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.