Initially switch `S` is opened and energy stored between the plates of capacitor is `E`. Now the switch `S` is closed. Work done by battery after the switch `S` is closed is `W`. Find `(2E)/W`.

Answer

3

Explanation

`E=1/2(C/2)V^(2) Q_(i)=(CV)/2 Q_(f)=(2CV)/3`
`Q_(f)-Q_(i)=(2CV)/3-(CV)/2=(CV)/6`
`E/W=(CV^(2)//4)/(CV^(2)//5)=3/2 implies(2E)/W=3`