Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1⁄2) 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 1⁄2.

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.,

FΔx = uAΔx

F= uA= (1/2  ε_{0 E}²) A

Electric intensity is given by,

E=V/d

Therefore F = 1/2  ε₀ (V/d)EA =1/2 ( ε₀ A V/d)E

However ,capacitance , C = ε₀ A/d

Therefore F = 1/2 (CV)E

Charge on capacitor is given by , Q=CV

Therefore F =1/2 QE

The physical origin of the factor, 1/2, 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.