A parallel plate capacitor of capacity 100 `mu`F is charged by a battery of 50 V. The battery remains connected and if the plates of the capacitor are separated so that the distance between them becomes double the original distance, the additional energy given by the battery to the capacitor in joule is

Options

A. `125/2 xx 10^(-3)`
B. `12.5 xx 10^(-3)`
C. `1.25 xx 10^(-3)`
D. `0.125 xx 10^(-3)`

Answer

(a)

Explanation

Initial Energy ,
`E_(1) = 1/2 cv^(2) = 1/2 xx (100 xx 10^(-6)) xx (50)^(2) = 125 xx 10^(-3) J`
`E_(2) = 1/2 c'v^(2) = 1/2 xx (50 xx 10^(-6)) xx (50)^(2) = 62.5 xx 10^(-3)J`
Additional energy
`= E_(1) - E_(2)`
`=62.5 xx 10^(-3)`
`= 125/2 xx 10^(-3)` J