Chapter 24 Problem 50

Problem

(a) Show that the force on a plate of a parallel-plate capacitor is $F={\frac {1}{2}}{\frac {Q^{2}}{\epsilon _{0}A}}$ by calculating $dW/dx$

(b) Why $F=QE$, is wrong?

Solution

(a)

Moving the plates of a charged and unconnected capacitor, does not change the amount of charge stored in it, thus we can use

$U={\frac {1}{2}}{\frac {Q^{2}}{C}}={\frac {1}{2}}{\frac {Q^{2}x}{\epsilon _{0}A}}$

The work done to move the plates infinitesimally apart is

$dW=Fdx=dU={\frac {1}{2}}{\frac {Q^{2}dx}{\epsilon _{0}A}}$

thus the Force being exerted is simply

$F={\frac {1}{2}}{\frac {Q^{2}}{\epsilon _{0}A}}$

(b)

Self interaction issue: The charges can not exert a force onto themselves, so on the plate, we have to ignore the field due to that particular plate. Consequently yhe electric field should've been the half of that.

$F_{\textrm {onplate}}=Q{\frac {1}{2}}E={\frac {1}{2}}{\frac {Q^{2}}{\epsilon _{0}A}}$