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Question

Show that the force on each plate of a parallel plate capacitor has amagnitude equal to (½) QE, where Q is the charge on the capacitor,and E is the magnitude of electric field between the plates. Explainthe origin of the factor ½.

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Solution

It is given that the charge on the parallel plate capacitor is Q and the magnitude of

electric field between the plates is E.

Let the force on each plate capacitor be F.

The work done by the force is,

W=FΔd

The potential energy of the capacitor will increase by an amount of uAΔd.

The work done should be equal to the increase in the potential energy.

FΔd=uAΔd F=uA =( 1 2 ε E 2 )A (1)

The magnitude of electrical energy is,

E= V d

Substitute the value of E in equation (1).

F= 1 2 ε( V d )EA = 1 2 ( εA V d )E

Substitute the value of capacitance, C= εA d in the above equation.

F= 1 2 ( CV )E

Substitute the value of Q in the above equation.

F= 1 2 QE


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