A battery of 200 V charges a capacitor to \( 10 \times 10^{-6}\,\mathrm{C} \). Now the battery is disconnected, and the plate separation is doubled. The potential of the capacitor will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 400\,\mathrm{V} \)
Once the battery is disconnected, the charge on the plates stays constant. The capacitance of a parallel-plate capacitor is
\( \displaystyle C = \frac{\varepsilon_0 A}{d} \)
so doubling the separation \( d \) halves the capacitance. The potential is \( V = Q/C \), so it doubles.
\( \displaystyle V' = \frac{Q}{C'} = \frac{Q}{C/2} = 2\,\frac{Q}{C} = 2 \times 200\,\mathrm{V} = 400\,\mathrm{V} \)