Show that the force on each plate of a parallel plate capacitor has a magnitude equal to $\frac{1}{2} Q E$,where $Q$ is the charge on the capacitor,and $E$ is the magnitude of the electric field between the plates. Explain the origin of the factor $\frac{1}{2}$.

(N/A) Let $F$ be the force applied to separate the plates of a parallel plate capacitor by a distance of $\Delta x$.
The work done by the external force is $W = F \Delta x$.
This work increases the potential energy of the capacitor by an amount equal to the change in energy density multiplied by the change in volume: $\Delta U = u A \Delta x$,where $u = \frac{1}{2} \varepsilon_0 E^2$ is the energy density,$A$ is the area of the plates,and $\Delta x$ is the change in separation.
Equating work done to the change in potential energy:
$F \Delta x = (\frac{1}{2} \varepsilon_0 E^2) A \Delta x$
$F = \frac{1}{2} \varepsilon_0 E^2 A$
Since the electric field $E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{A \varepsilon_0}$,we can write $\varepsilon_0 A = \frac{Q}{E}$.
Substituting this into the force equation:
$F = \frac{1}{2} (\frac{Q}{E}) E^2 = \frac{1}{2} Q E$.
The factor $\frac{1}{2}$ arises because the electric field $E$ between the plates is the sum of the fields produced by both plates. Each plate experiences a force only due to the field produced by the other plate,which is $\frac{E}{2}$. Thus,the force on a plate with charge $Q$ is $F = Q \times (\frac{E}{2}) = \frac{1}{2} Q E$.