# 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 the force applied to separate the plates of a parallel plate capacitor by a distance of x be F. Hence, work done = F x.

The increase in potential energy of the capacitor = u v = u A x

Where, u = Energy density, A = area of each plate and v = volume between plates of capacitors.

Since, work done will be equal to the increase in the potential energy i.e.,

work done = increase in potential energy of the capacitor

F x = u A x

F = u A Since, u = and = E (Electric intensity or Electric field)   and Therefore, F = 1/2 (CV)E and Q = CV

Thus, F = QE

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

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