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 Held between the plates. Explain the origin of the factor `1//2`

Explanation

Let area of plate is A and surface charge density is `sigma` .
Charge on capacitor q = `sigma`A and electric field `E= (sigma)/(in_(0))`
The work done for increasing the distance `Delta`x between two plates of capacitor opposite to the force,
W = `FDeltax `
If `rho_(E)` is the energy density of capacitor then potential energy of capacitor = density `xx` increase in volume
`F.Deltax = rho_(E)xxA Deltax`
`:. F = rho_(E) xxA `
`:. F = (1)/(2) in_(0) E^(2) A [ because rho_(E)= (2) in_(0)E^(2)]`
`=(1)/(2) in_(0)ExxEA`
`=(1)/(2) sigmaA xxE [because E = (sigma)/(in_(0))implies sigma = E in_(0)]`
`:. F = (1)/(2) qE [ because sigma = (q)/(A) implies q = sigma A ]`
The physical origin of the factor `(1)/(2)` in the force formula lies because just outside the conductor,field is E and inside it is zero. Hence, it is the average value `(E)/(2)` of the field that contribute to the force.