Question

A parallel plate air capacitor is connected to a battery. If plates of the capacitor are pulled further apart, then which of the following statements are correct?

• A. Strength of electric field inside the capacitor remain unchanged, if battery is disconnected before pulling the plate.
• B. During the process, work is done by an external force applied to pull the plates whether battery is disconnected or it remain connected.
• C. Potential energy in the capacitor decreases if the battery remains connected during pulling plates apart.
• D. Potential energy in the capacitor decreases if the battery is disconnected before pulling plates apart.

Answer 1

Answer: (A), (B), (C)

Potential energy in the capacitor decreases if the battery remains connected during pulling plates apart.

If battery is disconnected and plates are pulled apart, then charge will remain constant.

Electric field inside the capacitor, $E=\frac{Q}{2A{ϵ}_0}\times 2=\frac{Q}{A{ϵ}_0}$
therefore $E$ remain same.
Option (A) is correct.

Work is done against attarctive force by external force to move the plates apart. So option (B) is orrect.

Capacitance, $C=\frac{{ϵ}_{0}A}{d}$
As plates are pulled apart, $d\uparrow ,C\downarrow$

Now, Potential energy, $U=\frac{1}{2}C{V}^2$
$V=$constant [as battery is connected]
$U$ decraese, option (C) is correct.

When battery is disconnected, $U=\frac{Q^2}{2C}$.
$Q$ remains constant and C decreases, so $U$ increases.