Question

A parallel-plate capacitor of plate area $A$ and plate separation $d$ is charged to a potential difference and then the battery is disconnected. A slab of dielectric constant $K$ is then inserted between the plates of the capacitor so as to fill the whole space between the plates. Find the work done on the system the process of inserting the slab.

• A. $\frac{{\epsilon }_{0}A{V}^2}{2d}\left(1-\frac{1}{K}\right)$
• B. $\frac{{\epsilon }_{0}A{V}^2}{d}(\frac{1}{K}-1)$
• C. $\frac{{\epsilon }_{0}A{V}^2}{2d}(\frac{1}{K}+1)$
• D. $\frac{{\epsilon }_{0}A{V}^2}{d}(\frac{1}{K}+1)$

Answer 1

The correct option is A $\frac{{\epsilon }_{0}A{V}^2}{2d}\left(1-\frac{1}{K}\right)$

after inserting slab

$newcapacitance{C}^{\prime }=KC=\frac{K{ϵ}_{o}A}{d}$

$newpotentialdifference{V}^{\prime }=\frac{V}{K}$

$newelectricfield{E}^{\prime }=\frac{V^{\prime }}{d}=\frac{V}{kd}$

$newcharge{Q}^{\prime }{C}^{\prime }=\frac{{ϵ}_{o}AV}{d}$

$work=final-initial$

$=\frac{1}{2}{C}^{\prime }{V}^{\prime }-\frac{1}{2}{CV}^2$

$\frac{1}{2}\left(KC\right)=(1-\frac{1}{K})$

$\left[\omega \right]=\frac{{ϵ}_o{AV}^2}{2d}(1-\frac{1}{K})$