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. 0.5 N
B. 0.005N
C. 0.05 N
D. 0.0005N

Answer

C

Explanation

`E = 10^(5) V//m, q = 1 mu C = 10^(-6)C`
`F = (q^(2))/(2epsilon_(0)A)` and `E = (q)/(epsilon_(0)A) :. F = (qE)/(2) = (10^(-6) xx 10^(5))/(2) = 0.05 N`