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. The plates are pulled apart an additional distance $dx$. What is the change in the stored energy?

• A. $\left(\frac{Q^2}{2{ϵ}_{o}A}\right)\cdot dx$
• B. $\left(\frac{Q^2}{{ϵ}_{o}A}\right)\cdot dx$
• C. $\left(\frac{2Q^2}{{ϵ}_{o}A}\right)\cdot dx$
• D. $\left(\frac{Q^2}{4{ϵ}_{o}A}\right)\cdot dx$

Answer 1

The correct option is A $\left(\frac{Q^2}{2{ϵ}_{o}A}\right)\cdot dx$

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 $=U^{\prime }-U=\frac{Q^2}{2A{ϵ}_0}dx$