Question

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

Answer 1

Correct Answer - A
Let `E = (1)/(2)C_(0)V_(0)^(2)` then `E_(1) = 2E` and `E_(2) = (E)/(2)`
So `(E_(1))/(E_(2)) = (4)/(1)`