Question

A parallel plate capacitor with a slab of dielectric constant $3$ filling the whole space between the plates is charged to certain potential and isolated. Then the slab is drawn out and another slab of equal thickness but dielectric constant $2$ is introduced between the plates. The ratio of the energy stored in the capacitor later to that stored initially is?

• A. $2:3$
• B. $3:2$
• C. $4:9$
• D. $9:4$

Answer 1

Answer: (A)

$2:3$

Capacity of plate when K=3

$C_{k=3}=\frac{E_{0}3A}{d}$-------(1)

Again capacity of the plate when K=2

$C_{k=2}=\frac{E_{0}2A}{d}$-------(2)

Dividing 2 and 1 we get

The ratio of the energy stored in the capacitor later to stored initially$=\frac{\frac{E_{0}2A}{d}}{\frac{E_{0}3A}{d}}=\frac{2}{3}$