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:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 11.1%, Increase
The source stays connected, so the potential difference \( V \) is fixed. The capacitance of a parallel plate capacitor is \( C = \varepsilon_0 A/d \), and the stored energy is \( U = CV^{2}/2 \), so \( U \propto C \propto 1/d \).
\( \displaystyle d' = d - 0.1\,d = 0.9\,d \;\Rightarrow\; C' = \frac{C}{0.9} \)
\( \displaystyle \frac{U'}{U} = \frac{C'}{C} = \frac{1}{0.9} = 1.111 \)
\( \displaystyle \text{Percentage change} = (1.111 - 1) \times 100\% = 11.1\% \)
\( U' \gt U \), so the energy increases by 11.1%.