Capacitors and Dielectrics — Competition Zone

A capacitance of 2 µF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 µF capacitors are available which 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

32

Reason —

Given,

• Required capacitance = 2 µF, across V = 1.0 kV = 1000 V
• Each available capacitor = 1 µF, withstanding at most 300 V

To withstand 1000 V, the capacitors must be joined in series. The number in each row must be

$\text n = \dfrac{1000}{300} \\[1em] \text n = 3.33$

so at least n = 4 capacitors are needed in a row. The capacitance of one row is

$\text C_\text{row} = \dfrac{1}{4}\ \mu\text F$

If m such rows are joined in parallel to give 2 µF,

$\text m \times \dfrac{1}{4} = 2 \\[1em] \text m = 8$

Hence the minimum number of capacitors is

$\text n \times \text m = 4 \times 8 \\[1em] \text n \times \text m = 32$

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 between the plates
3. proportional to the square root of the distance between the plates
4. linearly proportional to the distance between the plates

Answer

independent of the distance between the plates

Reason —

Each plate lies in the field produced by the other plate alone, which is

$\text E = \dfrac{\sigma}{2\epsilon_0} \\[1em] \text E = \dfrac{\text Q}{2\epsilon_0 \text A}$

Hence the force on a plate carrying charge Q is

$\text F = \text{QE} \\[1em] \text F = \dfrac{\text Q^2}{2\epsilon_0 \text A}$

The capacitor is isolated, so Q is fixed. This expression contains no distance term, hence the electrostatic force is independent of the separation between the plates.

A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant $\text K = \dfrac{5}{3}$ is inserted between the plates, the magnitude of the induced charge will be :

1. 2.4 nC
2. 0.9 nC
3. 1.2 nC
4. 0.3 nC

Answer

1.2 nC

Reason —

Given,

• C = 90 pF, V = 20 V, K = 5/3
• The battery remains connected, so V is constant

With the dielectric inserted, the charge on the plates becomes

$\text Q = (\text{KC})\text V \\[1em] = \left(\dfrac{5}{3} \times 90 \times 10^{-12}\right)(20) \\[1em] = 3 \times 10^{-9}\ \text C \\[1em] \text Q = 3\ \text{nC}$

The charge induced on the faces of the dielectric is

$\text Q_\text{ind} = \text Q\left(1 - \dfrac{1}{\text K}\right)$

Substituting the values,

$\text Q_\text{ind} = 3\left(1 - \dfrac{3}{5}\right) \\[1em] = 3 \times \dfrac{2}{5} \\[1em] \text Q_\text{ind} = 1.2\ \text{nC}$

A parallel plate capacitor of capacitance 20 µF is being charged by a voltage source whose potential is changing at the rate of 3 V/s. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively :

1. 60 µA, zero
2. Zero, zero
3. Zero, 60 µA
4. 60 µA, 60 µA

Answer

60 µA, 60 µA

Reason —

Given,

• C = 20 µF = 20 × 10^-6 F
• $\dfrac{\text{dV}}{\text{dt}} = 3$ V s^-1

The charge on the capacitor is q = CV. Differentiating with respect to time,

$\dfrac{\text{dq}}{\text{dt}} = \text C\dfrac{\text{dV}}{\text{dt}} \\[1em] \text I_c = (20 \times 10^{-6})(3) \\[1em] = 60 \times 10^{-6}\ \text A \\[1em] \text I_c = 60\ \mu\text A$

The displacement current through the gap between the plates is always equal to the conduction current in the wires, so

$\text I_d = \text I_c \\[1em] \text I_d = 60\ \mu\text A$

Figure shows charge(q) versus voltage(V) graph for series and parallel combination of two given capacitors. The capacitance are :

1. 50 µF and 30 µF
2. 40 µF and 10 µF
3. 20 µF and 30 µF
4. 60 µF and 40 µF

Answer

40 µF and 10 µF

Reason —

From the graph, the slope q/V gives the capacitance. The line A of greater slope is the parallel combination and the line B the series combination,

$\text C_p = \dfrac{500}{10} \\[1em] \text C_p = 50\ \mu\text F \\[1em] \text C_s = \dfrac{80}{10} \\[1em] \text C_s = 8\ \mu\text F$

If the two capacitances are C₁ and C₂,

$\text C_1 + \text C_2 = 50 \qquad \text{and} \qquad \dfrac{\text C_1 \text C_2}{\text C_1 + \text C_2} = 8$

so that

$\text C_1 \text C_2 = 8 \times 50 \\[1em] \text C_1 \text C_2 = 400$

Hence C₁ and C₂ are the roots of

$\text x^2 - 50\text x + 400 = 0 \\[1em] \text x = \dfrac{50 \pm 30}{2} \\[1em] \text x = 40\ \text{or}\ 10$

Voltage rating of a parallel plate capacitor is 500 V. Its dielectric can withstand a maximum electric field of 10⁶ V/m. The plate area is 10^-4 m². What is the dielectric constant if the capacitance is 15 pF?

(Given ε₀ = 8.86 × 10^-12 C²/Nm²)

1. 3.8
2. 4.5
3. 6.2
4. 8.5

Answer

8.5

Reason —

Given,

• V = 500 V, E_max = 10⁶ V m^-1
• A = 10^-4 m², C = 15 pF = 15 × 10^-12 F

The least permissible separation follows from E = V/d,

$\text d = \dfrac{\text V}{\text E_\text{max}} \\[1em] = \dfrac{500}{10^6} \\[1em] \text d = 5 \times 10^{-4}\ \text m$

For a capacitor filled with a dielectric,

$\text C = \dfrac{\text K \epsilon_0 \text A}{\text d} \\[1em] \text K = \dfrac{\text{Cd}}{\epsilon_0 \text A}$

Substituting the values,

$\text K = \dfrac{(15 \times 10^{-12})(5 \times 10^{-4})}{(8.86 \times 10^{-12})(10^{-4})} \\[1em] \text K = \dfrac{7.5 \times 10^{-15}}{8.86 \times 10^{-16}} \\[1em] \text K \approx 8.5$

Determine the charge on the capacitor in the following circuit :

1. 2 µC
2. 60 µC
3. 200 µC
4. 10 µC

Answer

200 µC

Reason —

Given, emf = 72 V, resistances 6 Ω, 4 Ω, 2 Ω and 10 Ω, and C = 10 µF.

In the steady state no current flows through the branch containing the capacitor, so the capacitor is simply connected across the 10 Ω resistor.

The 4 Ω resistor is in parallel with the series pair (2 Ω + 10 Ω = 12 Ω),

$\text R' = \dfrac{4 \times 12}{4 + 12} \\[1em] \text R' = 3\ \Omega$

The total resistance of the circuit is

$\text R = 6 + 3 \\[1em] \text R = 9\ \Omega \\[1em] \text I = \dfrac{72}{9} \\[1em] \text I = 8\ \text A$

The potential difference across the parallel section is

$\text V' = \text{IR}' \\[1em] = 8 \times 3 \\[1em] \text V' = 24\ \text V$

The current in the 12 Ω branch is

$\text I' = \dfrac{24}{12} \\[1em] \text I' = 2\ \text A$

so the potential difference across the 10 Ω resistor, and hence across the capacitor, is

$\text V_C = 2 \times 10 \\[1em] \text V_C = 20\ \text V$

Therefore

$\text Q = \text{CV}_C \\[1em] = (10)(20) \\[1em] \text Q = 200\ \mu\text C$

A capacitor with capacitance 5 µF is charged to 5 µC. If the plates are pulled apart to reduce the capacitance to 2 µF, how much work is done?

1. 3.75 × 10^-6 J
2. 2.55 × 10^-6 J
3. 2.16 × 10^-6 J
4. 6.25 × 10^-6 J

Answer

3.75 × 10^-6 J

Reason —

Given,

• C₁ = 5 µF, C₂ = 2 µF, charge q = 5 µC = 5 × 10^-6 C

The charge remains constant while the plates are pulled apart, so the work done equals the increase in the stored energy,

$\text W = \dfrac{\text q^2}{2\text C_2} - \dfrac{\text q^2}{2\text C_1} \\[1em] \text W = \dfrac{\text q^2}{2}\left(\dfrac{1}{\text C_2} - \dfrac{1}{\text C_1}\right)$

Substituting the values,

$\text W = \dfrac{(5 \times 10^{-6})^2}{2}\left(\dfrac{1}{2 \times 10^{-6}} - \dfrac{1}{5 \times 10^{-6}}\right) \\[1em] = \dfrac{25 \times 10^{-12}}{2}\left(5 \times 10^5 - 2 \times 10^5\right) \\[1em] = (12.5 \times 10^{-12})(3 \times 10^5) \\[1em] \text W = 3.75 \times 10^{-6}\ \text J$

The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery of voltage, V. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is :

1. $\dfrac{\text V}{\text K + \text n}$
2. V
3. $\dfrac{(\text n + 1)\text V}{(\text K + \text n)}$
4. $\dfrac{\text {nV}}{\text K + \text n}$

Answer

$\dfrac{(\text n + 1)\text V}{(\text K + \text n)}$

Reason —

The two capacitors in parallel have a total capacitance C + nC = (n + 1)C, so the total charge stored is

$\text Q = (\text n + 1)\text{CV}$

The battery is then removed, so this charge remains constant. On introducing the dielectric in the first capacitor, its capacitance becomes KC, and the total capacitance is

$\text C' = \text{KC} + \text{nC} \\[1em] \text C' = (\text K + \text n)\text C$

Hence the new common potential difference is

$\text V' = \dfrac{\text Q}{\text C'} \\[1em] = \dfrac{(\text n + 1)\text{CV}}{(\text K + \text n)\text C} \\[1em] \text V' = \dfrac{(\text n + 1)\text V}{(\text K + \text n)}$

A simple pendulum of length L is placed between the plates of a parallel plate capacitor having electric field E, as shown in figure. Its bob has mass m and charge q. The time-period of the pendulum is given by :

1. $\text {2π}\sqrt{\dfrac{\text L}{\sqrt{\text g^2 - \dfrac{\text q^2\text E^2}{\text m^2}}}}$
2. $\text {2π}\sqrt{\dfrac{\text L}{\left(\text g + \dfrac{\text {qE}}{\text m}\right)}}$
3. $\text {2π}\sqrt{\dfrac{\text L}{\left(\text g - \dfrac{\text {qE}}{\text m}\right)}}$
4. $\text {2π}\sqrt{\dfrac{\text L}{\sqrt{\text g^2 + \left(\dfrac{\text {qE}}{\text m}\right)^2}}}$

Answer

$\text {2π}\sqrt{\dfrac{\text L}{\sqrt{\text g^2 + \left(\dfrac{\text {qE}}{\text m}\right)^2}}}$

Reason —

Given,

• Length of the pendulum = L
• Mass of the bob = m, charge on the bob = q
• Electric field between the plates = E, directed horizontally

Forces acting on the bob : the plates of the capacitor are vertical, so the field E between them is horizontal. The bob therefore experiences

(i) its weight mg, acting vertically downwards, and

(ii) the electric force qE, acting horizontally, along the direction of the field.

Resultant acceleration : the two forces are mutually perpendicular, so their resultant is obtained by the theorem of parallelogram of vectors,

$\text F = \sqrt{(\text{mg})^2 + (\text{qE})^2}$

The effective acceleration of the bob is this resultant force divided by the mass,

$\text g_\text{eff} = \dfrac{\text F}{\text m} \\[1em] \text g_\text{eff} = \dfrac{\sqrt{(\text{mg})^2 + (\text{qE})^2}}{\text m}$

Taking m inside the radical,

$\text g_\text{eff} = \sqrt{\dfrac{(\text{mg})^2 + (\text{qE})^2}{\text m^2}} \\[1em] \text g_\text{eff} = \sqrt{\text g^2 + \left(\dfrac{\text{qE}}{\text m}\right)^2}$

Time period : the bob now oscillates about the new equilibrium position, which is the line of the resultant force, and the time period of a simple pendulum in an effective gravitational field g_eff is

$\text T = 2\pi\sqrt{\dfrac{\text L}{\text g_\text{eff}}}$

Substituting the value of g_eff,

$\text T = 2\pi\sqrt{\dfrac{\text L}{\sqrt{\text g^2 + \left(\dfrac{\text{qE}}{\text m}\right)^2}}}$

The options in which g and qE/m are simply added or subtracted would apply only if the electric force were vertical, that is, parallel or antiparallel to gravity. Here the field is horizontal, so the two accelerations must be combined vectorially.

A parallel plate capacitor has 1 µF capacitance. One of its two plates is given +2 µC charge and the other plate, +4 µC charge. The potential difference developed across the capacitor is :

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

Answer

1 V

Reason —

Given, C = 1 µF, q₁ = + 2 µC and q₂ = + 4 µC.

When two plates carry unequal charges, only the difference of the charges is responsible for the field between them; half of this difference resides as the effective charge of the capacitor. Hence

$\text V = \dfrac{\text q_1 - \text q_2}{2\text C}$

Substituting the values,

$|\text V| = \dfrac{|2 - 4|}{2 \times 1} \\[1em] = \dfrac{2}{2} \\[1em] |\text V| = 1\ \text V$

In the circuit shown, find C if the effective capacitance of the whole circuit is to be 0.5 µF. All values in the circuit are in µF.

1. $\dfrac{6}{5}$ µF
2. 4 µF
3. $\dfrac{7}{10}$ µF
4. $\dfrac{7}{11}$ µF

Answer

$\dfrac{7}{11}$ µF

Reason —

Let P be the junction just after C, and let the lower rail be B.

Left branch from P : the 2 µF vertical capacitor is in series with the parallel pair of 2 µF and 2 µF at the bottom,

$2 + 2 = 4\ \mu\text F \\[1em] \dfrac{2 \times 4}{2 + 4} = \dfrac{4}{3}\ \mu\text F$

Right branch from P : the two 2 µF capacitors of the right-hand group are in series, giving 1 µF, and this is in parallel with the 1 µF capacitor, giving 2 µF. This is reached from P through the top 2 µF capacitor,

$\dfrac{2 \times 2}{2 + 2} = 1\ \mu\text F$

Between P and B : the two branches are in parallel,

$\text C_\text{PB} = \dfrac{4}{3} + 1 \\[1em] \text C_\text{PB} = \dfrac{7}{3}\ \mu\text F$

Finally C is in series with this,

$\dfrac{1}{0.5} = \dfrac{1}{\text C} + \dfrac{3}{7} \\[1em] 2 - \dfrac{3}{7} = \dfrac{1}{\text C} \\[1em] \dfrac{11}{7} = \dfrac{1}{\text C} \\[1em] \text C = \dfrac{7}{11}\ \mu\text F$

The charge on a capacitor plate in a circuit as a function of time is shown in the figure.

What is the value of current at t = 4 s?

1. 2 µA
2. 1.5 µA
3. Zero
4. 3 µA

Answer

Zero

Reason —

The current is the rate of change of charge,

$\text I = \dfrac{\text{dq}}{\text{dt}}$

which is given by the slope of the q–t graph. Between t = 2 s and t = 6 s the graph is a horizontal straight line, that is, the charge stays constant at 3 µC. Hence at t = 4 s the slope is zero and so the current is zero.

A parallel plate capacitor having capacitance 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 plates. The work done by the capacitor on the slab is :

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

Answer

508 pJ

Reason —

Given, C = 12 pF, V = 10 V and K = 6.5.

The initial energy stored is

$\text U_i = \dfrac{1}{2}\text{CV}^2 \\[1em] = \dfrac{1}{2}(12)(10)^2 \\[1em] \text U_i = 600\ \text{pJ}$

The battery is disconnected, so the charge remains constant while the capacitance becomes KC. Hence

$\text U_f = \dfrac{\text U_i}{\text K} \\[1em] = \dfrac{600}{6.5} \\[1em] \text U_f = 92.3\ \text{pJ}$

The work done by the capacitor in drawing the slab in is the decrease in its stored energy,

$\text W = \text U_i - \text U_f \\[1em] \text W = 600 - 92.3 \\[1em] \text W \approx 508\ \text{pJ}$

A parallel plate capacitor is of area 6 cm² and a separation 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 which give same capacitance when fully inserted in above capacitor, would be :

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

Answer

12

Reason —

From the figure the three dielectrics occupy three equal parts of the plate area, each of area A/3, with the full separation d. They therefore behave as three capacitors joined in parallel,

$\text C = \dfrac{\epsilon_0 (\text A/3)}{\text d}(\text K_1 + \text K_2 + \text K_3) \\[1em] = \dfrac{\epsilon_0 \text A}{3\text d}(10 + 12 + 14) \\[1em] \text C = \dfrac{36}{3}\dfrac{\epsilon_0 \text A}{\text d} \\[1em] \text C = \dfrac{12\thinspace\epsilon_0 \text A}{\text d}$

If a single material of dielectric constant K fully filled the capacitor, then C = Kε₀A/d. Comparing the two,

$\text K = 12$

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. $\dfrac{3}{4}\cdot\dfrac{\text q^2}{\text C}$
2. $\dfrac{5}{8}\cdot\dfrac{\text q^2}{\text C}$
3. $\dfrac{1}{8}\cdot\dfrac{\text q^2}{\text C}$
4. $\dfrac{3}{8}\cdot\dfrac{\text q^2}{\text C}$

Answer

$\dfrac{3}{8}\cdot\dfrac{\text q^2}{\text C}$

Reason —

With the switch at A, the capacitor C is charged to a potential V = q/C and carries the total charge q. On turning the switch to B, this capacitor is joined to the uncharged capacitor 3C.

The loss of energy on sharing charge between two capacitors is

$\Delta \text U = \dfrac{\text C_1 \text C_2}{2(\text C_1 + \text C_2)}(\text V_1 - \text V_2)^2$

Here C₁ = C, C₂ = 3C, V₁ = q/C and V₂ = 0. Substituting,

$\Delta \text U = \dfrac{\text C(3\text C)}{2(4\text C)}\left(\dfrac{\text q}{\text C}\right)^2 \\[1em] \Delta \text U = \dfrac{3\text C}{8}\cdot\dfrac{\text q^2}{\text C^2} \\[1em] \Delta \text U = \dfrac{3}{8}\cdot\dfrac{\text q^2}{\text C}$

Two isolated conducting spheres S₁ and S₂ of radius $\dfrac{2}{3}$ R and $\dfrac{1}{3}$ R have 12 µC and −3 µC charges, respectively, and are at a large distance from each other. They are now connected by a conducting wire. A long time after this is done the charges on S₁ and S₂ are respectively :

1. 4.5 µC on both
2. +4.5 µC and −4.5 µC
3. 6 µC and 3 µC
4. 3 µC and 6 µC

Answer

6 µC and 3 µC

Reason —

The total charge is conserved,

$\text Q = 12 + (-3) \\[1em] \text Q = 9\ \mu\text C$

After joining by a wire both spheres attain a common potential, so the charges are shared in the ratio of their capacitances, that is, in the ratio of their radii,

$\dfrac{\text Q_1}{\text Q_2} = \dfrac{2\text R/3}{\text R/3} \\[1em] \dfrac{\text Q_1}{\text Q_2} = \dfrac{2}{1}$

Hence

$\text Q_1 = \dfrac{2}{3} \times 9 \\[1em] \text Q_1 = 6\ \mu\text C \\[1em] \text Q_2 = \dfrac{1}{3} \times 9 \\[1em] \text Q_2 = 3\ \mu\text C$

Two capacitors of capacitances C and 2C are charged to potential differences V and 2V, respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is :

1. $\dfrac{3}{2}\text {CV}^2$
2. $\dfrac{9}{2}\text {CV}^2$
3. zero
4. $\dfrac{25}{6}\text {CV}^2$

Answer

$\dfrac{3}{2}\text {CV}^2$

Reason —

The initial charges are

$\text Q_1 = \text{CV} \\[1em] \text Q_2 = (2\text C)(2\text V) \\[1em] \text Q_2 = 4\text{CV}$

Since the positive plate of one is joined to the negative plate of the other, the charges partly neutralise, so the net charge is

$\text Q = 4\text{CV} - \text{CV} \\[1em] \text Q = 3\text{CV}$

The equivalent capacitance of the parallel combination is

$\text C_\text{eq} = \text C + 2\text C \\[1em] \text C_\text{eq} = 3\text C$

Hence the final energy is

$\text U = \dfrac{\text Q^2}{2\text C_\text{eq}} \\[1em] = \dfrac{(3\text{CV})^2}{2(3\text C)} \\[1em] = \dfrac{9\text C^2\text V^2}{6\text C} \\[1em] \text U = \dfrac{3}{2}\text{CV}^2$

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

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

Answer

8 : 5

Reason —

The original capacitance is C₀ = ε₀A/d. With a slab of thickness t = d/2 and dielectric constant K = 4 inserted,

$\text C = \dfrac{\epsilon_0 \text A}{\text d - \text t + \dfrac{\text t}{\text K}} \\[1em] \text C = \dfrac{\epsilon_0 \text A}{\text d - \dfrac{\text d}{2} + \dfrac{\text d}{8}}$

Simplifying the denominator,

$\text d - \dfrac{\text d}{2} + \dfrac{\text d}{8} = \dfrac{8\text d - 4\text d + \text d}{8} \\[1em] \text d - \dfrac{\text d}{2} + \dfrac{\text d}{8} = \dfrac{5\text d}{8} \\[1em] \text C = \dfrac{\epsilon_0 \text A}{\dfrac{5\text d}{8}} \\[1em] \text C = \dfrac{8}{5}\dfrac{\epsilon_0 \text A}{\text d}$

Hence

$\dfrac{\text C}{\text C_0} = \dfrac{8}{5}$

A capacitor C is fully charged with voltage V₀. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance $\dfrac{\text C}{2}$ . The energy loss in the process after the charge is distributed between the two capacitors is :

1. $\dfrac{1}{4}\text {CV}_0^2$
2. $\dfrac{1}{3}\text {CV}_0^2$
3. $\dfrac{1}{6}\text {CV}_0^2$
4. $\dfrac{1}{2}\text {CV}_0^2$

Answer

$\dfrac{1}{6}\text {CV}_0^2$

Reason —

The loss of energy when two capacitors at potentials V₁ and V₂ are joined in parallel is

$\Delta \text U = \dfrac{\text C_1 \text C_2}{2(\text C_1 + \text C_2)}(\text V_1 - \text V_2)^2$

Here C₁ = C, C₂ = C/2, V₁ = V₀ and V₂ = 0. Substituting,

$\Delta \text U = \dfrac{\text C\left(\dfrac{\text C}{2}\right)}{2\left(\text C + \dfrac{\text C}{2}\right)}\text V_0^2 \\[1em] \Delta \text U = \dfrac{\dfrac{\text C^2}{2}}{2\left(\dfrac{3\text C}{2}\right)}\text V_0^2 \\[1em] \Delta \text U = \dfrac{\text C^2/2}{3\text C}\text V_0^2 \\[1em] \Delta \text U = \dfrac{1}{6}\text{CV}_0^2$

A parallel plate capacitor has plate of length 'l', width 'w' and separation of plates is 'd'. It is connected to a battery of emf V. A dielectric slab of the same thickness 'd' and of dielectric constant k = 4 is being inserted between the plates of the capacitor. At what length of the slab inside plates, will the energy stored in the capacitor be two times the initial energy stored?

1. $\dfrac{l}{4}$
2. $\dfrac{l}{2}$
3. $\dfrac{2l}{3}$
4. $\dfrac{l}{3}$

Answer

$\dfrac{l}{3}$

Reason —

The battery remains connected, so V is constant and U = ½CV² is proportional to the capacitance. For the energy to be doubled, the capacitance must be doubled.

Let a length x of the slab be inside the plates. The filled and the empty portions form two capacitors in parallel,

$\text C = \dfrac{\epsilon_0 \text w(l - \text x)}{\text d} + \dfrac{\text k\thinspace\epsilon_0 \text w\thinspace\text x}{\text d} \\[1em] = \dfrac{\epsilon_0 \text w}{\text d}\left[(l - \text x) + 4\text x\right] \\[1em] \text C = \dfrac{\epsilon_0 \text w(l + 3\text x)}{\text d}$

The initial capacitance is C₀ = ε₀wl/d. Setting C = 2C₀,

$l + 3\text x = 2l \\[1em] \text x = \dfrac{l}{3}$

In the circuit shown, charge on the 5 µF capacitor is :

1. 16.36 µC
2. 18.00 µC
3. 5.45 µC
4. 10.90 µC

Answer

16.36 µC

Reason —

Let the potential of the common lower point O be zero. Both the 6 V cells raise the potentials of the two outer top corners to + 6 V each.

Let V be the potential of the junction to which the 2 µF, the 4 µF and the 5 µF capacitors are joined. Since this junction is isolated, the total charge on the plates meeting there is zero,

$2(\text V - 6) + 4(\text V - 6) + 5(\text V - 0) = 0 \\[1em] 2\text V - 12 + 4\text V - 24 + 5\text V = 0 \\[1em] 11\text V = 36 \\[1em] \text V = \dfrac{36}{11}\ \text V$

Hence the charge on the 5 µF capacitor is

$\text q = 5 \times \dfrac{36}{11} \\[1em] = \dfrac{180}{11} \\[1em] \text q = 16.36\ \mu\text C$

Polar molecules are the molecules :

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

Answer

having a permanent electric dipole moment

Reason —

In a polar molecule the centre of gravity of the positive charges does not coincide with that of the negative charges, so the molecule behaves as an electric dipole even in the absence of any external field, that is, it has a permanent electric dipole moment. Water and HCl are examples. A non-polar molecule, on the other hand, acquires a dipole moment only when an external field displaces its charges.

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

1. $\dfrac{\text R_1^{\thinspace2}}{\text R_2^{\thinspace2}}$
2. $\dfrac{\text R_1}{\text R_2}$
3. $\dfrac{\text R_2}{\text R_1}$
4. $\sqrt{\left(\dfrac{\text R_1}{\text R_2}\right)}$

Answer

$\dfrac{\text R_2}{\text R_1}$

Reason —

On connecting by a wire, both spheres come to the same potential V,

$\dfrac{\text Q_1}{4\pi\epsilon_0 \text R_1} = \dfrac{\text Q_2}{4\pi\epsilon_0 \text R_2} \\[1em] \dfrac{\text Q_1}{\text Q_2} = \dfrac{\text R_1}{\text R_2}$

In terms of the surface charge densities, Q = 4πR²σ, so

$\dfrac{\text Q_1}{\text Q_2} = \dfrac{\text R_1^2 \sigma_1}{\text R_2^2 \sigma_2}$

Equating the two expressions,

$\dfrac{\text R_1^2 \sigma_1}{\text R_2^2 \sigma_2} = \dfrac{\text R_1}{\text R_2} \\[1em] \dfrac{\sigma_1}{\sigma_2} = \dfrac{\text R_2}{\text R_1}$

A parallel plate capacitor has a uniform electric field '$\vec{\textbf E}$' 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 (ε₀ = permittivity of free space) :

1. $\dfrac{\text E^2\text {Ad}}{\epsilon_0}$
2. $\dfrac{1}{2}\epsilon_0\text E^2$
3. $\epsilon_0\text {EAd}$
4. $\dfrac{1}{2}\epsilon_0\text E^2\text {Ad}$

Answer

$\dfrac{1}{2}\epsilon_0\text E^2\text {Ad}$

Reason —

The energy stored per unit volume of the field, that is, the energy density, is

$\text u = \dfrac{1}{2}\epsilon_0 \text E^2$

The field exists only in the space between the plates, whose volume is Ad. Hence the total energy stored is

$\text U = \text u \times (\text{Ad}) \\[1em] \text U = \dfrac{1}{2}\epsilon_0 \text E^2 \text{Ad}$

The equivalent capacitance of the combination shown in the figure is :

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

Answer

2C

Reason —

The two capacitors marked C in the upper and the lower arms are joined between the same pair of points, so they are in parallel, giving 2C.

The third capacitor C is connected across the two ends of the same right-hand junction, that is, both of its plates are joined to the same point by the connecting wires. It is therefore short-circuited and carries no charge (both sides of the vertical capacitor are at the same potential), so it does not contribute to the equivalent capacitance.

$\text C_\text{eq} = \text C + \text C \\[1em] \text C_\text{eq} = 2\text C$

The material filled between the plates of a parallel plate capacitor has resistivity 200 Ω m. The value of capacitance of the capacitor is 2 pF. If a potential difference of 40 V is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is : (Given the value of relative permittivity of material is 50.)

1. 0.9 mA
2. 9.0 mA
3. 9.0 µA
4. 0.9 µA

Answer

0.9 mA

Reason —

Given, ρ = 200 Ω m, C = 2 pF = 2 × 10^-12 F, V = 40 V and K = 50.

For the capacitor,

$\text C = \dfrac{\text K \epsilon_0 \text A}{\text d} \\[1em] \dfrac{\text A}{\text d} = \dfrac{\text C}{\text K \epsilon_0}$

The resistance offered by the material between the plates is

$\text R = \dfrac{\rho \text d}{\text A} \\[1em] \text R = \dfrac{\rho\thinspace\text K \epsilon_0}{\text C}$

Substituting the values,

$\text R = \dfrac{(200)(50)(8.85 \times 10^{-12})}{2 \times 10^{-12}} \\[1em] = \dfrac{8.85 \times 10^{-8}}{2 \times 10^{-12}} \\[1em] \text R = 4.425 \times 10^4\ \Omega$

Hence the leakage current is

$\text I = \dfrac{\text V}{\text R} \\[1em] = \dfrac{40}{4.425 \times 10^4} \\[1em] \text I = 9.04 \times 10^{-4}\ \text A \\[1em] \text I \approx 0.9\ \text{mA}$

A parallel plate capacitor with plate area A has separation d between the plates. Two dielectric slabs of dielectric constant K₁ and K₂ of same area A/2 and thickness d/2 are inserted in the space between the plates. The capacitance of the capacitor will be given by :

1. $\dfrac{\epsilon_0\text A}{\text d}\left(\dfrac{1}{2} + \dfrac{\text K_1\text K_2}{\text K_1 + \text K_2}\right)$
2. $\dfrac{\epsilon_0\text A}{\text d}\left(\dfrac{1}{2} + \dfrac{2(\text K_1 + \text K_2)}{\text K_1\text K_2}\right)$
3. $\dfrac{\epsilon_0\text A}{\text d}\left(\dfrac{1}{2} + \dfrac{\text K_1 + \text K_2}{\text K_1\text K_2}\right)$
4. $\dfrac{\epsilon_0\text A}{\text d}\left(\dfrac{1}{2} + \dfrac{2\text K_1 \times \text K_2}{(\text K_1 + \text K_2)}\right)$

Answer

$\dfrac{\epsilon_0\text A}{\text d}\left(\dfrac{1}{2} + \dfrac{\text K_1\text K_2}{\text K_1 + \text K_2}\right)$

Reason —

The arrangement is equivalent to two capacitors in parallel, each of plate area A/2.

Left half (air throughout the gap d) :

$\text C_1 = \dfrac{\epsilon_0 (\text A/2)}{\text d} \\[1em] \text C_1 = \dfrac{\epsilon_0 \text A}{2\text d}$

Right half : here the two slabs, each of thickness d/2, lie one above the other, so they form two capacitors in series,

$\text C' = \dfrac{\text K_1 \epsilon_0 (\text A/2)}{\text d/2} \\[1em] \text C' = \dfrac{\text K_1 \epsilon_0 \text A}{\text d} \\[1em] \text C'' = \dfrac{\text K_2 \epsilon_0 \text A}{\text d} \\[1em] \text C_2 = \dfrac{\text C'\text C''}{\text C' + \text C''} \\[1em] \text C_2 = \dfrac{\epsilon_0 \text A}{\text d}\left(\dfrac{\text K_1 \text K_2}{\text K_1 + \text K_2}\right)$

Adding the two parallel parts,

$\text C = \text C_1 + \text C_2 \\[1em] \text C = \dfrac{\epsilon_0 \text A}{\text d}\left(\dfrac{1}{2} + \dfrac{\text K_1 \text K_2}{\text K_1 + \text K_2}\right)$

Three capacitors C₁ = 2 µF, C₂ = 6 µF and C₃ = 12 µF are connected as shown in figure. Find the ratio of the charges on capacitors C₁, C₂ and C₃ respectively.

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

Answer

1 : 2 : 2

Reason —

From the figure, C₂ and C₃ are joined in series between A and B, and C₁ is connected directly between A and B. The battery of emf V is applied across A and B.

Charge on C₁ : it has the full potential difference V,

$\text q_1 = \text C_1 \text V \\[1em] \text q_1 = 2\text V$

Charges on C₂ and C₃ : their series capacitance is

$\text C' = \dfrac{6 \times 12}{6 + 12} \\[1em] \text C' = 4\ \mu\text F$

and being in series they carry equal charges,

$\text q_2 = \text q_3 = \text C'\text V \\[1em] \text q_2 = \text q_3 = 4\text V$

Hence

$\text q_1 : \text q_2 : \text q_3 = 2\text V : 4\text V : 4\text V \\[1em] \text q_1 : \text q_2 : \text q_3 = 1 : 2 : 2$

A capacitor is connected to a 20 V battery through a resistance of 10 Ω. It is found that the potential difference across the capacitor rises to 2 V in 1 µs. The capacitance of the capacitor is ............... µF.

Given $\ln\left(\dfrac{10}{9}\right) = 0.105$

1. 0.95
2. 9.52
3. 1.85
4. 0.105

Answer

0.95

Reason —

Given, V₀ = 20 V, R = 10 Ω, V = 2 V at t = 1 µs = 10^-6 s.

During charging through a resistance, the potential difference across the capacitor grows as

$\text V = \text V_0\left(1 - \text e^{-\text t/\text{RC}}\right)$

Substituting the values,

$2 = 20\left(1 - \text e^{-\text t/\text{RC}}\right) \\[1em] \text e^{-\text t/\text{RC}} = 1 - 0.1 \\[1em] \text e^{-\text t/\text{RC}} = 0.9$

Taking the natural logarithm of both sides,

$\dfrac{\text t}{\text{RC}} = \ln\left(\dfrac{1}{0.9}\right) \\[1em] = \ln\left(\dfrac{10}{9}\right) \\[1em] \dfrac{\text t}{\text{RC}} = 0.105 \\[1em] \text{RC} = \dfrac{10^{-6}}{0.105} \\[1em] \text{RC} = 9.52 \times 10^{-6}\ \text s$

Hence

$\text C = \dfrac{9.52 \times 10^{-6}}{10} \\[1em] \text C = 0.952 \times 10^{-6}\ \text F \\[1em] \text C \approx 0.95\ \mu\text F$

A parallel plate capacitor with plate area 'A' and distance of separation 'd' is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as :

$\epsilon(\text x) = \epsilon_0 + \text {kx}, \text{ for } \left(0 \le \text x \le \dfrac{\text d}{2}\right)$

$\epsilon(\text x) = \epsilon_0 + \text k(\text d - \text x), \text{ for } \left(\dfrac{\text d}{2} \le \text x \le \text d\right)$

1. $\dfrac{\text {kA}}{2}\ln\left(\dfrac{2\epsilon_0}{2\epsilon_0 - \text {kd}}\right)$
2. $\left(\epsilon_0 + \dfrac{\text {kd}}{2}\right)^{2/\text {kA}}$
3. $\dfrac{\text {kA}}{2\ln\left(\dfrac{2\epsilon_0 + \text {kd}}{2\epsilon_0}\right)}$
4. 0

Answer

$\dfrac{\text {kA}}{2\ln\left(\dfrac{2\epsilon_0 + \text {kd}}{2\epsilon_0}\right)}$

Reason —

Consider a thin layer of thickness dx at a distance x from one plate. It behaves as an elementary capacitor of capacitance

$\text{dC} = \dfrac{\epsilon(\text x)\text A}{\text{dx}}$

All such layers are in series, so the reciprocals add,

$\dfrac{1}{\text C_1} = \int_0^{\text d/2}\dfrac{\text{dx}}{\text A(\epsilon_0 + \text{kx})}$

Putting u = ε₀ + kx, so that du = k dx,

$\dfrac{1}{\text C_1} = \dfrac{1}{\text{kA}}\left[\ln(\epsilon_0 + \text{kx})\right]_0^{\text d/2} \\[1em] \dfrac{1}{\text C_1} = \dfrac{1}{\text{kA}}\ln\left(\dfrac{\epsilon_0 + \dfrac{\text{kd}}{2}}{\epsilon_0}\right) \\[1em] \dfrac{1}{\text C_1} = \dfrac{1}{\text{kA}}\ln\left(\dfrac{2\epsilon_0 + \text{kd}}{2\epsilon_0}\right)$

By the symmetry of the given permittivity about x = d/2, the second half gives an equal contribution, so that

$\dfrac{1}{\text C_2} = \dfrac{1}{\text C_1}$

The two halves are themselves in series, hence

$\dfrac{1}{\text C} = \dfrac{1}{\text C_1} + \dfrac{1}{\text C_2} \\[1em] \dfrac{1}{\text C} = \dfrac{2}{\text{kA}}\ln\left(\dfrac{2\epsilon_0 + \text{kd}}{2\epsilon_0}\right) \\[1em] \text C = \dfrac{\text{kA}}{2\ln\left(\dfrac{2\epsilon_0 + \text{kd}}{2\epsilon_0}\right)}$

Two capacitors of capacities 2C and C are joined in parallel and charged up to potential V. The battery is removed and the capacitor of capacity C is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be :

1. $\dfrac{3\text V}{\text K + 2}$
2. $\dfrac{3\text V}{\text K}$
3. $\dfrac{\text V}{\text K + 2}$
4. $\dfrac{\text V}{\text K}$

Answer

$\dfrac{3\text V}{\text K + 2}$

Reason —

In parallel the total capacitance is 2C + C = 3C, so the total charge stored is

$\text Q = 3\text{CV}$

The battery is then removed, so this charge is conserved. On filling the capacitor of capacity C with the dielectric, its capacitance becomes KC, and the total capacitance becomes

$\text C' = 2\text C + \text{KC} \\[1em] \text C' = (\text K + 2)\text C$

Hence the new common potential difference is

$\text V' = \dfrac{\text Q}{\text C'} \\[1em] = \dfrac{3\text{CV}}{(\text K + 2)\text C} \\[1em] \text V' = \dfrac{3\text V}{\text K + 2}$

A simple pendulum of mass m, length l and charge '+q' suspended in the electric field produced by two conducting parallel plates as shown. The value of deflection of pendulum in equilibrium position will be :

1. $\tan^{-1}\left[\dfrac{\text q}{\text {mg}} \times \dfrac{\text C_2(\text V_1 + \text V_2)}{(\text C_1 + \text C_2)(\text d - \text t)}\right]$
2. $\tan^{-1}\left[\dfrac{\text q}{\text {mg}} \times \dfrac{\text C_1(\text V_1 + \text V_2)}{(\text C_1 + \text C_2)(\text d - \text t)}\right]$
3. $\tan^{-1}\left[\dfrac{\text q}{\text {mg}} \times \dfrac{\text C_2(\text V_2 - \text V_1)}{(\text C_1 + \text C_2)(\text d - \text t)}\right]$
4. $\tan^{-1}\left[\dfrac{\text q}{\text {mg}} \times \dfrac{\text C_1(\text V_2 - \text V_1)}{(\text C_1 + \text C_2)(\text d - \text t)}\right]$

Answer

$\tan^{-1}\left[\dfrac{\text q}{\text {mg}} \times \dfrac{\text C_2(\text V_1 + \text V_2)}{(\text C_1 + \text C_2)(\text d - \text t)}\right]$

Reason —

The plates are at potentials + V₂ and − V₁, so the total potential difference between them is V₁ + V₂.

The gap of width d contains a dielectric slab of thickness t and an air region of width (d − t). These two portions form two capacitors in series; let C₁ be the capacitance of the air portion, in which the bob hangs, and C₂ that of the dielectric portion. In a series combination the potential difference divides in the inverse ratio of the capacitances, so the part appearing across the air region is

$\text V_\text{air} = (\text V_1 + \text V_2)\dfrac{\text C_2}{\text C_1 + \text C_2}$

The field acting on the bob is therefore

$\text E = \dfrac{\text V_\text{air}}{\text d - \text t} \\[1em] \text E = \dfrac{\text C_2(\text V_1 + \text V_2)}{(\text C_1 + \text C_2)(\text d - \text t)}$

In the equilibrium position the string makes an angle θ with the vertical such that the horizontal electric force balances the horizontal component of the tension,

$\tan\theta = \dfrac{\text{qE}}{\text{mg}} \\[1em] \tan\theta = \dfrac{\text q}{\text{mg}} \times \dfrac{\text C_2(\text V_1 + \text V_2)}{(\text C_1 + \text C_2)(\text d - \text t)}$

Note: The figure does not label which of C₁ and C₂ belongs to which region. The printed answer corresponds to taking C₁ as the capacitance of the air portion and C₂ as that of the dielectric portion.

For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\dfrac{3}{4}\text d$, where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance (C₀) is given by the following relation.

1. $\text C' = \dfrac{3 + \text K}{4\text K}\text C_0$
2. $\text C' = \dfrac{4}{3 + \text K}\text C_0$
3. $\text C' = \dfrac{4 + \text K}{3}\text C_0$
4. $\text C' = \dfrac{4\text K}{\text K + 3}\text C_0$

Answer

$\text C' = \dfrac{4\text K}{\text K + 3}\text C_0$

Reason —

With a slab of thickness t and dielectric constant K inserted,

$\text C' = \dfrac{\epsilon_0 \text A}{\text d - \text t + \dfrac{\text t}{\text K}}$

Putting t = 3d/4,

$\text C' = \dfrac{\epsilon_0 \text A}{\text d - \dfrac{3\text d}{4} + \dfrac{3\text d}{4\text K}} \\[1em] = \dfrac{\epsilon_0 \text A}{\text d\left(\dfrac{1}{4} + \dfrac{3}{4\text K}\right)} \\[1em] = \dfrac{\epsilon_0 \text A}{\text d}\cdot\dfrac{4\text K}{\text K + 3} \\[1em] \text C' = \dfrac{4\text K}{\text K + 3}\text C_0$

An electron with kinetic energy K₁ enters between parallel plates of a capacitor at an angle 'α' with the plates. It leaves the plates at angle 'β' with kinetic energy K₂. Then the ratio of kinetic energies K₁ : K₂ will be :

1. $\dfrac{\sin^2\beta}{\cos^2\alpha}$
2. $\dfrac{\cos\beta}{\cos\alpha}$
3. $\dfrac{\cos\beta}{\sin\alpha}$
4. $\dfrac{\cos^2\beta}{\cos^2\alpha}$

Answer

$\dfrac{\cos^2\beta}{\cos^2\alpha}$

Reason —

The electric field between the plates is perpendicular to the plates, so the electron experiences a force only in that direction. The component of its velocity parallel to the plates therefore remains unchanged.

If v₁ and v₂ are the speeds at entry and exit, then

$\text v_1\cos\alpha = \text v_2\cos\beta \\[1em] \dfrac{\text v_1}{\text v_2} = \dfrac{\cos\beta}{\cos\alpha}$

Since the kinetic energy is proportional to the square of the speed,

$\dfrac{\text K_1}{\text K_2} = \left(\dfrac{\text v_1}{\text v_2}\right)^2 \\[1em] \dfrac{\text K_1}{\text K_2} = \dfrac{\cos^2\beta}{\cos^2\alpha}$

A parallel plate capacitor is formed by two plates each of area 30π cm², separated by 1 mm. A material of dielectric strength 3.6 × 10⁷ Vm^-1 is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is 7 × 10^-6 C, the value of dielectric constant of the material is :

$\left[\text{Use, } \dfrac{1}{4\pi\epsilon_0} = 9 \times 10^9\ \text{Nm}^2\ \text C^{-2}\right]$

1. 1.66
2. 1.75
3. 2.25
4. 2.33

Answer

2.33

Reason —

Given,

• A = 30π cm² = 30π × 10^-4 m², d = 1 mm = 10^-3 m
• Dielectric strength, E_max = 3.6 × 10⁷ V m^-1
• Q_max = 7 × 10^-6 C

The maximum permissible potential difference is

$\text V_\text{max} = \text E_\text{max}\thinspace\text d \\[1em] = (3.6 \times 10^7)(10^{-3}) \\[1em] \text V_\text{max} = 3.6 \times 10^4\ \text V$

Hence the capacitance is

$\text C = \dfrac{\text Q_\text{max}}{\text V_\text{max}} \\[1em] = \dfrac{7 \times 10^{-6}}{3.6 \times 10^4} \\[1em] \text C = 1.944 \times 10^{-10}\ \text F$

Now C = Kε₀A/d, so

$\text K = \dfrac{\text{Cd}}{\epsilon_0 \text A}$

With ε₀ = 8.85 × 10^-12 C² N^-1 m^-2,

$\epsilon_0 \text A = (8.85 \times 10^{-12})(30\pi \times 10^{-4}) \\[1em] \epsilon_0 \text A = 8.34 \times 10^{-14} \\[1em] \text K = \dfrac{(1.944 \times 10^{-10})(10^{-3})}{8.34 \times 10^{-14}} \\[1em] \text K = 2.33$

If the charge on a capacitor is increased by 2 C, the energy stored in it increases by 44%. The original charge on the capacitor is (in C) :

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

Answer

10

Reason —

The energy stored is U = q²/2C, so for a given capacitor U ∝ q².

Let the original charge be q. When it becomes q + 2, the energy becomes 1.44 times its original value,

$\dfrac{(\text q + 2)^2}{\text q^2} = 1.44$

Taking the square root of both sides,

$\dfrac{\text q + 2}{\text q} = 1.2 \\[1em] \text q + 2 = 1.2\thinspace\text q \\[1em] 0.2\thinspace\text q = 2 \\[1em] \text q = 10\ \text C$

Two metallic plates form a parallel plate capacitor. The distance between the plates is 'd'. A metal sheet of thickness $\dfrac{\text d}{2}$ and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?

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

Answer

2 : 1

Reason —

For a slab of thickness t and dielectric constant K,

$\text C = \dfrac{\epsilon_0 \text A}{\text d - \text t + \dfrac{\text t}{\text K}}$

For a metal sheet K = ∞, so t/K = 0. With t = d/2,

$\text C = \dfrac{\epsilon_0 \text A}{\text d - \dfrac{\text d}{2}} \\[1em] = \dfrac{2\epsilon_0 \text A}{\text d} \\[1em] \text C = 2\text C_0$

Hence the new capacitance is twice the original one, that is, 2 : 1.

Two capacitors having capacitance C₁ and C₂ respectively are connected as shown in figure. Initially, capacitor C₁ is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor C₁ is now connected to uncharged capacitor C₂ by closing the switch S. The amount of charge on the capacitor C₂ after equilibrium, is :

1. $\dfrac{\text C_1\text C_2}{(\text C_1 + \text C_2)}\text V$
2. $\dfrac{(\text C_1 + \text C_2)}{\text C_1\text C_2}\text V$
3. $(\text C_1 + \text C_2)\text V$
4. $(\text C_1 - \text C_2)\text V$

Answer

$\dfrac{\text C_1\text C_2}{(\text C_1 + \text C_2)}\text V$

Reason —

The initial charge on C₁ is Q = C₁V. The battery is removed, so on closing the switch this charge is merely shared between the two capacitors, which are then in parallel. The common potential difference is

$\text V' = \dfrac{\text Q}{\text C_1 + \text C_2} \\[1em] \text V' = \dfrac{\text C_1 \text V}{\text C_1 + \text C_2}$

Hence the charge on C₂ is

$\text Q_2 = \text C_2 \text V' \\[1em] \text Q_2 = \dfrac{\text C_1 \text C_2}{(\text C_1 + \text C_2)}\text V$

A condenser of 2 µF capacitance is charged steadily from 0 to 5 C. Which of the following graph represents correctly the variation of potential difference (V) across its plates with respect to the charge (Q) on the condenser?

Answer

Graph (a)

Reason —

For a capacitor of fixed capacitance,

$\text V = \dfrac{\text Q}{\text C}$

so V is directly proportional to Q, and the graph must be a straight line passing through the origin. When Q reaches its final value of 5 C,

$\text V = \dfrac{5}{2 \times 10^{-6}} \\[1em] \text V = 2.5 \times 10^6\ \text V$

The only graph which is a straight line through the origin ending at 2.5 × 10⁶ V for Q = 5 C is graph (a).

Note: The answer printed in the textbook is (d), which shows a curved graph saturating at 5 × 10⁶ V. This contradicts the relation V = Q/C, which requires a straight line reaching 2.5 × 10⁶ V. The correct option is (a).

Capacitance of an isolated conducting sphere of radius R₁ becomes n times when it is enclosed by a concentric conduction sphere of radius R₂ connected to earth. The ratio of their radii $\left(\dfrac{\text R_2}{\text R_1}\right)$ is :

1. $\dfrac{\text n}{\text n - 1}$
2. $\dfrac{2\text n}{2\text n + 1}$
3. $\dfrac{\text n + 1}{\text n}$
4. $\dfrac{2\text n + 1}{\text n}$

Answer

$\dfrac{\text n}{\text n - 1}$

Reason —

The capacitance of the isolated sphere is

$\text C_0 = 4\pi\epsilon_0 \text R_1$

When it is enclosed by a concentric earthed sphere of radius R₂, the arrangement becomes a spherical capacitor of capacitance

$\text C = 4\pi\epsilon_0 \dfrac{\text R_1 \text R_2}{\text R_2 - \text R_1}$

It is given that C = nC₀, so

$4\pi\epsilon_0 \dfrac{\text R_1 \text R_2}{\text R_2 - \text R_1} = \text n(4\pi\epsilon_0 \text R_1) \\[1em] \dfrac{\text R_2}{\text R_2 - \text R_1} = \text n \\[1em] \text R_2 = \text n\text R_2 - \text n\text R_1 \\[1em] \text n\text R_1 = (\text n - 1)\text R_2 \\[1em] \dfrac{\text R_2}{\text R_1} = \dfrac{\text n}{\text n - 1}$

A source of potential difference V is connected to the combination of two identical capacitors as shown in the figure. When key K is closed, the total energy stored across the combination is E₁. Now key K is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is E₂. The ratio E₁/E₂ will be :

1. $\dfrac{1}{10}$
2. $\dfrac{2}{5}$
3. $\dfrac{5}{13}$
4. $\dfrac{5}{26}$

Answer

$\dfrac{5}{13}$

Reason —

With the key K closed, both identical capacitors of capacitance C are in parallel across the source, so

$\text E_1 = 2\left(\dfrac{1}{2}\text{CV}^2\right) \\[1em] \text E_1 = \text{CV}^2$

and the charge on each is CV.

When K is opened, one capacitor remains connected to the source while the other is isolated with its charge CV. The dielectric of K = 5 is then introduced in both, so each capacitance becomes 5C.

Capacitor still connected to the source : the potential difference stays V,

$\text U_1 = \dfrac{1}{2}(5\text C)\text V^2 \\[1em] \text U_1 = \dfrac{5}{2}\text{CV}^2$

Isolated capacitor : the charge stays CV,

$\text U_2 = \dfrac{(\text{CV})^2}{2(5\text C)} \\[1em] \text U_2 = \dfrac{\text{CV}^2}{10}$

Hence

$\text E_2 = \dfrac{5}{2}\text{CV}^2 + \dfrac{1}{10}\text{CV}^2 \\[1em] = \dfrac{25 + 1}{10}\text{CV}^2 \\[1em] \text E_2 = \dfrac{26}{10}\text{CV}^2 \\[1em] \dfrac{\text E_1}{\text E_2} = \dfrac{\text{CV}^2}{\dfrac{26}{10}\text{CV}^2} \\[1em] = \dfrac{10}{26} \\[1em] \dfrac{\text E_1}{\text E_2} = \dfrac{5}{13}$

Two capacitors, each having capacitance 40 µF are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant K such that the equivalence capacitance of the system became 24 µF. The value of K will be :

1. 1.5
2. 2.5
3. 1.2
4. 3

Answer

1.5

Reason —

Given, C₁ = 40 µF and C₂ = 40K µF after filling one of them with the dielectric, and C_eq = 24 µF.

For a series combination,

$\dfrac{1}{\text C_\text{eq}} = \dfrac{1}{\text C_1} + \dfrac{1}{\text C_2} \\[1em] \dfrac{1}{24} = \dfrac{1}{40} + \dfrac{1}{40\text K} \\[1em] \dfrac{1}{40\text K} = \dfrac{1}{24} - \dfrac{1}{40} \\[1em] = \dfrac{5 - 3}{120} \\[1em] = \dfrac{2}{120} \\[1em] \dfrac{1}{40\text K} = \dfrac{1}{60} \\[1em] 40\text K = 60 \\[1em] \text K = 1.5$

Two identical thin metal plates has charge q₁ and q₂ respectively such that q₁ > q₂. The plates were brought close to each other to form a parallel plate capacitor of capacitance C. The potential difference between them is :

1. $\dfrac{(\text q_1 + \text q_2)}{\text C}$
2. $\dfrac{(\text q_1 - \text q_2)}{\text C}$
3. $\dfrac{(\text q_1 - \text q_2)}{2\text C}$
4. $\dfrac{2(\text q_1 - \text q_2)}{\text C}$

Answer

$\dfrac{(\text q_1 - \text q_2)}{2\text C}$

Reason —