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`, the force on each capacitor plate is

Options

A. 0.5 N
B. 0.05 N
C. 0.005 N
D. 5 N

Answer

B

Explanation

For a capacitor `E=(1)/(2)CV^(2)" but "E=W=Fxxd`
`therefore Fd=(1)/(2)CV^(2)`
`therefore F=(1)/(2)(CV^(2))/(d)=(1)/(2)(CV)((V)/(d))`
`because Q=CV and E=(V)/(d)`
`therefore F=(1)/(2)QE" "thereforeF=(1)/(2)xx10^(-6)xx10^(5)=0.05N`