JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Capacitance

A medium having dielectric constant

K > 1

fills the space between the plates of a parallel plate capacitor. The plates have large area, and the distance between them is

d

. The capacitor is connected to a battery of voltage

V

, as shown in Figure (a). Now, both the plates are moved by a distance of

\frac{d}{2}

from their original positions, as shown in Figure (b).

In the process of going from the configuration depicted in Figure (a) to that in Figure (b), which of the following statement(s) is(are) correct?

(A)The electric field inside the dielectric material is reduced by a factor of $2 K$ .
(B)The capacitance is decreased by a factor of $\frac{1}{K + 1}$ .✓ Correct
(C)The voltage between the capacitor plates is increased by a factor of $(K + 1)$ .
(D)The work done in the process DOES NOT depend on the presence of the dielectric material.

Answer: (B)

Solution

In configuration (a):

C = \frac{K ε _{0} A}{d}

In configuration (b), the capacitor has three regions in series:

Air gap of $d /2$ on one side
Dielectric of thickness $d / K$ in the middle
Air gap of $(d - d /2 - d / K) = d /2 - d / K$ on the other side

Effective separation:

d_{eff} = 2 d - d + d / K = d + d / K = d (K + 1) / K

New capacitance:

C^{'} = \frac{K ε _{0} A}{d ( K + 1 )} = \frac{C}{K + 1}

So the capacitance decreases by a factor of

\frac{1}{K + 1}

.

Since the battery remains connected, voltage stays constant at

V

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JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Capacitance

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