Q.
A parallel-plate capacitor of area $A$, plate separation d and capacitance $C$ is filled with four dielectric materials having dielectric constants $k_1, k_2, k_3 $ and $ k_4$ as shown in the figure below. If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor, then its dielectric constant $k$ is given by
Solution:
$\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} $
$ \frac{d}{kA \in_{0} } = \frac{3d}{2\left(k_{1} + k_{2}+k_{3}\right)A \in_{0}} + \frac{d}{2k_{4}A \in_{0}} $
$ \frac{d}{kA \in_{0}} = \frac{d}{A \in_{0}} \left[\frac{3}{3\left(k_{1} + k_{2} +k_{3}\right)} + \frac{1}{2k_{4}}\right] $
$ \frac{2}{k} = \frac{3}{k_{1} + k_{2} +k_{3}} = \frac{1}{k_{4} }$
