Which quantity will change on increasing the separation between the plates of a charged parallel plate capacitor?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Potential difference between the plates of the capacitor
A charged capacitor that is not connected to a battery is isolated: its charge has nowhere to go, so \( Q \) stays the same when the plates are moved apart.
\( \displaystyle E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A} \)
does not contain \( d \), so the field is unchanged. But the potential difference is the field times the separation:
\( \displaystyle V = E\,d \)
so \( V \) increases in proportion to \( d \).
Option (1) Energy density between the plates — Incorrect. The energy density \( u = \frac{1}{2}\varepsilon_0 E^{2} \) depends only on \( E \), which does not change.
Option (2) Charge on the capacitor — Incorrect. The capacitor is isolated, so its charge is conserved.
Option (3) Potential difference between the plates of the capacitor — Correct. \( V = Ed \) with \( E \) fixed, so \( V \) grows as the plates move apart.
Option (4) Electric field between the plates of the capacitor — Incorrect. \( E = Q/(\varepsilon_0 A) \) does not depend on the separation.
NCERT (2026-27):
The Parallel Plate Capacitor – “… Now for uniform electric field, potential difference is simply the electric field times the distance between the plates, that is,
\( \displaystyle V = E\,d = \frac{1}{\varepsilon_0}\frac{Qd}{A} \qquad (2.42) \)”