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 )

Answers (1)

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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