A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor? If another capacitor of 6pF is connected in series with it with the same battery connected across the combination, find the charge stored and potential difference across each capacitor.

Explanation

Electrostatic energy stored in the capacitor,
`U= (1)/(2)CV^2 = (1)/(2)xx12xx10^(-12)xx(50)^2= 1.5xx10^(-8) J`
When 6 pF is connected in series with 12 pF, charge stored across each capacitor,
`Q= (C_1C_2)/(C_1+C_2)V= (12xx6xx10^(-24))/((12+6)xx10^(-12))xx50 = 200 pC`
Potential difference across 12 pF is, `= (Q)/(C_1)= (200xx10^(-12))/(12xx10^(-12)) = 16.67 V`
Potential difference across 6 pF is,
`= (Q)/(C_2) = (200xx10^(-12))/(6xx10^(-12))= 33.33 V`.