A capacitor with capacitance `5mu F` is charged to `5 mu C` . If the plates are pulled apart to reduce the capacitance to `2 mu F` , how much work is done?

Options

A. `2.55xx10^(-6)J`
B. `3.75xx10^(-6)J`
C. `6.25xx10^(-6)J`
D. `2.16xx10^(-6)J`

Answer

B

Explanation

`W=|DeltaU|=(Q^(2))/(2C_(1))-(Q^(2))/(2C_(2))=3.75xx10^(-6)J`.