Question

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

Answer 1

Force $=\left(\frac{1}{2}\right)QE$

Work done $=\frac{1}{2}\times V\times Q$

$F\times d=\frac{1}{2}\times Ed\times Q$

$F=\frac{1}{2}QE$

The physical origion of the factor in the force formula lies in the fact that just outside the conductor, field is $E$ and inside it is zero. SO the average value contributes to the force.