Show that the force on each plate of a parallel plate capacitor has a magnitude equal to `(½) 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 `½`.

Explanation

Hint: Suppose we increase the separation of the plates by `Deltax` . Work done (by external agency) = F =`Deltax` . This goes to increase the potential energy of the capacitor by u a`Deltax` where u is energy density. Therefore, F = u a which is easily seen to be (1/2) QE, using `u = (1//2) epsilon_(0)E^(2)`. The physical origin of the factor 1/2 in the force formula lies in the fact that just outside the conductor, field is E, and inside it is zero. So, the average value E/2 contributes to the force.