Question

A parallel-plate vacuum capacitor with plate area $A$ and separation $x$ has charges $+Q$ and $-Q$ on its plates. The capacitor is disconnected from the source of charge, so the charge on each plate remains fixed. If $F$ is the force with which the plates attract each other, then the change in the stored energy must equal the work done $dW=Fdx$ in pulling the plates apart. Find an expression for $F$.

• A. $\frac{Q^2}{4{ϵ}_{o}A}$
• B. $\frac{Q^2}{{ϵ}_{o}A}$
• C. $\frac{Q^2}{2{ϵ}_{o}A}$
• D. $\frac{2Q^2}{{ϵ}_{o}A}$

Answer 1

Answer: (C)

$\frac{Q^2}{2{ϵ}_{o}A}$

The capacitance of parallel plate capacitor with vacuum is $C=\frac{A{ϵ}_0}{x}$

As the charges on plates is constant so the energy stored in the capacitor is

$U=\frac{Q^2}{2C}=\frac{Q^{2}x}{2A{ϵ}_0}$

When the plates are pulled apart additional distance $dx$ so effective separation is $x+dx$

Now the energy is $U^{\prime }=\frac{Q^2(x+dx)}{2A{ϵ}_0}$

Thus the change in stored energy $=U^{\prime }-U=\frac{Q^2}{2A{ϵ}_0}dx$

Now, $dW=Fdx=U^{\prime }-U$

or $Fdx=\frac{Q^2}{2A{ϵ}_0}dx$

or $F=\frac{Q^2}{2A{ϵ}_0}$