A capacitor of capacitor of capacitance `2.0muF` is charge to a potential diffrence of `12V` It is then connected of uncharged capacitor of capacitance `4.0muF` as shown in figure . Find (a ) the charge on each of the two capacitors after the connection, (b ) the electrostatic energy stored in each of the two capacitors.

Explanation

Initail charge stored
` C xx V = 12 xx 2 xx 10^(-6)`
` = 24 xx 10^(-6) C `
Let the charged on `2 mu F` and `4 muf` capacitance be ` q_1` and `q_2` respectively
Here , `q_2 = 2 q_1`
or , q_1 + q_2 = 24 xx 10^(-6) `
`rArr 3 q_1 = 24 xx 10^(-6) C `
` q_1 = 8 xx 10^(-6) C = 8 mu C `
`q_1 = 2 q_1 = 16 mu C`
`E_1 = 1 /2 xx C_1 v_2^(2) = 16 mu J`
`E_2 = 1 / 2 xx C_2 v_2^(2) = 32 mu J`.