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

Let the distance between the plates be increased by a very small distance `Deltax`. The force on each plate is F. the amount of work done in increasing the separation by `Deltax`=force`xx`increased distance=`F.Deltax` . . (i)
Increase in volume of capacitor=Area of plates`xx`increased distance=`A.Deltax`
u=energy density`=("Energy")/("Volume")`
Increase in energy=`uxx`volume=`u.A.Deltax` . .. (ii)
As, energy=work done
`F.Deltax=u.A.Deltax` [`because` from eqs. (i) and (ii)]
`=(1)/(2)epsi_(0).(V^(2))/(d^(2)).A=((epsi_(0)A)/(d).V)(V)/(d)xx(1)/(2)" "[becauseC=(epsi_(0)A)/(d),CV=Q], =(1)/(2)E.C.V=(1)/(2)QE`.