Prove that the force acting on one plate due to the other in a parallel plate capacitor is ` F =(1)/(2) (CV^(2))/(d)` .

Explanation

The electric field due to one plate is
`E_(1) = (sigma)/(2 epsilon_(0))`
A second plate having charge aA is present in the above electric field.
`:.` The force acting on second plate `F= F_(1) (sigmaA)`
Substituting the value of `E_(1)` from (1) we get
`F = (sigma^(2)A)/(2 epsilon_(0))` but `sigma= (Q)/(A)`
`:.F =((Q^(2))/(A^(2)).A)/(2epsilon_(0))=(Q^(2))/(2epsilon_(0)A)=((Q^(2))/(d))/(2epsilon_(0)(A)/(d))=(Q^(2))/(2dC)`
`[because (epsilon_(0)A)/(d) =C]`
`:. F =(1)/(2) (CV^(2))/(d ) " " ( because Q = CV)`