Question

A parallel plate 100 µF capacitor is charged to 500V. If the distance between the plates is halved, what will be the new potential difference between the plates and what will be the change in the new stored energy?

Answer 1

C = 100µF = 100 × 10^–6F = 10^–4F; V = 500 volts

When plate separation is decreased to half, the new capacitance C′ becomes twice i.e. C′ = 2C. Since the capacitor is not connected to the battery, the charge on the capacitor remains the same. The potential difference between the plates must decrease to maintain the same charge.