Question

A parallel plate capacitor of plate area $5{\text{cm}}^2$ and plate separation $20\text{mm}$ is charged to a potential difference of $10\text{V}$ and then the battery is disconnected. A slab of dielectric constant $5$ is then inserted between the plates to fill the space between the plates completely. What is the work done on the system in the process ?

• A. $8.85\times {10}^{-12}\text{J}$
• B. $-8.85\times {10}^{12}\text{J}$
• C. $8.85\times {10}^{12}\text{J}$
• D. $-8.85\times {10}^{-12}\text{J}$

Answer 1

The correct option is D $-8.85\times {10}^{-12}\text{J}$

When the capacitor is charged, total charge on plate is given by

$Q=CV$

$\Rightarrow Q=\frac{A{ϵ}_{0}V}{d}$

Work done on the system is a change in potential energy. So, work done will be

$W=U_f-U_i$

$\Rightarrow W=\frac{Q^2}{2KC}-\frac{Q^2}{2C}$

[After dielectric is inserted, the capacitance increases by factor $K$]

$\Rightarrow W=\frac{{ϵ}_{0}A{V}^2}{2d}(\frac{1}{K}-1)$

Given:
$A=5\times {10}^{-4}{\text{m}}^2;V=10\text{Volt};d=20\times {10}^{-3}\text{m};K=5$

So, $W=\frac{8.85\times {10}^{-12}\times 5\times {10}^{-4}\times 100}{2\times 2\times {10}^{-2}}(\frac{1}{5}-1)$

$\therefore W=-8.85\times {10}^{-12}\text{J}$

Hence, option $\left(d\right)$ is correct.

Key-concept :- Work done by external forces on the dielectric when it is inserted in a parallel plate capacitor with the battery removed.