The capacitance of capacitor increases when :(A) its charge increases…

The correct answer is Option (3) → (C) and (D)

The capacitance of a capacitor is the charge stored per unit potential difference, \( C = Q/V \). Its value depends only on the size and shape of the plates, their separation and the medium between them, not on the charge or voltage given to it. For a parallel plate capacitor,

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

(A) its charge increases — Incorrect. When the charge increases, the potential difference rises in the same proportion, so the ratio \( Q/V \) stays the same.

(B) its supply voltage decreases — Incorrect. A smaller voltage only puts a smaller charge on the plates; \( C \) does not depend on \( V \).

(C) common area of the plates increases — Correct. \( C \propto A \), so a larger overlapping area gives a larger capacitance.

(D) separation between the plates decreases — Correct. \( C \propto 1/d \), so bringing the plates closer increases the capacitance.

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.”