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 across the combination, find the charge stored and potential difference across each capacitor.

Explanation

Here, `C_(1) = 12 pF = 12xx10^(-12) F`.
V = 50 volt.
Electrostaic energy stored `U = (1)/(2) C_(1) V^(2)`
`= (1)/(2)xx12xx10^(-12) (50)^(2)`
`U = 1.5xx10^(-8) J`
Now, `C_(2) = 6 pF = 6xx10^(-12) F`
`(1)/(C_(s)) = (1)/(C_(1)) + (1)/(C_(2)) = (1)/(12) + (1)/(6) = (3)/(12) = (1)/(4)`
`C_(s) = 4 pF = 4xx10^(-12) F`
Charge on each condenser,
`q_(1) = q_(2) = C_(s) V = 4xx10^(-12) xx50`
`= 2xx10^(-10) C`
`V_(1) = (q_(1))/(C_(1)) = (2xx10^(-10))/(12xx10^(-12)) = (50)/(3)` volt
`V_(2) = (q_(2))/(C_(2)) = (2xx10^(-10))/(6xx10^(-12)) = (100)/(3)` volt