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

If F is the force on each plate of a parallel plate capacitor, then work done in increasing the separation between the plates by `Delta x = F. Delta x`.
This must be the increase in potential energy of the capacitor.
Now, increase in volume of capacitor `= A. Delta x`
If u = energy density = energy stored/ volume, then increase in pot, energy `= u.A Delta x :. F Delta x = u. A Delta x`
`F = u.A = ((1)/(2) in_(0) E^(2)) A = (1)/(2) (in_(0) AE)E , F = (1)/(2) (in_(0) A(V)/(d))E (( :. C = (in_(0) A)/(d))`
`F = (1)/(2) (CV) E = (1)/(2) QE`
This origin of factor `1//2` in force can be explained by the fact that inside the conductor, field is zero and outside the conductor, the field is E. Therefore the average value of the field `(i.e. E//2)` constribuites to the force.