Ncert Exercises & Solutions

Question 2.28:

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 the electric field between the plates. Explain the Origin Of the factor 1/2.

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

∆

x. Hence, work is done by the force to do so = F

∆

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

∆

x. Where,

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.,

\begin{array}{l} FΔx = uAΔx \\ F = uA = (\frac{1}{2} ϵ_{0} E^{2}) A \end{array}

Electric intensity is given by,

\begin{array}{l} E = \frac{V}{d} \\ ∴ F = \frac{1}{2} ϵ_{0} (\frac{V}{d}) EA = \frac{1}{2} (ϵ_{0} A \frac{V}{d}) E \end{array}

However, capacitance,

C = \frac{ϵ_{0} A}{d}

∴ F = \frac{1}{2} (CV) E

Charge on the capacitor is given by, Q = CV

∴ F = \frac{1}{2} QE

The physical origin of the factor, 1/2 in the force formula lies in the fact that just outside the conductor, the field is E, and inside it is zero. Hence, it is the average value, E/2, of the field that contributes to the force.

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