Keeping the voltage of the charging source constant, the separation between the plates of the capacitor is decreased by 10%. The percentage change in the energy stored in the capacitor will be:

11.1%, Increase

The correct answer is Option (1) → 11.1%, Increase

Energy stored in a capacitor: $U = \frac{1}{2} C V^2$

Capacitance of a parallel plate capacitor: $C = \frac{\varepsilon_0 A}{d}$

Given: voltage $V$ constant, separation decreases by 10%

New separation: $d' = 0.9 d$

New capacitance: $C' = \frac{\varepsilon_0 A}{d'} = \frac{\varepsilon_0 A}{0.9 d} = \frac{C}{0.9} \approx 1.111 C$

New energy: $U' = \frac{1}{2} C' V^2 = 1.111 \cdot \frac{1}{2} C V^2 = 1.111 U$

Percentage change in energy: $\frac{U' - U}{U} \times 100\% = (1.111 - 1) \times 100\% = 11.1\%$

Answer: Energy stored increases by 11.1%