Question

A parallel plate capacitor having capacity \(12\) pF is charged by a battery of \(10\) V between its plates. The battery is now disconnected and a slab of dielectric constant \(6\) is kept between the plates. The work done by the capacitor on the slab is

• A. \(1\times 10^{-10}\) J
• B. \(2\times 10^{-10}\) J
• C. \(5\times 10^{-10}\) J
• D. \(6\times 10^{-10}\) J

Hint:

With the battery disconnected, charge is constant; work done equals the fall in stored energy.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
With the battery removed, the charge stays fixed at \(Q=CV\). Energy stored is \(U=\dfrac{Q^{2}}{2C}\).

Step 2: Initial and Final Energy
\(U_{i}=\dfrac{1}{2}CV^{2}=\dfrac{1}{2}\times 12\times 10^{-12}\times 100=6\times 10^{-10}\) J
Capacitance becomes \(KC\), so \(U_{f}=\dfrac{U_{i}}{K}=1\times 10^{-10}\) J.

Step 3: Work Done
Work done by the capacitor on the slab \(=U_{i}-U_{f}=6\times 10^{-10}-1\times 10^{-10}\)
\(=5\times 10^{-10}\) J

Step 4: Compare the options
This matches option (C).

Final Answer:
The work done by the capacitor on the slab is \(5\times 10^{-10}\) J. \[ \boxed{\text{(C) }5\times 10^{-10}\text{ J}} \]