Question

The plate separation in a parallel plate condenser is $d$ and plate area is A. If it is charged to $V$ volt and battery is disconnected then the work done in increasing the plate separation to $2d$ will be-

• A. $\frac{3}{2}\frac{{\epsilon }_{0}A{V}^2}{d}$
• B. $\frac{{\epsilon }_{0}A{V}^2}{d}$
• C. $\frac{2{\epsilon }_{0}A{V}^2}{d}$
• D. $\frac{{\epsilon }_{0}A{V}^2}{2d}$

Answer 1

Answer: (D)

As battery is disconnected, charge remains constant in the work process
Work done $=$ final potential energy - initial potential energy
$=\frac{Q^2}{2C^{\prime }}-\frac{Q^2}{2C^o}$
$=\left\{\frac{1}{C^{\prime }}-\frac{1}{C^o}\right\}$
Where $Q=CV=\frac{A{\epsilon }_{0}V}{d}$
$C^{\prime }=\frac{A{\epsilon }_0}{d}{C}^0=\frac{A{\epsilon }_0}{2d}$
Now, work done $=\frac{{\epsilon }_{0}A{V}^2}{2d}$