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A parallel plate capacitor of capacity C...

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 doubled. find the ratio `E_(1)//E_(2)`.

A

4

B

`3//2`

C

2

D

`1//2`

Text Solution

Verified by Experts

The correct Answer is:
A
`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`
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