A parallel plate air capacitor P is charged by a battery of emf V and…

The correct answer is Option (4) → (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

The energy of a capacitor is \( U=\frac{1}{2}CV^{2} \), which also equals \( Q^{2}/2C \). With the battery connected, \( V \) stays fixed. With the battery removed, \( Q \) stays fixed. Halving the gap doubles \( C=\varepsilon_0A/d \).

(A) Battery kept connected and the plate separation halved — This matches (III) \( 2U \). The voltage stays \( V \) and the capacitance becomes \( 2C \), so the energy \( \frac{1}{2}(2C)V^{2} \) is \( 2U \).

(B) Battery removed and the plate separation halved — This matches (II) \( U/2 \). The charge stays \( Q \) and the capacitance becomes \( 2C \), so the energy \( Q^{2}/2(2C) \) is \( U/2 \).

(C) Capacitor charged again by a battery of emf \( 2V \) — This matches (IV) \( 4U \). The energy goes as \( V^{2} \), so doubling the voltage makes it four times.

(D) Battery removed and P joined across an identical uncharged capacitor — This matches (I) \( U/4 \). The charge is shared equally, so each capacitor keeps \( Q/2 \). The energy of P becomes \( (Q/2)^{2}/2C \), which is \( U/4 \). The two capacitors together hold \( U/2 \), but List-II gives the energy of P only.

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

    \( C=\frac{\varepsilon_0A}{d} \)\( \qquad (2.43) \)

Energy Stored in a Capacitor – “We can write the final result, in different ways”

    \( W=\frac{Q^{2}}{2C}=\frac{1}{2}CV^{2}=\frac{1}{2}QV \)\( \qquad (2.69) \)

Example 2.10 – “… charge on each capacitor is then Q′ = CV′. By charge conservation, Q′ = Q/2. This implies V′ = V/2. …”