The energy of a parallel plate capacitor when connected to a battery is E. With the battery still in connection, if the plates of the capacitor are separated so that the distance between them is twice the original distance, then the electrostatic energy becomes

Options

A. 2E
B. `E/4`
C. `E/2`
D. 4E

Answer

( c)

Explanation

`E = 1/2 CV^(2) = 1/2( epsilon_(0)A/d)V^(2)`
If the distance is doubled then
`E' = 1/2 epsilon_(0)A/2d V^(2) = e/2`