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. `6.25 xx 10^(-6)` J
B. `3.75 xx 10^(-6)`J
C. `2.16 xx 10^(-6)`J
D. `2.55 xx 10^(-6)` J

Answer

b

Explanation

`W= U_(f)-U_(i)= (q^(2))/(2)((1)/(c_(f))-(1)/(C_(i)))( because U= (q^(2))/(2C))`
`= ((5 xx 10^(-6))^(2))/(2)((1)/(2)-(1)/(5)) xx 10^(6)`
` 3.75 xx 10^(-6)J`