JEE Main 2019 · Physics · Capacitors · medium

A parallel plate capacitor, charged and then disconnected from the battery, has plate separation d and capacitance C0. A dielectric slab of constant…

Single correct option

A parallel plate capacitor, charged and then disconnected from the battery, has plate separation

d

and capacitance

C_0

. A dielectric slab of constant

K=4

and thickness

d/2

, with the same area as the plates, is slid in between the plates, parallel to them. The ratio of the final energy stored to the initial energy is

Show answer and solution

Answer: (D) $\dfrac58$

With the slab the capacitance is

C=\dfrac{\varepsilon_0A}{(d-t)+t/K}

with

t=d/2

C=\dfrac{\varepsilon_0 A}{d/2+d/8}=\dfrac{8}{5}C_0

. The capacitor is isolated, so the charge

Q

stays constant and

U=Q^2/2C

. Therefore

U_f/U_i = C_0/C = 5/8

. If the battery were kept connected, the potential would be constant and the ratio would be

U\propto C

, i.e.

8/5

, which is the common trap.

Original question written in the JEE Main 2019 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake