A parallel plate capacitor of capacity `C_(0)` is charged to a potential `V_(0)`. (i) The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is `E_(1)`.(ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is `E_(2)`.Then, `E_(1)/E_(2),` value is

Options

A. 4
B. `3//2`
C. 2
D. `1//2`

Answer

A

Explanation

`E_1=Q^2/(2(C_0//2))=(C_0^2C_0^2)/(C_0)=C_0V_0^2`
`E_2=1/2(C_0/2)V_0^2=C_0/4V_0^2 `
`E_1/E_2=4`