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

Options

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

Answer

A

Explanation

`E_(1)=(1)/(2)(Q^(2))/(C_(0)//2)=C_(0)V_(0)^(2), E_(2)=(1)/(2)(C_(0))/(2).V_(0)^(2)=(1)/(4)C_(0)V_(0)^(2), (E_(1))/(E_(2))=(4)/(1)`