A parallel plate capacitor has an electric field of `10^(5)V//m` between the plates. If the charge on the capacitor plate is `1muC`, then force on each capacitor plate is-

Options

A. 0.5N
B. 0.05 N
C. 0.005 N
D. None of these

Answer

B

Explanation

Net charge on the pararllel plate capacitor is `1muc` electric field near one of the plates due to the other plate is
`E=(10^(5))/(2)Vm^(-1)`
Now force is given by f=qE
substiuting `q=1muc=1xx10^(-6C)`
we get `F=0.05 N