A parallel plate capacitor has plate area 25.0 cm2 and a separation of 2.0 mm between the plates. The capacitor is connected to a battery of 12.0 V (a) Find the charge on the capacitor (b) The plate separation is decreased to 1.00 mm. Find the extra charge given by the battery to the positive plate.
Here, A = 25.0 cm2
= 25 × 10-4 m2
d = 2.0 mm = 2 × 10-2 m
V = 12 volt.
(b) When plate separation is decreased to half, capacity becomes twice. The charge (q’ = CV) becomes twice. Hence extra charge given by the battery.
= q’ – q = 2q – q = 1.35 × 10-10C