. n2/3 V
2. n1/3 V
3. nV
4. Vn/3

Solution: 1. n2/3 V

Question 34. Two conducting spheres of radii R1 and respectively are charged and joined by a wire. The ratio of electric fields of spheres is

1. \(\frac{\mathrm{R}_2^2}{\mathrm{R}_1^2}\)
2. \(\frac{R_1^2}{R_2^2}\)
3. \(\frac{R_2}{R_1}\)
4. \(\frac{R_1}{R_2}\)

Solution: 3. \(\frac{R_2}{R_1}\)

Question 35. Charge on a sphere of radius R is q and on the sphere of radius 2R is –2q. If these spheres are connected through a conducting wire then, the amount of charge flown through the wire will be :

1. \(-\frac{q}{3}\)
2. \(\frac{2 q}{3}\)
3. q
4. \(\frac{4 q}{3}\)

Solution: 4. \(\frac{4 q}{3}\)

Question 36. Two identical conducting spheres R and S have negative charges Q1and Q2respectively, but Q1 Q2. The spheres are brought to touch each other and then kept in their original positions, now the force between them is

1. Greater than that before the spheres touched
2. Less than that before the spheres touched
3. Same as before the spheres
4. Zero

Solution: 1. Greater than that before the spheres touched

Question 37. 27 smaller drops combine to form a bigger drop if the potential on a smaller drop is v then the potential on a bigger drop will be

1. 9V
2. 3V
3. 27V
4. 1/3V

Solution: 1. 9V

Question 38. A thin spherical conducting shell of radius R has a charge q. Another charge Q is placed at the center of the shell. The electrostatic potential at a point P at a distance R/2 from the center of the shell is :

1. \(\frac{2 \mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}}\)
2. \(\frac{2 Q}{4 \pi \varepsilon_0 R}-\frac{2 q}{4 \pi \varepsilon_0 R}\)
3. \(\frac{2 \mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}}+\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{R}}\)
4. \(\frac{(q+Q)}{4 \pi \varepsilon_0 R} \frac{2}{R}\)

Solution: 3. \(\frac{2 \mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}}+\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{R}}\)

Question 39. A charged oil drop is suspended in a uniform field of 3 × 104 V/m so that it neither falls nor rises. The charge on the drop will be : (take the mass of the charge = 9.9 × 10^-15 kg, g = 10m/sec2)

1. 3.3 × 10^-18 C
2. 3.2 × 10^-18 C
3. 1.6 × 10^-18 C
4. 4.8 × 10^-18 C

Solution: 1. 3.3 × 10^-18 C

Question 40. Two thin wire rings, each having a radius R are placed at a distance d apart with their axes coinciding. The charges on the two rings are + q and –q. The potential difference between the centers of the two rings is:

1. zero
2. \(\frac{q}{4 \pi \varepsilon_0}\left[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}\right]\)
3. \(\frac{q R}{4 \pi \varepsilon_0 d^2}\)
4. \(\frac{q}{2 \pi \varepsilon_0}\left[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}\right]\)

Solution: 4. \(\frac{q}{2 \pi \varepsilon_0}\left[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}\right]\)

Question 41. An electric charge of 10–3µC is placed at the origin (0, 0) of the X–Y coordinate system. Two points A and B are situated at will be (V₂, V₂ ) and (2, 0) respectively. The potential difference between the points A and B

1. 9 volt
2. zero
3. 2 volt
4. 4.5 volt

Solution: 2. zero

Question 42. Charges are placed on the vertices of a square as shown. Let Ebe be the electric field and V the potential at the center. If the charges on A and B are interchanged with those on D and C respectively, then

1. \(\text { E }\) remains unchanged, V changes
2. Both\(\text { E }\) and V change
3. \(\text { E }\) and V remain unchanged
4. \(\text { E }\) changes, V remains unchanged

Solution: 4. \(\text { E }\) changes, V remains unchanged

Question 43. A hollow uniformly charged sphere has a radius of r. If the potential difference between its surface and a point at a distance 3r from the center is V, then the electric field intensity at a distance 3r from the center is:

1. V/6r
2. V/4r
3. V/3r
4. V/2r

Solution: 1. V/6r

Question 44. A hollow sphere of radius 5 cm is uniformly charged such that the potential on its surface is 10 volts then the potential at the center of the sphere will be :

1. Zero
2. 10 volt
3. Same as at a point 5 cm away from the surface
4. Same as at a point 25 cm away from the center

Solution: 2. 10 volt

Section (4): Electric Potential Energy Of A Particle

Question 1. A nucleus has a charge of + 50e. A proton is located at a distance of 10^-12 m. The potential at this point in volt will be –

1. 14.4 x 10⁴
2. 7.2 x 10⁴
3. 7.2 x 10^-12
4. 14.4 x 10⁸

Solution: 2. 7.2 x 10⁴

Question 2. Under the influence of charge, a point charge q is carried along different paths from point A to point B, then work done will be –

1. Maximum for path four.
2. Maximum for path one.
3. Equal for all paths
4. Minimum for path three.

Solution: 3. Equal for all paths

Question 3. An electron moving in an electric potential field V1enters a higher electric potential field V2, then the change in kinetic energy of the electron is proportional to –

1. (V₂ — V₁)1/2
2. V₂ — V₁
3. (V₂ — V₁)₂
4. \(\frac{\left(V_2-V_1\right)}{V_2}\)

Solution: 2. V₂— V₁

Question 4. In the electric field of charge Q, another charge is carried from A to B. A to C, A to D, and A to E, then work done will be –

1. Minimum along path AB.
2. Minimum along path AD.
3. Minimum along path AE.
4. Zero along all the paths.

Solution: 4. Zero along all the paths.

Question 5. The work done to take an electron from rest where the potential is – 60 volts to another point where the potential is – 20 volts is given by –

1. 40 eV
2. –40 eV
3. 60 eV
4. –60 eV

Solution: 2. –40 eV

Question 6. If a charge is shifted from a low-potential region to high high-potential region. the electrical potential energy:

1. Increases
2. Decreases
3. Remains constant
4. May increase or decrease.

Solution: 4. May increase or decrease.

Question 7. A particle A has a charge +q and particle B has a charge + 4q with each of them having the same mass m. When allowed to fall from rest through the same electrical potential difference, the ratio of their speed vA: vB will be :

1. 2: 1
2. 1: 2
3. 4: 1
4. 1: 4

Solution: 2. 1: 2

Question 8. In an electron gun, electrons are accelerated through a potential difference of V volt. Taking electronic charge and mass to be respectively e and m, the maximum velocity attained by them is :

1. \({\frac{2 \mathrm{eV}}{\mathrm{m}}}\)
2. \(\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}\)
3. 2 m/eV
4. (√2/2em)

Solution: 2. \(\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}\)

Question 9. In a cathode ray tube, if V is the potential difference between the cathode and anode, the speed of the electrons, when they reach the anode is proportional to : (Assume initial velocity = 0)

1. V
2. 1/√-5
3. √V
4. √2

Solution: 3. √V

Question 10. An electron of mass m and charge e is accelerated from rest through a potential difference V in a vacuum. The final speed of the electron will be –

1. \(\mathrm{V} \sqrt{\mathrm{e} / \mathrm{m}}\)
2. \(\sqrt{\mathrm{eV} / \mathrm{m}}\)
3. \(\sqrt{2 \mathrm{eV} / \mathrm{m}}\)
4. \(2 \mathrm{eV} / \mathrm{m}\)

Solution: 3. \(\sqrt{2 \mathrm{eV} / \mathrm{m}}\)

Question 11. Positive and negative point charges of equal magnitude are kept \(\left(0,0, \frac{a}{2}\right)\)and \(\left(0,0, \frac{-a}{2}\right)\) respectively. The work done by the electric field when another positive point charge is moved from (–a, 0, 0) to (0, a, 0) is

1. Positive
2. Negative
3. Zero
4. Depends on the path connecting the initial and final positions.

Solution: 3. Zero

Question 12. If a positive charge is shifted from a low potential region to a high potential region, then electric potential energy

1. Decreases
2. Increases
3. Remains the same
4. May increase or decrease

Solution: 2. Increases

Question 13. An electron is accelerated by 1000 volts, potential difference, and its final velocity is :

1. 3.8 × 10⁷ m/s
2. 1.9 × 10⁶ m/s
3. 1.9 × 10⁷ m/s
4. 5.7 × 10⁷ m/s

Solution: 3. 1.9 × 10⁷ m/s

Question 14. As per this diagram, a point charge +q is placed at the origin O. Work done in taking another point charge –Q from the point A [co-ordiantes (o, a)] to another point B [co-ordinates(a,o)] along the straight path AB is :

1. Zero
2. \(\left(\frac{-q Q}{4 \pi \varepsilon_0} \frac{1}{\mathrm{a}^2}\right) \sqrt{2 \mathrm{a}}\)
3. \(\left(\frac{\mathrm{qQ}}{4 \pi \varepsilon_0} \frac{1}{\mathrm{a}^2}\right) \cdot \frac{\mathrm{a}}{\sqrt{2}}\)
4. \(\left(\frac{\mathrm{qQ}}{4 \pi \varepsilon_0} \frac{1}{\mathrm{a}^2}\right) \sqrt{2 a}\)

Solution: 1. Zero

Question 15. Two charges q₁ and q₂ are placed 30 cm apart, as shown in the figure. A third charge q3is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is \(\frac{\mathrm{q}_3}{4 \pi \varepsilon_0}\)k, where k is :

1. 8q₂
2. 8q₁
3. 6q₂
4. 6q₁

Solution: 1. 8q₂

Question 16. Charges +q and –q are placed at points A and B respectively which are a distance 2 L apart, C is the midpoint between A and B. The work done in moving a charge +Q along the semicircle CRD is :

1. \(\frac{\mathrm{qQ}}{4 \pi \varepsilon_0 \mathrm{~L}}\)
2. \(\frac{\mathrm{qQ}}{2 \pi \varepsilon_0 \mathrm{~L}}\)
3. \(\frac{\mathrm{qQ}}{6 \pi \varepsilon_0 \mathrm{~L}}\)
4. \(-\frac{\mathrm{qQ}}{6 \pi \varepsilon_0 \mathrm{~L}}\)

Solution: 4. \(-\frac{\mathrm{qQ}}{6 \pi \varepsilon_0 \mathrm{~L}}\)

Question 17. A charged particle ‘q’ is shot towards another charged particle ‘Q’, which is fixed, with a speed ‘v’. It approaches ‘Q’ up to the closest distance r and then returns. If q were given a speed of ‘2v’, the closest distance of approach would be :

1. r
2. 2r
3. r²
4. r⁴

Solution: 4. r⁴

Question 18. Two insulating plates are both uniformly charged in such a way that the potential difference between them is V₂– V₁= 20 V. (i.e. plate 2 is at a higher potential). The plates are separated by d = 0.1 m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1. What is its speed when it hits plate 2? (e = 1.6 × 10^-19 C, me= 9.11 × 10^-31 kg)

1. 1.87 × 10⁶ m/s
2. 32 × 10^-19m/s
3. 2.65 × 10⁶ m/s
4. 7.02 × 10¹² m/s

Solution: 3. 2.65 × 10⁶ m/s

Question 19. A particle of mass 2 g and charge 1μC is held at rest on a frictionless horizontal surface at a distance of 1 m from a fixed charge of 1 mC. If the particle is released it will be repelled. The speed of the particle when it is at a distance of 10 m from the fixed charge is:

1. 100 m/s
2. 90 m/s
3. 60 m/s
4. 45 m/s

Solution: 2. 90 m/s

Question 20. On moving a charge of 20 coulombs by 2 cm, 2 J of work is done, then the potential difference between the points is :

1. 0.1 V
2. 8 V
3. 2 V
4. 0.5 V

Solution: 1.0.1 V

Question 21. For an infinite line of charge having charge density λ lying along the x-axis, the work required in moving charge q from C to A along arc CA is :

1. \(\frac{\mathrm{q} \lambda}{\pi \varepsilon_0} \log _{\mathrm{e}} \sqrt{2}\)
2. \(\frac{\mathrm{q} \lambda}{4 \pi \varepsilon_0} \log _e \sqrt{2}\)
3. \(\frac{\mathrm{q} \lambda}{4 \pi \varepsilon_0} \log _{\mathrm{e}} 2\)
4. \(\frac{q \lambda}{2 \pi \varepsilon_0} \log _e \frac{1}{2}\)

Solution: 1. \(\frac{\mathrm{q} \lambda}{\pi \varepsilon_0} \log _{\mathrm{e}} \sqrt{2}\)

Question 22. A flat circular fixed disc has a charge +Q uniformly distributed on the disc. A charge +q is thrown with kinetic energy K, towards the disc along its axis. The charge q :

1. May hit the disc at the centre
2. May return along its path after touching the disc
3. May return along its path without touching the disc
4. Any of the above three situations is possible depending on the magnitude of K

Solution: 4. Any of the above three situations is possible depending on the magnitude of K

Section (5): Potential Energy Of A System Of Point Charge

Question 1. In the H atom, an electron is rotating around the proton in an orbit of radius r. Work done by an electron in moving once around the proton along the orbit will be –

1. ke/r
2. ke2/r²
3. 2πre
4. zero

Solution: 4. zero

Question 2. You are given an arrangement of three point charges q, 2q, and xq separated by equal finite distances so that the electric potential energy of the system is zero. Then the value of x is :

1. \(-\frac{2}{3}\)
2. \(-\frac{1}{3}\)
3. \(\frac{2}{3}\)
4. \(\frac{3}{2}\)

Solution: 1. \(-\frac{2}{3}\)

Question 3. You are given an arrangement of three point charges q, 2q, and xq separated by equal finite distances so that the electric potential energy of the system is zero. Then the value of x is :

1. \(-\frac{2}{3}\)
2. \(-\frac{1}{3}\)
3. \(\frac{2}{3}\)
4. \(\frac{3}{2}\)

Solution: 1. \(-\frac{2}{3}\)

Question 4. If a charge q is placed at the center of the line joining two equal charges Q each such that the system is in equilibrium, then the value of q is :

1. Q / 2
2. –Q/2
3. Q / 4
4. –Q/4

Solution: 4. –Q/4

Section (6): Self Energy And Energy Density

Question 1. A sphere of radius 1 cm has a potential of 8000 V. The energy density near the surface of the sphere will be:

1. 64 × 10⁵ J/m³
2. 8 × 10³ J/m³
3. 32 J/m³
4. 2.83 J/m³

Solution: 4. 2.83 J/m³

Question 2. If ‘ n ‘ identical water drops assumed spherical each charged to a potential energy U coalesce to a single drop, the potential energy of the single drop is(Assume that drops are uniformly charged):

1. n₁/3 U
2. n₂/3 U
3. n₄/3 U
4. n₅/3 U

Solution: 4. n₅/3 U

Question 3. Four charges equal to –Q each are placed at the four corners of a square and a charge q is at its center. If the system is in equilibrium, the value of q is:

1. \(-\frac{Q}{4}(1+2 \sqrt{2})\)
2. \(\frac{Q}{4}(1+2 \sqrt{2})\)
3. \(-\frac{Q}{2}(1+2 \sqrt{2})\)
4. \(\frac{Q}{2}(1+2 \sqrt{2})\)

Solution: 2. \(\frac{Q}{4}(1+2 \sqrt{2})\)

Section (7): Questions Based On Relation Between \(\overrightarrow{\mathrm{E}}\) And V :

Question 1. A family of equipotential surfaces is shown. The direction of the electric field at point A is along- –

1. AB
2. AC
3. AD
4. AF

Solution: 4. AF

Question 2. Some equipotential surfaces are shown in the figure. The magnitude and direction of the electric field is-

1. 100 V/m making angle 1200 with the x-axis
2. 100 V/m making angle 600 with the x-axis
3. 200 V/m making angle 1200 with the x-axis
4. None of the above

Solution: 3. 200 V/m making angle 1200 with the x-axis

Question 3. The variation of potential with distance r from a fixed point is shown in Figure. The electric field at r = 5 cm, is :

1. (2.5) V/cm
2. (–2.5) V/cm
3. (–2/5) cm
4. (2/5) V/cm

Solution: 1. (2.5) V/cm

Question 4. The electric field and the electric potential at a point are E and V respectively

1. If E = 0, V must be zero
2. If V = 0, E must be zero
3. If E ≠ 0, V cannot be zero
4. None of these

Solution: 4. None of these

Question 5. The electric field in a region is directed outward and is proportional to the distance r from the origin. Taking the electric potential at the origin to be zero, the electric potential at a distance r :

1. Is uniform in the region
2. Is proportional to r
3. Is proportional to r²
4. Increases as one goes away from the origin.

Solution: 3. Is proportional to r²

Question 6. V = any, then the electric field at a point will be proportional to :

1. r
2. r^-1
3. r^-2
4. r²

Solution: 1. r

Question 7. A point charge is located at O. There is a point P at a distance r from it. The electric field at point P is 500 V/m and has a potential of 3000 V. Then the value of r is

1. 6 m
2. 12 m
3. 24 m
4. 36 m

Solution: 1. 6 m

Question 8. Figure shows three points A, B, and C in a region of uniform electric field \(\overrightarrow{\mathrm{E}}\). The line AB is perpendicular and BC is parallel to the field lines. Then which of the following holds good? 

1. V_A= V_B= V_C
2. V_A= V_B> V_C
3. V_A= V_B< V_C
4. V_A> V_B= V_C

where V_A> V_Band represents the electric potential at points A, B, and C respectively.

Solution: 2. V_A= V_B> V_C

Question 9. The variation of potential with distance r from a fixed point is shown in the figure. The electric field at r = 3 cm is :

1. Zero
2. –2.5 V/cm
3. +2.5 V/cm
4. +5 V/cm

Solution: 1. Zero

Question 10. Electric potential at any point is V = –5x + 3y + √15z, then the magnitude of the electric field is –

1. 3 √2
2. 4 √2
3. 5√2
4. 7

Solution: 4. 7

Question 11. The potential at a point x (measured in µm) due to some charges situated on the x-axis is given by V(x) = 20/(x² – 4) volts. The electric field E at x = 4 μm is given by :

1. 5/3 volt/µm and in the –ve x direction
2. 5/3 volt/µm and in the +ve x direction
3. 10/9 volt/µm and in the –ve x direction
4. 10/9 volt/µm and in the +ve x direction

Solution: 4. 10/9 volt/µm and in the +ve x direction

Question 12. The electric potential V as a function of distance x (in meters) is given by V = (5x² + 10x ^-9) volt.

The value of the electric field at x = 1 m would be :

1. – 20 volt/m
2. 6 volt/m
3. 11 volt/m
4. –23 volt/m

Solution: 1. – 20 volt/m

Question 13. A uniform electric field having a magnitude and direction along a positive X-axis exists. If the electric potential V is zero at x = 0, then its value at x = +x will be :

1. V_X= xE₀
2. V_X= –xE₀
3. V_X= x²E₀
4. V_X= –x² E₀

Solution: 2. V_X= –xE₀

Question 14. A uniform electric field pointing in a positive x-direction exists in a region. Let A be the origin, B be the point on the x-axis at x = + 1 cm and C be the point on the y-axis at y = + 1 cm. Then the potentials at points A, B, and C satisfy :

V_A< V_B

V_A> V_B

V_A< V_C

V_A> V_C

Solution: 2. V_A> V_B

Question 15. A 5-coulomb charge experiences a constant force of 2000 N when moved between two points separated by a distance of 2 cm in a uniform electric field. The potential difference between these two points is:

1. 8 V
2. 200 V
3. 800 V
4. 20,000 V

Solution: 1. 8 V

Question 16. An equipotential surface and an electric line of force :

1. Never intersect each other
2. Intersect at 45º
3. Intersect at 60º
4. Intersect at 90º

Solution: 4. Intersect at 90º

Section (8): Dipole

Question 1. If an electric dipole is kept in a non-uniform electric field, then it will experience –

1. Only torque
2. No torque
3. A resultant force and a torque
4. Only a force

Solution: 3. A resultant force and a torque

Question 2. The force on a charge situated on the axis of a dipole is F. If the charge is shifted to double the distance, the acting force will be –

1. 4F
2. F/2
3. F/4
4. F/8

Solution: 4. F/8

Question 3. A dipole of dipole moment p is placed in an electric field \(\vec{E}\) and is in stable equilibrium. The torque required to rotate the dipole from this position by angle θ will be

1. pE cos θ
2. pE sin θ
3. pE tan θ
4. –pE cosθ

Solution: 2. pE sin θ

Question 4. The electric potential at a point due to an electric dipole will be –

1. \(\frac{k(\vec{p} \cdot \vec{r})}{r^3}\)
2. \(\frac{k(\vec{p} \cdot \vec{r})}{r^2}/\)
3. \(\frac{k(\vec{p} \times \vec{r})}{r}\)
4. \(\frac{k(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{r}})}{r^2}\)

Solution: 1. \(\frac{k(\vec{p} \cdot \vec{r})}{r^3}\)

Question 5. The ratio of electric fields due to an electric dipole on the axis and the equatorial line at an equal distance will be –

1. 4: 1
2. 1: 2
3. 2: 1
4. 1: 1

Solution: 3. 2: 1

Question 6. An electric dipole is made up of two equal and opposite charges of 2 x 10^-6 coulomb at a distance of 3 cm. This is kept in an electric field of 2 x 105 N/C, then the maximum torque acting on the dipole –

1. 12 x 10^-1 Nm
2. 12 x 10^-3 Nm
3. 24 x 10^-3 Nm
4. 24 x 10^-1 Nm

Solution: 2. 12 x 10^-3 Nm

Question 7. The distance between two singly ionized atoms is 1Å. If the charge on both ions is equal and opposite then the dipole moment in coulomb-metre is –

1. 1.6 × 10^-29
2. 0.16 × 10^-29
3. 16 × 10^-29
4. 1.6 × 10^-29/ 4πε₀

Solution: 1. 1.6 × 10^-29

Question 8. The electric potential in volts at a distance of 0.01 m on the equatorial line of an electric dipole of dipole moment p is –

1. \(\mathrm{p} / 4 \pi \epsilon_0 \times 10^{-4}\)
2. zero
3. \(4 \pi \epsilon_0 \mathrm{p} \times 10^{-4}\)
4. \(4 \pi \epsilon_0 / p \times 10^{-4}\)

Solution: 2. zero

Question 9. The electric potential in volts due to an electric dipole of dipole moment 2 x 10^-8 C-m at a distance of 3m on a line making an angle of 600 with the axis of the dipole is –

1. 0
2. 10
3. 20
4. 40

Solution: 2. 10

Question 10. A dipole of electric dipole moment P is placed in a uniform electric field of strength E. If θ is the angle between positive directions of P and E, then the potential energy of the electric dipole is largest when θ is :

1. zero
2. π/2
3. π
4. π/4

Solution: 3. π

Question 11. Potential due to an electric dipole at some point is maximum or minimum when the axis of the dipole and line joining point and dipole are at angles respectively :

1. 90° and 180°
2. 0° and 90°
3. 90° and 0°
4. 0° and 180°

Solution: 4. 0° and 180°

Question 12. The electric field on the axis of an electric dipole, at a distance of r from its center, is E. If the dipole is rotated through 90°; then the electric field intensity at the same point will be :

1. E
2. \(\frac{E}{4}\)
3. \(\frac{E}{2}\)
4. 2E

Solution: 3. \(\frac{E}{2}\)

Question 13. An electric dipole is placed along a North-South direction in a sphere filled with water. Which statement is true?

1. Electric flux is coming towards the sphere.
2. Electric flux is going out of the sphere
3. As much electric flux is going out of the sphere, as much is coming toward the sphere.
4. Water does not allow electric flux to come inside the sphere

Solution: 3. As much electric flux is going out of the sphere, as much is coming toward the sphere.

Question 14. At the equator of an electric dipole, the angle between the electric dipole moment and an electric field is :

1. 0°
2. 90°
3. 180°
4. None of these

Solution: 3. 180°

Question 15. The potential of the dipole at its axial position is proportional to distance r as :

1. r^-2
2. r^-1
3. r
4. r0

Solution: 1. r^-2

Question 16. An electric dipole has the magnitude of its charge as q and its dipole moment is p. It is placed in a uniform electric field E. If its dipole moment is along the direction of the field, the force on it and its potential energy are respectively :

1. 2qE and minimum
2. qe and pE
3. zero and minimum
4. qE and maximum

Solution: 3. zero and minimum

Question 17. An electric dipole of moment dipole by 90° is pis lying along a uniform electric field \(\vec{E}\). The work done in rotating the

1. √2pE
2. \(\frac{\mathrm{pE}}{2}\)
3. 2pE
4. pE

Solution: 4. pE

Question 18. Three point charges +q, –2q and + q are placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = a, y = 0, z = 0), respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are :

1. v2qa along + y direction
2. 2qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)
3. qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)
4. 2qa along + x direction

Solution: 2. 2qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)

Question 19. An electric dipole is placed at an angle of 30 to a non-uniform electric field. The dipole will experience

1. A torque as well as a translational force.
2. A torque only.
3. A translational force only in the direction of the field.
4. A translational force is only in a direction normal to the direction of the field.

Solution: 1. A torque as well as a translational force.

Question 20. Due to an electric dipole shown in fig., the electric field intensity is parallel to the dipole axis :

1. At P only
2. A Q only
3. Both at P and at Q
4. Neither at P nor at Q

Solution: 3. Both at P and at Q

Question 21. An electric dipole consists of two opposite charges each of magnitude 1.0 μC, separated by a distance of 2.0 cm. The dipole is placed in an external electric field of 1.0 × 10⁵ N/C. The maximum torque on the dipole is :

1. 0.2 × 10^-3 N-m
2. 1.0 × 10^-3 N-m
3. 2.0 × 10^-3 N-m
4. 4.0 × 10^-3 N-m

Solution: 3. 2.0 × 10^-3 N-m

Question 22. Two opposite and equal charges of magnitude 4 × 10^-8 coulomb each when placed 2 × 10^-5 cm apart form a dipole. If this dipole is placed in an external electric field of 4 × 10⁸ N/C, the value of maximum torque and the work required in rotating it through 180º from its initial orientation which is along the electric field will be : (Assume rotation of dipole about an axis passing through the center of the dipole):

1. 64 × 10^-4 N-m and 44 × 10^-4 J
2. 32 × 10^-4 N-m and 32 × 10^-4J
3. 64 × 10^-4 N-m and 32 × 10^-4 J
4. 32 × 10^-4 N-m and 64 × 10^-4 J

Solution: 4. 32 × 10^-4 N-m and 64 × 10^-4J

Question 23. At a point on the axis (but not inside the dipole and not at infinity) of an electric dipole

1. The electric field is zero
2. The electric potential is zero
3. Neither the electric field nor the electric potential is zero
4. The electric field is directed perpendicular to the axis of the dipole

Solution: 3. Neither the electric field nor the electric potential is zero

Section (9): Flux Calculation And Gauss’s Law

Question 1. For an electrostatic system which of the statements is always true :

1. Electric lines are parallel to metallic surfaces.
2. The electric field inside a metallic surface is zero.
3. Electric lines of force are perpendicular to the equipotential surface.
   1. (1) and (2) only
   2. (2) and (3) only
   3. (1) and (3) only
   4. (1), (2), and (3)

Solution: 3. (1) and (3) only

Question 2. Total flux coming out of some closed surface is :

1. q/ε₀
2. ε₀/q
3. qε₀
4. Vq/ ε0

Solution: 4. Vq/ ε₀

Question 3. Three charges q1= 1 × 10^-6, q2= 2 × 10^-6, q3= –3 × 10^-6 C have been placed, as shown in the figure, in four surfaces S₁, S₂, S₃and S₄electrical flux emitted from the surface S₂ in N–m2/C will be –

1. 36π × 10³
2. –36π × 10³
3. 36π × 10⁹
4. –36π × 10⁹

Solution: 2. –36π × 10³

Question 4. The intensity of an electric field at some point distant r from the axis of an infinitely long pipe having charges per unit length as q will be :

1. Proportional to r2
2. Proportional to r3
3. Inversely proportional to r.
4. Inversely proportional to r2.

Solution: 1. Proportional to r2

Question 5. Eight charges, 1μC, -7μC, -4μC, 10μC, 2μC, -5μC, -3μC and 6μC are situated at the eight corners of a cube of side 20 cm. A spherical surface of radius 80 cm encloses this cube. The center of the sphere coincides with the center of the cube. Then the total outgoing flux from the spherical surface (in units of volt meter) is-

1. 36π x 10³
2. 684π x 10³
3. zero
4. None of the above

Solution: 3. zero

Question 6. A closed cylinder of radius R and length L is placed in a uniform electric field E, parallel to the axis of the cylinder. Then the electric flux through the cylinder must be –

1. 2πR²E
2. (2πR² + 2πRL)E
3. 2πRLE
4. zero

Solution: 4. zero

Question 7. A charge q is placed at the center of the cubical vessel (with one face open) as shown in the figure. The flux of the electric field through the surface of the vessel is

1. zero
2. q/ε₀
3. \(\frac{\mathrm{q}}{4 \varepsilon_0}\)
4. 5q/6ε₀

Solution: 4. 5q/6ε₀

Question 8. Electric charge is uniformly distributed along a long straight wire of radius 1 mm. The charge per cm length of the wire is Q coulomb. Another cylindrical surface of radius 50 cm and length 1m symmetrically encloses the wire as shown in Fig. The total electric flux passing through the cylindrical surface is –

1. \(\frac{\mathrm{Q}}{\varepsilon_0}/\)
2. \(\frac{100 Q}{\varepsilon_0}\)
3. \(\frac{10 Q}{\left(\pi \varepsilon_0\right)}\)
4. \(\frac{100 Q}{\left(\pi \varepsilon_0\right)}\)

Solution: 2. \(\frac{100 Q}{\varepsilon_0}\)

Question 9. If the geometric axis of the cylinder is parallel to the electric field, the flux through the cylinder will be –

1. 2πr × B
2. πr² × B
3. 2πr² × B
4. 0

Solution: 4. 0

Question 10. A cubical box contains charge +Q at its center. The total electric flux emerging from the box is :

1. \(\frac{\mathrm{Q}}{\varepsilon_0}\)
2. \(\frac{\mathrm{Q}}{6 \varepsilon_0}\)
3. \(\frac{\mathrm{Q}}{4 \varepsilon_0}\)
4. \(\varepsilon_0 \mathrm{Q}\)

Solution: 1. \(\frac{\mathrm{Q}}{\varepsilon_0}\)

Question 11. If a square coil is making an angle of 60° with electric field E according to the figure, the electric flux passing through the square coil is (the side of the square is 4 cm) :

1. 853 E
2. 8E
3. 16 E
4. None of these

Solution: 2. 8E

Question 12. Four equal charges q are placed at the center of a conducting hollow sphere. If they are displaced 1.5 cm from the center, the change in flux will be (Radius of sphere> 1.5 cm) :

1. Doubled
2. Rripled
3. Constant
4. None of these

Solution: 4. None of these

Question 13. If the electric flux entering and leaving an enclosed surface respectively is φ1and φ2, the electric charge inside the surface will be

1. (φ2– φ1) ε₀
2. (φ1+ φ2)/ε₀
3. (φ2– φ1)/ ε₀
4. (φ1+ φ2) ε₀

Solution: 1. (φ2– φ1) ε₀

Question 14. A square surface of side L m is in the plane of the paper. A uniform electric field \(\overrightarrow{\mathrm{E}}\)(V/m), also in the plane of the paper, is limited only to the lower half of the square surface, (see figure). The electric flux in SI units associated with the surface is :

1. EL2/ (2ε₀)
2. EL2/ 2
3. zero
4. EL2

Solution: 3. zero

Question 15. If the electric flux entering and leaving an enclosed surface respectively is φ1and φ2, the electric charge inside the surface will be :

1. (φ2– φ1)ε₀
2. (φ1+ φ2)/ε₀
3. (φ2– φ1)/ε₀
4. (φ1+ φ2) ε₀

Solution: 1. (φ2– φ1)ε₀

Question 16. An electric dipole is placed at the center of a sphere. Mark the correct options.

1. The electric field is zero at every point of the sphere.
2. The flux of the electric field through the sphere is non-zero.
3. The electric field is zero on a circle on the sphere.
4. The electric field is not zero anywhere on the sphere.

Solution: 4. The electric field is not zero anywhere on the sphere.

Question 17. A charge Q is placed at a distance of 4R above the center of a disc of radius R. The magnitude of flux through the disc is φ. Now a hemispherical shell of radius R is placed over the disc such that it forms a closed surface. The flux through the curved surface (taking the direction of the area vector along outward normal as positive), is –

1. Zero
2. φ
3. – φ
4. 2φ

Solution: 3. – φ

Question 18. A charge q is placed at the corner of a cube of side a. The electric flux through the cube is :

1. \(\frac{q}{\varepsilon_0}\)
2. \(\frac{q}{3 \varepsilon_0}\)
3. \(\frac{q}{6 \varepsilon_0}\)
4. \(\frac{q}{8 \varepsilon_0}\)

Solution: 4. \(\frac{q}{8 \varepsilon_0}\)

Question 19. A charge qμC is placed at the center of a cube of a side 0.1 m, then the electric flux diverging from each face of the cube is :

1. \(\frac{\mathrm{q} \times 10^{-6}}{24 \varepsilon_0}\)
2. \(\frac{\mathrm{q} \times 10^{-4}}{\varepsilon_0}\)
3. \(\frac{\mathrm{q} \times 10^{-6}}{6 \varepsilon_0}\)
4. \(\frac{\mathrm{q} \times 10^{-4}}{12 \varepsilon_0}\)

Solution: 3. \(\frac{\mathrm{q} \times 10^{-6}}{6 \varepsilon_0}\)

Question 20. Gauss law is given by ∈0 \(\int \vec{E} \cdot \overrightarrow{d s}=q\) if net charge enclosed by Gaussian surface is zero then –

1. E on the surface must be zero
2. Incoming and outgoing electric lines are equal
3. There is a net incoming electric flux
4. None

Solution: 2. Incoming and outgoing electric lines are equal

Section (10): Conductor, Its Properties & Electric Pressure

Question 1. The electric field near the conducting surface of a uniform charge density σ will be –

1. σ / ∈0and parallel to the surface.
2. 2σ /∈0and parallel to surface.
3. σ / ∈0and perpendicular to the surface.
4. 2σ / ∈0 and perpendicular to the surface.

Solution: 3. σ / ∈0and perpendicular to the surface.

Question 2. An uncharged conductor A is brought close to another positive charged conductor B, then the charge on B –

1. Will increase but potential will be constant.
2. Will be constant but the potential will increase
3. Will be constant but the potential decreases.
4. The potential and charge on both are constant.

Solution: 3. Will be constant but potential decreases.

Question 3. The fig. shows lines of constant potential in a region in which an electric field is present. The value of the potential is written in brackets of the points A, B, and C, the magnitude of the electric field is greatest at point –

1. A
2. B
3. C
4. A and C

Solution: 2. B

Question 4. The electric charge in uniform motion produces –

1. An electric field only
2. A magnetic field only
3. Both electric and magnetic fields
4. Neither electric nor magnetic fields

Solution: 3. Both electric and magnetic fields

Question 5. Which of the following represents the correct graph for electric field intensity and the distance r from the center of a hollow charged metal sphere or solid metallic conductor of radius R :

Solution: 4.

Question 6. A neutral metallic object is placed near a finite metal plate carrying a positive charge. The electric force on the object will be :

1. Towards the plate
2. A way from the plate
3. Parallel to the plate
4. Zero

Solution: 1. Towards the plate

Question 7. The figure shows a thick metallic sphere. If it is given a charge +Q, then an electric field will be present in the region

1. r < R₁only
2. r > R₁and R₁< r < R₂
3. r ≥R₂only
4. r≤R₂only

Solution: 3. r ≥R2only

Question 8. An uncharged sphere of metal is placed in a uniform electric field produced by two large conducting parallel plates having equal and opposite charges, then lines of force look like

Solution: 3.

Question 9. You are traveling in a car during a thunderstorm, to protect yourself from lightning would you prefer to :

1. Remain in the car
2. Take shelter under a tree
3. Get out and be flat on the ground
4. Touch the nearest electrical pole

Solution: 1. Remain in the car

Question 10. The amount of work done in Joules in carrying a charge +q along the closed path PQRSP between the oppositely charged metal plates is (where E is the electric field between the plates)

1. zero
2. q
3. qE (PQ + QR + SR + SP)
4. q\ε₀

Solution: 1. zero

Question 11. The figure shows a closed surface that intersects a conducting sphere. If a positive charge is placed at the point P, the flux of the electric field through the closed surface

1. Will remain zero
2. Will become positive
3. Will become negative
4. Will become undefined

Solution: 2. Will become positive

Question 12. A charge ‘ q ‘ is placed at the center of a conducting spherical shell of radius R, which is given a charge Q. An external charge Q′ is also present at distance R′ (R′ > R) from ‘ q ‘. Then the resultant field will be best represented for region r < R by: [ where r is the distance of the point from q ]

Solution: 1.

Question 13. In the above question, if Q’ is removed then which option is correct :

Solution: 1.

Question 14. The net charge is given to an isolated conducting solid sphere:

1. must be distributed uniformly on the surface
2. may be distributed uniformly on the surface
3. must be distributed uniformly in the volume
4. may be distributed uniformly in the volume.

Solution: 1. must be distributed uniformly on the surface

Question 15. The net charge is given to a solid insulating sphere:

1. must be distributed uniformly in its volume
2. may be distributed uniformly in its volume
3. must be distributed uniformly on its surface
4. the distribution will depend upon whether other charges are present or not.

Solution: 2. may be distributed uniformly in its volume

Question 16. A charge Q is kept at the center of a conducting sphere of inner radius and outer radius R2. A point charge q is kept at a distance r (> R2) from the center. If q experiences an electrostatic force of 10 N then assuming that no other charges are present, the electrostatic force experienced by Q will be:

1. –10 N
2. 0
3. 20 N
4. None of these

Solution: 2. 0

Question 17. A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge –3Q, the new potential difference between the same two surfaces is :

1. V
2. 2V
3. 4V
4. –2V

Solution: 1. V

Question 18. A point charge ‘ q ‘ is placed at a point inside a hollow conducting sphere. Which of the following electric force patterns is correct?

Solution: 1.

Question 19. Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then,

1. Negative and distributed uniformly over the surface of the sphere
2. Negative and appears only at the point on the sphere closest to the point charge
3. Negative and distributed non-uniformly over the entire surface of the sphere
4. Zero

Solution: 4. Zero

Question 20. Three concentric metallic spherical shells of radii R, 2R, and 3R, are given charges Q₁, Q₂, and Q₃, respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells, Q₁: Q₂: Q₃, is

1. 1: 2 : 3
2. 1 : 3: 5
3. 1: 4: 9
4. 1: 8: 18

Solution: 2. 1 : 3: 5

Question 21. A positive point charge q is brought near a neutral metal sphere.

1. The sphere becomes negatively charged.
2. The sphere becomes positively charged.
3. The interior remains neutral and the surface gets non-uniform charge distribution.
4. The interior becomes positively charged and the surface becomes negatively charged.

Solution: 3. The interior remains neutral and the surface gets non-uniform charge distribution.

Question 22. Two small conductors A and B are given charges q₁ and respectively. Now they are placed inside a hollow metallic conductor (C) carrying a charge Q. If all the three conductors A, B, and C are connected by conducting wires as shown, the charges on A, B, and C will be respectively:

1. \( \frac{q_1+q_2}{2}, \frac{q_1+q_2}{2}, Q\)
2. \(\frac{Q+q_1+q_3}{3}, \frac{Q+q_1+q_2}{3}, \frac{Q+q_1+q_2}{3}\)
3. \(\frac{q_1+q_2+Q}{2}, \frac{q_1+q_2+Q}{2}, 0\)
4. 0, 0, Q + q₁ + q₂

Solution: 4. 0, 0, Q + q₁+ q₂

Question 23. A charge Q is kept at the center of a conducting sphere of inner radius R₁ and outer radius R₂. A point charge q is kept at a distance r (> R2) from the center. If q experiences an electrostatic force of 10 N then assuming that no other charges are present, the electrostatic force experienced by Q will be:

1. –10 N
2. 0
3. 20 N
4. None of these

Solution: 2. 0

Question 24. Some charge is being given to a conductor then its potential is :

1. Maximum at surface
2. Maximum at centre
3. The same throughout the conductor
4. Maximum somewhere between surface and center

Solution: 3. Same throughout the conductor

Question 25. A solid metallic sphere has a charge of +3Q. Concentric with this sphere is a conducting spherical shell having charge –Q. The radius of the sphere is a and that of the spherical shell is b(>a). What is the electric field at a distance r(a < r < b) from the center?

1. \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{r}}\)
2. \(\frac{1}{4 \pi \varepsilon_0} \frac{3 Q}{r}\)
3. \(\frac{1}{4 \pi \varepsilon_0} \frac{3 \mathrm{Q}}{\mathrm{r}^2}\)
4. \(\frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2}\)

Solution: 3. \(\frac{1}{4 \pi \varepsilon_0} \frac{3 \mathrm{Q}}{\mathrm{r}^2}\)

Question 26. Two charged spheres having radii a and b are joined with a wire then the ratio of electric field Ea/Ebon on their surface is –

1. a/b
2. b/a
3. a²/b²
4. b²/a²

Solution: 2. b/a

Question 27. A long hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of a larger radius. Both the cylinders are initially electrically neutral.

1. A potential difference appears between the two cylinders when a charge density is given to the inner cylinder.
2. A potential difference appears between the two cylinders when a charge density is given to the outer cylinder.
3. No potential difference appears between the two cylinders when a uniform line charge is kept along the axis of the cylinders.
4. No potential difference appears between the two cylinders when the same charge density is given to both cylinders.

Solution: 1. A potential difference appears between the two cylinders when a charge density is given to the inner cylinder.

Electrostatics Exercise – 2

Question 1. A dipole having dipole moment p is placed in front of a solid uncharged conducting sphere as shown in the diagram. The net potential at point A lying on the surface of the sphere is :

1. \(\frac{k p \cos \phi}{r^2}\)
2. \(\frac{\mathrm{kpcos}^2 \phi}{\mathrm{r}^2}\)
3. Zero
4. \(\frac{2 \mathrm{kp} \cos ^2 \phi}{\mathrm{r}^2}\)

Solution: 2. \(\frac{\mathrm{kpcos}^2 \phi}{\mathrm{r}^2}\)

Question 2. Two equal charges are separated by a distance d. A third charge placed on a perpendicular bisector at x distance will experience maximum coulomb force when –

1. x = d / √2
2. x = d/2
3. x = d/2√2
4. x = d/2 v3

Solution: 3. x = d/2 √2

Question 3. The work done in placing four charges at the corners of a square as shown in the figure, will be –

1. \( (4-\sqrt{2}) \frac{K q^2}{a}\)
2. \((4+\sqrt{2}) \frac{K q^2}{\mathrm{a}}\)
3. \((4-\sqrt{2}) \frac{K q^2}{a^2}\)
4. \((4+\sqrt{2}) \frac{K q^2}{\mathrm{a}^2}\)

Solution: 1. \( (4-\sqrt{2}) \frac{K q^2}{a}\)

Question 4. Six charges q,q,q, – q, –q, and –q to be arranged on the vertices of a regular hexagon PQRSTU such that the electric field at the center is double the field produced when only charge ‘q’ is placed at vertex R. The sequence of the charges from P to U is

1. q, –q, q, q, –q, –q
2. q, q, q, –q, –q, –q
3. –q, q, q, –q, –q, q
4. –q, q, q, q, –q, –q

Solution: 1. q, –q, q, q, –q, –q

Question 5. Which of the following groups do not have the same dimensions

1. Young’s modulus, pressure, stress
2. work, heat, energy
3. Electromotive force, potential difference, voltage
4. Electric dipole, electric flux, electric field

Solution: 4. Electric dipole, electric flux, electric field

Question 6. A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is

1. Zero everywhere
2. Is not zero but uniform
3. Nonuniform
4. Is zero at the center only

Solution: 2. Is not zero but uniform

Question 7. Statement -1: For practical purposes, the earth is used as a reference at zero potential in electrical circuits.

Statement -2: The electrical potential of a sphere of radius R with charge Q uniformly distributed on the surface is given by \(\)

1. Statement -1 is True, Statement -2 is True; Statement -2 is a correct explanation for Statement -1
2. Statement -1 is True, Statement -2 is True; Statement -2 is NOT a correct explanation for Statement -1
3. Statement -1 is True, Statement -2 is False
4. Statement -1 is False, Statement -2 is True.

Solution: 2. Statement -1 is True, Statement -2 is True; Statement -2 is NOT a correct explanation for Statement -1

Question 8. Which of the following statement(s) is/are correct?

1. If the electric field due to a point charge varies as r –2.5 instead of r –2, then the Gauss law will still be valid.
2. The Gauss law can be used to calculate the field distribution around an electric dipole.
3. If the electric field between two point charges is zero somewhere, then the sign of the two charges is the same.
4. The work done by the external force in moving a unit positive charge from point A at potential VA to point B at potential VBis (VB — VA).

Solution: 3. If the electric field between two point charges is zero somewhere, then the sign of the two charges is the same.

Question 9. Let E1= x ˆ i+ y ˆ j ,and E2= xy2 ˆ i+ x2y ˆ j, then :

1. Represents a constant electric field
2. Represents a constant electric field
3. Both represent a constant electric field
4. None of these

Solution: 4. None of these

Question 10. When a glass rod is rubbed with silk, the amount of positive charge acquired by the glass rod in magnitude is :

1. Less than the charge for silk
2. Greater than the charge on silk
3. Equal to the charge on silk
4. None of these

Solution: 3. Equal to the charge on silk

Question 11. A cube has point charges of magnitude – q at all its vertices. The electric field at the center of the cube is :

1. \(1) \frac{1}{4 \pi \varepsilon_0} \frac{6 q}{3 a^2}\)
2. \(\frac{1}{4 \pi \varepsilon_0} \frac{8 \mathrm{q}}{\mathrm{a}^2}\)
3.  zero
4. \(\frac{1}{4 \pi \varepsilon_0} \frac{-8 \mathrm{q}}{\mathrm{a}^2}\)

Solution: 3.

Question 12. Three point charges +q, – 2q and +q placed at points (x = 0, y = a, z = 0), (x = 0, y = 0, z = 0) and (x = a, y = 0, z = 0), respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are

1. √2qa along +y direction
2. √2qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0) (2)
3. qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0)
4. √2qa along +x direction

Solution: 2. v2qa along the line joining points (x = 0, y = 0, z = 0) and (x = a, y = a, z = 0) (2)

Question 13. An electric dipole is placed along the x-axis at the origin O. A point P is at a distance of 20 cm from this origin such that OP makes an angle π/3 with the x-axis. If the electric field at P makes an angle θ with the x-axis, the value of θ would be

1. \(\frac{\pi}{3}\)
2. \(\frac{\pi}{3}+\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
3. \(\frac{2 \pi}{3}\)
4. \(\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)

Solution: 2. \(\frac{\pi}{3}+\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)

Question 14. A Charged wire is bent in the form of a semi-circular arc of radius a. If charge per unit length is λ coulomb/meter, the electric field at the center O is :

1. Zero
2. \(\frac{\lambda}{2 \pi \mathrm{a}^2 \varepsilon_0}\)
3. \(\frac{\lambda}{4 \pi^2 \varepsilon_0 a}\)
4. \(\frac{\lambda}{2 \pi \varepsilon_0 \mathrm{a}}\)

Solution: 3. \(\frac{\lambda}{4 \pi^2 \varepsilon_0 a}\)

Question 15. The dimensions of \(\)(ε₀: permittivity of free space; E: electric field) are: 2

1. M L T^-1
2. M L² T^-2
3. M L T^-2
4. M L^-1 T^-2

Solution: 4. M L^-1T^-2

Question 16. Two non–non-conducting spheres of radii R₁ and drying uniform volume charge densities +ρ and – ρ, respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region :

1. The electrostatic field is zero
2. The electrostatic potential is constant
3. The electrostatic field is constant
4. The electrostatic field has the same magnitude only

Solution: 3. The electrostatic field is constant

Question 17. Charges Q₁, 2Q, and 4Q are uniformly distributed in three dielectric solid spheres 1, 2, and 3 of radii R/2, R, and 2R respectively, as shown in the figure. If magnitudes of the electric fields at point P at a distance R from the center of spheres 1, 2, and 3 are E1 and respectively, then

1. E₁> E₂> E₃
2. E₃> E₁> E₂
3. E₂> E₁> E₃
4. E₃> E₂> E₁

Solution: 3. E₂> E1> E₃

Question 18. Four charges Q₁, Q₂, Q₃, and Q₄, of the same magnitude, are fixed along the x-axis at x = –2a –a, +a, and +2a, respectively. A positive charge q is placed on the positive y-axis at a distance b > 0. Four options for the signs of these charges are given in List I. The direction of the forces on the charge q is given in List- II Match List-1 with List-II and select the correct answer using the code given below the lists.

List-I   List-II

P. Q₁,Q2,Q₃, Q₄, all positive 1. +x

Q. Q₁,Q₂positive Q₃,Q₄ negative 2. –x

R. Q₁,Q₄positive Q₂, Q₃negative 3. +y

S. Q₁,Q₃positive Q₂, Q₄negative 4. –y

Code :

1. P-3, Q-1, R-4,S-2
2. P-4, Q-2, R-3, S-1
3. P-3, Q-1, R-2,S-4
4. P-4, Q-2, R-1, S-3

Solution: 1. P-3, Q-1, R-4,S-2

Electrostatics Exercise – 3

Question 1. The electric potential at a point (x, y, z) is given by V = – x2 y – xz3 + 4 The electric field \(\text { है }\) at that point is

1. \(\vec{E}=\hat{i}\left(2 x y+z^3\right)+\hat{j} x^2+\hat{k} 3 x z^2\)
2. \(\vec{E}=\hat{i} 2 x y+\hat{j}\left(x^2+y^2\right)+\hat{k}\left(3 x z-y^2\right)\)
3. \(\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}} z^3+\hat{\mathrm{j}} x y z+\hat{k} z^2\)
4. \(\vec{E}=\hat{i}\left(2 x y-z^3\right)+\hat{j} x y^2+\hat{k} 3 z^2 x\)

Solution: 1. \(\vec{E}=\hat{i}\left(2 x y+z^3\right)+\hat{j} x^2+\hat{k} 3 x z^2\)

Question 2. The electric field at a distance \(\frac{3 R}{2}\)from the center of a charged conducting spherical shell of the radius is E. The electric field at a distance \(\frac{R}{2}\) from the centre of the sphere is

1. zero
2. E
3. E/2
4. E/3

Solution: 1. zero

Question 3. A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will :

1. Increase four times
2. Be reduced to half
3. Remain the same
4. Be doubled

Solution: 3. Remain the same

Question 4. Four electric charges +q, +q, –q, and –q are placed at the corners of a square of side 2L (see figure). The electric potential at point A, midway between the two charges +q and +q, is :

1. \(\frac{1}{4 \pi \epsilon_0} \frac{2 q}{L}(1+\sqrt{5})\)
2. \(\frac{1}{4 \pi \epsilon_0} \frac{2 \mathrm{q}}{\mathrm{L}}\left(1+\frac{1}{\sqrt{5}}\right)\)
3. \(\frac{1}{4 \pi \epsilon_0} \frac{2 q}{L}\left(1-\frac{1}{\sqrt{5}}\right)\)
4. Zero

Solution: 3. \(\frac{1}{4 \pi \epsilon_0} \frac{2 q}{L}\left(1-\frac{1}{\sqrt{5}}\right)\)

Question 5. The electric potential V at any point (x, y, z), all in meters in space is given by V = 4×2 volt. The electric field at the point (1, 0, 2) in volt/meter is :

1. 8 along the positive X-axis
2. 16 along the negative X-axis
3. 16 along the positive X-axis
4. 8 along the negative X-axis

Solution: 4. 8 along the negative X-axis

Question 6. Three charges, each +q, are placed at the corners of an isosceles triangle ABC of sides BC and AC, 2a. D and E are the midpoints of BC and CA. The work done in taking a charge Q from D to E is:

1. \(\frac{\mathrm{eqQ}}{8 \pi \epsilon_0 \mathrm{a}}\)
2. \(\frac{q Q}{4 \pi \epsilon_0 a}\)
3. zero
4. \(\frac{3 q Q}{4 \pi \epsilon_0 a}\)

Solution: 3. zero

Question 7. An electric dipole of the moment ´p´ is placed in an electric field of intensity ´E´. The dipole acquires a position such that the axis of the dipole makes an angle θ with the direction of the field. Assuming that the potential energy of the dipole is zero when θ = 90º, the torque and the potential energy of the dipole will respectively be :

1. p E sin θ, – p E cos θ
2. p E sin θ, – 2 p E cos θ
3. p E sin θ, 2 p Ecos θ
4. p E cos θ, – p Ecos θ

Solution: 1. p E sin θ, – p E cos θ

Question 8. Four point charges –Q, –q, 2q, and 2Q are placed, one at each corner of the square. The relation between Q and q for which the potential at the center of the square is zero is :

1. Q = – q
2. \(Q=-\frac{1}{q}\)
3. Q = q
4. \(Q=\frac{1}{q}\)

Solution: 1. Q = – q

Question 9. What is the flux through a cube of side ‘a’ if a point charge of q is at one of its corners:

1. \(\frac{2 \mathrm{q}}{\varepsilon_0}\)
2. \(\frac{q}{8 \varepsilon_0}\)
3. \(\frac{\mathrm{q}}{\varepsilon_0}\)
4. \(\frac{q}{2 \varepsilon_0} 6 a^2\)

Solution: 2. \(\frac{q}{8 \varepsilon_0}\)

Question 10. Two metallic spheres of radii 1 cm and 3 cm are given charges of –1×10^-2 C and 5×10^-2 C, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is :

1. 2×10^-2 C
2. 3×10^-2 C
3. 4×10^-2 C
4. 1×10^-2 C

Solution: 2. 3×10^-2 C

Question 11. A, B, and C are three points in a uniform electric field. The electric potential is :

1. Maximum at B
2. Maximum at C
3. Same at all three points A, B, and C
4. Maximum at A

Solution: 1. Maximum at B

Question 12. Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is r. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now becomes:

1. \(\left(\frac{r}{\sqrt[3]{2}}\right)\)
2. \(\left(\frac{2 r}{\sqrt{3}}\right)\)
3. \(\left(\frac{2 r}{3}\right)\)
4. \(\left(\frac{r}{\sqrt{2}}\right)^2\)

Solution: 1. \(\left(\frac{r}{\sqrt[3]{2}}\right)\)

Question 13. A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the center of the sphere respectively are:

1. \(\text { Zero } \& \frac{\mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}^2}\)
2. \(\frac{\mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}} \& \text { Zero }\)
3. \(\frac{\mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}} \ \frac{\mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}^2}\)
4. Both are zero.

Solution: 2. \(\frac{\mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}} \& \text { Zero }\)

Question 14. In a region the potential is represented by V(x, y, z) = 6x – 8xy –8y + 6yz, where V is in volts and x, y, and z, are in meters. The electric force experienced by a charge of 2 coulomb situated at point (1, 1,1) is :

1. 6 √5N
2. 30N
3. 24N
4. 4 √35N

Solution: 4. 4 v35N

Question 15. The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a sphere of radius ‘a’ centered at the origin of the field, will given by :

1. Aa ∈a²
2. 4π∈0 Aa³
3. ∈0 Aa³
4. 4π∈0 Aa³

Solution: 2. 4π∈0 Aa³

Question 16. If potential (in volts) in a region is expressed as V(x, y, z) = 6 xy – y + 2yz, the electric field (in N/C) at point (1, 1, 0) is :

1. \(-(6 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \hat{k})\)
2. \(-(2 \hat{i}+3 \hat{j}+\hat{k})\)
3. \(-(6 \hat{i}+9 \hat{j}+\hat{k})\)
4. \(-(3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})\)

Solution: 1. \(-(6 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \hat{k})\)

Question 17. Two identical charged spheres suspended from a common point by two mass-less strings of lengths l are initially at a distance d(d << l) apart because of their mutual repulsion. The charges begin to leak from both spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then v varies as a function of the distance x between the spheres, as :

1. v ∝ x–1
2. v ∝ x1/2
3. v ∝ x
4. v ∝ x–1/2

Solution: 4. v ∝ x–1/2

Question 18. When an α-particle of mass ‘m’ moving with velocity ‘ v ‘ bombards on a heavy nucleus of charge ‘Ze’ its distance of closet approach from the nucleus depends on m as : 

1. m
2. \(\frac{1}{m}\)
3. \(\frac{1}{\sqrt{m}}\)
4. \(\frac{1}{\mathrm{~m}^2}\)

Solution: 2. \(\frac{1}{m}\)

Question 19. An electric dipole is placed at an angle of 30º with an electric field intensity of 2 ×105 N/C. It experiences a torque equal to 4 N m. The charge on the dipole, if the dipole length is 2cm, is

1. μC
2. 8 mC
3. 2 mC
4. 5 mC

Solution: 3. 2 mC

Question 20. Suppose the charge of a proton and an electron differ slightly. One of them is – e, and the other is (e + Δe). If the net of electrostatic force and the gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order of [Given the mass of hydrogen mh = 1.67 × 10^-27 kg]

1. 10^-20 C
2. 10^-23 C
3. 10^-37 C
4. 10^-47 C

Solution: 3. 10–37 C

Question 21. The diagrams below show regions of equipotentials.

1. The positive charge is moved from A to B in each diagram
2. Maximum work is required to move q in Figure (c).
3. In all four cases, the work done is the same.
4. Minimum work is required to move q in Figure (a)
5. Maximum work is required to move q in Figure (b).

Solution: 2. In all four cases the work done is the same.

Question 22. An electron falls from rest through a vertical distance h in a uniform and vertically upward-directed electric field E. The direction of the electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is : 

1. Smaller
2. Equal
3. 10 times greater
4. 5 times greater

Solution: 1. Smaller

Question 24. Two point charges A and B, having charges +Q and –Q respectively, are placed at a certain distance apart, and the force acting between them is F. If 25% charge of A is transferred to B, then the force between the charges becomes:

1. \(\frac{\mathrm{4F}}{\mathrm{3}}\)
2. F
3. \(\frac{\mathrm{9F}}{\mathrm{16}}\)
4. \(\frac{\mathrm{16F}}{\mathrm{9}}\)

Solution: 3. \(\frac{\mathrm{9F}}{\mathrm{16}}\)

Question 25. Two parallel infinite line charges with linear charge densities +λ C/m and –λ C/m are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges?

1. \(\frac{\lambda}{2 \pi \varepsilon_0 R} \mathrm{R} / \mathrm{C}\)
2. zero
3. \(\frac{2 \lambda}{\pi \varepsilon_0 R} N / C\)
4. \(\frac{\lambda}{\pi \varepsilon_0 R} N / C\)

Solution: 4. \(\frac{\lambda}{\pi \varepsilon_0 R} N / C\)

Question 26. Two metal spheres, one of radius R and the other of radius 2R respectively have the same surface charge density σ. They are brought in contact and separated. What will be the new surface charge densities on them?

1. \(\sigma_1=\frac{5}{6} \sigma, \sigma_2=\frac{5}{6} \sigma\)
2. \(\sigma_1=\frac{5}{2} \sigma, \sigma_2=\frac{5}{6} \sigma/\)
3. \(\sigma_1=\frac{5}{2} \sigma, \sigma_2=\frac{5}{3} \sigma\)
4. \(\sigma_1=\frac{5}{3} \sigma, \sigma_2=\frac{5}{6} \sigma\)

Solution: 4. \(\sigma_1=\frac{5}{3} \sigma, \sigma_2=\frac{5}{6} \sigma\)

Question 27. A sphere encloses an electric dipole with charges ±3 × 10^-6 C. What is the total electric flux across the sphere?

1. 3 × 10^-6
2. Zero
3. 3 × 10^-6 Nm²/C
4. 6 × 10^-6 Nm²/C

Solution: 2. Zero

Question 28. The electric field at a point on the equatorial plane at a distance r from the center of a dipole having dipole moment is given by (r >> separation of two charges forming the dipole,∈ −0permittivity of free space)

1. \(\overrightarrow{\mathrm{E}}=\frac{\overrightarrow{\mathrm{p}}}{4 \pi \epsilon_0 \mathrm{r}^3}\)
2. \(\overrightarrow{\mathrm{E}}=\frac{2 \overrightarrow{\mathrm{p}}}{4 \pi \epsilon_0 \mathrm{r}^3}\)
3. \(\vec{E}=-\frac{\vec{p}}{4 \pi \epsilon_0 r^2}\)
4. \(\overrightarrow{\mathrm{E}}=-\frac{\overrightarrow{\mathrm{p}}}{4 \pi \epsilon_0 \mathrm{r^3}}\)

Solution: 4. \(\overrightarrow{\mathrm{E}}=-\frac{\overrightarrow{\mathrm{p}}}{4 \pi \epsilon_0 \mathrm{r^3}}\)

Question 29. The acceleration of an electron due to the mutual attraction between the electron and a proton when they are Aapart is, \(\left(\mathrm{m}_{\mathrm{e}} \times 10^{-31} \mathrm{~kg}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\right)\left(\text { Take } \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right)\)

1. 10²⁴ m/s²
2. 10²³ m/s²
3. 10²² m/s²
4. 10²⁵ m/s²

Solution: 3. 1022 m/s²

Question 30. A spherical conductor of radius 10cm has a charge of 3.2 × 10^-7 C distributed uniformly. What is the magnitude of the electric field at a point 15 cm from the center of the sphere? \(\left(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2\right)\)

1. 1.28 × 10⁷ N/C
2. 1.28 × 10⁴ N/C
3. 1.28 × 10⁵ N/C
4. 1.28 × 10⁶ N/C

Solution: 3. 1.28 × 10⁵ N/C

Question 31. Two points P and Q are maintained at the potentials of 10 V and –4 V respectively. The work done in moving 100 electrons from P to Q is :

1. 9.60 × 10^-17J
2. –2.24 × 10^-15 J
3. 2.24 × 10^-16 J
4. –9.60 × 10^-17 J

Solution: 3. 2.24 × 10^-16 J

Question 32. A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then Q/q equals:

1. –1
2. 1
3. \(-\frac{1}{\sqrt{2}}\)
4. \(-2 \sqrt{2}\)

Solution: 4. \(-2 \sqrt{2}\)

Question 33. Statement 1: For a charged particle moving from point P to point Q, the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q.

Statement 2: The net work done by a conservative force on an object moving along a closed loop is zero.

1. Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.
2. Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.
3. Statement-1 is false, Statement-2 is true.
4. Statement-1 is true, Statement-2 is false.

Solution: 1. Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

Question 34. A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field E at the center O is

1. \(\frac{q}{4 \pi^2 \varepsilon_0 r^2} \hat{j}\)
2. \(-\frac{q}{4 \pi^2 \varepsilon_0 r^2} \hat{j}\)
3. \(-\frac{q}{2 \pi^2 \varepsilon_0 r^2} \hat{j}\)
4. \(\frac{q}{2 \pi^2 \varepsilon_0 r^2} \hat{j}\)

Solution: 3. \(-\frac{q}{2 \pi^2 \varepsilon_0 r^2} \hat{j}\)

Question 35. Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 30º with each other. When suspended in a liquid of density 0.8 g cm–3, the angle remains the same. If the density of the material of the sphere is 1.6 g cm–3, the dielectric constant of the liquid is

1. 4
2. 3
3. 2
4. 1

Solution: 3. 2

Question 36. The electrostatic potential inside a charged spherical ball is given by φ = ar2 + b where r is the distance from the center; a,b are constants. Then the charge density inside the ball is :

1. –24π aε₀r
2. –6π aε₀r
3. –24π aε₀
4. –6 aε₀

Solution: 4. –6 aε₀

Question 37. Two identical charged spheres suspended from a common point by two massless strings of length l are initially a distance d(d < < l) apart because of their mutual repulsion. The charge begins to leak from both spheres at a constant rate. As a result, the charges approach each other with a velocity υ. Then as a function of distance x between them :

1. υ ∝ x–1/2
2. υ ∝ x–1
3. υ ∝ x1/2
4. υ ∝ x

Solution: 1. υ ∝ x–1/2

Question 38. Two positive charges of magnitude ‘q’ are placed at the ends of a side (side 1) of a square of side ‘2a’. Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charge Q moves from the middle of side 1 to the center of the square, its kinetic energy at the center of the square is :

1. zero
2. \(\frac{1}{4 \pi \varepsilon_0} \frac{2 \mathrm{qQ}}{\mathrm{a}}\left(1+\frac{1}{\sqrt{5}}\right)\)
3. \(\frac{1}{4 \pi \varepsilon_0} \frac{2 q \mathrm{Q}}{\mathrm{a}}\left(1-\frac{2}{\sqrt{5}}\right)\)
4. \(\frac{1}{4 \pi \varepsilon_0} \frac{2 q Q}{a}\left(1-\frac{1}{\sqrt{5}}\right)\)

Solution: 4. \(\frac{1}{4 \pi \varepsilon_0} \frac{2 q Q}{a}\left(1-\frac{1}{\sqrt{5}}\right)\)

Question 39. In a uniformly charged sphere of total charge Q and radius R, the electric field E is plotted as a function of distance from the center. The graph that would correspond to the above will be :

Solution: 3.

Question 40. This question has statement 1 and statement 2. Of the four choices given after the statements, choose the one that best describes the two statements. An insulating solid sphere of radius R has a uniformly positive charge density ρ. As a result of this uniform charge distribution, there is a finite value of the electric potential at the center of the sphere, at the surface of the sphere, and also at a point outside the sphere. The electric potential at infinite is zero.

Statement-1: When a charge ‘q’ is taken from the center of the surface of the sphere its potential energy changes by \(\frac{\mathrm{q} \rho}{3 \varepsilon_0}\)

Statement-2: The electric field at a distance r (r < R) from the center of the sphere is \(\frac{\rho r}{3 \varepsilon_0}\)

1. Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of statement-1.
2. Statement 1 is true Statement 2 is false.
3. Statement 1 is false Statement 2 is true.
4. Statement 1 is true, Statement 2 is true, and Statement 2 is the correct explanation of Statement 1.

Solution: 3. Statement 1 is false Statement 2 is true.

Question 41. Two charges, each equal to q, are kept at x = – a and x = a on the x-axis. A particle of mass m and charge \(q_0=\frac{q}{2}\) is placed at the origin. If charge q0is given a small displacement (y <<a) along the y-axis, the net force acting on the particle is proportional to :

1. y
2. –y
3. \(\frac{1}{y}\)
4. \(-\frac{1}{y}\)

Solution: 1. y

Question 42. A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying at distance L from the end A is :

1. \(\frac{Q}{8 \pi \in_0 L}\)
2. \(\frac{3 Q}{4 \pi \epsilon_0 L}\)
3. \(\frac{Q}{4 \pi \in_0 L \ln 2}\)
4. \(\frac{Q \ln 2}{4 \pi \epsilon_0 L}\)

Solution: 4. \(\frac{Q \ln 2}{4 \pi \epsilon_0 L}\)

Question 43. Assume that an electric field \(\overrightarrow{\mathrm{E}}=30 \mathrm{x}^2 \hat{\mathrm{i}}\) exists in space. Then the potential difference VA– VO, where VO is the potential at the origin and the potential at x = 2 m is :

1. 120 V
2. –120 V
3. – 80 V
4. 80 V

Solution: 3. – 80 V

Question 44. A long cylindrical shell carries a positive surface charge σ in the upper half and a negative surface charge – σ in the lower half. The electric field lines around the cylinder will look like the figure given in : (figures are schematic and not drawn to scale)

Solution: 1.

Question 45. A uniformly charged solid sphere of radius R has potential V0(measured concerning ∞) on its surface. For this sphere the equipotential surfaces with potentials \(\frac{3 \mathrm{~V}_0}{2}, \frac{5 \mathrm{~V}_0}{4}, \frac{3 \mathrm{~V}_0}{4} \text { and } \frac{\mathrm{V}_0}{4}\) have radius R₁, R2, R3and R4respectively. Then

1. R₁= 0 and R₂> (R₄– R₃)
2. R₁ ≠0 and (R₂– R₁) > (R₄– R₃)
3. R₁= 0 and R₂< (R₄– R₃)
4. 2R < R₄

Solution: 3. R₁= 0 and R₂< (R₄– R₃)

Question 46. The region between two concentric spheres of radii ‘a’ and ‘b’, respectively (see figure), has volume charge density Ar ρ=\(\), where A is a constant and r is the distance from the center. At the center of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is :

1. \(\frac{Q}{2 \pi\left(b^2-a^2\right)}\)
2. \(\frac{2 Q}{\pi\left(a^2-b^2\right)}\)
3. \(\frac{2 Q}{\pi a^2}\)
4. \(\frac{Q}{2 \pi a^2}\)

Solution: 4. \(\frac{Q}{2 \pi a^2}\)

Question 47. An electric dipole has a fixed dipole moment \(\overrightarrow{\mathrm{p}}\), which makes angle θ concerning the x-axis. When subjected to an electric field \(\overrightarrow{\mathrm{E}}_1=\mathrm{E} \hat{\mathbf{i}},\), it experiences a torque \(\overrightarrow{\mathrm{T}}_1=\tau \hat{\mathrm{k}}\). When subjected to another \(\overrightarrow{\mathrm{E}}_2=\sqrt{3} \mathrm{E}_1 \hat{\mathrm{j}}\) electric field ˆit experiences a torque \(\overrightarrow{\mathrm{T}}_2=-\overrightarrow{\mathrm{T}}_1\). The angle θ is :

1. 90°
2. 30°
3. 45°
4. 60°

Solution: 4. 60°

Question 48. Three concentric metal shells A, B, and C of respective radii a,b, and c (a < b < c) have surface charge densities +σ, –σ, and +σ respectively. The potential of shell B is :

1. \(\frac{\sigma}{\varepsilon_0}\left[\frac{b^2-c^2}{b}+a\right]\)
2. \(\frac{\sigma}{\varepsilon_0}\left[\frac{\mathrm{b}^2-\mathrm{c}^2}{\mathrm{c}}+\mathrm{a}\right]\)
3. \(\frac{\sigma}{\varepsilon_0}\left[\frac{\mathrm{a}^2-\mathrm{b}^2}{\mathrm{a}}+\mathrm{c}\right]\)
4. \(\frac{\sigma}{\varepsilon_0}\left[\frac{\mathrm{a}^2-\mathrm{b}^2}{\mathrm{~b}}+\mathrm{c}\right]\)

Solution: 4. \(\frac{\sigma}{\varepsilon_0}\left[\frac{\mathrm{a}^2-\mathrm{b}^2}{\mathrm{~b}}+\mathrm{c}\right]\)

Question 49. Three charges +Q, q, +Q are placed respectively, at distance, 0, d/2, and d from the origin, on the x-axis. If the net force experienced by +Q, placed at x = 0, is zero, then the value of q is :

1. +Q/2
2. +Q/4
3. –Q/2
4. –Q/4

Solution: 4. –Q/4

Question 50. For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance of h from its center. The value of h is :

1. \(\frac{\mathrm{R}}{\sqrt{2}}\)
2. R
3. \(\frac{\mathrm{R}}{\sqrt{5}}\)
4. R 2

Solution: 1. \(\frac{\mathrm{R}}{\sqrt{2}}\)

Question 51. Two point charges \(\rho(r)=\frac{A}{r^2} e^{-2 r / a}\)and q2( −25μCa )re placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is,\(\left[\text { take }=\frac{1}{4 \pi \mathrm{g} \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right]\)

1. \((-81 \hat{i}+81 \hat{j}) \times 10^2\)
2. \((81 \hat{\mathrm{i}}-81 \hat{\mathrm{j}}) \times 10^2\)
3. \((-63 \hat{\mathbf{i}}+27 \hat{\mathbf{j}}) \times 10^2\)
4. \((63 \hat{\mathbf{i}}-27 \hat{\mathrm{j}}) \times 10^2\)

Solution: 4. \((63 \hat{\mathbf{i}}-27 \hat{\mathrm{j}}) \times 10^2\)

Question 52. Charge is distributed within a sphere of radius R with a volume charge density \(\), where A 2e r and a are constant. If Q is the total charge of this charge distribution, the radius R is :

1. \(\frac{Q}{12 \pi \epsilon_0} \frac{a b+b c+c a}{a b c}\)
2. \(\frac{a}{2} \log \left(\frac{1}{1-\frac{Q}{2 \pi a A}}\right)\)
3. \(\frac{a}{2} \log \left(1-\frac{Q}{2 \pi a A}\right)\)
4. \(\mathrm{a} \log \left(1-\frac{\mathrm{Q}}{2 \pi \mathrm{aA}}\right)\)

Solution: 2. \(\frac{Q}{12 \pi \epsilon_0} \frac{a b+b c+c a}{a b c}\)

Question 53. A charge Q is distributed over three concentric spherical shells of radii a, b, c (a < b < c) such that their surface charge densities are equal. The total potential at a point at distance r from their common center, where r < a, would be

1. \(a \log \left(\frac{1}{1-\frac{Q}{2 \pi a A}}\right)\)
2. \(\frac{Q}{4 \pi \in_0(a+b+c)} \)
3. \(\frac{Q(a+b+c)}{4 \pi \epsilon_0\left(a^2+b^2+c^2\right)}\)
4. \(\frac{Q\left(a^2+b^2+c^2\right)}{4 \pi \epsilon_0\left(a^3+b^3+c^3\right)}\)

Solution: 3. \(a \log \left(\frac{1}{1-\frac{Q}{2 \pi a A}}\right)\)

Question 54. Two electric dipoles, A, B with respective dipole moments A d \(\overrightarrow{\mathrm{d}}_{\mathrm{A}}=-4 q a \hat{i} \text { and } \overrightarrow{\mathrm{d}}_{\mathrm{B}}=-2 q a \hat{i}\) is placed on x-axis with a separation R, as shown in the figure.

The distance from A at which both of them produce the same potential is :

1. \(\frac{R}{\sqrt{2}-1}\)
2. \(\frac{\sqrt{2} R}{\sqrt{2}+1}\)
3. \(\frac{\sqrt{2} R}{\sqrt{2}-1}\)
4. \(\frac{R}{\sqrt{2}+1}\)

Solution: 3. \(\frac{\sqrt{2} R}{\sqrt{2}-1}/\)

Question 55. Four equal point charges Q each are placed in the xy-plane at (0, 2), (4, 2), (4, –2), and (0, –2). The work required to put a fifth charge Q at the origin of the coordinate system will be

1. \(\frac{Q^2}{4 \pi \in_0}\left(1+\frac{1}{\sqrt{3}}\right)\)
2. \(\frac{Q^2}{4 \pi \epsilon_0}\left(1+\frac{1}{\sqrt{5}}\right)\)
3. \(\frac{Q^2}{4 \pi \epsilon_0}\)
4. \(\frac{Q^2}{2 \sqrt{2} \pi \epsilon_0}\)

Solution: 2. \(\frac{Q^2}{4 \pi \epsilon_0}\left(1+\frac{1}{\sqrt{5}}\right)\)

Question 56. Charge –q and +q located at A and B, respectively, constitute an electric dipole. The distance AB = 2a, O is the mid-point of the dipole and OP is perpendicular to AB. A charge Q is placed at P where y and y >> 2a. The charge Q experiences an electrostatic force F. If Q is now moved along the equatorial line y, the force on Q will be close to : \(\left(\frac{y}{3}>2 a\right)\)

1. 3F
2. 27F
3. 9F
4. F/3

Solution: 2. 27F

Question 57. The given graph shows a variation (with distance r from center) of :

1. The electric field of a uniformly charged sphere
2. Potential of a uniformly charged spherical shell
3. The potential of a uniformly charged sphere
4. The electric field of a uniformly charged spherical shell

Solution: 2. Potential of a uniformly charged spherical shell

Question 58. Three charges Q, +q, and +q and placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of Q is :

1. +q
2. \(\frac{-\sqrt{2} q}{\sqrt{2}+1}\)
3. -2q
4. \(\frac{-q}{1+\sqrt{2}}\)

Solution: 2. \(\frac{-\sqrt{2} q}{\sqrt{2}+1}\)

Question 59. A particle of mass m and charge q is in an electric and magnetic field given by \(\overrightarrow{\mathrm{E}}=2 \hat{i}+3 \hat{j}; \vec{B}=4 \hat{j}+6 \hat{k}\) The charged particle is shifted from the origin to the point P(x = 1; y = 1) along a straight path. The magnitude of the total work done is :

1. 5q
2. (2.5)q
3. (0.35) q
4. (0.15)q

Solution: 1. 5q

Question 60. An electric field of 1000 V/m is applied to an electric dipole at an angle of 45°. The value of the electric dipole moment is 10–29 C.m. What is the potential energy of the electric dipole?

1. –9 × 10^-20 J
2. –7 × 10^-27 J
3. –10 × 10^-29 J
4. –20 × 10^-18 J

Solution: 2. –7 × 10^-27 J

Question 61. Determine the electric dipole moment of the system of three charges, placed on the vertices of an equilateral triangle, as shown in the figure :

1. \(-\sqrt{3} q \ell \hat{j}\)
2. \(\left(q \ell \ell \frac{\hat{i}+\hat{j}}{\sqrt{2}}\right.\)
3. \(2 q \ell \hat{j}\)
4. \(\sqrt{3} \mathrm{q} \ell \frac{\hat{\mathrm{j}}-\hat{\mathrm{i}}}{\sqrt{2}}\)

Solution: 1. \(-\sqrt{3} q \ell \hat{j}\)

Question 62. There is a uniform spherically symmetric surface charge density at a distance R0 from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed V(R(t)) of the distribution as a function of its instantaneous radius R(t) is:

Solution: 4.

Question 63. A sphere of radius 1 cm has a potential of 8000 V. The energy density near the surface of the sphere will be:

1. 64 × 105 J/m³
2. 8 × 103 J/m³
3. 2 J/m³
4. 2.83 J/m³

Solution: 4. 2.83 J/m³

Question 64. In the above question, the electric force acting on a point charge of 2 C placed at the origin will be :

1. 2 N
2. 500 N
3. –5 N
4. –500 N

Solution: 4. –500 N

Question 65. The figure shows two large cylindrical shells having uniform linear charge densities + λ and – λ. The radius of the inner cylinder is ‘a’ and that of the outer cylinder is ‘b’. A charged particle of mass m, charge q revolves in a circle of radius r. Then, its speed ‘v’ is : (Neglect gravity and assume the radii of both the cylinders to be very small in comparison to their length.)

1. \(\sqrt{\frac{\lambda q}{2 \pi \epsilon_0 m}}\)
2. \(\sqrt{\frac{2 \lambda q}{\pi \epsilon_0 m}}\)
3. \(\sqrt{\frac{\lambda q}{\pi \epsilon_0 m}}\)
4. \(\sqrt{\frac{\lambda \mathrm{q}}{4 \pi \varepsilon_0 \mathrm{~m}}}\)

Solution: 1. \(\sqrt{\frac{\lambda q}{2 \pi \epsilon_0 m}}\)

Question 66. A charge q is uniformly distributed over a large plastic plate. The electric field at point P close to the center and just above the surface of the plate is 50 V/m. If the plastic plate is replaced by a copper plate of the same geometrical dimensions and carrying the same uniform charge q, the electric field at point P will become:

1. zero
2. 25 V/m
3. 50 V/m
4. 100 V/m

Solution: 3. 50 V/m

Question 67. A point charge q is brought from infinity (slowly so that heat developed in the shell is negligible) and is placed at the center of a conducting neutral spherical shell of inner radius a and outer radius b, then work done by the external agent is:

1. 0
2. \(\frac{k q^2}{2 b}\)
3. \(\frac{k q^2}{2 b}-\frac{k q^2}{2 a}\)
4. \(\frac{k q^2}{2 a}-\frac{k q^2}{2 b}\)

Solution: 3. \(\frac{k q^2}{2 b}-\frac{k q^2}{2 a}\)

Question 68. The magnitude of the electric force on 2 μ c charge placed at the center O of two equilateral triangles each of side 10 cm, as shown in the figure is P. If charge A, B, C, D, E, and F are 2 μ c, 2 μ c, 2 μ c, -2 μc, – 2 μ c, – 2μ c respectively, then P is:

1. 21.6 N
2. 64.8 N
3. 0
4. 43.2 N

Solution: 4. 43.2 N

Question 69. A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field 81 5 of strength \(\frac{81 \pi}{7} \times 10^5 \mathrm{Vm}^{-1}\). When the field is switched off, the drop is observed to fall with terminal velocity 2 × 10–3 m s–1. Given g = 9.8 m s^-2, the viscosity of the air = 1.8 × 10–5 Ns m^-2, and the density of oil = 900 kg m–3, the magnitude of q is :

1. 1.6 × 10^-19 C
2. 3.2 × 10^-19 C
3. 4.8 × 10^-19 C
4. 8.0 × 10^-19 C

Solution: 4. 8.0 × 10^-19 C

Question 70. Identical charges (–q) are placed at each corner of a cube of side b, then the electrostatic potential energy of charge (+q) placed at the center of the cube will be :

1. \(-\frac{4 \sqrt{2} q^2}{\pi \varepsilon_0}\)
2. \(\frac{8 \sqrt{2} q^2}{\pi \varepsilon_0 \mathrm{~b}}\)
3. \(-\frac{4 q^2}{\sqrt{3} \pi \varepsilon_0 b}\)
4. \(\frac{8 \sqrt{2} q^2}{4 \pi \varepsilon_0 b}\)

Solution: 3. \(-\frac{4 q^2}{\sqrt{3} \pi \varepsilon_0b}\)

Question 71. Three charges Q, + q, and + q are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if Q is equal to :

1. \(\frac{-q}{1+\sqrt{2}}\)
2. \(\frac{-2 q}{2+\sqrt{2}}\)
3. -2q
4. +q

Solution: 2. \(\frac{-2 q}{2+\sqrt{2}}\)

Question 72. Six-point charges are kept at the vertices of a regular hexagon of side L and center O, as shown in the 1 q K figure. Given that \(\mathrm{K}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{L}^2}\), which of the following statement (s) is incorrect?

1. The electric field at O is 6K along OD
2. The potential at O is zero
3. The potential at all points on the line PR is the same
4. The potential at all points on the line ST is the same.

Solution: 4. The potential at all points on the line ST is the same.

Question 73. Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities ρ₁and ρ₂ respectively, touch each other. The net electric field at a distance 2R from the center of the smaller ρ₁ sphere, along the line joining the centers of the spheres, is zero. The ratio ρcan be ;

1. –4
2. 2
3. 32/25
4. 4

Solution: 4. 4

Question 74. Let E₁(r), E₂(r), and E₃(r) be the respective electric fields at a distance r from a point charge Q, an infinitely long wire with constant linear charge density λ, and an infinite plane with uniform surface charge density σ. if E₁(r0) = E₂(r0) = E₃(r0) at a given distance r0, then

1. \(\mathrm{Q}=4 \sigma \pi \mathrm{r}_0^2\)
2. \(r_0=\frac{\lambda}{2 \pi \sigma}\)
3. \(E_1\left(r_0 / 2\right)=2 E_2\left(r_0 / 2\right)\)
4. \(E_2\left(r_0 / 2\right)=4 E_3\left(r_0 / 2\right)\)

Solution: 3. \(E_1\left(r_0 / 2\right)=2 E_2\left(r_0 / 2\right)\)

Capacitance

1. Capacitance Introduction

A capacitor Can Store Energy In The Form Of Potential Energy In An Electric Field. In This Chapter, We Will Discuss The Capacity of conductors to hold charge and energy.

2. Capacitance Of An Isolated Conductor

When a conductor is charged its potential increases, it is found that for an isolated conductor (the conductor should be of finite dimension, so that the potential of infinity can be assumed to be zero), the potential of the conductor is proportional to the charge given to it.

q = charge on a conductor

V = potential of the conductor

q ∞ V

⇒ q = CV

Where C is the proportionality constant called the capacitance of the conductor.

2.1 Definition of capacitance :

The capacitance of a conductor is defined as the charge required to increase the potential of a conductor by one unit.

2.2 Important points about the capacitance of an isolated conductor :

1. It is a scalar quantity.
2. Unit of capacitance is farad in SI units and its dimensional formula is M –1 L–2 μ2 T4
3. 1 Farad: 1 Farad is the capacitance of a conductor for which 1-coulomb charge increases potential by 1 volt.
   1. \(1 \text { Farad }=\frac{1 \text { Coulomb }}{1 \text { Volt }}\)
   2. 1 μF = 10–6 F, 1nF = 10–9 F or 1 pF = 10–12 F
4. The capacitance of an isolated conductor depends on the following factors :
   1. Shape and size of the conductor: On increasing the size, capacitance increases.
   2. On surrounding medium: With the increase in dielectric constant K, capacitance increases.
   3. Presence of other conductors: When a neutral conductor is placed near a charged conductor, the capacitance of conductors increases.
5. The capacitance of a conductor does not depend on
   1. Charge on the conductor
   2. The potential of the conductor
   3. The potential energy of the conductor.

3. Potential Energy Or Self Energy Of An Isolated Conductor

Work done in charging the conductor to the charge on it against its electric field or total energy stored in the electric field of the conductor is called self-energy or self-potential energy of the conductor.

3.1 Electric potential energy (Self Energy): Work done in charging the conductor

⇒ \(W=\int_0^q \frac{q}{c} d q=\frac{q^2}{2 c}\)

⇒ \(W=U=\frac{q^2}{2 c}=\frac{1}{2} C V^2=\frac{q V}{2} .\)

q = Charge on the conductor

V = Potential of the conductor

C = Capacitance of the conductor

3.2 Self-energy is stored in the electric field of the conductor with energy density (Energy per unit volume)

⇒ \(\frac{\mathrm{dU}}{\mathrm{dV}}=\frac{1}{2} \varepsilon_0 \mathrm{E}^2 \text { [The energy density in a medium is } \frac{1}{2} \varepsilon_0 \varepsilon_{\mathrm{r}} \mathrm{E}^2 \text { ] }\)

where E is the electric field at that point.

3.3 In the case of a charged conductor energy stored is only outside the conductor but in the case of charged insulating material it is outside as well as inside the insulator.

4. Capacitance Of An Isolated Spherical Conductor

Solved Examples

Example 1. Find out the capacitance of an isolated spherical conductor of radius R.
Solution :

Let there be charge Q on the sphere.

Potential \(V=\frac{K Q}{R}\)

Hence by the formula: Q = CV

⇒ \(Q=\frac{C K Q}{R}\)

C = 4πε₀R (∴ CEarth = 711 μF)

The capacitance of an isolated spherical conductor

C = 4πε₀R

If the medium around the conductor is vacuum or air.

C_Vaccum = 4πε₀R

R = Radius of spherical conductor. (maybe solid or hollow.)

If the medium around the conductor is a dielectric of constant K from the surface of a sphere to infinity.

C_Medium = 4πε₀KR

⇒ \(\frac{\mathrm{C}_{\text {medium }}}{\mathrm{C}_{\text {air/vaccum }}}=\mathrm{K}=\text { dielectric constant. }\)

5. Sharing Of Charges On Joining Two Conductors (By A Conducting Wire) :

Whenever there is a potential difference, there will be movement of charge.

If released, the charge always tends to move from high potential energy to low potential energy.

If released, positive charge moves from high potential to low potential [if only electric force acts on charge].

If released, the negative charge moves from low potential to high potential [if only electric force acts on charge].

The movement of charge will continue till there is a potential difference between the conductors (finally potential difference = 0).

Formulae related to redistribution of charges :

⇒ \(V=\frac{Q_1^{\prime}}{\mathrm{C}_1}=\frac{\dot{Q}_2^{\prime}}{\mathrm{C}_2}\)

⇒ \(\frac{Q_1^{\prime}}{Q_2^{\prime}}=\frac{C_1}{C_2}\)

But,Q’₁ +Q’₂ = Q₁ + Q₂

⇒ \( V=\frac{Q_1+Q_2}{C_1+C_2}=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}\)

⇒ \(Q_1^{\prime}=\frac{C_1}{C_1+C_2}\left(Q_1+Q_2\right)\)

⇒ \(Q_2^{\prime}=\frac{C_2}{C_1+C_2}\left(Q_1+Q_2\right)\)

Heat loss during redistribution : \(\Delta H=\frac{1}{2} \frac{C_1 C_2}{C_1+C_2}\left(V_1-V_2\right)^2\)

The loss of energy is in the form of Joule heating in the wire.

Note : Always put Q₁, Q₂, V₁ and V₂ with sign.

Solved Examples

Example 2. A and B are two isolated conductors (that means they are placed at a large distance from each other). When they are joined by a conducting wire:

1. Find out the final charges on A and B.
2. Find out the heat produced during the process of the flow of charges.
3. Find out common potential after joining the conductors by conducting wires.

Solution :

⇒ \(Q_A^{\prime}=\frac{3}{3+6}(6+3)=3 \mu C, Q_B^{\prime}=\frac{6}{3+6}(6+3)=6 \mu C\)

⇒ \(\Delta H=\frac{1}{2} \cdot \frac{3 \mu F \cdot 6 \mu F}{(3 \mu F+6 \mu F)} \cdot\left(2-\frac{1}{2}\right)^2=\frac{1}{2} \cdot(2 \mu F)\left(\frac{3}{2}\right)^2 \cdot=\frac{9}{4} \mu \mathrm{J}\)

⇒ \(V_c=\frac{3 \mu \mathrm{C}+6 \mu \mathrm{C}}{3 \mu \mathrm{F}+6 \mu \mathrm{F}}=1 \text { volt. }\)

Example 3. When 30μC charge is given to an isolated conductor of capacitance 5μF. Find out the following

1. The potential of the conductor
2. Energy stored in the electric field of the conductor
3. If this conductor is now connected to another isolated conductor by a conducting wire (at a very large distance) of a total charge of 50 μC and capacity of 10 μF then
   1. Find out the common potential of both the conductors.
   2. Find out the heat dissipated during the process of charge distribution.
   3. Find out the ratio of final charges on conductors.
   4. Find out the final charges on each conductor.

Solution:

Q₁ = 30μC, C₁ = 5μF

⇒ \(\text { (i) } V_1=\frac{Q_1}{C_1}=\frac{30}{5}=6 \mathrm{~V}\)

⇒ \(U=\frac{1}{2} \frac{Q^2}{C}=\frac{1}{2} \frac{\left(30 \times 10^{-6}\right)^2}{\left(5 \times 10^{-6}\right)}=90 \mu \mathrm{J}\)

⇒ \(\mathrm{Q}_2=50 \mu \mathrm{C}, \mathrm{C}_2=10 \mu \mathrm{F}, \quad \mathrm{V}_2=\frac{\mathrm{Q}_2}{\mathrm{C}_2}=\frac{50}{10}=5 \mathrm{~V}\)

1. Common potential \(V=\frac{Q_1+Q_2}{C_1+C_2}=\frac{30+50}{5+10}=\frac{16}{3} V\)
2. \(\Delta H=\frac{1}{2} \frac{C_1 C_2}{C_1+C_2}\left(V_1-V_2\right)=\frac{1}{2} \frac{5 \times 10}{5+10}(6-5)^2=\frac{5}{3} \mathrm{~mJ}\)
3. \(\frac{Q_1^1}{Q_2^1}=\frac{C_1}{C_2}=\frac{5}{10}=\frac{1}{2}\)
4. \(Q_1{ }^1=C_1 V=5 \times \frac{16}{3}=\frac{80}{3} \mathrm{mC}\)

⇒ \(\mathrm{Q}_1{ }^1=\mathrm{C}_1 \mathrm{~V}=5 \times \frac{16}{3}=\frac{80}{3} \mathrm{mC}\)

⇒ \( \mathrm{Q}_2{ }^2=\mathrm{C}_2 \mathrm{~V}=10 \times \frac{16}{3}=\frac{160}{3} \mu \mathrm{C}\)

6. Capacitor :

A capacitor or condenser consists of two conductors separated by an insulator or dielectric.

1. When an uncharged conductor is brought near to a charged conductor, the charge on conductors remains the same but its potential decreases increasing capacitance.
2. In a capacitor, two conductors have equal but opposite charges.
3. The conductors are called the plates of the capacitor. The name of the capacitor depends on the shape of the capacitor.
4. Formulae related to capacitors

⇒ \(Q=C V \Rightarrow \quad C=\frac{Q}{V}=\frac{Q_A}{V_A-V_B}=\frac{Q_B}{V_B-V_A}\)

Q = Charge of a positive plate of the capacitor.

V = Potential difference between positive and negative plates of capacitor

C = Capacitance of capacitor.

Energy stored in the capacitor

Initially charge = 0

Intermediate

Finally,

⇒ \(W=\int d W=\int_0^Q \frac{q}{C} d q=\frac{Q^2}{C}\)

Energy stored in the capacitor = \(=U=\frac{Q^2}{2 C}=\frac{1}{2} C V^2=\frac{1}{2} Q V \text {. }\)

This energy is stored inside the capacitor in its electric field with energy density

⇒ \(\frac{d U}{d V}=\varepsilon \mathrm{E}^2 \frac{1}{2} \text { or } \frac{1}{2} \varepsilon_0 \varepsilon_r E^2\)

5. The capacitor is represented as follows:

6. Based on the shape and arrangement of capacitor plates there are various types of capacitors.

1. Parallel plate capacitor.
2. Spherical capacitor.
3. Cylindrical capacitor.

7. The capacitance of a capacitor depends on

1. Area of plates.
2. Distance between the plates.
3. The dielectric medium between the plates.

Solved Examples

Example 4. Find out the capacitance of the parallel plate capacitor of plate area A and plate separation d.
Solution :

⇒ \(\mathrm{E}=\frac{\mathrm{Q}}{\mathrm{A} \varepsilon_0}\)

⇒ \(V_A-V_B=E. d .=\frac{Q d}{A \varepsilon_0}=\frac{Q}{C}\)

⇒ \(C=\frac{\varepsilon_0 A}{d}\)

where A = area of the plates.

d = distance between plates.

Electric field intensity between the plates of capacitors (air-filled )

E =σ/ε₀ = V/d

The force experienced by any plate of the capacitor

⇒ \(F=\frac{q^2}{2 A \varepsilon_0}\)

7. Circuit Solution For R–C Circuit At t = 0 (Initial State) AND At t − ∞ (Final State)

Note :

1. The charge on the capacitor does not change instantaneously or suddenly if there is a resistance in the path (series) of the capacitor.

2. When an uncharged capacitor is connected with a battery then its charge is zero initially hence potential difference across it is zero initially. At this time the capacitor can be treated as a conducting wire

3. The current will become zero finally (that means in a steady state) in the branch that contains the capacitor.

Solved Examples

Example 5. Find out the current in the circuit and charge on capacitor which is initially uncharged in the following situations.

1. Just after the switch is closed.
2. After a long time, the switch was closed.

Solution :

For just after closing the switch: potential difference across capacitor = 0

Q_C = 0

⇒ \(i=\frac{10}{2}=5 A\)

After a long time at steady state current, i = 0

and potential difference across capacitor = 10 V

∴ Q_C = 3 × 10 = 30 C

Example 6. Find out current I₁, I₂, I₃, the charge on the capacitor and \(\frac{\mathrm{dQ}}{\mathrm{dt}}\) of capacitor in the circuit which is initially uncharged in the following situations.

1. Just after the switch is closed
2. After a long time, the switch is closed.

Solution :

Initially, the capacitor is uncharged so its behavior is like a conductor Let potential at A be zero so at B and C also zero, and at F it is ε. Let potential at E is x so at D also x. Apply Kirchhoff’s 1st law at point E :

⇒ \(\frac{x-\varepsilon}{R}+\frac{x-0}{R}+\frac{x-0}{R}=0 \quad \Rightarrow \quad \frac{3 x}{R}=\frac{\varepsilon}{R}\)

⇒ \(x=\frac{\varepsilon}{3}\)

Q_C = 0

⇒ \(\mathrm{I}_1=\frac{-\varepsilon / 3+\varepsilon}{\mathrm{R}}=\frac{2 \varepsilon}{3 \mathrm{R}} \quad \Rightarrow \quad \mathrm{I}_2=\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\varepsilon}{3 \mathrm{R}} \Rightarrow \quad \mathrm{I}_3=\frac{\varepsilon}{3 \mathrm{R}}\)

Alternatively

⇒ \(\mathrm{i}_1=\frac{\varepsilon}{\mathrm{R}_{\text {eq }}}=\frac{\varepsilon}{\mathrm{R}+\frac{\mathrm{R}}{2}}=\frac{2 \varepsilon}{3 \mathrm{R}} \quad \Rightarrow \quad \mathrm{i}_2=\mathrm{i}_3=\frac{\mathrm{i}_1}{2}=\frac{\varepsilon}{3 \mathrm{R}}\)

At t = ∞ (finally)

the capacitor is completely charged so there will be no current through it.

⇒ \(\mathrm{I}_2=0, \quad \mathrm{I}_1=\mathrm{I}_3=\frac{\varepsilon}{2 \mathrm{R}}\)

⇒ \(\mathrm{V}_{\mathrm{E}}-\mathrm{V}_{\mathrm{B}}=\mathrm{V}_{\mathrm{D}}-\mathrm{V}_{\mathrm{C}}=(\varepsilon / 2 \mathrm{R}) \mathrm{R}=\varepsilon / 2\)

⇒ \(\mathrm{Q}_{\mathrm{c}}=\frac{\varepsilon \mathrm{C}}{2}, \quad \frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{I}_2=0\)

Example 7. A capacitor of capacitance C which is initially uncharged is connected to a battery. Find out heat dissipated in the circuit during the process of charging.
Solution:

Final status

Let potential at point A be 0, so at B also 0, and at C and D it is ε.

Finally, charge on the capacitor

⇒ \(\mathrm{Q}_{\mathrm{C}}=\varepsilon \mathrm{C}, \mathrm{U}_{\mathrm{i}}=0, \mathrm{U}_{\mathrm{f}}=\frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \mathrm{C}^2\)

(Now onwards remember that w.d. by battery = εQ if Q has flown out of the cell from high potential and w.d. on battery is εQ if Q has flown into the cell through high potential)

Heat produced = W = (U f – U i ) = \(\varepsilon^2 \mathrm{C}-\frac{1}{2} \varepsilon^2 \mathrm{C}=\frac{\mathrm{C} \varepsilon^2}{2} .\)

Example 8. A capacitor of capacitance C which is initially charged up to a potential difference ε is connected with a battery of emf ε such that the positive terminal of the battery is connected with a positive plate of the capacitor. Find out heat loss in the circuit during the process of charging.
Solution :

Since the initial and final charge on the capacitor is the same before and after connection. Here no charge will flow in the circuit so heat loss = 0

Example 9. A capacitor of capacitance C which is initially charged up to a potential difference ε is connected with a battery of emf ε/2 such that the positive terminal of the battery is connected with a positive plate of the capacitor. After a long time

1. Find out the total charge flow through the battery
2. Find out the total work done by the battery
3. Find out heat dissipated in the circuit during the process of charging.

Solution: Let the potential of A is 0 so at B it is \(\frac{\varepsilon}{2}.\)So the final charge on the capacitor = Cε/2

Charge flow through the capacitor = (Cε/2 – Cε) = –Cε/2

So the charge enters the battery.

Finally,

Change in energy of capacitor = U final – U initial

⇒ \(=\frac{1}{2} C\left(\frac{\varepsilon}{2}\right)^2-\frac{\varepsilon^2 C}{2}=\frac{1}{8} \varepsilon^2 C-\frac{1}{2} \varepsilon^2 C=-\frac{3 \varepsilon^2 C}{8}\)

Work done by battery \(=\frac{\varepsilon}{2} \times\left(-\frac{\varepsilon C}{2}\right)=-\frac{\varepsilon^2 C}{4}\)

Work done by battery = Change in energy of capacitor + Heat produced
Heat produced\(\frac{3 \varepsilon^2 C}{8}-\frac{\varepsilon^2 C}{4}=\frac{\varepsilon^2 C}{8}\)

Example 10. A capacitor of capacitance C, a resistor of resistance R, and a battery of emf ε are connected in series at t = 0. What is the maximum value of

1. The potential difference across the resistor,
2. The current in the circuit,
3. The potential difference across the capacitor,
4. The energy stored in the capacitors.
5. The power delivered by the battery and
6. The power is converted into heat.

Solution:

At t = 0 C is replaced by wire.

(1) V_rmax = ε

V_C = ε

⇒ \(\mathrm{U}_{\mathrm{c}}=\frac{1}{2} \mathrm{C} \varepsilon^2\)

⇒ \(\mathrm{P}_{\text {battery }}=\mathrm{i} . \mathrm{v} .=\frac{\varepsilon}{\mathrm{R}} \cdot \varepsilon=\frac{\varepsilon^2}{\mathrm{R}}\)

⇒ \(\Delta H=\frac{\varepsilon^2}{R} .\)

8. Distribution Of Charges On Connecting Two Charged Capacitors:

When two capacitors C₁ and C₂ are connected as shown in the figure

Common potential :

By charge conservation of plates A and C before and after connection.

Q₁ + Q₂ = C₁V + C₂V

⇒ \(V=\frac{Q_1+Q_2}{C_1+C_2}=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}=\frac{\text { Total charge }}{\text { Total capacitance }}\)

⇒ \(Q_1^{\prime}=C_1 V=\frac{C_1}{C_1+C_2}\left(Q_1+Q_2\right)\)

Heat Loss During Redistribution :

⇒ \(\Delta H=U_i-U_f=\frac{1}{2} \frac{C_1 C_2}{C_1+C_2}\left(V_1-V_2\right)^2\)

The loss of energy is in the form of Joule heating in the wire.

Note :

When plates of similar charges are connected (+ with + and – with –) then put all values (Q₁, Q₂, V₁, V₂) with a positive sign.

When plates of opposite polarity are connected (+ with –) then take charge and the potential of one of the plates to be negative.

Derivation Of The Above Formulae :

Let the potential of B and D be zero and the common potential on capacitors is V, then at A and C it will be C₁V + C₂V = C₁V₁ + C₂V₂

⇒ \(V=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}\)

⇒ \(H=\frac{1}{2} C_1 V_1^2+\frac{1}{2} C_2 V_2^2-\frac{1}{2}\left(C_1+C_2\right) V^2\)

⇒ \(=\frac{1}{2} C_1 V_1^2+\frac{1}{2} C_2 V_2^2-\frac{1}{2} \frac{\left(C_1 V_1+C_2 V_2\right)^2}{\left(C_1+C_2\right)}\)

⇒ \(=\frac{1}{2}\left[\frac{\mathrm{C}_1^2 \mathrm{~V}_1^2+\mathrm{C}_1 \mathrm{C}_2 \mathrm{~V}_1^2+\mathrm{C}_2 \mathrm{C}_1 \mathrm{~V}_2^2+\mathrm{C}_2^2 \mathrm{~V}_2^2-\mathrm{C}_1^2 \mathrm{~V}_1^2-\mathrm{C}_2 \mathrm{~V}_2^2-2 \mathrm{C}_1 \mathrm{C}_2 \mathrm{~V}_1 \mathrm{~V}_2}{\mathrm{C}_1+\mathrm{C}_2}\right]\)

⇒ \(=\frac{1}{2} \frac{C_1 C_2}{C_1+C_2}\left(V_1-V_2\right)^2\)

⇒ \(H=\frac{1}{2} \frac{C_1 C_2}{C_1+C_2}\left(V_1-V_2\right)^2\)

When oppositely charge terminals are connected then

C₁V + C₂V = C₁V₁ – C₂V₂

⇒ \(V=\frac{C_1 V_1-C_2 V_2}{C_1+C_2} ; \quad H=\frac{1}{2} \frac{C_1 C_2}{C_1+C_2}\left(V_1+V_2\right)^2\)

Solved Examples

Example 11 Find out the following if A is connected with C and B is connected with D.

1. How much charge flows in the circuit?
2. How much heat is produced in the circuit?

Solution:

Let the potential of B and D be zero and the common potential on capacitors is V, then at A and C, it will be V.

By charge conservation,

3V + 2V = 40 + 30

5V = 70

V = 14 volt

Charge flow

= 40 – 28

= 12 μC

Now final charges on each plate are shown in the figure

Heat produced \(=\frac{1}{2} \times 2 \times(20)^2+\frac{1}{2} \times 3 \times(10)^2-\frac{1}{2} \times 5 \times(14)^2\)

= 400 + 150 – 490

= 550 – 490 = 60 μJ

Note:

1. When capacitor plates are joined then the charge remains conserved.
2. We can also use the direct formula of redistribution as given above.

Example 12. Repeat the above question if A is connected with D and B is connected with C.

Solution: Let the potential of B and C be zero and the common potential on capacitors be V, then at A and D it will be V

2V + 3V = 10

⇒ V = 2 volt

Now charge on each plate is shown in the figure

Heat produced = \(400+150-\frac{1}{2} \times 5 \times 4\)

= 550 – 10

= 540 μJ

Example 13. Three capacitors as shown by capacitance 1 μF, 2μF, and 2μF are charged up to a potential difference of 30 V, 10 V, and 15 V respectively. If terminal A is connected with D, C is connected with E, and F is connected with B Then find out the charge flow in the circuit and find the final charges on capacitors.

Solution: Let charge flow is q.

Now applying Kirchhoff’s voltage low

⇒ \(-\frac{(q-20)}{2}-\frac{(30+q)}{2}+\frac{30-q}{1}=0\)

–2q = – 25

q = 12.5 μC

Final Charges on plates

9. Combination Of Capacitors :

9.1 Series Combination :

When initially uncharged capacitors are connected as shown in the combination is called a series combination.

All capacitors will have the same charge but different potential differences across them.

We can say that \(V_1=\frac{Q}{C_1}\)

V₁ = potential across C₁

Q = charge on positive plate of C₁

C₁ = capacitance of capacitor similarly

⇒ \(V_2=\frac{Q}{C_2}, V_3=\frac{Q}{C_3} \ldots \ldots .\)

⇒ \(V_1: V_2: V_3=\frac{1}{C_1}: \frac{1}{C_2}: \frac{1}{C_3}\)

We can say that the potential difference across capacitors is inversely proportional to their capacitance in a series combination.

⇒ \(V \propto \frac{1}{C}\)

⇒ \(\mathrm{V}_1=\frac{\frac{1}{\mathrm{C}_1}}{\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}+\ldots \ldots} \mathrm{V}\)

⇒ \(V_2=\frac{\frac{1}{C_2}}{\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\ldots . .} V\)

⇒ \(V_3=\frac{\frac{1}{C_3}}{\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\ldots . .} V\)

Where V = V₁ + V₂ + V₃

Equivalent Capacitance: Equivalent capacitance of any combination is that capacitance which when connected in place of the combination stores the same charge and energy that of the combination.

In series :

⇒ \(\frac{1}{\mathrm{C}_{\text {eq }}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3} \ldots \ldots .\)

Note: In series combination equivalent is always less than the smallest capacitor of the combination.

Energy stored in the combination

⇒ \(\mathrm{U}_{\text {combination }}=\frac{\mathrm{Q}^2}{2 \mathrm{C}_1}+\frac{\mathrm{Q}^2}{2 \mathrm{C}_2}+\frac{\mathrm{Q}^2}{2 \mathrm{C}_3}\)

⇒ \(\mathrm{U}_{\text {combination }}=\frac{\mathrm{Q}^2}{2 \mathrm{C}_{\mathrm{eq}}}\)

The energy supplied by the battery in charging the combination

⇒ \(\mathrm{U}_{\text {battery }}=\mathrm{Q} \times \mathrm{V}=\mathrm{Q} \cdot \frac{\mathrm{Q}}{\mathrm{C}_{\mathrm{eq}}}=\frac{\mathrm{Q}^2}{\mathrm{C}_{\text {eq }}} \Rightarrow \quad \frac{\mathrm{U}_{\text {combination }}}{\mathrm{U}_{\text {battery }}}=\frac{1}{2}\)

μ Half of the energy supplied by the battery is stored in the form of electrostatic energy and half of the energy is converted into heat through resistance.

Derivation of Formulae :

Meaning Of Equivalent Capacitor

⇒ \(C_{e q}=\frac{Q}{V}\)

Now, Initially, the capacitor has no charge.

Applying Kirchhoff’s voltage law

⇒ \(\frac{-Q}{C_1}+\frac{-Q}{C_2}+\frac{-Q}{C_3}+V=0\)

⇒ \(V=Q\left[\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right]\)

⇒ \(\frac{V}{Q}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\)

⇒ \(\frac{1}{C_{e q}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\) in general \(\frac{1}{C_{\text {eq }}}=\sum_{n=1}^n \frac{1}{C_n}\)

Solved Examples

Example 14. Three initially uncharged capacitors are connected in series as shown in the circuit with a battery of EMF 30V. Find out the following:-

1. Charge flow through the battery,
2. The potential energy in 3 μF capacitor.
3. U total in capacitors
4. The heat produced in the circuit

Solution:

⇒ \(\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3+2+1}{6}=1\)

C_eq = 1 μF.

Q = C_eq V = 30μC.

charge on 3μF capacitor = 30μC  energy = \(\frac{Q^2}{2 C}=\frac{30 \times 30}{2 \times 3}=150 \mu \mathrm{J}\)

⇒ \(\mathrm{U}_{\text {total }}=\frac{30 \times 30}{2} \mu \mathrm{J}=450 \mu \mathrm{J}\)

Heat produced = (30 μC) (30) – 450 μJ = 450 μJ.

Example 15. Two capacitors of capacitance 1 μF and 2μF are charged to potential differences of 20V and 15V as shown in the figure. If now terminals B and C are connected to terminal A with a positive battery and D with a negative terminal of the battery of emf 30 V. Then find out the final charges on both the capacitor

Solution: Now apply the Kirchoff voltage law

⇒ \(\frac{-(20+q)}{1}-\frac{30+q}{2}+30=0\)

– 40 – 2q – 30 – q = – 60

3q = –10

Charge flow = –10/3 μC.

50 Charge on capacitor of capacitance \(1 \mu F=20+q=\frac{50}{3}\)

Charge on capacitor of capacitance \(2 \mu F=30+q=\frac{80}{3}\)

9.2 Parallel Combination :

When one plate of each capacitor (more than one) is connected and the other plate of each capacitor is connected, such a combination is called a parallel combination.

All capacitors have the same potential difference but different charges.

We can say that :

Q₁ = C₁V

Q₁ = Charge on capacitor C₁

C₁ = Capacitance of capacitor C₁

V = Potential across capacitor C₁

Q₁ : Q₂ : Q₃ = C₁ : C₂ : C₃

The charge on the capacitor is proportional to its capacitance Q μ C

⇒ \(Q_1=\frac{C_1}{C_1+C_2+C_3} Q \quad Q_2=\frac{C_2}{C_1+C_2+C_3} Q \quad Q_3=\frac{C_3}{C_1+C_2+C_3} Q\)

Where Q = Q₁ + Q₂ + Q₃ ……

Note: Maximum charge will flow through the capacitor of the largest value.

Equivalent capacitance of parallel combination

C_eq = C₁ + C₂ + C₃

Note: Equivalent capacitance is always greater than the largest capacitor of combination.

Energy stored in the combination :

⇒ \(V_{\text {combination }}=\frac{1}{2} C_1 V^2+\frac{1}{2} C_2 V^2+\ldots .=\frac{1}{2}\left(C_1+C_2+C_3 \ldots . .\right) V^2\)

⇒ \(=\frac{1}{2} C_{\text {eq }} \mathrm{V}^2\)

⇒ \(\mathrm{U}_{\text {battery }}=\mathrm{QV}=\mathrm{CV}^2\)

⇒ \(\frac{U_{\text {combination }}}{U_{\text {battery }}}=\frac{1}{2}\)

Note: Half of the energy supplied by the battery is stored in the form of electrostatic energy and half of the energy is converted into heat through resistance.

Formulae Derivation for parallel combination :

Q = Q₁ + Q₂ + Q₃

= C₁V + C₂V + C₃V

= V(C₁ + C₂ + C₃)

⇒ \(\frac{Q}{V}=C_1+C_2+C_3\)

C_eq = C₁ + C₂ + C₃

In general \(C_{e q}=\sum_{n=1}^n C_n\)

Solved Examples

Example 16. Three initially uncharged capacitors are connected to a battery of 10 V in parallel combination out the following

1. Charge flow from the battery
2. Total energy stored in the capacitors
3. The heat produced in the circuit
4. The potential energy in the 3μF capacitor.

Solution :

Q = (30 + 20 + 10)μC = 60μC

⇒ \(U_{\text {total }}=\frac{1}{2} \times 6 \times 10 \times 10=300 \mu \mathrm{J}\)

heat produced = 60 × 10 – 300 = 300 μJ

9.3 Mixed Combination :

The combination that contains a mixing of a series of parallel combinations or other complex combinations falls into the mixed category.

There are two types of mixed combinations

1. Simple
2. Complex.

Solved Examples

Example 17. In the given circuit find out the charge on the 6μF and 1 μF capacitors.

Solution: It can be simplified as

⇒ \(C_{\text {eq }}=\frac{18}{9}=2 \mu \mathrm{F}\)

charge flow through the cell = 30 × 2 μC

Q = 60 μC

Now charge on 3μF = Charge on 6μF= 60 μC

Potential difference across 3μF = 60/ 3= 20 V

∴Charge on 1 μF = 20 μC.

Example 18. Comprehension: In the arrangement of the capacitors shown in the figure, each C₁ capacitor has a capacitance of 3μF and each C₂ capacitor has a capacitance of 2μF Then,

1. Equivalent capacitance of the network between the points a and b is :

1. 1μF
2. 2μF
3. 4μC
4. \(\frac{3}{2} \mu F\)

2. If V_ab = 900 V, the charge on each capacitor nearest to the points ‘a’ and ‘b’ is :

1. 300 μC
2. 600 μC
3. 450 μC
4. 900 μC

3. If V_ab = 900 V, then the potential difference across points c and d is :

1. 60 V
2. 100 V
3. 120 V
4. 200 V

Solution :

1.  \(\frac{1}{C_1^1}=\frac{1}{C_1}-\frac{1}{C_1}+\frac{1}{C_2} \quad \Rightarrow \quad C_1^1=1 \mu F\)

C₂1 = C₂ + C₂ 1 = 3μF C_eq = 1 μF

2.  C_eq= 1 μF Q = C_eq V = 900μF

charge on nearest capacitor = 900μF

3. From the point of potential method

V_C – V_d = 100V

Example 19. Comprehension: Capacitor C₃ in the circuit is a variable capacitor (its capacitance can be varied). The graph is plotted between the potential difference V₁ (across capacitor C₁) versus C₃. Electric potential V₁ approaches on the asymptote of 10 V as C₃ → ∞.

1. EMF of the battery is equal to :

1. 10 V
2. 12 V
3. 16 V
4. 20 V

2. The capacitance of the capacitor C₁ has value :

1. 2 μ F
2. 6 μ F
3. 8 μ F
4. 12 μ F

3. The capacitance of C₂ is equal to :

1. 2 μ F
2. 6 μ F
3. 8 μ F
4. 12 μ F

Solution:

When C₃ = μ, there will be no charge on C₁

As V₁ = 10 V therefore V = 10 V

From graph when C₃ = 10 μ F, V₁ = 6 V

Charge on C₁= Charge on C₂ + Charge on C₃

6C₁ = 4C₂ + 40 μ C …. (1)

Also when C₃ = 6 μ F, V₁ = 5V

Again using the charge equation

5C₁ = 5C₂ + 30 μ C ….(2)

Solving (1) and (2)

C₁ = 8 μ F

C₂ = 2 μF.

10. Charging And Discharging Of A Capacitor

10.1 Charging of a condenser: In the following circuit. If key 1 is closed then the condenser gets charged. Finite time is taken in the charging process. The quantity of charge at any instant of time t is given by q = q₀[1 – e^-(t/RC)]

Where q₀ = maximum final value of charge at t = τ.

According to this equation, the quantity of charge on the condenser increases exponentially with the increase of time.

If t = RC = τ then

q = q₀ [1 – e^-(t/RC)] = \(q_0\left[1-\frac{1}{e}\right]\)

q = q₀ (1 – 0.37) = 0.63 q₀

= 63% of q₀

Time t = RC is known as the time constant.

The time constant is the time during which the charge rises on the condenser plates to 63% of its maximum value.

The potential difference across the condenser plates at any instant of time is given by V = V₀[1 – e^-(t/RC)] volt

The potential curve is also similar to that of charge. During the charging process, an electric current flows in the circuit for a small interval of time which is known as the transient current.

The value of this current at any instant of time is given by I = I ₀[e^-(t/RC))] ampere

According to this equation, the current falls in the circuit exponentially (Fig.).

If t = RC = τ = Time constant

⇒ \(\mathrm{I}=\mathrm{I}_0 \mathrm{e}^{(-{RC/RC})}=\frac{\mathrm{I}_0}{\mathrm{e}}=0.37 \mathrm{I}_0=37 \% \text { of } \mathrm{I}_0\)

The time constant is the time during which the current in the circuit falls to 37% of its maximum value.

Derivation of formulae for charging of capacitor

It is given that initially capacitor is uncharged.

Let at any time Apply Kirchoff voltage law

⇒ \(\varepsilon-i R-\frac{q}{C}=0 \Rightarrow i R=\frac{\varepsilon C-q}{C} \quad \Rightarrow \quad i=\frac{\varepsilon C-q}{C R} \quad \Rightarrow \quad \frac{d q}{d t}=\frac{\varepsilon C-q}{C R}\)

⇒ \(\frac{d q}{d t}=\frac{\varepsilon C-q}{C R} \Rightarrow \frac{C R}{\varepsilon C-q} \cdot d q=d t.\)

⇒ \(\int_0^q \frac{d q}{\varepsilon C-q}=\int_0^t \frac{d t}{R C}\)

⇒ \(-\ln (\varepsilon C-q)+\ln \varepsilon C=\frac{t}{R C}\)

⇒ \(\ln \frac{\varepsilon C}{\varepsilon C-q}=\frac{t}{R C}\)

RC = time constant of the RC series circuit.

After One Time Constant

⇒ \(q=\varepsilon C\left(1-\frac{1}{e}\right)=\varepsilon C(1-0.37)=0.63 \varepsilon C\)

Current At Any Time t

⇒ \(\mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}}=\varepsilon C\left(-\mathrm{e}^{-\mathrm{t} / \mathrm{RC}}\left(-\frac{1}{\mathrm{RC}}\right)\right)=\frac{\varepsilon}{\mathrm{R}} \mathrm{e}^{-\mathrm{t} / \mathrm{RC}}\)

The voltage across the capacitor after one-time constant V = 0.63 ε

Q = CV, V_C = ε(1 – e^-t/RC)

The voltage across the resistor

VR = iR = εe^-(t/RC)

By energy conservation,

Heat dissipated = work done by battery – ΔU capacitor

⇒ \(=C \varepsilon(\varepsilon)-\left(\frac{1}{2} C \varepsilon^2-0\right)=\frac{1}{2} C \varepsilon^2\)

Alternatively :

⇒ \(\text { Heat }=H=\int_0^{\infty} i^2 R d t=\int_0^{\infty} \frac{\varepsilon^2}{R^2} e^{-\frac{2 t}{R C}} R d t=\frac{\varepsilon^2}{R} \int_0^{\infty} e^{-2 t / R C} d t \quad=\frac{\varepsilon^2}{R}\left[\frac{e^{-\frac{2 t}{R C}}}{-2 / R C}\right]_0^{\infty}\)

In the figure time constant of (2) is more than (1)

⇒ \(=-\frac{\varepsilon^2 R C}{2 R}\left[e^{-\frac{2 t}{R C}}\right]_0^{\infty}=\frac{\varepsilon^2 C}{2}\)

Solved Examples

Example 20: A capacitor is connected to a 12 V battery through a resistance of 10μ. It is found that the potential difference across the capacitor rises to 4.0 V in 1 μs. Find the capacitance of the capacitor.
Solution :

The charge on the capacitor during charging is given by Q = Q₀(1 –e^-(t/RC)).

Hence, the potential difference across the capacitor is V = Q/C = Q₀/C (1 – e^-(t/RC)).

Here, at t = 1 μs, the potential difference is 4V whereas the steady potential difference is

Q₀/C = 12V. So,  4V = 12V(1 – e^-(t/RC))

⇒ \(\text { or } 1-\mathrm{e}^{-\mathrm{URC}}=\frac{1}{3} \text { or } \mathrm{e}^{-\mathrm{URC}}=\frac{2}{3} \text { or } \frac{\mathrm{t}}{\mathrm{RC}}=\ln \left(\frac{3}{2}\right)=0.405\)

⇒ \(\mathrm{RC}=\frac{\mathrm{t}}{0.405}=\frac{1 \mu \mathrm{s}}{0.45}=2.469 \mu \mathrm{s} \text { or } \mathrm{C}=\frac{2.469 \mu \mathrm{s}}{10 \Omega}=0.25 \mu \mathrm{F}\)

Method For Objective :

In any circuit when there is only one capacitor then \(q=Q_{s t}\left(1-e^{-t / \tau}\right); Q_{s t}\) = steady-state charge on capacitor (has been found in article 6 in this sheet)

τ = R_eff

Reflective is the resistance between the capacitor when the battery is replaced by its internal resistance.

10.2 Discharging of a condenser :

In the above circuit (in article 8.1) if key 1 is opened and key 2 is closed then the condenser gets discharged.

The quantity of charge on the condenser at any instant of time t is given by q = q₀e^-(t/RC) i.e. the charge falls exponentially.

If t = RC = τ = time constant, then

⇒ \(q=\frac{q_0}{e}=0.37 q_0=37 \% \text { of } q_0\)

The time constant is the time during which the charge on the condenser plate discharge process falls to 37%

The dimensions of RC are those of time i.e. MºLºT¹ and the dimensions of \(\frac{1}{\mathrm{RC}}\) frequency i.e. MºLºT^-1

The potential difference across the condenser plates at any instant of time t is given by V = V₀e^-(t/RC) Volt.

The transient current at any instant of time is given by the I = –I₀e^-(t/RC)ampere. i.e. the current in the circuit decreases exponentially but its direction is opposite to that of the charging current.

Derivation of equation of discharging circuit :

Applying K.V.L.

⇒ \(+\frac{q}{C}-i R=0, \quad i=\frac{q}{C R}\)

⇒ \(\int_Q^q \frac{-d q}{q}=\int_0^t \frac{d t}{C R}-\ln \frac{q}{Q}=+\frac{t}{R C}\)

⇒ \(q=Q \cdot e^{-t / R C} \Rightarrow \quad i=-\frac{d q}{d t}=\frac{Q}{R C} e^{-t / R C}\)

Solved Examples

Example 21. Two parallel conducting plates of a capacitor of capacitance C containing charges Q and –2Q at a distance d apart. Find out the potential difference between the plates of capacitors.
Solution :

Capacitance = C

Electric field =\(=\frac{3 Q}{2 A \varepsilon_0}\)

⇒ \(V=\frac{3 Q d}{2 A \varepsilon_0} \Rightarrow V=\frac{3 Q}{2 C}\)

11. Capacitors With Dielectric

In the Absence Of Dielectric

⇒ \(\mathrm{E}=\frac{\sigma}{\varepsilon_0}\)

When a dielectric fills the space between the plates then molecules having dipole moment align themselves in the direction of electric field.

σb = induced charge density (called bound charge because it is not due to free electrons).

• For polar molecules dipole moment  of 0
• For non-polar molecules dipole moment = 0

Capacitance in the presence of dielectric

⇒ \(C=\frac{\sigma \mathrm{A}}{\mathrm{V}}=\frac{\sigma \mathrm{A}}{\frac{\sigma}{\mathrm{K} \varepsilon_0} \cdot \mathrm{d}}=\frac{\mathrm{AK} \varepsilon_0}{\mathrm{~d}}=\frac{\mathrm{AK} \varepsilon_0}{\mathrm{~d}}\)

Here capacitance is increased by a factor K.

⇒ \(C=\frac{A K \varepsilon_0}{d}\)

Polarisation Of Material :

When a nonpolar substance is placed in the electric field then a dipole moment is induced in the molecule. This induction of dipole moment is called the polarisation of material. The induced charge also produces an electric field.

Ib = induced (bound) charge density.

⇒ \(\mathrm{E}_{\mathrm{in}}=\mathrm{E}-\mathrm{E}_{\mathrm{ind}} \quad=\frac{\sigma}{\varepsilon_0}-\frac{\sigma_{\mathrm{b}}}{\varepsilon_0}\)

The ratio of the electric field between the plates in the absence of a dielectric and the presence of

a dielectric is constant for a dielectric material. This ratio is called the ‘Dielectric constant’ of that
material. It is represented by r or k.

⇒ \(E_{\text {in }}=\frac{\sigma}{\mathrm{K} \varepsilon_0} \quad \Rightarrow \quad \sigma_{\mathrm{b}}=\sigma\left(1-\frac{1}{\mathrm{~K}}\right)\)

If the medium does not fill between the plates completely then the electric field will be as shown in the figure

Case: 1

The total electric field produced by a bound-induced charge on the dielectric outside the slab is zero because they cancel each other.

Case: 2

Comparison of E (electric field),  (surface charges density), Q (charge ), C (capacitance), and before and after inserting a dielectric slab between the plates of a parallel plate capacitor.

⇒ \(C=\frac{\varepsilon_0 A}{d}\)

Q = CV

Here is the potential difference between the plates,

Ed = V

⇒ \(E=\frac{V}{d}\)

⇒ \(\frac{\mathrm{V}}{\mathrm{d}}=\frac{\sigma}{\varepsilon_0}\)

⇒ \(C^{\prime}=\frac{A \varepsilon_0 K}{d}\)

Q’ = C’V

⇒ \(\mathrm{E}^{\prime}=\frac{\sigma’}{\mathrm{K} \varepsilon_0}=\frac{\mathrm{C}}{\mathrm{A} \varepsilon_0}\)

⇒ \(=\frac{\mathrm{V}}{\mathrm{d}} \text { also }\)

I’d = V

⇒ \(E^{\prime}=\frac{V}{d}\)

⇒ \(\frac{V}{d}=\frac{\sigma^{\prime}}{K \varepsilon_0}\)

Equating both

⇒ \(\frac{\sigma}{\varepsilon_0}=\frac{\sigma}{K \varepsilon_0}\)

σ’ = Kσ

In the presence of dielectric, i.e. in case II, the capacitance of a capacitor is higher.

Energy density in a dielectric = \(\frac{1}{2} \varepsilon_0 \varepsilon_{\mathrm{r}} \mathrm{E}^2\)

Solved Examples

Example 22. If a dielectric slab of thickness t and area A is inserted in between the plates of a parallel plate capacitor of plate area A and the distance between the plates d (d > t) then find out the capacitance of the system. What do you predict about the dependence of capacitance on the location of the slab?
Solution :

⇒ \(C=\frac{Q}{V}=\frac{\sigma A}{V}\)

⇒ \(\mathrm{V}=\frac{\sigma \mathrm{t}_1}{\varepsilon_0}+\frac{\sigma \mathrm{t}}{\mathrm{K} \varepsilon_0}+\frac{\sigma \mathrm{t}_2}{\varepsilon_0}\)

⇒ \(\left(t_1+t_2=d-t\right)=\frac{\sigma}{\varepsilon_0}\left[t_1+t_2+\frac{t}{k}\right]\)

⇒ \(V=\frac{\sigma}{\varepsilon_0}\left[\mathrm{~d}-\mathrm{t}+\frac{\mathrm{t}}{\mathrm{k}}\right]=\frac{\mathrm{Q}}{\mathrm{C}}=\frac{\sigma \mathrm{A}}{\mathrm{C}}\)

⇒ \(C=\frac{\varepsilon_0 A}{d-t+t / K}\)

Note:

Capacitance does not depend upon the position of the dielectric (it can be shifted up or down but capacitance does not change).

If the slab is of metal then: C \(C=\frac{A \varepsilon_0}{d-t}\)

Solved Examples

Example 23. A dielectric of constant K is slipped between the plates of parallel plate condenser in half of the space as shown in the figure. If the capacity of the air condenser is C, then new capacitance between A and B will be

1. \(\frac{C}{2}\)
2. \(\frac{\mathrm{C}}{2 \mathrm{~K}}\)
3. \(\frac{C}{2}[1+K]\)
4. \(\frac{2[1+K]}{C}\)

Solution: This system is equivalent to two capacitors in parallel with area of each plate

⇒ \(\frac{\mathrm{A}}{2}\)

⇒\(\mathrm{C}^{\prime}=\mathrm{C}_1+\mathrm{C}_2=\frac{\varepsilon_0 \mathrm{~A}}{2 \mathrm{~d}}+\frac{\varepsilon_0 \mathrm{AK}}{2 \mathrm{~d}}=\frac{\varepsilon_0 \mathrm{~A}}{2 \mathrm{~d}}[1+\mathrm{K}]=\frac{\mathrm{C}}{2}[1+\mathrm{K}]\)

Hence the correct answer will be (3).

Example 24. The parallel plates of a capacitor have an area of 0.2 m² and are 10^-2 m apart. The original potential difference between them is 3000 V, and it decreases to 1000 V when a sheet of dielectric is inserted between the plates filling the full space. Compute: (∈₀= 9 x 10^-12 S. I. units)

1. Original capacitance C₀.
2. The charge Q on each plate.
3. Capacitance C after insertion of the dielectric.
4. Dielectric constant K.
5. The permittivity ∈ of the dielectric.
6. The original field E₀ between the plates.
7. The electric field E after insertion of the dielectric.

Solution:

⇒ \(C_0=\frac{\epsilon_0 A}{d}=\frac{0.2 \epsilon_0}{10^{-2}}=20 \epsilon_0 \quad=20 \times 9 \times 10^{-12}=180 \mathrm{pF}\)

Q = C₀V = 180 × 10–12 × 3000 = 5.4 × 10–7 C

⇒ \(C_1=\frac{Q}{V_1}=\frac{5.4 \times 10^{-7}}{1000}=540 \mathrm{pF}\)

⇒ \(\mathrm{K}=\frac{\mathrm{C}_1}{\mathrm{C}_0}=\frac{540}{180}=3\)

⇒ \(\epsilon=\epsilon_{\mathrm{r}} \epsilon_0=K \epsilon_0\)

⇒ \(E_0=\frac{V}{d}=\frac{3000}{10^{-2}}=3 \times 10^5 \mathrm{~V} / \mathrm{m}\)

⇒ \(=\frac{V_1}{d}=\frac{1000}{10^{-2}}=1 \times 10^5 \mathrm{~V} / \mathrm{m}\)

12. Combination Of Parallel Plates

Example 25. Find out the equivalent capacitance between A and B

Solution: Put numbers on the plates The charges will be as shown in the figure.

V₁₂ = V₃₂ = V₃₄

so all the capacitors are in parallel combination.

C_eq = C₁ + C₂ + C₃

Example 26. Find out the equivalent capacitance between A and B.

These are only two capacitors. C_eq= C₁ + C₂

13. Other Types Of Capacitors

Spherical Capacitor :

This arrangement is known as a spherical capacitor.

⇒ \(V_1-V_2=\left[\frac{K Q}{a}-\frac{K Q}{b}\right]-\left[\frac{K Q}{b}-\frac{K Q}{b}\right]=\frac{K Q}{a}-\frac{K Q}{b}\)

⇒ \(\mathrm{C}=\frac{\mathrm{Q}}{V_1-V_2}=\frac{Q}{\frac{K Q}{a}-\frac{K Q}{b}}=\frac{a c}{K(b-a)}=\frac{4 \pi \varepsilon_0 a b}{b-a}\)

⇒ \(\mathrm{C}=\frac{4 \pi \varepsilon_0 \mathrm{ab}}{\mathrm{b}-\mathrm{a}}\)

If b >> a

C = 4πε₀a

⇒ \(\mathrm{C}=\frac{4 \pi \varepsilon_0 \varepsilon_{\mathrm{r}_2} \mathrm{ab}}{\mathrm{b}-\mathrm{a}}\)

Cylindrical Capacitor

There are two co-axial conducting cylindrical surfaces

where l >> a and l>> b

where a and b are the radius of cylinders.

Capacitance per unit length

⇒ \(\mathrm{C}=\frac{\lambda}{\mathrm{V}}=\frac{\lambda}{2 \mathrm{~K} \lambda \ell \mathrm{n} \frac{\mathrm{b}}{\mathrm{a}}}=\frac{4 \pi \varepsilon_0}{2 \ell \mathrm{n} \frac{\mathrm{b}}{\mathrm{a}}}=\frac{2 \pi \varepsilon_0}{\ell \ln \frac{\mathrm{b}}{\mathrm{a}}}\)

Capacitance per unit length = \(=\frac{2 \pi \varepsilon_0}{\ln \frac{\mathrm{b}}{\mathrm{a}}} \mathrm{F} / \mathrm{m}\)

Miscellaneous Solved Examples

Problem 1. Find out the capacitance of the earth. (Radius of the earth = 6400 km)
Solution :

⇒ \(\mathrm{C}=4 \pi \varepsilon_0 \mathrm{R}=\frac{6400 \times 10^3}{9 \times 10^9}=711 \mu \mathrm{F}\)

Problem 2. When two isolated conductors A and B are connected by a conducting wire positive charge will flow from

1. A to B
2. B to A
3. will not flow
4. Can not say.

Solution: Charge always flows from higher potential body to lower potential body

Hence \(V_A=\frac{30}{10}=3 \mathrm{~V} \Rightarrow V_B=\frac{20}{5}=4 \mathrm{~V} . \text { As } V_B>V_B\)

(2) is the correct Answer.

Problem 3. A conductor of capacitance 10μF connected to another conductor of capacitance 40 μF having equal charges 100 μC initially. Find out the final voltage and heat loss during the process.
Answer :

1. V = 4V
2. H = 225 μJ.

Solution : C₁ = 10µF C₂ = 40µF

Q₁ = 100 µC Q₂ = 100µC

V₁ = Q₁/C₁ = 10 V V₂ = Q₂/C₂ = 2.5

Final voltage (V)\(=\frac{C_1 V_1+C_2 V_2}{\left(C_1+C_2\right)}=\frac{Q_1+Q_2}{C_1+C_2}=\frac{200 \mu}{50 \mu}=4 V\)

Heat loss during the process \(=\frac{1}{2}\left[\mathrm{C}_1 \mathrm{~V}_1^2+\mathrm{C}_2 \mathrm{~V}_2^2\right]-\frac{1}{2} \mathrm{~V}^2\left(\mathrm{C}_1+\mathrm{C}_2\right)\)

⇒ \(=\frac{1}{2}\left[Q_1 V_1+Q_2 V_2\right]-\frac{1}{2} V^2\left(C_1+C_2\right)\)

⇒ \(=\frac{1}{2} \times 100 \mu[12.5]-\frac{1}{2} \times 16(50) \mu=225 \mu \mathrm{J}\)

Problem 4. In the above question, if the positive terminal of the battery is connected to the negative plate of the capacitor. Find out heat loss in the circuit during the process of charging.

1. Net charge flow through battery = 2εC
2. Work done by battery = ε × 2εC = 2ε²C
3. The heat produced = 2ε²C.

Solution: From figure

Net charge flow through battery = Qfinal – Qinitial = εC – (–εC) = 2εC

work done by battery (W) = Q × V = 2εC × ε = 2ε²C

or Heat produced = 2ε²C

Problem 5. Find out capacitance between A and B if two dielectric slabs of dielectric constant K₁ and K₂ of thickness d₁ and d₂ and each of area A are inserted between the plates of parallel plate capacitor of plate area A as shown in the figure.

Solution:

⇒ \(C=\frac{\sigma \mathrm{A}}{\mathrm{V}} ; V=\mathrm{E}_1 \mathrm{~d}_1+\mathrm{E}_2 \mathrm{~d}_2=\frac{\sigma \mathrm{d}_1}{\mathrm{~K}_1 \varepsilon_0}+\frac{\sigma \mathrm{d}_2}{\mathrm{~K}_2 \varepsilon_0}=\frac{\sigma}{\varepsilon_0}\left(\frac{\mathrm{d}_1}{\mathrm{k}_1}+\frac{\mathrm{d}_2}{\mathrm{k}_2}\right)\)

⇒ \(C=\frac{A \varepsilon_0}{\frac{d_1}{K_1}+\frac{d_2}{K_2}} \Rightarrow \frac{1}{C}=\frac{d_1}{A K_1 \varepsilon_0}+\frac{d_2}{A K_2 \varepsilon_0}\)

This formula suggests that the system between A and B can be considered as a series combination of two capacitors.

Problem 6. Find out capacitance between A and B if three dielectric slabs of dielectric constant K₁ of area A₁ and thickness d, K₂ of area A₂ and thickness d 1 and K₃ of area A₂ and thickness d₂ are inserted between the plates of parallel plate capacitor of plate area A as shown in the figure. (Given the distance between the two plates d =d 1+d₂)

Solution:

It is Equivalent \(C=C_1+\frac{C_2 C_3}{C_2+C_3}\)

⇒ \(C=\frac{A_1 K_1 \varepsilon_0}{d_1+d_2}+\frac{\frac{A_2 K_2 \varepsilon_0}{d_1} \cdot \frac{A_2 K_3 \varepsilon_0}{d_2}}{\frac{A_2 K_2 \varepsilon_0}{d_1}+\frac{A_2 K_3 \varepsilon_0}{d_2}}\)

⇒ \(=\frac{A_1 K_1 \varepsilon_0}{d_1+d_2}+\frac{A_2^2 K_2 K_3 \varepsilon_0^2}{A_2 K_2 \varepsilon_0 d_2+A_2 K_3 \varepsilon_0 d_1}\)

⇒ \(=\frac{\mathrm{A}_1 \mathrm{~K}_1 \varepsilon_0}{\mathrm{~d}_1+\mathrm{d}_2}+\frac{\mathrm{A}_2 \mathrm{~K}_2 \mathrm{~K}_3 \varepsilon_0}{\mathrm{~K}_2 \mathrm{~d}_2+\mathrm{K}_3 \mathrm{~d}_1}\)

Problem 7. Find out capacitance between A and B if two dielectric slabs of dielectric constant K₁ and K₂ of area A₁ and A₂ and each of thickness d are inserted between the plates of parallel plate capacitor of plate area A as shown in the figure.
Solution :

⇒ \(\mathrm{C}_1=\frac{\mathrm{A}_1 \mathrm{~K}_1 \varepsilon_0}{d}, \mathrm{C}_2=\frac{\mathrm{A}_2 \mathrm{~K}_2 \varepsilon_0}{\mathrm{~d}}\)

⇒\(E_1=\frac{V}{d}=\frac{\sigma_1}{\mathrm{~K}_1 \varepsilon_0}, E_2=\frac{V}{d}=\frac{\sigma_2}{\mathrm{~K}_2 \varepsilon_0}\)

⇒ \(\sigma_1=\frac{\mathrm{K}_1 \varepsilon_0 V}{\mathrm{~d}} \quad \sigma_2=\frac{\mathrm{K}_2 \varepsilon_0 V}{\mathrm{~d}}\)

⇒ \(C=\frac{Q_1+Q_2}{V}=\frac{\sigma_1 A_1+\sigma_2 A_2}{V}=\frac{K_1 \varepsilon_0 A_1}{d}+\frac{K_2 \varepsilon_0 A_2}{d}\)

The combination is equivalent to :

∴ C = C₁ + C₂

Problem 8. Find out the equivalent capacitance between A and B.

Solution:

⇒ \(C_{e q}=\frac{2 C}{3}\)

Other method :

⇒ \(C_{e q}=\frac{Q}{V}=\frac{2 x A}{V}\)

V = V₂ – V₄ = (V₂ – V₃) + (V₃ – V₄)

⇒ \(=\frac{x d}{\varepsilon_0}+\frac{2 x d}{\varepsilon_0}=\frac{3 x d}{\varepsilon_0}\)

⇒ \(C_{e q}=\frac{2 A x \varepsilon_0}{3 x d}=\frac{2 A \varepsilon_0}{3 d}=\frac{2 C}{3} .\)

Problem 9. Find out the equivalent capacitance between A and B.
Solution:

⇒ \(C=\frac{A \varepsilon_0}{d}\)

⇒ \(\frac{1}{C_{e q}}=\frac{1}{C}+\frac{2}{3 C}=\frac{5}{3 C}\)

⇒ \(C_{e q}=\frac{3 C}{5}=\frac{3 A \varepsilon_0}{5 d}\)

Alternative Method :

⇒ \(C=\frac{Q}{V}=\frac{x+y}{V_{A B}}\)

⇒ \(C=\frac{Q}{V}=\frac{x+y}{V_{A B}}\)

Potential of 1 and 4 is same \(\frac{y}{\mathrm{~A} \varepsilon_0}=\frac{2 \mathrm{x}}{\mathrm{A} \varepsilon_0}\)

⇒ \(V=\left(\frac{2 y+x}{A \varepsilon_0}\right) d\)

⇒ \(C=\frac{(x+2 x) A \varepsilon_0}{(5 x) d}=\frac{3 A \varepsilon_0}{5 d}\)

Problem 10. Five similar condenser plates, each of area A, are placed at equal distances d apart and are connected to a source of e.m.f. E as shown in the following diagram. The charge on the plates 1 and 4 will be

1. \(\frac{\varepsilon_0 A}{d}, \frac{-2 \varepsilon_0 A}{d}\)
2. \(\frac{\varepsilon_0 A V}{d}, \frac{-2 \varepsilon_0 A V}{d}\)
3. \(\frac{-\varepsilon_0 A V}{d}, \frac{-3 \varepsilon_0 A V}{d}\)
4. \(\frac{\varepsilon_0 A V}{d}, \frac{-4 \varepsilon_0 A V}{d}\)

Solution: Equivalent circuit diagram Charge on the first plate

Q = CV

⇒ \(Q=\frac{\varepsilon_0 A V}{d}\)

Charge on the fourth plate

Q” = C(–V)

⇒ \(Q^{\prime}=\frac{-\varepsilon_0 A V}{d}\)

A plate 4 is repeated twice, hence charge on 4 will be Q´´ = 2Q´

⇒ \(Q^{\prime \prime}=-\frac{2 \varepsilon_0 A V}{d}\)

Hence the correct answer will be (2).

key Concept

q ∞ V ⇒  q = CV

q: Charge on the positive plate of the capacitor

C: Capacitance of capacitor.

V: Potential difference between positive and negative plates.

Representation of capacitor:

It is a scalar quantity having dimensions [C] = [M^-1 L^-2 T⁴ A² ]

The S.I. Unit is Farad. (F).

Energy stored in the capacitor \(: U=\frac{1}{2} C V^2=\frac{Q^2}{2 C}=\frac{Q V}{2}\)

Energy density =\(\frac{1}{2} \epsilon_0 \in_{\mathrm{r}} \mathrm{E}^2=\frac{1}{2} \epsilon_0 \mathrm{KE}^2\)

K = ∈_r = Relative permittivity of the medium (Dielectric Constant) For vacuum, energy density = \(\frac{1}{2} \epsilon_0 \mathrm{E}^2\)

Types of Capacitors :

Parallel Plate Capacitor

⇒ \(C=\frac{\epsilon_0 \epsilon_r A}{d}=K \frac{\epsilon_0 A}{d}\)

A: Area of plates

d : distance between the plates( << size of plate )

Spherical Capacitor :

The capacitance of an isolated spherical Conductor (hollow or solid )

C = 4 π ∈₀∈_r R

R = Radius of the spherical conductor

The capacitance of a spherical capacitor

⇒ \(\mathrm{C}=4 \pi \epsilon_0 \frac{\mathrm{ab}}{(\mathrm{b}-\mathrm{a})}\)

⇒ \(C=\frac{4 \pi \epsilon_0 K_2 a b}{(b-a)}\)

⇒ \(C=\frac{4 \pi \epsilon_0 b^2}{(b-a)}\)

Cylindrical Capacitor : l >> {a,b}

Capacitance per unit length = \(\frac{2 \pi \epsilon_0}{\ln (\mathrm{b} / \mathrm{a})} \mathrm{F} / \mathrm{m}\)

The capacitance of the capacitor depends on

Area of plates

Distance between the plates

The dielectric medium between the plates.

Electric field intensity between the plates of capacitor \(E=\frac{\sigma}{\epsilon_0}=\frac{V}{d}\)

σ: Surface charge density

Force experienced by any plate of capacitor :\(F=\frac{q^2}{2 A \epsilon_0}\)

Distribution of Charges on Connecting two Charged Capacitors:

When two capacitors C₁ and C₂ are connected as shown in the figure

Common potential :\(V=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}=\frac{\text { Total charge }}{\text { Total capacitance }}\)

⇒ \(Q_1^{\prime}=C_1 V=\frac{C_1}{C_1+C_2}\left(Q_1+Q_2\right)\)

⇒ \(Q_2^{\prime}=C_2 V=\frac{C_2}{C_1+C_2}\left(Q_1+Q_2\right)\)

Heat Loss During Redistribution :

⇒ \(\Delta H=U_i-U_1=\frac{1}{2} \frac{C_1 C_2}{C_1+C_2}\left(V_1-V_2\right)^2\)

The loss of energy is in the form of Joule heating in the wire.

Note :

When plates of similar charges are connected (+ with + and – with –) then put all values (Q₁, Q₂, V₁, V₂) with positive sign.

When plates of opposite polarity are connected (+ with –) then take charge and the potential of one of the plates to be negative.

Combination of capacitors:

Series Combination

⇒ \(\frac{1}{C_{\text {eq }}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} ; V_1: V_2: V_3=\frac{1}{C_1}: \frac{1}{C_2}: \frac{1}{C_3}\)

Parallel Combination :

C_eq = C₁ + C₂ + C₃ + ……….

Q₁: Q₂ :Q₃ = C₁ : C₂ : C₃

Charging and Discharging of a capacitor :

Charging of Capacitor ( Capacitor initially uncharged ): q = q0 ( 1 – e– t /)

q₀ = Charge on the capacitor at steady

State  q₀ = CV

τ : Time constant = CR_eq

⇒ \(i=\frac{q_0}{\tau} e^{-t / \tau}=\frac{V}{R} e^{-t / \tau}\)

63% of the maximum charge is deposited in one time constant.

Discharging of Capacitor :

q = q₀ e^-t/τ

q₀= Initial charge on the capacitor

⇒ \(i=\frac{q_0}{\tau} e^{-t / \tau}\)

63% of discharging is complete in one time constant.

Capacitor with dielectric :

Capacitance in the presence of dielectric :

C₀ = Capacitance in the absence of dielectric.

If the thickness of the dielectric slab is t, then its capacitance

⇒\(C=\frac{\epsilon_0 A}{(d-t+t / k)}\) where k is the dielectric constant of the slab.

It does not depend on the position of the slab.

k = 1 for vacuum or air.

k = ∞ for metals.

⇒ \(E_{\text {in }}=E-E_{\text {ind }}=\frac{\sigma}{\epsilon_0}-\frac{\sigma_{\mathrm{b}}}{\epsilon_0}=\frac{\sigma}{\mathrm{K} \epsilon_0}=\frac{V}{d}\)

⇒ \(E=\frac{\sigma}{\epsilon_0}\) Electric field in the absence of dielectric

Find: Induced electric field

⇒ \(\sigma_{\mathrm{b}}=\sigma\left(1-\frac{1}{\mathrm{~K}}\right)\) (induced charge density)

Capacitance Exercise – 1

Section (1): Definition Of Capacitance

Question 1. The radii of two metallic spheres are 5 cm and 10 cm and both carry an equal charge of 75μC. If the two spheres are shorted then a charge will be transferred–

1. 25 μC from smaller to bigger
2. 25 μC from bigger to smaller
3. 50 μC from smaller to bigger
4. 50 μC from bigger to smaller

Answer: 1. 25 μC from smaller to bigger

Question 2. Two isolated charged metallic spheres of radii R₁ and R₂ having charges Q₁ and Q₂ respectively are connected, then there is: