Let $F$ be the force applied to separate the plates of a parallel plate capacitor by a distance of $\triangle x .$ Hence, work done by the force to do so $=F \triangle x$

As a result, the potential energy of the capacitor increases by an amount given as $u A \triangle x$

Where,

$u=$ Energy density

$A=$ Area of each plate

$d=$ Distance between the plates

$V=$ Potential difference across the plates

The work done will be equal to the increase in the potential energy i.e…

$F \triangle x=u A \triangle x$

$F=u A=\left(\frac{1}{2} \in_{0} E^{2}\right) A$

Electric intensity is given by, $E=\frac{V}{d}$

$\therefore F=\frac{1}{2} \in_{0}\left(\frac{V}{d}\right) E A=\frac{1}{2}\left(\in_{0} A \frac{V}{d}\right) E$

However, capacitance, $C=\frac{\epsilon_{0} A}{d}$

$\therefore F=\frac{1}{2}(C V) E$

Charge on the capacitor is given by, $Q=C V$

$\therefore F=\frac{1}{2} Q E$

The physical origin of the factor, $\frac{1}{2},$ in the force formula lies in the fact that just outside the conductor, field is $\mathrm{E}$ and inside it is zero. Hence, it is the average value,

$\frac{E}{2},$ of the field that contributes to the force.