Question

Two plates of a parallel plate capacitor carry charges q and -q are separated by a distance x from each other. The capacitor is connected to a constant voltage source $V_0$. The distance between the plates is changed to x+ dx. Then in steady state.

• A. Change in electrostatic energy stored in the capacitor is $\frac{-Udx}{x}$, where U is initial potential energy.
• B. Change in electrostatic energy stored in the capacitor is $-\frac{Ux}{dx}$
• C. Attraction force between the plates is $\frac{1}{2}qE$
• D. Attraction force between the plates is qE. (where E is electric field between the plates)

Answer 1

Answer: (A), (C)

The correct options are
A Change in electrostatic energy stored in the capacitor is $\frac{-Udx}{x}$, where U is initial potential energy.
C Attraction force between the plates is $\frac{1}{2}qE$

Initial stored energy $U_i=\frac{1}{2}\frac{{\epsilon }_{0}A{V}_0^2}{x}$
Final stored energy $U_f=\frac{1}{2}\frac{{\epsilon }_{0}A{V}_0^2}{(x+dx)}$
So $\Delta U=U_f-U_i=\frac{1}{2}{\epsilon }_{0}A{V}_0^2[\frac{1}{x+dx}-\frac{1}{x}]$
$=\frac{1}{2}{\epsilon }_{0}A{V}_0^2\left[\frac{x-x-dx}{x(x+dx)}\right]=-\frac{1}{2}\frac{{\epsilon }_{0}A{V}_0^2}{x^2}dx=-\frac{1}{2}\frac{{\epsilon }_{0}A{V}_0^2}{x}.\frac{dx}{x}=-\frac{Udx}{x}$

$F=-\frac{dY}{dx}=-\frac{U}{x}$
$F=-\frac{1}{2}\frac{{\epsilon }_{0}A}{x}.\frac{V_0^2}{x}=-\frac{1}{2}\left(\frac{{\epsilon }_{0}A}{x}{V}_0\right)\frac{V_0}{x}=-\frac{1}{2}qE$
So magnitude of attractive force is $\frac{1}{2}qE.$