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
Answer & explanation
Correct answer: option 1
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