The capacity of a parallel plate capacitor does not depend on:

The correct answer is Option (3) → the potential applied across the plates

The capacitance of a parallel plate capacitor depends only on its geometry and the medium between the plates.

    \( \displaystyle C = \frac{K\,\varepsilon_0\,A}{d} \)

Here \( K \) is the dielectric constant of the medium, \( A \) the area of each plate and \( d \) the separation. The potential does not appear. If the applied potential is doubled, the charge doubles too, so \( C = Q/V \) stays the same.

Option (1) the medium between the plates — Incorrect. The medium enters through \( K \); a dielectric raises \( C \) by the factor \( K \).

Option (2) the area of the plates — Incorrect. \( C \propto A \); larger plates hold more charge at the same potential.

Option (3) the potential applied across the plates — Correct. \( C \) is fixed by \( K \), \( A \) and \( d \); a change in \( V \) changes \( Q \) in the same ratio.

Option (4) the separation between the plates — Incorrect. \( C \propto 1/d \); bringing the plates closer raises the capacitance.

NCERT (2026-27):
The Parallel Plate Capacitor – “… The capacitance C of the parallel plate capacitor is then

    \( \displaystyle C = \frac{Q}{V} = \frac{\varepsilon_0 A}{d} \qquad (2.43) \)

which, as expected, depends only on the geometry of the system.”