Problem 17: A 6 µF capacitor is charged to a P.D 120 V and then connected to an uncharged 4 µF capacitor. Calculate the P.D across the capacitors.
When the capacitor is at a potential 120 V, the charge deposited on it is,
Q = CV ⇒ Q = 6 × 10-6 × 120 = 720 × 10-6F.
When it is connected to the uncharged capacitor, then the equivalent capacitance of the combination is Ce = C1 + C2.
Put the values
Ce = 6 + 4 = 10 µF.
The resultant voltage is then