Question

A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab having dielectric constant (relative permittivity) K, is now introduced between its two plates in order to occupy the space completely.

State, in terms of K, its effect on the following:

1. The capacitance of the capacitor.
2. The potential difference between its plates
3. The energy stored in the capacitor.

Answer 1

We have, C = `(epsilon_0 A)/d`,

V = `Q/C` and

U = `Q^2/(2 C)`

After inserting the dielectric constant K:

i. Capacitance (C') = `(epsilon_0 KA)/d`

= KC

⇒ Capacitance increases. It becomes K times the initial value.

ii. Potential difference between the plates (V') = `Q/(C')`

= `Q/(KC)`

= `V/K`

⇒ Potential difference between the plates decreases to `1/K` times the initial value.

iii. Energy stored (U') = `Q^2/(2 C')`

= `Q^2/(2 C K)`

= `U/K`

⇒ Energy stored decreases to `1/K` times the initial value.