The charge on a 48 `muF` capacitor is increased from 0.1 C to 0.5 C. the enerrgy stored in the capacitor increases by

Options

A. 250 J
B. 2500 J
C. `2.5xx10^(6)J`
D. `2.42xx10^(-6)J`

Answer

B

Explanation

Energy sotred in a capacitor `=(1)/(2)CV^(2)=(1)/(2)(Q^(2))/(C)`
`therefore DeltaE=E_(2)-E_(1)=(1)/(2C)[Q_(2)^(2)-Q_(1)^(2)]`
`=(1)/(2xx48xx10^(-6))[25xx10^(-2)-1xx10^(-2)]`
`therefore DeltaE=(24xx10^(-2))/(2xx48xx10^(-6))=2500J`