Question

Question 2.28: 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.

Class 12 - Physics - Electrostatic Potential And Capacitance Electrostatic Potential And Capacitance Page-90

Answer 1

magnitude of electric field between the plates of charged capacitor is $E=\frac{\sigma}{\epsilon_0}$
However, magnitude of electric field of one plates on the other plate of charged capacitor is $E_1=\frac{\sigma}{2\epsilon_0}$.so, force on the one plate of charged capacitor due to other plate is $\vec{F}=Q\vec{E_1}$
$F=Q.\frac{\sigma}{2\epsilon_0}=\frac{1}{2}\times Q\times\frac{\sigma}{\epsilon_0}\\F=\frac{1}{2}QE$

here, the factor 1/2 is because the electric Field of one plate on the other plate of charged capacitor is 1/2 of resultant electric field E between the plate of charged capacitor.

Answer 2

Answer:

Suppose we increase the separation of the plates by Ax.

Work done (by external agency) = F Ax.

This goes to increase the potential energy of the capacitor by u a Ax where u is energy density.

Therefore, F=ua which is easily seen to be (1/2) QE, using u= (1/2) e E².

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

So. the average value E/2 contributes to the force.