A 20 μF capacitor is charged to a potential difference of 1000 V . The terminals of this charged capacitor are then connected to those of an uncharged 5 μF capacitor. Calculate the final potential difference across each capacitor.

The charge on the 20 μ F capacitor: Q = C V = 20 μ F × 1000 V = 20 , 000 μC As the 20 μ F capacitor is connected to 5 μ F capacitor, the total charge on it will get distributed between the two capacitors. The charge will flow from 20 μ F capacitor until the potential difference across each capacitor become equal. Q 1 + Q 2 = 20 , 000 μ F ......(i) Q 1 20 μ C = Q 2 5 μ C ⇒ Q 1 = 4 Q 2 .....(ii) On solving (i) and (ii) we get: Q 2 = 4000 μ C , Q 1 = 16 , 000 μC Hence, the potential difference across each capacitor: V 1 = Q 1 C 1 = 16 , 000 μ C 20 μ F = 800 V V 2 = Q 2 C 2 = 4000 μ C 50 μ F = 800 V