Question 28

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) 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 ½.


Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of Δx.

Hence, work done by the force to do so = FΔx

As a result, the potential energy of the capacitor increases by an amount given as uAΔx.


u = Energy density

A = Area of each plate

d = Distance between the plates

V = Potential difference across the plates

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

Electric intensity is given by,

However, capacitance,

Charge on the capacitor is given by,

Q = CV

The physical origin of the factor, , in the force formula lies in the fact that just outside the conductor, 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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