Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of Δx. Hence, work done by the force to do so = FΔx
As a result, the potential energy of the capacitor increases by an amount given as uAΔx. 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.,
Electric intensity is given by,
The physical origin of the factor, ½, in the force formula lies in the fact that just outside the conductor, field is E and inside it is zero. Hence, it is the average value, E/2, of the field that contributes to the force.
