Question

A dielectic of dielectric constant $K$ completely fills the gap between the plates of a parallel plate capacitor whose capacitance is equal to $C_0$ , when dielectric is absent. Find the mechanical work that must be done against electric forces for extracting the dielectric out of the capacitor , if voltage $\left(V\right)$ of the capacitor is maintained constant.

• A. $\frac{1}{2}(\frac{1}{K}-1)C_0{V}^2$
• B. $\frac{1}{2}(K-1)C_0{V}^2$
• C. $\frac{K}{2}{C}_0{V}^2$
• D. $\frac{(2K-1)C_0{V}^2}{2}$

Answer 1

The correct option is B $\frac{1}{2}(K-1)C_0{V}^2$

In order to maintain the constant voltage, let a battery of emf $V\text{volt}$ is connected.
After extracting the dielectric, the change in capacitance will be,

$\Delta C=C_f-C_i=C_0-K{C}_0$

$\Rightarrow \Delta C=-(K-1)C_0$

Charge supplied by the battery.

$\Delta q=V\Delta C$

So, workdone by battery.

$W_B=V\Delta q$

$\Rightarrow W_B=V\times V\Delta C=V^2\Delta C$

Change in potential energy of capacitor,

$\Delta U=\frac{1}{2}{V}^2\Delta C$

Using work energy theorem,

$W_{\text{mechancial}}+W_B=\Delta K.E+\Delta U$

$(\because \Delta K.E=0)$

$\Rightarrow W_{\text{mechanical}}=0+\Delta U-W_B$

$\Rightarrow W_{\text{mechanical}}=\frac{1}{2}{V}^2\Delta C-V^2\Delta C=-\frac{1}{2}{V}^2\Delta C$

[Substituting $\Delta C=-(K-1)C_0]$

$\therefore W_{mechanical}=\frac{1}{2}(K-1)C_0{V}^2$

Hence, option (b) is the correct answer.

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: In this problem after the dielectric is extracted}\\ \text{the capcitance decreases, hence}\Delta C\text{is}-ve.\\ \text{Therefore charge is released by capcitor and workdone by}\\ \text{battery}\left(W_b\right)\text{will be}-ve.\end{array}$