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

Answer 1

Answer: (A)

$\frac{1}{2}{CV}^2$

Capacitance of a parallel plate capacitor is

$C=\frac{{\epsilon }_{0}A}{d}...\left(i\right)$

where $A$ is the area of each plate and $d$ is the distance between the plates.

Initial energy stored in the capacitor

$U_i=\frac{1}{2}C{V}^2...\left(ii\right)$

When the separation between plates is doubled, its capacitance becomes

$C^{\prime }=\frac{{\epsilon }_{0}A}{d}=\frac{1}{2}\frac{{\epsilon }_{0}A}{d}=\frac{C}{2}$ (Using (i))....(iii)

As the battery is disconnected, so charged capacitor becomes isolated and charge on it will remain constant,

$\therefore$ $Q^{\prime }=Q$

$C^{\prime }{V}^{\prime }=CV$ (As $Q=CV$)

$V^{\prime }=\left(\frac{C}{C^{\prime }}\right)V=\frac{C}{\left(\frac{C}{2}\right)}V=2V....\left(iv\right)$

Final energy stored in the capacitor

$U_f=\frac{1}{2}C{V}^2=\frac{1}{2}\left(\frac{C}{2}\right){\left(2V\right)}^2=C{V}^2...\left(v\right)$ (Using (iii) and (iv))

Required work done, $W=U_f-U_i=C{V}^2-\frac{1}{2}C{V}^2=\frac{1}{2}C{V}^2$