A parallel plate capacitor carries equal and opposite charge , hence they must attract each other with a force say F.

At any instant let the plate sepration be `x` , then

`C = (in_0 A)/x`

also

`U = Q^2/(2C)`

`U = (Q^2 /(2epsilonA))x`

Let the plates be removed towards each other throungh dx, such that the new separaion between the plates is (x - dx). If `U_f` is the potential energy, then

`U_f = Q^2/(2C') = Q^2/(2epsilonA)(x-dx)`

If dU is the charge in potential energy, then

`dU = U_f - U_i`

`dU = Q^2/(2epsilon_0A)(x-dx) - Q^2/(2epsilon_0A)x`

`dU = Q^2/(2epsilon_0A)dx`

Further since

`F = -(dU)/(dx)`

`F = Q^2/(2epsilon_0A) = sigma^2/(2epsilon_0)A = 1/2 (epsilon_0E^2)A` `(therefore Q = sigmaA, E = sigma/epsilon_0)`