Question

A parallel plate capacitor has capacity 'C' is fully charged and diconnected from the source. What will the effect on its capacity and energy stored in it if the separation between the plates increased ?​

Answer 1

Explanation:

A parallel plate capacitor is charged and then the battery is disconnected. When the plates of the capacitor are brought closer, then. d is decreased. Hence, C will increase.

Answer 2

If a parallel plate capacitor is disconnected from the source, the charge in the capacitor $\sf{Q}$ remains unchanged.

The capacitance of the capacitor is given by,

$\sf{\longrightarrow C=\dfrac{A\epsilon_0}{d}}$

This implies,

$\sf{\longrightarrow C\propto\dfrac{1}{d}\quad\quda\dots(1)}$

Hence if the separation between the plates is increased, capacitance will be decreased.

The energy stored in the capacitor is given by,

$\sf{\longrightarrow E=\dfrac{Q^2}{2C}}$

Since $\sf{Q}$ is constant,

$\sf{\longrightarrow E\propto\dfrac{1}{C}}$

From (1),

$\sf{\longrightarrow E\propto\dfrac{1}{\left(\dfrac{1}{d}\right)}}$

$\sf{\longrightarrow E\propto d}$

Hence the energy stored will also be increased if the separation is increased.