A battery of 200 V charges a capacitor to $10×10^{-6} C$. Now the battery is disconnected, and the plate separation is doubled. The potential of the capacitor will be

400 V

The correct answer is Option (1) → 400 V

Given:

Initial charge, Q = 10 × 10⁻⁶ C

Initial voltage, V = 200 V

Capacitance, C = Q / V = (10 × 10⁻⁶) / 200 = 5 × 10⁻⁸ F

When the battery is disconnected and plate separation is doubled:

→ Charge remains constant

→ Capacitance becomes half (since C ∝ 1/d)

New capacitance = 2.5 × 10⁻⁸ F

New potential, V = Q / C = (10 × 10⁻⁶) / (2.5 × 10⁻⁸) = 400 V