Question

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to $\frac{1}{2}$ QE, where Q is the charge on the capacitor, and E is the magnitude of the electric field between the plates. Explain the origin of the factor $\frac{1}{2}$.

Answer 1

Force = $\frac{1}{2}$ QE
Work done = $\frac{1}{2}$ V x Q
The physical origin of the factor $\frac{1}{2}$ in the force formula lies in the fact that just outside the conductor, the field is E, and inside it is zero. So the average value $\frac{E}{2}$ contributes to the force.