NEET Physics Class 12 Chapter 2 Capacitance Notes

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

\(\mathrm{i}=\frac{\varepsilon}{\mathrm{R}}\)

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

⇒ \(V=\frac{Q_1+Q_2}{C_1+C_2}=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}=\frac{\text { Total charge }}{\text { Total capacitance }}\)

⇒ \(Q_1^{\prime}=C_1 V=\frac{C_1}{C_1+C_2}\left(Q_1+Q_2\right)\)

\(Q_2^{\prime}=C_2 V=\frac{C_2}{C_1+C_2}\left(Q_1+Q_2\right)\)

Heat Loss During Redistribution :

⇒ \(\Delta H=U_i-U_f=\frac{1}{2} \frac{C_1 C_2}{C_1+C_2}\left(V_1-V_2\right)^2\)

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

\(U_{3 \mu \mathbb{F}}=\frac{1}{2} \times 3 \times 10 \times 10=150 \mu \mathrm{J}\)

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

\(\mathrm{E}=\frac{\sigma}{\varepsilon_0}=\frac{\mathrm{CV}}{\mathrm{A} \varepsilon_0}\) \(=\frac{\mathrm{V}}{d}\)

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 :

\(C=\frac{K \in \epsilon_0 A}{d}=K C_0\)

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:

1. No change in the electrical energy of the system
2. An increase in the electrical energy of the system
3. Always a decrease in the electrical energy of the system
4. A decrease in electrical energy of the system until Q₁ R₂ = Q2 R1

Answer: 4. A decrease in electrical energy of the system until Q₁ R₂ = Q₂ R₁

Question 3. A parallel plate capacitor is charged and the charging battery is then disconnected. The plates of the capacitor are now moved, farther apart. The following things happen :

1. The charge on the capacitor increases
2. The electrostatic energy stored in the capacitor increases
3. The voltage between the plates decreases
4. The capacitance increases.

Answer: 2. The electrostatic energy stored in the capacitor increases

Question 4. A parallel plate capacitor is charged and then isolated. On increasing the plate separation–

1. Charge Potential Capacitance
2. Remains constant remains constant decreases
3. Remains constant increases decreases
4. Remains constant decreases increases
5. Increases increase decreases

Answer: 2. Remains constant increases decreases

Question 5. The value of one farad in e.s.u. is-

1. 3 × 10¹⁰
2. 9 × 10¹¹
3. 1/9 × 10^-11
4. 1/3× 10^-10

Answer: 2. 9 × 10¹¹

Question 6. A parallel plate air capacitor is charged to a potential difference V. After disconnecting the battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates-

1. Increases
2. Decreases
3. Does not change
4. Becomes zero

Answer: 1. Increases

Question 7. The capacitance of a parallel plate capacitor is 10 µF when the distance between its plates is 8 cm. If the distance between the plates is reduced to 4 cm, its capacitance will be

1. 5 µF
2. 10 µF
3. 20 µF
4. 40 µF

Answer: 3. 20 µF

Question 8. Capacitance in the farad of a spherical conductor with a radius of 1 meter is-

1. 1.1 × 10^-10
2. 10^-6
3. 9 × 10^-9
4. 10^-3

Answer: 1. 1.1 × 10^-10

Question 9. The capacitance of a spherical conductor is given by

1. \(\mathrm{C}=\frac{1}{4 \pi \varepsilon_0 R}\)
2. C = 4πε₀ R
3. C = 4πε₀R₂
4. C = rπε₀R₃

Answer: 2. C = 4πε₀R

Question 10. Eight drops of mercury of equal radii each possessing the same charge combine to form a big drop. The capacitance of this big drop as compared to that of each smaller drop is-

1. 2 times
2. 4 times
3. 8 times
4. 16 times

Answer: 1. 2 times

Question 11. The capacity of a conductor depends upon the

1. Size of conductor
2. Thickness of conductor
3. Material of conductor
4. All of these

Answer: 1. Size of conductor

Question 12. A metallic sphere of radius R is charged to potential V. Then charge q is proportional to

1. V
2. R
3. Both
4. None

Answer: 3. Both

Question 13. A conducting sphere of radius 10 cm is charged with 10 μC. Another uncharged sphere of radius 20 cm is allowed to touch it for some time. After that, if the sphere is separated, then the surface density of charge on the spheres will be in the ratio of :

1. 1: 4
2. 1 : 3
3. 2: 1
4. 1: 1

Answer: 3. 2: 1

Question 14. Capacitance (in F) of a spherical conductor having a radius of 1m, is :

1. 1.1 × 10^-10
2. 10^-6
3. 9 × 10^-9
4. 10^-3

Answer: 1. 1.1 × 10^-10

Question 15. The capacitance of a parallel plate capacitor is 12μF. If the distance between its plates is reduced to half and the area of the plates is doubled, then the capacitance of the capacitor will become

1. 24μF
2. 12μF
3. 16μF
4. 48μF

Answer: 4. 48μF

Question 16. The radius of the circular plates of a parallel plate capacitor is R. Air is the dielectric medium between the plates. If the capacitance of the capacitor is equal to the capacitance of a sphere of radius R, then the distance between the plates is

1. R/4
2. R/2
3. R
4. 2R

Answer: 1. R/4

Section (2): Circuits With Capacitor And Use Of KCL And KVL

Question 1. The work done against electric forces in increasing the potential difference of a condenser from 20V to 40V is W. The work done in increasing its potential difference from 40V to 50V will be

1. 4W
2. \(\frac{3 W}{4}\)
3. 2W
4. \(\frac{W}{2}\)

Answer: 2.\(\frac{3 W}{4}\)

Question 2. The magnitude of charge in steady state on either of the plates of condenser C in the adjoining circuit is

1. Ce
2. \(\frac{\mathrm{CER}_2}{\left(\mathrm{R}_1+\mathrm{r}\right)}\)
3. \(\frac{\text { CER }_2}{\left(R_2+r\right)}\)
4. \(\frac{\text { CER }_1}{\left(R_2+r\right)}\)

Answer: 3. \(\frac{\text { CER }_2}{\left(R_2+r\right)}\)

Question 3. The plate separation in a parallel plate condenser is d and the plate area is A. If it is charged to V volt & battery is disconnected then the work done in increasing the plate separation to 2d will be–

1. \(\frac{\mathrm{C}_1 \mathrm{~V}}{\mathrm{C}_1+\mathrm{C}_2}\)
2. \(\frac{C_2 V}{C_1+C_2}\)
3. \(\frac{\left(C_1+C_2\right) V}{C_1+C_2}\)
4. \(\frac{\mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}\)

Answer: 4. \(\frac{\mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}\)

Question 5. If the charge on a body is increased by 2μC, the energy stored in it increases by 21%. The original charge on the body in micro-coulombs is

1. 10
2. 20
3. 30
4. 40

Answer: 2. 20

Question 6. What fraction of the energy drawn from the charging battery is stored in a capacitor

1. 100%
2. 75%
3. 50%
4. 25%

Answer: 3. 50%

Question 7. The plates of a parallel plate condenser are pulled apart with a velocity v. If at any instant mutual distance of separation is d, then the magnitude of the time of rate of change of electrostatic energy of the capacity depends on d as follows(potential difference between plates is kept constant)-

1. \(\frac{1}{d}\)
2. \(\frac{1}{d^2}\)
3. d2
4. d

Answer: 2. \(\frac{1}{d^2}\)

Question 8. 125 water drops of equal radius and equal capacitance C, coalesce to form a single drop of capacitance C´. The relation between C and C´ is-

1. C´ = 125 C
2. C´ = C
3. \(C^{\prime}=\frac{C}{125}\)
4. C´ = 5C

Answer: 4. C´ = 5C

Question 9. Energy per unit volume for a capacitor having area A and separation d kept at potential difference V is given by

1. \(\frac{1}{2} \varepsilon_0 \frac{V^2}{d^2}\)
2. \(\frac{1}{2 \varepsilon_0} \frac{\mathrm{V}^2}{\mathrm{~d}^2}\)
3. \(\frac{1}{2} \mathrm{CV}^2\)
4. \(\frac{Q^2}{2 C}\)

Answer: 1. \(\frac{1}{2} \varepsilon_0 \frac{V^2}{d^2}\)

Question 10. The mean electric energy density between plates of a charged capacitor is- (here q = charge on the capacitor and A = area of the capacitor plate)

1. \(\frac{q^2}{2 \varepsilon_0 A^2}\)
2. \(\frac{\mathrm{q}}{2 \varepsilon_0 \mathrm{~A}^2}\)
3. \(\frac{q^2}{2 \varepsilon_0 A}\)
4. None Of these

Answer: 1. \(\frac{q^2}{2 \varepsilon_0 A^2}\)

Question 11. A capacitor when charged by a potential difference of 200 volts, stores a charge of 0.1 C. By discharging, the energy liberated by the capacitor is-

1. –30 J
2. –15 J
3. 10 J
4. 20 J

Answer: 3. 10 J

Question 12. Work done in placing a charge of 8 × 10^-18 C on a condenser of capacity 100 microfarad is-

1. 16 × 10^-32 J
2. 3.1 × 10^-26 J
3. 4 × 10^-10 J
4. 32 × 10^-32 J

Answer: 4. 32 × 10^-32 J

Question 13. The work done in doubling the separation between plates of a parallel plate capacitor of capacity C and having charge Q is

1. \(\frac{\mathrm{Q}^2}{\mathrm{C}}\)
2. \(\frac{Q^2}{2 C}\)
3. \(\frac{Q^2}{4 C}\)
4. \(\frac{2 Q^2}{C}\)

Answer: 2. \(\frac{Q^2}{2 C}\)

Question 14. In the adjoining diagram, (assuming the battery to be ideal) the condenser C will be charged to potential V if

1. S₁ and S₂ are both open
2. S₁ and S₂ are both closed
3. S₁ is closed and S₂ is open
4. S₁ is open and S₂ is closed.

Answer: 3. S₁ is closed and S₂ is open

Question 15. A capacitor 4mF charged to 50 V is connected to another capacitor of 2mF charged to 100V with plates of like charges connected. The total energy before and after connection in multiples [10 J] is-

1. 1.5 and 1.33
2. 1.33 and 1.5
3. 3.0 and 2.67
4. 2.67 and 3.0

Answer: 1. 1.5 and 1.33

Question 16. A condenser of capacitance 10mF has been charged to 100 volts. It is now connected to another uncharged condenser in parallel. The common potential becomes 40 volts. The capacitance of another condenser is-

1. 15 mF
2. 5 mF
3. 10 mF
4. 16.6 mF

Answer: 2. 5 mF

Question 17. Two capacitors of capacitances 3μF and 6μF are charged to a potential of 12V each. They are now connected, with the positive plate of each joined to the negative plate of the other. The potential difference across each will be

1. 6V
2. 4V
3. 3V
4. Zero

Answer: 2. 4V

Question 18. In the following figure, the charge on each condenser in the steady state will be–

1. 3μC
2. 6μC
3. 9μC
4. 12μC

Answer: 4. 12μC

Question 19. Two parallel plate condensers of the capacity of 20µF and 30µF are charged to the potentials of 30V and 20V respectively. If likely charged plates are connected then the common potential difference will be

1. 100 V
2. 50 V
3. 24 V
4. 10 V

Answer: 3. 24 V

Question 20. The work done in placing a charge of 8 × 10^-18 coulomb on a condenser of capacity 100 micro-farad is :

1. 16 × 10^-32 joule
2. 3.1 × 10^-26 joule
3. 4 × 10^-10 joule
4. 32 × 10^-32 joule

Answer: 4. 32 × 10^-32 joule

Question 21. A capacitor is charged by connecting a battery across its plates. It stores energy U. Now the battery is disconnected and another identical capacitor is connected across it, then the energy stored by both capacitors of the system will be :

1. U
2. \(\frac{\mathrm{U}}{2}\)
3. 2U
4. \(\frac{\mathrm{3}}{2}U\)

Answer: 1. U

Question 22. In a parallel plate capacitor, the distance between the plates is d, and the potential difference across plates is V. Energy stored per unit volume between the plates of the capacitor is :

1. \(\frac{\mathrm{Q}^2}{2 \mathrm{~V}^2}\)
2. \(\frac{1}{2} \varepsilon_0 \frac{\mathrm{V}^2}{\mathrm{~d}^2}\)
3. \(\frac{\varepsilon_0^2 V^2}{d^2}\)
4. \(\frac{1}{2} \frac{\varepsilon_0^2 V^2}{d^2}\)

Answer: 4. \(\frac{1}{2} \frac{\varepsilon_0^2 V^2}{d^2}\)

Question 23. A capacitor of capacity C₁ is charged up to a potential V volt and then connected in parallel to an uncharged capacitor of capacity C₂. The final potential difference across each capacitor will be :

1. \($\frac{C_2 V}{C_1+C_2}\)
2. \(\frac{C_1 V}{C_1+C_2}\)
3. \(\left(1+\frac{C_2}{C_1}\right) V\)
4. \(\left(1-\frac{C_2}{C_2}\right) V\)

Answer: 3. \(\left(1+\frac{C_2}{C_1}\right) V\)

Question 24. A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates :

1. Decreases
2. Does not change
3. Becomes zero
4. Increases

Answer: 4. Increases

Question 25. Two condensers, one of capacity C and the other of capacity \(\frac{C}{2}\) is connected to a V-volt battery, as shown.

The work done in charging fully both the condensers is

1. 2CV₂
2. \(\frac{1}{4} \mathrm{CV}^2\)
3. \(\frac{3}{2} C V^2\)
4. \(\frac{1}{2} \mathrm{CV}^2\)

Answer: 3. \(\frac{3}{2} C V^2\)

Question 26. The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field between the plates is E, is

1. \(\frac{1}{2} \varepsilon_0 \mathrm{E}^2 / \mathrm{Ad}\)
2. _ε0E₂/Ad
3. _ε0E₂Ad
4. \(\frac{1}{2} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}\)

Answer: 3. ε₀E₂Ad

Question 27. A 40 μF capacitor in a defibrillator is charged to 3000 V. The energy stored in the capacitor is sent through the patient during a pulse of duration 2 ms. The power delivered to the patient is

1. 45 kW
2. 90 kW
3. 180 kW
4. 360 kW

Answer: 2. 90 kW

Question 28. If there are n capacitors of capacitance C in parallel connected to the V volt source, then the energy stored is equal to :

1. CV
2. \(\frac{1}{2} n C V^2\)
3. CV₂
4. \(\frac{1}{2 n} C V^2\)

Answer: 2. \(\frac{1}{2} n C V^2\)

Question 29. A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be

1. 1
2. 2
3. 1/4
4. 1/2

Answer: 4. 1/2

Question 30. A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separation between its plates is ‘d’. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant k₁ = 3 and thickness d/3 while the other one has dielectric constant k₂ = 6 and thickness 2d/3. The capacitance of the capacitor is now :

1. 45 pF
2. 40.5 pF
3. 20.25 pF
4. 1.8 pF

Answer: 2. 40.5 pF

Question 31. If the energy of a capacitor of capacitance 2μF is 0.16 joule, then its potential difference will be

1. 800 V
2. 400 V
3. 16 × 10⁴ V
4. 16 × 10^-4 V

Answer: 2. 400 V

Question 32. A capacitor of 6μF is charged to such an extent that the potential difference between the plates becomes 50 V. The work done in this process will be

1. 7.5 × 10^-2 J
2. 7.5 × 10^-3 J
3. 3 × 10^-6 J
4. 3 × 10^-3 J

Answer: 2. 7.5 × 10^-3 J

Question 33. Two identical capacitors have the same capacitance C. One of them is charged to potential V₁ and the other to V₂. The negative ends of the capacitors are connected. When the positive ends are also connected, the decrease in energy of the combined system is:

1. \(\frac{1}{4} C\left(V_1{ }^2-V_2{ }^2\right)\)
2. \(\frac{1}{4} C\left(V_1^2+V_2{ }^2\right)\)
3. \(\frac{1}{4} C\left(V_1-V_2\right)^2\)
4. \(\frac{1}{4} C\left(V_1+V_2\right)^2\)

Answer: 3. \(\frac{1}{4} C\left(V_1-V_2\right)^2\)

Question 34. A capacitor is connected to a cell of emf E having some internal resistance r. The potential difference across the :

1. The cell is < E
2. Cell is E
3. The capacitor is > E
4. The capacitor is < E

Answer: 2. cell is E

Question 35. A 10 μF capacitor is charged to a 1000-volt potential, then it is removed from the power supplied and connected to a 6 μF uncharged capacitor. Find potential differences across each capacitor.

1. 167 V
2. 100 V
3. 625 V
4. 250 V

Answer: 3. 625 V

Question 36. The uniform electric field in the space between the plates of a parallel plate condenser of plate separation d and plate areas A is E. The energy of this charged condenser is :

1. \(\frac{1}{2} \cdot \frac{\epsilon_0 \mathrm{E}^2}{\mathrm{~A} \cdot \mathrm{d}}\)
2. \(\epsilon_0 E^2 \mathrm{Ad}\)
3. \(\frac{1}{2} \epsilon_0 E^2 A d\)
4. \(\frac{1}{2} \cdot \frac{\epsilon_0 E^2}{\mathrm{Ad}}\)

Answer: 3. \(\frac{1}{2} \epsilon_0 E^2 A d\)

Question 37. In the given circuit with a steady current, the potential drop across the capacitor must be :

1. V
2. V/2
3. V/3
4. 2V/3

Answer: 3. V/3

Section (3): Combination Of Capacitors

Question 1. In the adjoining circuit, the capacity between the points A and B will be

1. C
2. 2C
3. 3C
4. 4C

Answer: 2. 2C

Question 2. The resultant capacity between the points A and B in the adjoining circuit will be

1. C
2. 2C
3. 3C
4. 4C

Answer: 3. 3C

Question 3. The effective capacity in the following figure between the points P and Q will be –

1. 3µF
2. 5µF
3. 2µF
4. 1µF

Answer: 4. 1µF

Question 4. The charge on the condenser of capacitance 2µF in the following circuit will be –

1. 4.5 µC
2. 6.0 µC
3. 7 µC
4. 30 µC

Answer: 2. 6.0 µC

Question 5. Five capacitors of 10µF capacity each are connected to a d.c. potential difference of 100 volts as shown in the figure. The equivalent capacitance between the points A and B will be equal to

1. 40µF
2. 20µF
3. 30µF
4. 10µF

Answer: 4. 10µF

Question 6. The equivalent capacitance between the terminals X and Y in the figure shown will be–

1. 100 pF
2. 200 pF
3. 300 pF
4. 400 pF

Answer: 2. 200 pF

Question 7. Three capacitors of capacity 1µF each, are connected in such a way, that the resultant capacity is 1.5µF, then:

1. All the capacitors are joined in a series
2. All the capacitors are joined in parallel
3. Two capacitors are in parallel, while the third is in series
4. Two capacitors are in series, while the third is in parallel

Answer: 4. Two capacitors are in series, while the third is in parallel

Question 8. n identical condensers are joined in parallel and are charged to potential V. Now they are separated and joined in series. Then the total energy and potential difference of the combination will be

1. Energy and potential differences remain the same
2. Energy remains the same and a potential difference is nV
3. Energy increases n times and potential differences are nV
4. Energy increases n times and potential difference remains the same

Answer: 2. Energy remains the same and a potential difference is nV

Question 9. A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is connected to another battery and is charged to a potential difference of 2V. The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

1. Zero
2. \(\frac{25 \mathrm{CV}^2}{6}\)
3. \(\frac{3 C V^2}{2}\)
4. \(\frac{9 \mathrm{CV}^2}{2}\)

Answer: 3. \(\frac{3 C V^2}{2}\)

Question 10. A 3 µF capacitor is charged to a potential of 300 V and a 2µF capacitor is charged to 200 V. The capacitors are then connected in parallel with plates of opposite polarity joined together. What amount of charge will flow when the plates are so connected-

1. 250µC
2. 600µC
3. 700µC
4. 1300µC

Answer: 2. 600µC

Question 11. The equivalent capacitance of the circuit shown, between points A and B will be

1. \($\frac{2}{3} \mu F\)
2. \(\frac{5}{3} \mu \mathrm{F}\)
3. \(\frac{8}{3} \mu \mathrm{F}\)
4. \(\frac{7}{3} \mu \mathrm{F}\)

Answer: 3. \(\frac{8}{3} \mu \mathrm{F}\)

Question 12. In the electric circuit given below, the capacitance of each capacitor is 1 µF. The effective capacitance between the points A and B is- (in µF)

1. \(\frac{2}{3}\)
2. \(\frac{3}{2}\)
3. \(\frac{1}{6}\)
4. 6

Answer: 1. \(\frac{2}{3}\)

Question 13. The equivalent capacitance of three capacitors of capacitance C₁, C₂, and C₃ connected in parallel is 12 units and the product C₁. C₂ . C₃ = 48. When the capacitors C₁ and C₂ are connected in parallel the equivalent capacitance is 6 units. Then the capacitance is-

1. 2, 4, 6
2. 1, 5, 6
3. 1.5, 2.5, 8
4. 2, 3, 7

Answer: 1. 2, 4, 6

Question 14. When a potential difference of 10³ V is applied between A and B, a charge of 0.75 mC is stored in the system of capacitors. The value of C is (µ F)

1. 1
2. 3
3. 2.5
4. 2

Answer: 4. 2

Question 15. What is the effective capacitance between points X and Y in fig.-

Answer: 1.

Question 16. Ten capacitors are joined in parallel and charged with a battery up to a potential V. They are then disconnected from the battery and joined again in series the potential of this combination will be

1. V
2. 10 V
3. 5 V
4. 2 V

Answer: 2. 10 V

Question 17. The energy stored in the capacitor is U when it is charged with a battery. After disconnecting the battery another capacitor of the same capacity is connected in parallel with it, then the energy stored in each capacitor is

1. \(\frac{U}{6}\)
2. \(\frac{U}{4}\)
3. 9U
4. 8U

Answer: 2. \(\frac{U}{4}\)

Question 18. Minimum numbers of 8µF and 250V capacitors are used to make a combination of 16µF and 1000V are-

1. 4
2. 32
3. 8
4. 3

Answer: 2. 32

Question 19. A condenser of capacity 50 µF is charged to 10V. The energy stored is-

1. 1.25 × 10–3 J
2. 2.5 × 10–3 J
3. 3.75 × 10–3 J
4. 5 × 10–3 J

Answer: 2. 2.5 × 10–3 J

Question 20. The capacitors A and B are connected in series with a battery as shown in the figure. When the switch S is closed and the two capacitors get charged fully, then-

1. The potential difference across the plates of A is 4V and across the plates of B is 6V
2. The potential difference across the plates of A is 6V and across the plates of B is 4V
3. The ratio of electric energies stored in A and B is 2 : 3
4. The ratio of charges on A and B is 3: 2

Answer: 2. The potential difference across the plates of A is 6V and across the plates of B is 4V

Question 21. The potential difference across capacitance of 4.5 µF-

1. 8 V
2. 4 V
3. 2 V
4. 6 V

Answer: 1. 8 V

Question 22. The equivalent capacitance in the circuit shown in fig. will be

1. \(\frac{10}{3} \mu F\)
2. 4 µF
3. 6 µF
4. 8 µF

Answer: 4. 8 µF

Question 23. Three condensers of capacity C each are joined first in series and then in parallel. The capacity becomes n times, where n is-

1. 3
2. 6
3. 9
4. 12

Answer: 3. 9

Question 24. Two spherical conductors A and B of radius a and b (b > a) are placed in air concentrically. B is given charge +Q coulomb and A is grounded. The equivalent capacitance of these will be

1. \(4 \pi \varepsilon_0 \frac{\mathrm{ab}}{\mathrm{b}-\mathrm{a}}\)
2. \(4 \pi \varepsilon_0(\mathrm{a}+\mathrm{b})\)
3. \(4 \pi \varepsilon_0 \mathrm{~b}\)
4. \(4 \pi \varepsilon_0 \frac{b^2}{\mathrm{~b}-\mathrm{a}}\)

Answer: 4. \(4 \pi \varepsilon_0 \frac{b^2}{\mathrm{~b}-\mathrm{a}}\)

Question 25. Two capacitors C1 = 2µF, and C2 = 6µF are in series order, connected in parallel to a third capacitor of 4µF. This combination is connected to a 2-volt battery, in charging these capacitors energy consumed by the battery is-

1. 22 × 10^-6 joule
2. 11 × 10^-6 joule (3)
3. \(\frac{32}{3} \times 10^{-6} joule\)
4. \(\frac{16}{3} \times 10^{-6}joule\)

Answer: 1. 22 × 10^-6 joule

Question 26. A capacitor of capacity C1 = 1 µF can withstand maximum voltage V1 = 6 kV and another capacitor C2 = 3µF can withstand maximum voltage V2 = 4 kV. If these are connected in series, then the combined system can withstand a maximum voltage at-

1. 4 kV
2. 6 kV
3. 8 kV
4. 10 kV

Answer: 3. 8 kV

Question 27. The conducting sphere of radius R1 is covered by a concentric sphere of radius R2. The capacity of this combination is proportional to

1. \(\frac{R_2-R_1}{R_1 R_2}\)
2. \(\frac{R_2+R_1}{R_1 R_2}\)
3. \(\frac{R_1 R_2}{R_1+R_2}\)
4. \(\frac{R_1 R_2}{R_2-R_1}\)

Answer: 4. \(\frac{R_1 R_2}{R_2-R_1}\)

Question 28. Two capacitors when connected in series have a capacitance of 3 µF, and when connected in parallel have a capacitance of 16 µF. Their capacities are-

1. 1 µF, 2 µF
2. 6 µF, 2 µF
3. 12 µF, 4 µF
4. 3 µF, 16 µF

Answer: 3. 12 µF, 4 µF

Question 29. A parallel plate capacitor is formed by placing n plates in alternate series one over another. If the capacity between any two consecutive plates is C, then the total capacity of the capacitor is-

1. C
2. C
3. (n – 1) C
4. (n + 1) C

Answer: 3. (n – 1) C

Question 30. Consider the situation shown in Fig. The capacitor A has a charge q on it whereas B is uncharged. The charge appearing on the capacitor B a long time after the switch is closed is

1. Zero
2. \(\frac{q}{2}\)
3. q
4. 2q

Answer: 1. Zero

Question 31. Two capacitors are joined as shown in the figure. Potentials at points A and B are V1 and V2 respectively. The potential of point D is

1. \(\frac{1}{2}\left(V_1+V_2\right)\)
2. \(\frac{C_1 V_2+C_2 V_1}{C_1+C_2}\)
3. \(\frac{C_1 V_1+C_2 V_2}{C_1+C_2}\)
4. \(\frac{C_2 V_1-C_1 V_2}{C_1+C_2}\)

Answer: 3. \(\frac{C_1 V_1+C_2 V_2}{C_1+C_2}\)

Question 32. Two capacitors of 2µF and 3µF are connected in series. The potential at point A is 1000 volts and the outer plate of the 3µF capacitor is earthed. The potential at point B is

1. 300 volt
2. 500 volt
3. 600 volt
4. 400 volt

Answer: 4. 400 volt

Question 33. How the seven condensers, each of capacity 2µF, should be connected to obtain a resultant 10 capacitance of \(\frac{10}{11} \mu \mathrm{F}?\)

Answer: 3.

Question 34. Five capacitors, each of capacitance value C are connected as shown in the figure. The ratio of capacitance between P and R, and the capacitance between P and Q is –

1. 3: 1
2. 5: 2
3. 2 : 3
4. 1: 1

Answer: 3. 2 : 3

Question 35. In the given figure, the capacitors C1, C3, C4 C5 have a capacitance of 4µF each. If the capacitor C2 has a capacitance of 10 µF, then the effective capacitance between A and B will be :

1. 2µF
2. 4µF
3. 6µF
4. 8µF

Answer: 2. 4µF

Question 36. Three capacitors each of capacity 4µF are to be connected in such a way that the effective capacitance is 6 µF. This can be done by :

1. Connecting two in series and one in parallel
2. Connecting two in parallel and one in series
3. Connecting all of them in a series
4. Connecting all of them in parallel

Answer: 1. Connecting two in series and one in parallel

Question 37. A network of four capacitors of capacity equal to C₁ = C, C₂ = 2C, C₃ = 3C, and C₄ = 4C are connected to a battery as shown in the figure. The ratio of the charges on C₂ and C₄ is :

1. \(\frac{22}{3}\)
2. \(\frac{3}{22}\)
3. \(\frac{7}{4}\)
4. \(\frac{4}{7}\)

Answer: 2. \(\frac{3}{22}\)

Question 38. A parallel plate capacitor is made by stacking n equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is ‘C’, then the resultant capacitance is :

1. (n – 1)C
2. (n + 1) C0
3. C
4. C

Answer: 1. (n – 1)C

Question 39. Two spheres of capacitances 3μF and 5μF are charged to 300 V and 500 V respectively and are connected. The common potential in steady state will be

1. 400 V
2. 425 V
3. 350 V
4. 375 V

Answer: 2. 425 V

Question 40. In the combination shown in the figure, the ideal voltmeter reading will be

1. 4.5 V
2. \(\left(\frac{18}{11}\right) V\)
3. 3V
4. 2V

Answer: 4. 2V

Question 41. In the given network capacitance C₂ = 10 µF, C₁ = 5 µF and C₃ = 4 µF. The resultant capacitance between P and Q will be :

1. 4.7 µF
2. 1.2 µF
3. 3.2 µF
4. 2.2 µF

Answer: 3. 3.2 µF

Question 42. If the equivalent capacitance between A and B is 1µF, then the value of C will be

1. 2µF
2. 4µF
3. 3µF
4. 6µF

Answer: 1. 2µF

Question 43. The equivalent capacitance between points A and B is

1. C/4
2. C/2
3. C
4. 2C

Answer: 4. 2C

Question 44. A combination arrangement of the capacitors is shown in the figure

1. C₁ = 3 µF, C₂ = 6 µF and C₃ = 2 µF then equivalent capacitance between ‘a’ and ‘b’ is :

1. 4 µF
2. 6 µF
3. 1 µF
4. 2 µF

Answer: 1. 4 µF

2. If a potential difference of 48 V is applied across points a and b, then the charge on the capacitor C3 at steady state condition will be :

1. 8 µC
2. 16 µC
3. 32 µC
4. 64 µC

Answer: 4. 64 µC

Question 45. Two spherical conductors A₁ and A₂ of radii r₁ and r₂ are placed concentrically in air. The two are connected by a copper wire as shown in the figure. Then the equivalent capacitance of the system is

1. \(\frac{4 \pi \varepsilon_0 k r_1 r_2}{r_2-r_1}\)
2. 4π∈₀ (r₁ + r₂)
3. 4π∈₀r₂
4. 4π∈₀r₁

Answer: 3. 4π∈₀r₂

Question 46. Eight drops of mercury of the same radius and having the same charge coalesce to form a big drop. The capacitance of a big drop relative to that of a small drop will be

1. 16 times
2. 8 times
3. 4 times
4. 2 times

Answer: 4. 2 times

Question 47. In the figure given, the effective capacitance between A and B will be

1. C
2. C/2
3. 2C
4. 3C

Answer: 3. 2C

Question 48. If the shown figure is a combination of four capacitors their values are given. Charge and potential difference across 4 µF capacitors will be –

1. 600 µC ; 150 Volt
2. 300 µC ; 75 Volt
3. 800 µC ; 200 Volt
4. 580 µC ; 145 Volt

Answer: 4. 580 µC; 145 Volt

Section (4): Equation Of Charging And Discharging

Question 1. The plates of a capacitor of capacitance 10 µF, charged to 60 µC, are joined together by a wire of resistance 10 µ at t = 0, then

1. The charge on the capacitor in the circuit at t = 0 is :

1. 120 µC
2. 60 µC
3. 30 µC
4. 44 µC

Answer: 2. 60 µC

2. The charge on the capacitor in the circuit at t = 100 µs is :

120 µC

60 µC

22 µC

18 µC

Answer: 3. 22 µC

3. The charge on the capacitor in the circuit at t = 1.0 ms is :

1. 0.003 µC
2. 60 µC
3. 44 µC
4. 18 µC

Answer: 1. 0.003 µC

Question 2. An uncharged capacitor of capacitance 8.0 μF is connected to a battery of emf 6.0 V through a resistance of 24 µ, then

1. The current in the circuit just after the connections are made is :

1. 0.25 A
2. 0.5 A
3. 0.4 A
4. 0 A

Answer: 1. 0.25 A

2. The current in the circuit at one time constant after the connections are made is :

1. 0.25 A
2. 0.09 A
3. 0.4 A
4. 0 A

Answer: 2. 0.09 A

Question 3. An uncharged capacitor of capacitances 12.0 µF is connected to a battery of emf 6.00 V and internal resistance 1.00 µ through resistanceless leads. At 12.0 µs after the connections are made :

1. The current in the circuit is :

1. 4.42 A
2. 6 A
3. 2.21 A
4. 0 A

Answer: 3. 2.21 A

2. The power spent by the battery is :

1. 26.4 W
2. 13.2 W
3. 4.87 W
4. 0

Answer: 2. 13.2 W

3. The power dissipated in heat is :

1. 26.4 W
2. 13.2 W
3. 4.87 W
4. 0

Answer: 4. 0

4. The rate at which energy stored in the capacitor is increasing is :

1. 26.4 W
2. 13.2 W
3. 4.87 W
4. 8.37 W

Answer: 3. 4.87 W

Question 4. The charge on each of the capacitors is 0.20 ms after switch S is closed in the figure:

1. 24 µC
2. 16.8 µC
3. 10.37 µC
4. 4.5 µC

Answer: 3. 10.37 µC

Question 5. The time constant of a series R-C circuit is

1. +RC
2. –RC
3. R/C
4. C/R

Answer: 1. +RC

Question 6. If a current, that charges a capacitor, is constant, then the graph representing the change in voltage across the capacitor with time t is

Answer: 3.

Question 7. In the adjoining diagram, (assuming the battery to be ideal) the condenser C will be fully charged to potential V if

1. S₁ and S₂ are both open
2. S₁ and S₂ are both closed
3. S₁ is closed and S₂ is open
4. S₁ is open and S₂ is closed.

Answer: 3. S₁ is closed and S₂ is open

Question 8. In the following figure, the charge on each condenser in the steady state will be–

1. 3µC
2. 6µC
3. 9µC
4. 12µC

Answer: 4. 12µC

Question 9. The plates of a capacitor of capacitance 10 µF, charged to 60 µC, are joined together by a wire of resistance 10 µ at t = 0, then

1. The charge on the capacitor in the circuit at t = 0 is :

1. 120 µC
2. 60 µC
3. 30 µC
4. 4 µC

Answer: 2. 60 µC

2. The charge on the capacitor in the circuit at t = 100 µs is :

1. 120 µC
2. 60 µC
3. 22 µC
4. 18 µC

Answer: 3. 22 µC

3. The charge on the capacitor in the circuit at t = 1.0 ms is : (take e10 = 20000)

1. 0.003 µC
2. 60 µC
3. 44 µC
4. 18 µC

Answer: 1. 0.003 µC

Question 10. The dotted line represents the charging of a capacitor with resistance X. If resistance is made 2X then which will be the graph of charging

1. P
2. Q
3. R
4. S

Answer: 2. Q

Question 11. n resistances each of resistance R is joined with capacitors of capacity C (each) and a battery of emf E as shown in the figure. In steady-state conditions ratio of the charge stored in the first and last capacitors is

1. n: 1
2. (n – 1) : (n + 1)
3. (n2 + 1) : (n2 – 1)
4. 1: 1

Answer: 4. 1: 1

Section (5): Capacitor With Dielectric

Question 1. The distance between the plates of a parallel plate condenser is d. If a copper plate of the same area but d thickness 2 is placed between the plates then the new capacitance will become-

1. Half
2. Double
3. One fourth
4. Unchanged

Answer: 2. Double

Question 2. On placing a dielectric slab between the plates of an isolated charged condenser it– Capacitance Charge Potential Difference Energy stored Electric field

1. Decreases remain decreases increases unchanged
2. Increases remain increases decreases unchanged
3. Increases remain decreases decreases decreases unchanged
4. Decreases remain decreases increases remain unchanged unchanged

Answer: 3. Increases remain decreases decreases decrease unchanged

Question 3. A parallel plate condenser is connected to a battery of e.m.f. 4 volts If a plate of dielectric constant 8 is inserted into it, then the potential difference on the condenser will be

1. 1/2 V
2. 2V
3. 4V
4. 32V

Answer: 3. 4V

Question 4. In the above problem if the battery is disconnected before inserting the dielectric, then the potential difference will be

1. 1/2 V
2. 2V
3. 4V
4. 32V

Answer: 1. 1/2 V

Question 5. A parallel plate condenser with plate separation d is charged with the help of a battery so that U 0 energy is stored in the system. A plate of dielectric constant K and thickness d is placed between the plates of the condenser while the battery remains connected. The new energy of the system will be

1. KU0
2. K2U0
3. \(\frac{\mathrm{U}_0}{\mathrm{~K}}\)
4. \(\frac{U_0}{K^2}\)

Answer: 1. KU0

Question 6. In the above problem if the battery is disconnected before placing the plate, then new energy will be–

1. K2U0
2. \(\frac{U_0}{K^2}\)
3. \(\frac{\mathrm{U}_0}{\mathrm{~K}}\)
4. KU0

Answer: 3. \(\frac{\mathrm{U}_0}{\mathrm{~K}}\)

Question 7. A parallel plate capacitor is first charged and then disconnected from the battery and then a dielectric slab is introduced between the plates. The quantity that remains unchanged is

1. Charge Q
2. Potential V
3. Capacity C
4. Energy U

Answer: 1. Charge Q

Question 8. A parallel capacitor of capacitance C is charged and disconnected from the battery. The energy stored in it is E. If a dielectric slab of dielectric constant 6 is inserted between the plates of the capacitor then energy and capacitance will become

1. 6E, 6C
2. E, C
3. \(\frac{E}{6}, 6 C\)
4. E, 6C

Answer: 3. \(\frac{E}{6}, 6 C\)

Question 9. When a dielectric material is introduced between the plates of a charged condenser, the electric field between the plates

1. Decreases
2. Increases
3. Does not change
4. May increase or decrease

Answer: 1. Decreases

Question 10. A condenser is charged and then the battery is removed. A dielectric plate is put between the plates of the condenser, then the correct statement is

1. Q constant V and U decrease
2. Q constant V increases U decreases
3. Q increases V decreases U increases
4. None of these

Answer: 1. Q constant V and U decrease

Question 11. While a capacitor remains connected to a battery, a dielectric slab is slipped between the plates. Then

1. The energy stored in the capacitor decreases
2. The electric field between the plates increases
3. Charges flow from the battery to the capacitor
4. The potential difference between the plates is changed

Answer: 3. Charges flow from the battery to the capacitor

Question 12. In a parallel plate capacitor of capacitance C, a metal sheet is inserted between the plates, parallel to them. The thickness of the sheet is half of the separation between the plates. The capacitance now becomes

1. C/4
2. C/2
3. 2C
4. 4C

Answer: 3. 2C

Question 13. The plates of parallel plate capacitor are charged up to 100 V. A 2 mm thick plate is inserted between the plates. Then to maintain the same potential difference, the distance between the plates increases by 1.6 mm. The dielectric constant of the plate is-

1. 5
2. 1.25
3. 4
4. 2.5

Answer: 1. 5

Question 14. Between the plates of parallel plate condenser, a plate of thickness t1 and dielectric constant K1 is placed. In the rest of the space, there is another plate of thickness t2 and dielectric constant k2. The potential difference across the condenser will be

1. \(\frac{\mathrm{Q}}{\mathrm{A} \varepsilon_0}\left(\frac{\mathrm{t}_1}{\mathrm{k}_1}+\frac{\mathrm{t}_2}{\mathrm{k}_2}\right)\)
2. \(\frac{\varepsilon_0 \mathrm{Q}}{\mathrm{A}}\left(\frac{\mathrm{t}_1}{\mathrm{k}_1}+\frac{\mathrm{t}_2}{\mathrm{k}_2}\right)\)
3. \(\frac{Q}{A \varepsilon_0}\left(\frac{k_1}{t_1}+\frac{k_2}{t_2}\right)\)
4. \(\frac{\varepsilon_0 Q}{A}\left(k_1 t_1+k_2 t_2\right)\)

Answer: 1. \(\frac{\mathrm{Q}}{\mathrm{A} \varepsilon_0}\left(\frac{\mathrm{t}_1}{\mathrm{k}_1}+\frac{\mathrm{t}_2}{\mathrm{k}_2}\right)\)

Question 15. A sheet of aluminum is inserted in the air gap of a parallel plate capacitor, without touching any of the two plates of the capacitor. The capacitance of the capacitor is-

1. An invariant for all positions of the sheet
2. Maximum when the sheet is midway between the 2 plates
3. Maximum when the sheet is just near the +ve plate.
4. Maximum when the sheet is just near the –ve plate.

Answer: 1. Invariant for all positions of the sheet

Question 16. The value of a capacitor formed by a thin metallic foil is 2 µF. The foil is folded with a layer of paper having a thickness of 0.015 mm. The dielectric constant of the paper is 2.5 and its breadth is 40 mm. The length of the foil used is-

1. 0.34 m
2. 1.33 m
3. 13.4 mm
4. 33.9 m

Answer: 4. 33.9 m

Question 17. A parallel plate air capacitor is charged by connecting its plates to a battery. Without disconnecting the battery, a dielectric is introduced between its plates. As a result-

1. P.D. between the plates increases
2. Charge on the plates decreases
3. The capacitance of the capacitor decreases
4. None of the above

Answer: 4. None of the above

Question 18. A capacitor is charged using a battery, and the battery is withdrawn later on. Now a dielectric slab is introduced between the capacitor plates then the correct statement is-

1. Q increase, V decrease, U increase
2. Q remains constant, V increases, U decreases
3. Q remains constant, and V and U both decrease
4. None of these

Answer: 3. Q remains constant, and V and U both decrease

Question 19. A dielectric slab of thickness d is inserted in a parallel plate capacitor whose negative plate is at X = 0 and positive plate is at X = 3d. The slab is equidistant from the plates. The capacitor is given some charge. As X goes from 0 to 3 d-

1. The electric potential increases at first, then decreases, and again increases.
2. The electric potential increases continuously.
3. The direction of the electric field remains the same
4. The magnitude of the electric field remains the same
   1. a, b
   2. b, c
   3. b, d
   4. a, b, d

Answer: 2. b, c

Question 20. A parallel plate condenser with oil between the plates (dielectric constant of oil K = 2) has a capacitance C. If the coil is removed, then the capacitance of the capacitor becomes

1. \(\frac{C}{2}\)
2. \(\frac{C}{\sqrt{2}}\)
3. 2C
4. √2C

Answer: 1. \(\frac{C}{2}\)

Question 21. The plate separation of a 15µF capacitor is 2 mm. A dielectric slab (K = 2) of thickness 1 mm is inserted between the plates. Then new capacitance is given by

1. 15 µF
2. 20 µF
3. 30 µF
4. 25 µF

Answer: 2. 20 µF

Question 22. The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is 2 µF. The separation is reduced to half and it is filled with a substance dielectric of value 2.8. The new capacity of the capacitor is-

1. 11.2 µF
2. 15.6 µF
3. 19.2 µF
4. 22.4 µF

Answer: 1. 11.2 µF

Question 23. Effective capacitance if Cair = 10 µF-

1. 30 µF
2. 15 µF
3. 5 µF
4. 10 µF

Answer: 1. 30 µF

Question 24. While a capacitor remains connected to a battery, a dielectric slab is slipped between the plates-

1. The electric field between the plates increases
2. The energy stored in the capacitor decreases
3. The potential difference between the plates is changed
4. Charges flow from the battery to the capacitor.

Answer: 4. Charges flow from the battery to the capacitor.

Question 25. A parallel plate condenser is filled with two dielectrics as shown in fFig The area of each plate is A m 2 and the separation is d metre. The dielectric constants are K1 and K2 respectively. Its capacitance in farad will be

1. \(\frac{A \varepsilon_0\left(K_1 \times K_2\right)}{d\left(K_1+K_2\right)} A\)
2. \(\frac{A \varepsilon_0\left(K_1-K_2\right)}{d}\)
3. \(\frac{\mathrm{A} \varepsilon_0 \mathrm{~K}_1 \mathrm{~K}_2}{\left(\mathrm{~K}_1+\mathrm{K}_2\right)}\)
4. \(\frac{A \varepsilon_0\left(K_1+K_2\right)}{2 d}\)

Answer: 4. \(\frac{A \varepsilon_0\left(K_1+K_2\right)}{2 d}\)

Question 26. The capacity of the air capacitor (parallel plate) is 10µC. Now a dielectric of dielectric constant 4 is filled in the half-space between the plates, then new capacity will be an 

1. 25 µF
2. 20 µF
3. 40 µF
4. 5 µF

Answer: 1. 25 µF

Question 27. A copper plate of thickness b fills a parallel plate capacitor. The new capacity will be

1. \(\frac{\varepsilon_0 A}{2 d-b}\)
2. \(\frac{\varepsilon_0 A}{d-b}\)
3. \(\frac{\varepsilon_0 A}{d-b / 2}\)
4. \(\frac{\varepsilon_0 A}{d}\)

Answer: 2. \(\frac{\varepsilon_0 A}{d-b}\)

Question 28. Putting a dielectric substance between two plates of a condenser, the capacity, potential, and potential energy respectively-

1. Increases, decreases, decreases
2. Decreases, increases, increases
3. Increases, increases, increases
4. Decreases, decreases, decreases

Answer: 1. Increases, decreases, decreases

Question 29. A parallel plate capacitor of plate area A, separation d is filled with dielectrics as shown in fFig The dielectric constants are K1 and K2. Net capacitance is

1. \(\frac{\varepsilon_0 A}{d}\left(K_1+K_2\right)\)
2. \(\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\left(\frac{\mathrm{K}_1+\mathrm{K}_2}{\mathrm{~K}_1 \mathrm{~K}_2}\right)\)
3. \(\frac{2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\left(\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_1+\mathrm{K}_2}\right)\)
4. \(\frac{2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\left(\frac{\mathrm{K}_1+\mathrm{K}_2}{\mathrm{~K}_1 \mathrm{~K}_2}\right)\)

Answer: 3. \(\frac{2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\left(\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_1+\mathrm{K}_2}\right)\)

Question 30. Two parallel plates capacitors of value C and 2C are connected in parallel and are charged to a potential difference V. If the battery is disconnected and a medium of dielectric constant K is introduced between the plates of the capacitor C, then the potential between the plates of the capacitor will become-

1. 3V [K + 2]
2. \(\frac{[\mathrm{K}+2]}{3 \mathrm{~V}}\)
3. \(\frac{3 \mathrm{~V}}{[\mathrm{~K}+2]}\)
4. \(\frac{3[\mathrm{~K}+2]}{V}\)

Answer: 3. \(\frac{3 \mathrm{~V}}{[\mathrm{~K}+2]}\)

Question 31. As shown in the figure half the space between the plates of a capacitor is filled with an insulator material of dielectric constant K, if the initial capacity was C then the new capacity is

1. \(\frac{\mathrm{C}}{2}(\mathrm{~K}+1)\)
2. \(\frac{C}{2(K+1)}\)
3. \(\frac{(\mathrm{K}+1)}{2 \mathrm{C}}\)
4. C (K + 1)

Answer: 1. \(\frac{\mathrm{C}}{2}(\mathrm{~K}+1)\)

Question 32. Two materials of dielectric constant k₁ and k₂ are filled between two parallel plates of a capacitor as shown in the figure. The capacity of the capacitor is :

1. \(\frac{A \in_0\left(k_1+k_2\right)}{2 d}\)
2. \(\frac{2 A \epsilon_0}{d}\left(\frac{k_1 k_2}{k_1+k_2}\right)\)
3. \(\frac{A \in_0}{d}\left(\frac{k_1 k_2}{k_1+k_2}\right)\)
4. \(\frac{A \epsilon_0}{2 d}\left(\frac{k_1+k_2}{k_1 k_2}\right)\)

Answer: 1. \(\frac{A \in_0\left(k_1+k_2\right)}{2 d}\)

Question 33. A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capacity of C and is charged to a potential of V volts. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is

1. \(\frac{1}{2}(\mathrm{~K}-1) \mathrm{CV}^2\)
2. CV2(K – 1)/K
3. (K – 1)CV2
4. zero

Answer: 4. zero

Question 34. After charging a capacitor the battery is removed. Now by placing a dielectric slab between the plates

1. The potential difference between the plates and the energy stored will decrease but the charge on the plates will remain the same
2. The charge on the plates will decrease and the potential difference between the plates will increase
3. The potential difference between the plates will increase and the energy stored will decrease but the charge on the plates will remain the same
4. The potential difference, energy stored and the charge will remain unchanged

Answer: 1. The potential difference between the plates and the energy stored will decrease but the charge on the plates will remain the same

Question 35. A parallel plate capacitor is filled with two dielectrics as shown in the figure. If A is the area of each plate, then the effective capacitance between X and Y is

1. \(\frac{\varepsilon_0 \mathrm{~A}}{d}\left(\frac{\mathrm{K}_1+\mathrm{K}_2}{2}\right)\)
2. \(\frac{2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\left(\mathrm{~K}_1+\mathrm{K}_2\right)\)
3. \(\frac{2 \varepsilon_0 A}{d}\left(\frac{K_1+K_2}{K_1 K_2}\right)\)
4. \(\frac{2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\left(\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_1+\mathrm{K}_2}\right)\)

Answer: 4. \(\frac{2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\left(\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_1+\mathrm{K}_2}\right)\)

Question 36. An insulator plate is passed between the plates of a capacitor. Then current

1. First flows from A to B and then from B to A
2. First flows from B to A and then from A to B
3. Always flows from B to A
4. Always flows from A to B

Answer: 2. First flows from B to A and then from A to B

Question 37. A sheet of aluminum foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor.

1. Decreases
2. Remains unchanged
3. Becomes infinite
4. Increases

Answer: 2. Remains unchanged

Question 38. In the adjoining diagram, two geometrically identical capacitors A and B are connected to a battery. Air is filled between the plates of C₁ and a dielectric is filled between the plates of C₂, then –

1. q₁ < q₂
2. q₁ > q₂
3. q₁ = q₂
4. None of these

Answer: 1. q₁ < q₂

Question 39. The electric field between two parallel plates of a capacitor is 2.1 × 10^-5. If a medium is inserted between the plates then the electric field becomes 1.0 × 10^-5. Now, the value of the dielectric will be

1. 2
2. 3
3. 4
4. 5

Answer: 1. 2

Question 40. The capacity of a parallel plate condenser without any dielectric is C. If the distance between the plates is doubled and the space between the plates is filled with a substance of dielectric constant 3, the capacity of the condenser becomes :

1. \(\frac{3}{4} \mathrm{C}\)
2. \(\frac{9}{2} C\)
3. \(\frac{2}{3} C\)
4. \(\frac{3}{2} C\)

Answer: 4. \(\frac{3}{2} C\)

Capacitance Exercise – 2

Question 1. In the figure given, the potential difference between A and B in a steady state will be

1. 20 V
2. 25 V
3. 75 V
4. 100 V

Answer: 2. 25 V

Question 2. In the above question, the potential difference between B and C in a steady state will be

1. 20 V
2. 25 V
3. 50 V
4. 75 V

Answer: 4. 75 V

Question 3. Three capacitors of the same capacitance are connected in parallel. When they are connected to a cell of 2 volts, a total charge of 1.8µC is accumulated on them. Now they are connected in series and then charged by the same cell. The total charge stored in them will be

1. 1.8µC
2. 0.9µC
3. 0.6µC
4. 0.2µC

Answer: 4. 0.2µC

Question 4. Each edge of the cube contains a capacitance C. The equivalent capacitance between the points A and B will be –

1. \(\frac{6 C}{5}\)
2. \(\frac{5 C}{6}\)
3. \(\frac{12 C}{7}\)
4. \(\frac{7 C}{12}\)

Answer: 1. \(\frac{6 C}{5}\)

Question 5. A capacitor of capacitance 500μF is charged at the rate of 100μC/s. The time in which the potential difference will become 20 V, is

1. 100 s
2. 50 s
3. 20 s
4. 10 s

Answer: 1. 100 s

Question 6. A network of uncharged capacitors and resistances is shown

Current through the battery immediately after key K is closed and after a long time interval is :

1. \(\frac{E}{R_1}, \frac{E}{R_1+R_3}\)
2. \(\frac{E}{R_1+R_3}, \frac{E}{R_1+\frac{R_2 R_3}{R_2+R_3}}\)
3. \(\text { zero, } \frac{E}{R_1}\)
4. \(\frac{E}{R_1+\frac{R_2 R_3}{R_2+R_3}}, \frac{E}{R_1}\)

Answer: 1. \(\frac{E}{R_1}, \frac{E}{R_1+R_3}\)

Question 7. In the given circuit, a charge of +80 µC is given to the upper plate of the 4µF capacitor. Then in the steady state, the charge on the upper plate of the 3µF capacitor is :

1. +32 µC
2. +40 µC
3. +48 µC
4. +80 µC

Answer: 3. +48 µC

Capacitance Exercise – 3

Question 1. Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitance and breakdown voltage of the combination will be

1. \(\frac{\mathrm{C}}{3}, \frac{\mathrm{V}}{3}\)
2. \(3 \mathrm{C}, \frac{\mathrm{V}}{3}\)
3. \(\frac{C}{3}, 3 V\)
4. 3C ,3V

Answer: 3. \(\frac{C}{3}, 3 V\)

Question 2. A series combination of n₁ capacitors, each of value C₁ is charged by a source of potential difference 4V. When another parallel combination of n₂ capacitors, each of value C₂, is charged by a source of potential difference V, it has the same (total) energy stored in it, as the first combination has. The value of C₂, in terms of C₁, is then

1. \(\frac{2 C_1}{n_1 n_2}\)
2. \(16 \frac{n_2}{n_1} C_1\)
3. \(2 \frac{n_2}{n_1} C_1\)
4. \(\frac{16 C_1}{n_1 n_2}\)

Answer: 4. \(\frac{16 C_1}{n_1 n_2}\)

Question 3. A parallel plate condenser has a uniform electric field E(V/m) in the space between the plates. If the distance between the plates is d(m) and the area of each plate is A(m²) the energy (joules) stored in the condenser is :

1. E²Ad/∈₀
2. \(\frac{1}{2} \epsilon_0 \mathrm{E}^2\)
3. ∈₀ EAd
4. \(\frac{1}{2} \epsilon_0 E^2 \mathrm{Ad}\)

Answer: 4. \(\frac{1}{2} \epsilon_0 E^2 \mathrm{Ad}\)

Question 4. A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and the area of each plate is A, the energy stored in the capacitor is :

1. \(\frac{1}{2} \varepsilon_0 \mathrm{E}^2\)
2. E² Ad/ε₀
3. \(\frac{1}{2} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}\)
4. ε₀EAd

Answer: 3. \(\frac{1}{2} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}\)

Question 5. Two thin dielectric slabs of dielectric constants K₁ and K₂ (K₁ < K₂) are inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field ‘E’ between the plates with distance ‘d’ as measured from plate P is correctly shown by:

Answer: 3.

Question 6. A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?

1. The energy stored in the capacitor decreases K times.
2. The chance in energy stored is\(\frac{1}{2} C V^2\left(\frac{1}{K}-1\right)[/laex]
3. The charge on the capacitor is not conserved.
4. The potential difference between the plates decreases K times.

Answer: 3. The charge on the capacitor is not conserved.

Question 7. A capacitor of 2µF is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is :

1. 80%
2. 0%
3. 20%
4. 75%

Answer: 1. 80%

Question 8. A parallel plate capacitor of area A, plate separation d, and capacitance C is filled with four dielectric materials having dielectric constants k₁, k₂, k₃, and k₄ as shown in the figure below. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by

1. 1/K= 1/K₁+2/K₂+1/K₃
2. k = k₁ + k₂ + k₃ + 3k₄
3. K = 2/3 (K₁+K₂+K₃)
4. 2/k= 3/(K₁+K₂+K₃)+1/K₄

Answer: 4. 2/k= 3/(K₁+K₂+K₃)+1/K₄

Question 9. A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system :

1. Increases by a factor of 4
2. Decreases by a factor of 2
3. Remains the same
4. Increases by a factor of 2

Answer: 2. Decreases by a factor of 2

Question 10. The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is :

1. Independent of the distance between the plates
2. Inversely proportional to the distance between the plates
3. Proportional to the square root of the distance the plates
4. Linearly proportional to the distance between the plates

Answer: 1. Independent of the distance between the plates

Question 11. Two identical capacitors C1 and C2 of equal capacitance are connected as shown in the circuit. Terminals a and b of the key k are connected to charge capacitor C1 using a battery of emf V volt. Now disconnecting a and b the terminals b and c are connected. Due to this, what will be the percentage loss of energy?

1. 75%
2. 0%
3. 50%
4. 25%

Answer: 3. 50%

Question 12. The variation of electrostatic potential with radial distance r from the center of a positively charged metallic thin shell of radius R is given by the graph.

Answer: 4.

Question 13. A parallel plate capacitor having cross-sectional area A and separation d has air in between the plates. Now an insulating slab of the same area but thickness d/2 is inserted between the plates as shown in the figure having dielectric constant K(= 4). The ratio of new capacitance to its original capacitance will be,

1. 2: 1
2. 8: 5
3. 6: 5
4. 4: 1

Answer: 2. 8: 5

Question 14. A short electric dipole has a dipole moment of. The electric potential due to the dipole at a point at a distance of 0.6m from the center of the dipole situated on a line making an angle of 600 with the 9 2 2 dipole axis is [latex]\left(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 / \mathrm{c}^2\right)\)

1. zero
2. 50 V
3. 200 V
4. 400 V

Answer: 3. 200 V

Question 15. In a certain region of space with a volume of 0.2 m3 the electric potential is found to be 5V throughout. The magnitude of the electric field in this region is

1. 5 N/C
2. zero
3. 0.5 N/C
4. 1 N/C

Answer: 2. zero

Question 16. The capacitance of a parallel plate capacitor with air as medium is. With the introduction of a 30 Fμ dielectric medium, the capacitance becomes. The permittivity of the medium is \(\left(\epsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)\)

1. \(5.00 \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\)
2. \(0.44 \times 10^{-13} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\)
3. \(1.77 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\)
4. \(0.44 \times 10^{-10} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\)

Answer: 4. \(0.44 \times 10^{-10} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\)

Question 17. Polar molecules are the molecules:

1. Acquire a dipole moment only in the presence of an electric field due to the displacement of charges.
2. Acquire a dipole moment only when the magnetic field is absent.
3. Having a permanent electric dipole moment
4. Having zero dipole moment.

Answer: 3. Having a permanent electric dipole moment

Question 18. Two charged spherical conductors of radius R₁ and R₂ are connected by a wire. Then the ratio of surface(σ₁ / σ₂) charge densities of the spheres is

1. \(\frac{\mathrm{R}_2}{\mathrm{R}_1}\)
2. \(\sqrt{\left(\frac{R_2}{R_1}\right)}\)
3. \(\frac{R_1^2}{R_2^2}\)
4. \(\frac{R_1}{R_2}\)

Answer: 1. \(\frac{\mathrm{R}_2}{\mathrm{R}_1}\)

Question 19. A dipole is placed in an electric field as shown. In which direction will it move?

1. Towards the right as its potential energy will decrease
2. Towards the left as its potential energy will decrease
3. Towards the right as its potential energy will increase
4. Towards the left as its potential energy will increase

Answer: 1. Towards the right as its potential energy will decrease

Question 20. A parallel plate capacitor has a uniform electric field ‘ ’ in the space between the plates. If the distance ε₀ = between the plates is ‘d’ and the area of each plate is ‘A’ the energy stored in the capacitor is: ( permittivity of free space)

1. ε₀ EAd
2. \(\frac{1}{2} \varepsilon_0 E^2 A d\)
3. \(\frac{E^2 A d}{\varepsilon_0}\)
4. \(\frac{1}{2} \varepsilon_0 E^2\)

Answer: 2. \(\frac{1}{2} \varepsilon_0 E^2 A d\)

Question 21. The equivalent capacitance of the combination shown in the figure is :

1. 2C
2. C/2
3. 3C/2
4. 3C

Answer: 1. 2C

Question 22. Twenty-seven drops of the same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop

1. 1320 V
2. 1520 V
3. 1980 V
4. 660 V

Answer: 3. 1980 V

Question 23. Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t1 is the time taken for the energy stored in the capacitor to reduce to half its initial value and t2 is the time taken for the charge to reduce to one-fourth of its initial value. Then the ratio t1/t2 will be

1. 1
2. 1/2
3. 1/4
4. 2

Answer: 3. 1/4

Question 24. A resistor ‘R’ and 2μF capacitor in series are connected through a switch to a 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5s after the switch has been closed. (log 102.5 = 0.4)

1. 1.3 × 10⁴ Ω
2. 1.7 × 10⁵ Ω
3. 2.7 × 10⁶ Ω
4. 3.3 × 10⁷ Ω

Answer: 4. 3.3 × 10⁷ Ω

Question 25. A combination of two identical capacitors, a resistor R, and a dDCvoltagedc voltage voltage 6V is used in an experiment on a (C – R) circuit. It is found that for a parallel combination of the capacitor, the time in which the voltage of the fully charged combination reduces to half its original voltage is 10 seconds. For series combination, the time needed for reducing the voltage of the fully charged series combination by half is :

1. 10 second
2. 5 second
3. 2.5 second
4. 20 second

Answer: 4. 20 second

Question 26. The figure shows an experimental plot discharging of a capacitor in an RC circuit. The time constant τ of this circuit lies between :

1. 150 sec and 200 sec
2. 0 and 50 sec
3. 50 sec and 100 sec
4. 100 sec and 150 sec

Answer: 3. 50 sec and 100 sec

Question 27. Two capacitors C₁ and C₂ are charged to 120 V and 200 V respectively. It is found that by connecting them the potential of each one can be made zero. Then :

1. 5C₁ = 3C₂
2. 3C₁ = 5C₂
3. 3C₁ + 5C₂ = 0
4. 9C₁ = 4C₂

Answer: 3. 3C₁ + 5C₂ = 0

Question 28. A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 10 4 V/m, the charge density of the positive plate will be close to :

1. 6 × 10–7C/m²
2. 3 × 10–7C/m²
3. 3 × 104C/m²
4. 6 × 104C/m

Answer: 3. 3 × 104C/m²

Question 29. In the given circuit, charge Q₂ on the 2μF capacitor changes as C is varied from 1 μF to 3μF. Q₂ as a function of ‘C’ is given properly by : (figures are drawn schematically and are not to scale)

Answer: 2.

Question 30. A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4μF and 9μF capacitors), at a point distance 30 m from it, would equal :

1. 360 N/C
2. 420 N/C
3. 480 N/C
4. 240 N/C

Answer: 2. 420 N/C

Question 31. A capacitance of 2μF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 μF capacitors are available that can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is : 

1. 32
2. 2
3. 16
4. 24

Answer: 1. 32

Question 32. In the given circuit diagram when the current reaches a steady state in the circuit, the charge on the capacitor of capacitance C will be :

1. \({CE} \frac{r_1}{\left(r_1+r\right)}\)
2. CE
3. \(C E \frac{r_1}{\left(r_2+r\right)}\)
4. \(C E \frac{r_2}{\left(r+r_2\right)}\)

Answer: 4. \(C E \frac{r_2}{\left(r+r_2\right)}\)

Question 33. A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20V. If a dielectric mate- 5 rials of dielectric constant K = \(\frac{5}{3}\) is inserted between the plates, the magnitude of the induced charge will be :

1. 2.4 n C
2. 0.9 n C
3. 1.2 n C
4. 0.3 n C

Answer: 3. 1.2 n C

Question 34. A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants K₁, K₂, K₃, and K₄ arranged as shown in the figure. The effective dielectric constant K will be :

1. \(\mathrm{K}=\frac{\left(\mathrm{K}_1+\mathrm{K}_4\right)\left(\mathrm{K}_2+\mathrm{K}_3\right)}{2\left(\mathrm{~K}_1+\mathrm{K}_2+\mathrm{K}_3+\mathrm{K}_4\right)}\)
2. \(\mathrm{K}=\frac{\left(\mathrm{K}_1+\mathrm{K}_3\right)\left(\mathrm{K}_2+\mathrm{K}_4\right)}{\mathrm{K}_1+\mathrm{K}_2+\mathrm{K}_3+\mathrm{K}_4}\)
3. \(\mathrm{K}=\frac{\left(\mathrm{K}_1+\mathrm{K}_2\right)\left(\mathrm{K}_3+\mathrm{K}_4\right)}{2\left(\mathrm{~K}_1+\mathrm{K}_2+\mathrm{K}_3+\mathrm{K}_4\right)}\)
4. \(\mathrm{K}=\frac{\left(\mathrm{K}_1+\mathrm{K}_2\right)\left(\mathrm{K}_3+\mathrm{K}_4\right)}{\mathrm{K}_1+\mathrm{K}_2+\mathrm{K}_3+\mathrm{K}_4}\)

Answer: 4. \(\mathrm{K}=\frac{\left(\mathrm{K}_1+\mathrm{K}_2\right)\left(\mathrm{K}_3+\mathrm{K}_4\right)}{\mathrm{K}_1+\mathrm{K}_2+\mathrm{K}_3+\mathrm{K}_4}\)

Question 35. A parallel plate capacitor is of area 6 cm² and a separation of 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants K₁ = 10, K₂ = 12, and K₃ = 14. The dielectric constant of a material that when fully inserted in the above capacitor, gives the same capacitance would be :

1. 4
2. 14
3. 36
4. 12

Answer: 4. 12

Question 36. A parallel plate capacitor having a capacitance of 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plate. The work done by the capacitor on the slab is

1. 692 pJ
2. 600 pJ
3. 508 pJ
4. 560 pJ

Answer: 3. 508 pJ

Question 37. Seven capacitors, each of capacitance 2pF. are to be connected in a configuration to obtain an effect tive capacitance of \(\left(\frac{6}{13}\right) \mu F\). Which of the combinations, shown in the figures below, will achieve the desired value?

Answer: 3.

Question 38. In the figure shown below, the charge on the left plate of the 10 µF capacitor is –30 µC. The charge on the right plate of the 6 µF capacitor is :

1. –12 µC
2. +12 µC
3. –18 µC
4. + 18 µC

Answer: 4. + 18 µC

Question 39. A parallel plate capacitor with pates of area 1 m 2 each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate. (Take ∈₀ = 8.85 × 10–12 \(\frac{C^2}{N-m^2}\))

1. 6.85 × 10–10 C
2. 7.85 × 10–10 C
3. 9.85 × 10–10 C
4. 8.85 × 10–10 C

Answer: 4. 8.85 × 10–10 C

Question 40. In the figure shown, after the switch ‘S’ is turned from position ‘A’ to position ‘B’, the energy dissipated in the circuit in terms of capacitance ‘C’ and total charge ‘Q’ is:

1. \(\frac{3}{8} \frac{Q^2}{C}\)
2. \(\frac{5}{8} \frac{Q^2}{C}\)
3. \(\frac{1}{8} \frac{Q^2}{C}\)
4. \(\frac{3}{4} \frac{Q^2}{C}\)

Answer: 1. \(\frac{3}{8} \frac{Q^2}{C}\)

Question 41. A spherical condenser has 10 cm and 12 cm as the radii of inner and outer spheres. The space between the two spheres is filled with a dielectric of dielectric constant 5. The capacity when;

1. The outer sphere is earthed.

1. \(\frac{2}{3} \times 10^{-10} \mathrm{~F}\)
2. \(\frac{8}{3} \times 10^{-10} \mathrm{~F}\)
3. \(\frac{10}{3} \times 10^{-10} \mathrm{~F}\)
4. \(\frac{16}{3} \times 10^{-10} \mathrm{~F}\)

Answer: 3. \(\frac{10}{3} \times 10^{-10} \mathrm{~F}\)

2. The inner sphere is earthed.

1. \(\frac{104}{30} \times 10^{-10} \mathrm{~F}\)
2. \(\frac{52}{30} \times 10^{-10} \mathrm{~F}\)
3. \(\frac{26}{30} \times 10^{-10} \mathrm{~F}\)
4. 6 × 10^-11 F

Answer: 1. \(\frac{104}{30} \times 10^{-10} \mathrm{~F}\)

Question 42. A parallel plate condenser of capacity C is connected to a battery and is charged to potential V. Another condenser of capacity 2C is connected to another battery and is charged to potential 2V. The charging batteries are removed and now the condensers are connected in such a way that the positive plate of one is connected to the negative plate of another. The final energy of this system is–

1. zero
2. \(\frac{25 C V^2}{6}\)
3. \(\frac{3 C V^2}{2}\)
4. \(\frac{9 C V^2}{2}\)

Answer: 3. \(\frac{3 C V^2}{2}\)

Question 43. An uncharged capacitor of capacitance C is connected to a battery of emf ε at t = 0 through a resistance R, then

1. The maximum rate at which energy is stored in the capacitor is :

1. \(\frac{\varepsilon^2}{4 R}\)
2. \(\frac{\varepsilon^2}{2R}\)
3. \(\frac{\varepsilon^2}{R}\)
4. \(\frac{2\varepsilon^2}{R}\)

Answer: 1. \(\frac{\varepsilon^2}{4 R}\)

2. The time at which the rate has this maximum value is

1. 2CR ln2
2. 1/2 CR ln2
3. CR ln2
4. 3CR ln2

Answer: 3. CR ln2

Question 44. The V versus x plot for six identical metal plates of cross-sectional area A is as shown. The equivalent capacitance between 2 and 5 is (Adjacent plates are placed at a separation d) :

1. \(\frac{2 \epsilon_0 A}{d}\)
2. \(\frac{\epsilon_0 A}{d}\)
3. \(\frac{3 \epsilon_0 A}{d}\)
4. \(\frac{2 \epsilon_0 A}{2d}\)

Answer: 2. \(\frac{\epsilon_0 A}{d}\)

Question 45. In the circuit shown in the figure, the capacitors are initially uncharged. The current through resistor PQ just after closing the switch is :

1. 2A from P to Q
2. 2A from Q to P
3. 6A from P to Q
4. zero

Answer: 4. zero

Question 46. The plates of a parallel plate condenser are being moved away with a constant speed v. If the plate separation at any instant of time is d then the rate of change of capacitance with time is proportional to–

1. \(\frac{1}{\mathrm{~d}}\)
2. \(\frac{1}{d^2}\)
3. d2
4. d

Answer: 2. \(\frac{1}{d^2}\)

Question 47. A parallel plate capacitor of capacitance C is as shown. A thin metal plate A is placed between the plates of the given capacitor in such a way that its edges touch the two plates as shown. The capacity now becomes.

1. 0
2. 3C
3. 4C
4. ∞

Answer: 4. ∞

Question 48. A capacitor of capacitance C₀ is charged to a potential V0 and then isolated. A capacitor C is then charged from C₀, discharged, and charged again; the process is repeated n times. Due to this, the potential of the larger capacitor is decreased to V, then the value of C is :

1. C₀ [V₀/V]^1/n
2. C₀[(V₀/V)^1/n – 1]
3. C₀ [(V₀/V) – 1]
4. C₀ [(V/V₀)n + 1]

Answer: 2. C₀[(V₀/V)1/n – 1]

Question 49. A 3 mega ohm resistor and an uncharged 1 μF capacitor are connected in a single loop circuit with a constant source of 4 volts. At one second after the connection is made what are the rates at which;

1. The charge on the capacitor is increasing.

1. \(4\left(1-\mathrm{e}^{-1 / 3}\right) \mu \mathrm{C} / \mathrm{s}\)
2. \(4 e^{-1 / 3} \mu \mathrm{C} / \mathrm{s}\)
3. \(\frac{4}{3} e^{-1 / 3} \mu \mathrm{C} / \mathrm{s}\)
4. \(\frac{4}{3}\left(1-\mathrm{e}^{-1 / 3}\right) \mu \mathrm{C} / \mathrm{s}\)

Answer: 3. \(\frac{4}{3} e^{-1 / 3} \mu \mathrm{C} / \mathrm{s}\)

2. Energy is being stored in the capacitor.

1. \(\frac{16}{3}\left(1-\mathrm{e}^{-1 / 3}\right) \mathrm{e}^{-1 / 3} \mu \mathrm{J} / \mathrm{s}\)
2. \(\frac{1}{2} e^{-1 / 3} \mu \mathrm{J} / \mathrm{s}\)
3. \(\frac{16}{3} \mathrm{e}^{-2 / 3} \mu \mathrm{J} / \mathrm{s}\)
4. None Of These

Answer: 1. \(\frac{16}{3}\left(1-\mathrm{e}^{-1 / 3}\right) \mathrm{e}^{-1 / 3} \mu \mathrm{J} / \mathrm{s}\)

3. Joule heat appears in the resistor.

1. \(\frac{16}{3} e^{-1 / 3} \mu \mathrm{J} / \mathrm{s}\)
2. \(\frac{1}{2} e^{-1 / 3} \mu \mathrm{J} / \mathrm{s}\)
3. \(\frac{16}{3}\left(\mathrm{e}^{-2 / 3}\right) \mu \mathrm{J} / \mathrm{s}\)
4. \(\frac{16}{3}\left(1-\mathrm{e}^{-2 / 3}\right) \mu \mathrm{J} / \mathrm{s}\)

Answer: 3. \(\frac{16}{3}\left(\mathrm{e}^{-2 / 3}\right) \mu \mathrm{J} / \mathrm{s}\)

4. Energy is being delivered by the source.

1. \(16\left(1-\mathrm{e}^{-1 / 3}\right) \mu \mathrm{J} / \mathrm{s}\)
2. 16μ J/s
3. \(\frac{16}{3} e^{-1 / 3} \mu \mathrm{J} / \mathrm{s}\)
4. \(\frac{16}{3}\left(1-\mathrm{e}^{-1 / 3}\right) \mu \mathrm{J} / \mathrm{s}\)

Answer: 3. \(\frac{16}{3} e^{-1 / 3} \mu \mathrm{J} / \mathrm{s}\)

Question 50. An uncharged capacitor of capacitance 100μF is connected to a battery of emf 20V at t = 0 through a resistance 10μ, then

1. The maximum rate at which energy is stored in the capacitor is :

1. 10J/s
2. 20 J/s
3. 40J/s
4. 5J/s

Answer: 1. 10J/s

2. The time at which the rate has this maximum value is

1. (4 ln 2) ms
2. (2 ln 2) ms
3. (ln 2) ms
4. (3 ln 2) ms

Answer: 3. (ln 2) ms

Question 51.

1. A 3μF capacitor is charged up to 300 volts and 2μF is charged up to 200 volts. The capacitors are connected so that the plates of the same polarity are connected. The final potential difference between the plates of the capacitor after they are connected is :

1. 220 V
2. 160 V
3. 280 V
4. 260 V

Answer: 4. 260 V

2. If instead of this, the plates of opposite polarity were joined together, then the amount of charge that flows is :

1. 6 × 10^-4 C
2. 1.5 × 10^-4 C
3. 3 × 10^-4 C
4. 7.5 × 10^-4 C

Answer: 1. 6 × 10^-4 C

Question 52. An uncharged capacitor of capacitance 4µF, a battery of EMF 12 volt, and a resistor of 2.5 M Ω are connected in series. The time after which V_C = 3VR is (take ln2 = 0.693)

1. 6.93 seconds
2. 13.86 seconds
3. 7 seconds
4. 14 seconds

Answer: 2. 13.86 seconds

Question 53. A circuit is connected as shown in the figure with the switch S open. When the switch is closed, the total amount of charge that flows from Y to X is

1. 0
2. 54 μC
3. 27 μC
4. 81 μC

Answer: 3. 27 μC

Question 54. In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switch S₁ is pressed first to fully charge the capacitor C₁ and then released. The switch S₂ is then pressed to charge the capacitor C₂. After some time, S₂ is released and then S3 is pressed. After some time.

1. The charge on the upper plate of C₁ is 2CV₀
2. The charge on the upper plate of C₁ is CV₀
3. The charge on the upper plate of C₂ is 0
4. The charge on the upper plate of C₂ is –2CV₀

Answer: 2. The charge on the upper plate of C₁ is CV₀

Question 55. A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the area of its plates, as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is C₁. When the capacitor is charged, the plate area covered by the dielectric gets charged Q₁, and the rest of the area gets charged Q₂. Choose the correct option/options, ignoring edge effects.

1. \(\frac{E_1}{E_2}=1 / 2\)
2. \(\frac{E_1}{E_2}=\frac{1}{K}\)
3. \(\frac{Q_1}{Q_2}=\frac{3}{K}\)
4. \(\frac{\mathrm{C}}{\mathrm{C}_1}=\frac{2+\mathrm{K}}{\mathrm{K}}\)

Answer: 4. \(\frac{\mathrm{C}}{\mathrm{C}_1}=\frac{2+\mathrm{K}}{\mathrm{K}}\)

Question 56. A parallel plate capacitor having plates of area S and plate separation d has capacitance C₁ in the air. When two dielectrics of different relative permittivities (ε₁ = 2 and ε₂ = 4) are introduced between the two 2 plates as shown in the figure, the capacitance becomes C₂. The ratio \(\frac{C_2}{C_1}\) is

1. 6/5
2. 5/3
3. 7/5
4. 7/3

Answer: 4. 7/3

Question 57. A fully charged capacitor has a capacitance of ‘C’. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity ‘s’ and mass ‘m’. If the temperature of the block is raised by ‘ΔT’, the potential difference ‘V’ across the capacitance is :

1. \(\sqrt{\frac{2 m C \Delta T}{s}}\)
2. \(\frac{\mathrm{mC} \Delta \mathrm{T}}{\mathrm{s}}\)
3. \(\frac{\mathrm{ms} \Delta \mathrm{T}}{\mathrm{C}}\)
4. \(\sqrt{\frac{2 m s \Delta T}{C}}\)

Answer: 4. \(\sqrt{\frac{2 m s \Delta T}{C}}\)