A 12 pF capacitor is connected to a 50 V battery . How much electrostatic energy is stored in the capacitor ? If another capacitor of 6 pF 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

Here , `C_1 = 12 pF = 12 xx 10^(-12) F , V = 50 V ` and `C_(2) = 6 pF = 6 xx 10^(-12) F`
`therefore` Electrostatic energy in first capacitor `U_(1) = (1)/(2) C_(1) V^2 = (1)/(2) xx (12 xx 10^(-12)) xx (50)^2 = 15 xx 10^(-9) J = 15 nJ`
When the two capacitors are joined in series across same battery , then resultant capacitance C is given by
`1/C = (1)/(C_(1)) + (1)/(C_(2)) = (1)/(12) + (1)/(6) = (1)/(4) implies C = 4 pF`
`therefore` Charge stored on 1st capacitor = Charge stored on 2nd capacitor
`Q = CV = (4 pF) xx 50 V = 200 pC`
`therefore` Potential across 1st capacitor `V_1 = (Q)/(C_1) = (200 pC)/(12 pF) = 16.67 V`
and potential difference across 2nd capacitor `V_2 = (Q)/(C_(2)) (200 pC)/(6pF) = 33.33 V`