Question

A parallel plate capacitor has area of each plate A, the separation between the plates is d. It is charged to a potential V and then disconnected from the battery. The amount of work done in filling the capacitor completely with a dielectric constant k is :

• A. $\frac{1}{2}\frac{{\epsilon }_{0}A{V}^2}{d}[1-\frac{1}{k^2}]$
• B. $\frac{1}{2}\frac{V^2{\epsilon }_{0}A}{k^{2}d}$
• C. $\frac{1}{2}\frac{{\epsilon }_{0}A{V}^2}{d}[1-\frac{1}{k}]$
• D. $\frac{1}{2}\frac{V^2{\epsilon }_{0}A}{kd}$

Answer 1

Answer: (C)

When dielectric was not filled $C=\frac{{\epsilon }_{0}A}{d.}$

It was changed till V $\therefore Q=\frac{{\epsilon }_{0}AV}{d.}$

energy stored $E_1=\frac{1}{2}QV=\frac{1}{2}\frac{{\epsilon }_{0}A{V}^2}{d}$

Now when battery is disconnected Q remains constant. When it is filled with dielectric K

We get $C=\frac{K{\epsilon }_{0}A}{d.}$ as Q is constant

V decreases by k

$\therefore V_1=\frac{1}{2}\cdot \frac{K{\epsilon }_{0}A}{d}(\frac{V}{K}{)}^2=1\frac{{\epsilon }_{0}A{V}^2}{2dk}$

$\therefore workdonebyus=-(V_2-V_1)=\frac{{\epsilon }_{0}A{V}^2}{2d}(1-\frac{1}{k})$