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A parallel-plate capacitor has plate area A and separation d. A dielectric slab of constant K=3 and thickness d/2, with the same area as the plates…

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A parallel-plate capacitor has plate area

A

and separation

d

. A dielectric slab of constant

K=3

and thickness

d/2

, with the same area as the plates, is placed flush against one plate. The capacitor is charged to

V

by a battery, the battery is removed, and the slab is then slowly pulled out. The work done by the external agent is

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Answer: (C) $\dfrac{3\varepsilon_0AV^2}{8d}$

With the slab:

\dfrac1C=\dfrac{d/2}{\varepsilon_0A}+\dfrac{d/2}{3\varepsilon_0A}=\dfrac{2d}{3\varepsilon_0A}

, so

C_1=\dfrac{3\varepsilon_0A}{2d}

. Charge

Q=C_1V

stays constant after the battery is removed. Initial energy

\tfrac12C_1V^2=\tfrac{3\varepsilon_0AV^2}{4d}

. Without the slab

C_0=\varepsilon_0A/d

, so

U_f=\dfrac{Q^2}{2C_0}=\dfrac{9\varepsilon_0AV^2}{8d}

. Work

=U_f-U_i=\dfrac{9-6}{8}\dfrac{\varepsilon_0AV^2}{d}=\dfrac{3\varepsilon_0AV^2}{8d}

Original question written in the JEE Main 2023 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake