Question

A parallel-plate capacitor of area $A$, plate separation 𝑑, and capacitance $C$ is filled with four dielectric material having dielectric constants $k_1,k_2,k_3$ and $k_4$ as shown in the figure. 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

• A. $k=k_1+k_2+k_3+3k_4$
• B. $K=\frac{2}{3}(k_1+k_2+k_3)+2k_4$
• C. $\frac{2}{k}=\frac{3}{k_1+k_2+k_3}+\frac{1}{k_4}$
• D. $\frac{1}{k_4}=\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}+\frac{3}{2k_4}$

Answer 1

The correct option is C $\frac{2}{k}=\frac{3}{k_1+k_2+k_3}+\frac{1}{k_4}$


The 3 di-electrics $k_1,k_2$ and $k_3$ of capacitor in zone A can be considered as shown

We know general capacitance with di-electric is given by
$C=\frac{k{\epsilon }_{0}A}{d}$
$C_{k_1}=\frac{k_1{\epsilon }_0\frac{A}{3}}{\frac{d}{2}}=\frac{2k_1{\epsilon }_{0}A}{3d}$
Similarly
$C_{k_2}=\frac{2k_2{\epsilon }_{0}A}{3d}$
$C_{k_3}=\frac{2k_3{\epsilon }_{0}A}{3d}$
Since all these capacitors are in parllel
$\therefore C_{zoneA}=C_{k_1}+C_{k_2}+C_{k_3}\Rightarrow C_{zoneA}=\frac{2{\epsilon }_{0}A}{3d}(k_1+k_2+k_3)$
Now taking zone B
$C_{k_4}=\frac{k_4{\epsilon }_{0}A}{\frac{d}{2}}=\frac{2k_4{\epsilon }_{0}A}{d}$
Since zone A and zone B are in series
$C_{eq}=\frac{C_{zoneA}\times C_{k_4}}{C_{zoneA}+C_{k_4}}$
$\Rightarrow \frac{k{\epsilon }_{0}A}{d}=\frac{\frac{2{\epsilon }_{0}A}{3d}(k_1+k_2+k_3)\times \frac{2k_4{\epsilon }_{0}A}{d}}{\frac{2{\epsilon }_{0}A}{3d}(k_1+k_2+k_3)+\frac{2k_4{\epsilon }_{0}A}{d}}$
$\Rightarrow \frac{2}{k}=\frac{3}{k_1+k_2+k_3}+\frac{1}{k_4}$