The separation between the plates of a charged parallel plate capacitor is increased. Which of the following quantities will change?
(A) charge on the capacitor plates
(B) potential difference across the capacitor
(C) energy of the capacitor
(D) energy density between the plates of the capacitor
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B) and (C) only
For a charged capacitor (disconnected from battery):
- (A) Charge on the plates remains constant → Does not change
- (B) Potential difference $V = Q/C$ increases as separation increases → Changes
- (C) Energy $U = Q^2/(2C)$ increases as capacitance decreases → Changes
- (D) Energy density $u = U/(Ad) = Q^2/(2 \epsilon_0 A d)$ → For parallel plate capacitor, $u = \frac{1}{2} \epsilon_0 E^2 = \frac{1}{2} \epsilon_0 (V/d)^2$; with V increasing proportionally to d, $E = V/d$ remains constant → Energy density does not change
∴ Quantities that change: B and C