Question

A parallel plate capacitor is charged by a battery and after charging the battery is removed. Now the distance between the plates is reduced. Choose the correct statement

• A. Electric field is not constant
• B. Potential difference is increased
• C. Capacitance is decreased
• D. Electrostatic potential energy is decreased

Answer 1

The correct option is D Electrostatic potential energy is decreased

The capacitance of a parallel plate capacitor having plates of area $A$ and $d$ distance between the plates is given by
$C=\frac{{ϵ}_{o}A}{d}$

With a decrease in $d$, $C$ increases. Hence, option (c) is incorrect.

As the battery is removed after charging the capacitor, the charge on the capacitor $‘q^{\prime }$ will not change.

Now, $q=CV$. As $‘C^{\prime }$ is increasing, $‘V^{\prime }$ will decrease to keep $‘q^{\prime }$ constant. Hence, option (b) is incorrect.

Electric field is given by $E=\frac{q}{A{ϵ}_o}$, which will not change as $‘q^{\prime }$ is constant. Hence, option (a) is incorrect.

Electrostatic potential energy is given by $U=\frac{q^2}{2C}$. As $‘q^{\prime }$ is constant and $‘C^{\prime }$ is increasing, $‘U^{\prime }$ will decrease. Hence, option (d) is correct.

Key concept - To compare the stored potential energy in a capacitor when charge on its plates remains the same, use $U=\frac{q^2}{2C}$.