Question 1

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 (\varepsilon_0 = permittivity of free space)

Accepted Answer

\frac{1}{2}\varepsilon_0 E^2 Ad. U = \frac{1}{2}CV^2 = \frac{1}{2}\frac{\varepsilon_0 A}{d}(Ed)^2 = \frac{1}{2}\varepsilon_0 E^2 Ad

Question 2

Two identical capacitors C_1 and C_2 of equal capacitance are connected as shown in the circuit. Terminals a and b of the key are connected to charge capacitor C_1 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?

Accepted Answer

50%. Initial energy U = \frac{1}{2}CV^2. On sharing equally with the identical uncharged capacitor, final energy = \frac{1}{4}CV^2.

Loss = \frac{1}{4}CV^2, i.e. \frac{1/4}{1/2} = 50\%

Question 3

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

Accepted Answer

decreases by a factor of 2. Charge is conserved (q = CV). Connecting an identical uncharged capacitor doubles the capacitance to 2C, so the new energy = \frac{q^2}{2(2C)} = \frac{1}{2}\cdot\frac{q^2}{2C} — half the original. The energy decreases by a factor of 2. (This matches the identical question Q51/CBSE 2000, answer U/2.)

Question 4

A capacitor of 2 µF is charged (switch S at position 1, across the battery). When the switch S is turned to position 2, connecting it across an 8 µF capacitor, the percentage of its stored energy dissipated is

Accepted Answer

80%. When the 2 µF (charged to V) shares charge with the 8 µF, the energy lost fraction = \frac{C_2}{C_1 + C_2} = \frac{8}{2 + 8} = \frac{8}{10} = 80\%

Question 5

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?

Accepted Answer

The charge on the capacitor is not conserved. With the battery disconnected, the charge stays constant (q is conserved). So the statement "the charge on the capacitor is not conserved" is incorrect — the other three statements are all correct.

Question 6

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 area of each plate is A (m²), the energy (joule) stored in the condenser is

Accepted Answer

\frac{1}{2}\varepsilon_0 E^2 Ad. U = \frac{1}{2}CV^2 = \frac{1}{2}\frac{\varepsilon_0 A}{d}(Ed)^2 = \frac{1}{2}\varepsilon_0 E^2 Ad

Question 7

A series combination of n_1 capacitors, each of value C_1, is charged by a source of potential difference 4V. When another parallel combination of n_2 capacitors, each of value C_2, 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 C_2, in terms of C_1, is then

Accepted Answer

\frac{16C_1}{n_1 n_2}. Series: U_s = \frac{1}{2}\frac{C_1}{n_1}(4V)^2. Parallel: U_p = \frac{1}{2}(n_2 C_2)V^2.

Setting equal: \frac{16C_1}{n_1} = n_2 C_2 \Rightarrow C_2 = \frac{16C_1}{n_1 n_2}

Question 8

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

Accepted Answer

\frac{C}{3}, 3V. In series C_s = \frac{C}{3}; the breakdown voltages add, so the combination withstands 3V.

Question 9

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

Accepted Answer

\varepsilon_0 E^2 Ad. The energy stored is \frac{1}{2}CV^2 = \frac{1}{2}\varepsilon_0 E^2 Ad; among the given options this corresponds to (c) \varepsilon_0 E^2 Ad (the only choice with the correct dimensions of energy).

Question 10

Two condensers, one of capacity C and the other of capacity \frac{C}{2}, are connected to a V volt battery in parallel. The work done in charging fully both the condensers is

Accepted Answer

\frac{3}{4}CV^2. Parallel: C' = C + \frac{C}{2} = \frac{3C}{2}.

W = \frac{1}{2}C'V^2 = \frac{1}{2}\cdot\frac{3C}{2}\cdot V^2 = \frac{3}{4}CV^2

Question 11

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

Accepted Answer

connecting two in series and one in parallel. Two in series give \frac{4 	imes 4}{4 + 4} = 2 µF; this in parallel with the third: 2 + 4 = 6 µF.

Question 12

A capacitor of capacity C_1 is charged up to potential V volt and then connected in parallel to an uncharged capacitor of capacity C_2. The final potential difference across each capacitor will be

Accepted Answer

\frac{C_1 V}{C_1 + C_2}. Common potential V_{eq} = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} with V_1 = V, V_2 = 0:

V_{eq} = \frac{C_1 V}{C_1 + C_2}

Question 13

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

Accepted Answer

\frac{1}{2}\frac{\varepsilon_0 V^2}{d^2}. Energy density u = \frac{1}{2}\varepsilon_0 E^2 = \frac{1}{2}\varepsilon_0\left(\frac{V}{d}\right)^2 = \frac{1}{2}\frac{\varepsilon_0 V^2}{d^2}

Question 14

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

Accepted Answer

\frac{U}{2}. Charge conserved. Connecting an identical capacitor doubles capacitance to 2C, so U' = \frac{q^2}{2(2C)} = \frac{U}{2}

Question 15

Five 6 µF capacitors are arranged as a Wheatstone bridge between points X and Y — four capacitors in the four arms (X–A–Y and X–C–Y paths) and the fifth as the bridge between the two middle nodes. The effective capacitance between points X and Y is

Accepted Answer

6 µF. The bridge is balanced (\frac{C_{AB}}{C_{BD}} = \frac{C_{AC}}{C_{CD}}), so the middle capacitor carries no charge. Each arm has two 6 µF in series = 3 µF, and the two arms in parallel give 3 + 3 = 6 µF.

Question 16

If the potential of a capacitor having capacity 6 µF is increased from 10 V to 20 V, then increase in its energy will be

Accepted Answer

9 	imes 10^{-4} J. \Delta U = \frac{1}{2}C(V_2^2 - V_1^2) = \frac{1}{2}	imes 6	imes 10^{-6}	imes(400 - 100) = 9 	imes 10^{-4} J

Question 17

Four capacitors, each of 25 µF, are connected as shown, and a DC voltmeter reads 200 V (the potential difference across each capacitor is 200 V). The charge on each plate of a capacitor is

Accepted Answer

\pm 5 	imes 10^{-3} C. Q = CV = 25 	imes 10^{-6} 	imes 200 = 5 	imes 10^{-3} C

Question 18

A 4 µF capacitor is charged to 400 V and then its plates are joined through a resistance of 1 kΩ. The heat produced in the resistance is

Accepted Answer

0.32 J. All the stored energy is dissipated as heat: \frac{1}{2}CV^2 = \frac{1}{2}	imes 4	imes 10^{-6}	imes(400)^2 = 0.32 J

Combination of Capacitors and Energy Stored in a Capacitor — NEET Physics PYQ MCQs with Solutions

Questions with solutions