If the distance between the plates of a charged parallel plate capacitor is doubled then the potential difference across the capacitor
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → is doubled
A capacitor that is already charged and not connected to a battery keeps its charge, so Q stays constant while the plates are moved apart. The capacitance depends on the plate separation.
\( \displaystyle C = \frac{\varepsilon_0 A}{d} \)
Doubling d halves the capacitance:
\( \displaystyle C' = \frac{C}{2} \)
\( \displaystyle V' = \frac{Q}{C'} = \frac{2Q}{C} \)
\( \displaystyle V' = 2V \)
Equivalently, the field between the plates \( E = \sigma/\varepsilon_0 \) does not change, and \( V = Ed \) doubles when d doubles.
Note: The question does not say whether the capacitor is still connected to the battery. A “charged” capacitor is taken to be disconnected, so its charge stays constant and the potential difference doubles, which gives the answer given by NTA. If the battery were kept connected, the potential difference would remain the same (Option 4) and the charge on the plates would halve instead.