Question

A parallel plate capacitor consists of two plates, each with area $A$ and separated by a distance $d$. What will be the work done against electrostatic force in increasing the separation between the plates from $l_1$ to $l_2$, while the potential difference $V$ across the capacitor is kept constant ?

• A. $\frac{{ϵ}_{0}A{V}^2}{2}\left(\frac{l_1}{l_1+l_2}\right)$
• B. $\frac{{ϵ}_{0}A{V}^2}{2}\left(\frac{l_1}{l_1-l_2}\right)$
• C. $\frac{{ϵ}_{0}A{V}^2}{2}\left(\frac{l_1+l_2}{l_1{l}_2}\right)$
• D. $\frac{{ϵ}_{0}A{V}^2}{2}\left(\frac{l_2-l_1}{l_1{l}_2}\right)$

Answer 1

The correct option is D $\frac{{ϵ}_{0}A{V}^2}{2}\left(\frac{l_2-l_1}{l_1{l}_2}\right)$

Using relation between energy density and workdone,

$U=\frac{dW}{dv}=\frac{1}{2}{\epsilon }_0{E}^2$

Where,
$v=$ volume between the plates
$E=$ electric field inside the plates


Let $dW$ is the work done in increasing separation between the plates by $dx$ is,

$dW=\frac{1}{2}{\epsilon }_0{E}^{2}dv$

$\because E=\frac{V}{x}\text{and}dv=Adx$

$dW=\frac{1}{2}{\epsilon }_0{\left(\frac{V}{x}\right)}^{2}Adx=\frac{{\epsilon }_{0}A{V}^2}{2}\frac{dx}{x^2}$

Integrating the above equation with proper limits, we get,

$W=\int dW=\frac{{\epsilon }_{0}A{V}^2}{2}{\int }_{l_1}^{l_2}{x}^{-2}dx$

$W=\frac{{\epsilon }_{0}A{V}^2}{2}{\left[\frac{x^{-1}}{-1}\right]}_{l_1}^{l_2}=\frac{{\epsilon }_{0}A{V}^2}{2}(\frac{1}{l_1}-\frac{1}{l_2})$

$\therefore W=\frac{{\epsilon }_{0}A{V}^2}{2}\left(\frac{l_2-l_1}{l_1{l}_2}\right)$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.

Alternate solution: Since, electrostatic force $\left(F\right)$ is a conservative force, so work done against this force in PPC will be $W=Q_f-Q_i$ $W=\frac{1}{2}{C}_i{V}^2-\frac{1}{2}{C}_f{V}^2=\frac{1}{2}{V}^2(C_i-C_f)$ $W=\frac{1}{2}{V}^2(\frac{A{ϵ}_0}{l_1}-\frac{A{ϵ}_0}{l_2})=\frac{{ϵ}_{0}A{V}^2}{2}\left(\frac{l_2-l_1}{l_1{l}_2}\right)$