Question

A parallel plate capacitor having plate area ' \(A\) ' and separation ' \(d\) ' is charged to a potential difference ' \(V\) '. The charging battery is disconnected and the plates are pulled apart to four times the initial separation. The work required to increase the distance between the plates is ( \(\varepsilon_0=\) permittivity of free space)

• A. \(\frac{\epsilon _{0}AV^{2}}{3d}\)
• B. \(\frac{\epsilon _{0}AV^{2}}{4d}\)
• C. \(\frac{2\epsilon _{0}AV^{2}}{d}\)
• D. \(\frac{3\epsilon _{0}AV^{2}}{2d}\)

Hint:

With the battery disconnected the charge is constant, so the energy stored increases in proportion to the plate separation.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
After disconnection, charge \(Q\) stays constant. Energy \(U=\frac{Q^2}{2C}\) and \(C=\frac{\epsilon_0A}{d}\), so \(U\propto d\).

Step 2: Initial energy
\(U_1=\frac{1}{2}\frac{\epsilon_0A}{d}V^2\).

Step 3: Final energy
With separation \(4d\), \(U_2=4U_1=\frac{2\epsilon_0AV^2}{d}\).

Step 4: Work done
\(W=U_2-U_1=3U_1=\frac{3\epsilon_0AV^2}{2d}\), option (D).

Final Answer:
The work required is \(\frac{3\epsilon_0AV^2}{2d}\). \[ \boxed{\text{(D) }\dfrac{3\epsilon _{0}AV^{2}}{2d}} \]