Derivation and Description→
Consider, A parallel plate capacitor with a charge $+q$ on one of its plates and $-q$ on the other plate. Let initially the plates of the capacitor are almost, but not quite touching. Due to opposite polarity, there is an attractive force $F$ between the plates. Now If these plates are gradually pulling and apart to a distance $d$, in such a way that $d$ is still small compared to the linear dimension of the plates, then the approximation of a uniform field between the plates is maintained and thus the force remains constant.
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| The force between the charged parallel plates of the capacitor |
Now the work done in separating the plates from near $0$ to $d$,
$W=F.d \qquad(1)$
This work done $(W)$ is stored as electrostatic potential energy $(U)$ between the plates, i.e.
$U=\frac{1}{2}q V $
But $V=E.d$, then the above equation can be written as
$U=\frac{1}{2}q \: E \: d \qquad(2)$
But the equation $(1)$ and equation $(2)$ both are equal then
$F.d=\frac{1}{2}q \: E \: d$
$ F=\frac{1}{2}q \: E $
The factor $\frac{1}{2}$ arises because just outside the conducting plates the field is $E$ and inside the plates, the field is zero. So the average value $\frac{E}{2}$ contributes to the force.
