A capacitor of `6mu F` is charged to such an extent that the potential difference between the plates becomes `50V`. The work done in this process will be

Options

A. `7.5 xx 10^(-2)J`
B. `7.5 xx 10^(-3)J`
C. `3 xx 10^(-6)J`
D. `3 xx 10^(-3)J`

Answer

B

Explanation

`W=(1)/(2) CV^(1)= (1)/(2) xx 6 xx 10^(-6) xx (50)^(2)`
`=3 xx 10^(-4) xx 25`
`=75 xx 10^(-4)J`
`=7.5 xx 10^(-3)J`