Two capacitors of capacitances `20 .0pF` and `50.0pF ` are connected in series with a `6.00V` battery . Find (a )the potential difference across each capacitor and (b) the energy stored in each capacitor .

Explanation

`C_1 = 20.0 pF, `
`C_2 = 50.0 pF`
(a) Charge on both capacitor is eaqual
So, ` `q = (C_1 xx C_2 / C_1 + C_2) xx V`
`= (20 xx 50 / 20 + 50) xx 6.0 pC `
`= V_1 = q / C_1 = 20 / 70 xx 50 xx 20 xx 6.0`
`= 1.71 V `
`= 1.71 V `
`V_2 = (6.00 - 1.71 ) V = 4.29 V ` (b) `E_1 = q_2 / 2 C_1 `
= [ 20 xx 50 / 70 xx 6.0 ]^(2) xx 1 / 2 xx 20 `
= 184 pJ `
`E_2 = q_2 / 2C_1`
`= [20 xx 50 / 70 xx 6.0] ^(2) xx 1 / 2 xx 50 `
`= 73.5 pJ`.