Each plate of a parallel -plate air capacitor has an area `S`. What amount of work has to be performed to slowly increases teh distance between the plates from `x_(1)` to `x_(2)` if
(a) Capacitance of the capacitor which is equal to `q`, or (b) the voltage across the capacitor, which is equal to `V`, is kept constant in the process?

Explanation

(a) Sought work is equivalent to the work is equavlaent to the work performed against the electric field created by one plate, holding at rest and to bring the other plate away. Therefore the requried work,
`A_("agent") = q E(x_(2) - x_(1))`.
where `E = (sigma)/(2 epsilon_(0))` is the intensity of the field created by one plate at the location of other.
So, `A_("gent") = q (sigma)/(2 epsilon_(0)) (x_(2) - x_(1)) = (q^(2))/(2 epsilon_(0) S) (x_(2) - x_(1))`
Alternate : `A_("ext") = Delta U` (as field potential)
`= (q^(2))/(2 epsilon_(0)S) x_(2) = (q^(2))/(2 epsilon_(0) S) x = (q^(2))/(2 epsilon_(0) S) (x_(2) - x_(1))`
(b) When voltage is kept const, the force acting on each plate acting on each plate of capacitor will depend on the distance between the plates.
So, elementary work done by agent, is it displacement over a distanace `dx`, relative to the other,
`dA = -F_(x) dx`
But, `F_(x) = - ((sigma (x))/(2 epsilon_(0))) S sigma(x)` and `sigma(x) = (epsilon_(0)V)/(x)`
Hence, `A = int DA = int_(x_(1))^(x_(2)) (1)/(2) epsilon_(0) (S V^(2))/(x^(2)) dx = (epsilon_(0) S V^(2))/(2) [(1)/(x_(1)) - (1)/(x_(2))]`
Alternate : From energy Conservation,
`U_(f) - U_(i) = A_(cell) + A_("agent")`
or `(1)/(2) (epsilon_(0) S)/(x_(2)) V^(2) - (1)/(2) (epsilon_(0) S)/(x_(1)) V^(2) = [(epsilon_(0) S)/(x_(2)) - (epsilon_(0) S)/(x_(1))] V^(2) + A_("agent")`
`(as A_("cell") = (q_(f) - q_(i)) V = (C_(f) - C_(i)) V^(2))`
So `A_("agent") = (epsilon_(0) S V^(2))/(2) [(1)/(x_(1)) - (1)/(x_(2))]`