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 $2 d$ will be-

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

Answer 1

Answer: (D)

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