Target Exam

CUET

Subject

Physics

Chapter

Electrostatic Potential and Capacitance

Question:

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:

Options:

(A), (B) and (D) only

(A), (C) and (D) only

(B) and (C) only

(B), (C) and (D) only

Correct Answer:

(B) and (C) only

Explanation:

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