# 2.28 Show that the force on each plate of a parallel plate capacitor has a magnitude equal to $\left (\frac{1}{2} \right )$ 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 $\left (\frac{1}{2} \right )$

Let

The surface charge density of the capacitor = $\sigma$

Area of the plate = $A$

NOW,

As we know,

$Q=\sigma A\:and \: E=\frac{\sigma}{\epsilon _0}$

When the separation is increased by $\Delta x$,

work done by external force= $F\Delta x$

Now,

Increase in potential energy :

$\Delta u=u*A\Delta x$

By work-energy theorem,

$F\Delta x=u*A\Delta x$

$F=u*A=\frac{1}{2}\epsilon _0E^2A$

putting the value of $\epsilon _0$

$F=\frac{1}{2}\frac{\sigma}{E}E^2A=\frac{1}{2}\sigma AE=\frac{1}{2}QE$

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

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