First slide

Capacitance

Question

In an isolated charged parallel plate capacitor the electrostatic potential energy stored in it is 12µJ. Find the work done in increasing the separation between the plates to 2 times its original value.

Moderate

A

6µJ
B

18µJ
C

8µJ
D

12µJ

Solution

$\begin{array}{l} C_{i} = \frac{ε_{0} A}{d} and C_{f} = \frac{ε_{0} A}{2 d} = \frac{C_{i}}{2} \\ ∴ U_{i} = \frac{Q^{2}}{2 C_{i}} and U_{f} = \frac{Q^{2}}{2 C_{f}} = \frac{Q^{2}}{2 (\frac{C_{i}}{2})} = 2 U_{i} \\ ∴ W = U_{f} - U_{i} = 2 U_{i} - U_{i} = U_{i} = 12 μJ \end{array}$

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