Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of ∆Δx. Hence, work is done by the force to do so = F∆ΔX
As a result, the potential energy of the capacitor increases by an amount given as uA∆Δx. 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 increase in the potential energy i.e.,
FΔx=uAΔxF=uA=(12ϵ0E2)AFΔx=uAΔxF=uA=(12ϵ0E2)A
Electric intensity is given by,
E=Vd∴F=12ϵ0(Vd)EA=12(ϵ0AVd)E
However, capacitance, C=ϵ0Ad
∴F=12(CV)E
Charge on the capacitor is given by, Q = CV
∴F=12QE
The physical origin of the factor, 1/2 in the force formula lies in the fact that just outside the conductor, the field is E, and inside it is zero. Hence, it is the average value, E/2, of the field that contributes to the force.
