In a parallel plate capacitor with plate area A and charge Q, the force on one plate because of the charge on the other is equal to

Options

A. `(Q^(2))/(epsilon_(0)A^(2))`
B. `(Q^(2))/(2epsilon_(0)A^(2))`
C. `(Q^(2))/(epsilon_(0))`
D. `(Q^(2))/(2epsilon_(0)A)`

Answer

D

Explanation

The electric field on one plate due to the charge on the other is `E = (Q)/(2 A epsilon_(0))`
`:.` The force on one plate due to the charge on the other is
`F = QE = Q((Q)/(2A epsilon_(0))) = (Q^(2))/(2A epsilon_(0))`