Question

A parallel-plate 100 uF capacitor is charged to 500 V
and the charging battery is removed. If the distance
between its plates is halved, then what will be the
new potential difference between the plates and what
will be the change in the stored energy?​

Answer 1

Answer:

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.