The capacitance of a parallel plate capacitor increases, when

The correct answer is Option (4) → the area of the plates is increased.

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

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

It does not depend on the charge or the voltage; C = Q/V only tells how much charge is stored per volt.

Option (1) the charge on the plates is decreased. — Incorrect. If Q decreases, V decreases in the same ratio and C stays the same.

Option (2) the voltage across the plates is increased. — Incorrect. A larger voltage stores proportionally more charge, but the ratio Q/V is unchanged.

Option (3) the distance between the plates is increased. — Incorrect. C is inversely proportional to d, so increasing the distance decreases the capacitance.

Option (4) the area of the plates is increased. — Correct. C is directly proportional to the plate area A.

NCERT (2026-27):
Capacitors and Capacitance – “… The constant C is called the capacitance of the capacitor. C is independent of Q or V, as stated above. The capacitance C depends only on the geometrical configuration (shape, size, separation) of the system of two conductors.”

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.”