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

Explanation

Consider a parallel plate capacitor with one plate having + Q charge and the other having - Q charge. Let F be the force between them. If the plates be displaced by a small distance dl, the work done dW = Fdl. This work results in the change in electrostatic potential energy of the capacitor.
`:.` Energy density of capacitor `u = 1/2 epsi_0 E^2`
and charge in volume = A.dl
`:.` Change in potential energy `du = 1/2 epsi_0 E^2. Adl`
Thus, we have `Fdl = 1/2 epsi_0 E^2 Adl`
`rArr F = 1/2 epsi_0 E^2 A = 1/2 epsi_0 E.A.E = 1/2 (epsi_0 (sigma)/(epsi_0)A) E = 1/2 (sigmaA)E = 1/2 QE`
The physical origin of the factor `1/2 ` in the above formula for force lies in the fact that just between the plates of capacitor electric field is E but just outside E = 0. Hence, average value `E/2` contributes to the force.