Question

A parallel plate capacitor without any dielectric within its plates, has a capacitance $C$ and is connected to a battery of emf $V$. The battery is disconnected and the plates of the capacitor are pulled apart until the separation between the plates is doubled. What is the work done by the agent pulling the plates apart, in this process?

• A. $\frac{1}{2}C{V}^2$
• B. $\frac{3}{2}C{V}^2$
• C. $-\frac{3}{2}C{V}^2$
• D. $C{V}^2$

Answer 1

Answer: (A)

Capacitance of capacitor
$C=\frac{{\epsilon }_{0}A}{d}$
Initial energy stored in the capacitor
$U_i=\frac{1}{2}C{V}^2$
When the separation between the plates is doubled, its capacitance becomes
$C^{\prime }=\frac{{\epsilon }_{0}A}{2d}=\frac{1}{2}\frac{{\epsilon }_{0}A}{d}=\frac{C}{2}$
As the battary is disconnected, so charged capacitor becomes isolated and charge on it will remain constant
$O^{\prime }=Q$
$C^{\prime }{V}^{\prime }=CV$
$V^{\prime }=\left(\frac{C}{C^{\prime }}\right)V=\frac{C}{\frac{C}{2}}=2V$
Final energy stored in the capacitor
$U_f=\frac{1}{2}{C}^{\prime }{V}^2=\frac{1}{2}\left(\frac{C}{2}\right)(2V{)}^2$
$=C{V}^2$
Work done, $W=U_f-U_i$
$=C{V}^2-\frac{1}{2}C{V}^2$
$=\frac{1}{2}C{V}^2$