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 F be the force applied to separate the plates of a parallel plate capacitor by a distance of `Deltax`. Hence , work done by the force to do so =`FDeltaX` ltBrgt As a result, the potential energy of the capacitros increases by an amount given as uA`Deltax`.
Where, u=energy density
A=area of each plate
d=distance between the plates
V=potential difference across the plates
the work done will be equal to the increases in the potential energy i.e,.
`F Detlax=uA Deltax `
`F=uA=(1/2 in_(0)E^(2))A`
Electric intensity is given by
`E=V/d`
`:. F=1/2 in_(0)(V/d)EA=1/2 (in_(0)AV/d)E`
however, capacitance, `C=(in_(0)A)/d`
`:. F=1/2 (CV)E`
charge on the capacitor is given by
`Q=CV`
`:. F=1/2 QE`
The physical origin of the factors , `1/2` , in the force formula lies in the fact that just outside
the conductor, field is E and inside it is zero. it is the avearge value, `F/2`, of the field that contributes to the force.