Question

A parallel plate capacitor of area $A$, plate separation $d$ and capacitance $C$ is filled with three different dielectric materials having dielectric constants $k_1,k_2$ and $k_3$ as shown. 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. $\frac{1}{k}=\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{2k_3}$
• B. $\frac{1}{k}=\frac{1}{k_1+k_2}+\frac{1}{2k_3}$
• C. $k=\frac{k_1{k}_2}{k_1+k_2}+2k_3$
• D. $k=k_1+k_2+2k_3$

Answer 1

Answer: (B)

$C_1=\frac{k_1{\epsilon }_0\frac{A}{2}}{\left(\frac{d}{2}\right)}=\frac{k_1{\epsilon }_{0}A}{d}$
$C_2=\frac{k_2{\epsilon }_0\frac{A}{2}}{\left(\frac{d}{2}\right)}=\frac{k_2{\epsilon }_{0}A}{d}$
and $C_3=\frac{k_3{\epsilon }_{0}A}{\left(\frac{d}{2}\right)}=\frac{2k_3{\epsilon }_{0}A}{d}$
$\frac{1}{C_{eq}}=\frac{1}{C_1+C_2}+\frac{1}{C_3}$
$=\frac{1}{\frac{{\epsilon }_{0}A}{d}\left(k_1+k_2\right)}+\frac{1}{\frac{{\epsilon }_{0}A}{d}\times 2k_3}$
$\frac{1}{C_{eq}}=\frac{d}{{\epsilon }_{0}A}\left[\frac{1}{k_1+k_2}+\frac{1}{2k_3}\right]$
$C_{eq}={\left[\frac{1}{k_1+k_2}+\frac{1}{2k_3}\right]}^{-1}\cdot \frac{{\epsilon }_{0}A}{d}$
So $k_{eq}={\left[\frac{1}{k_1+k_2}+\frac{1}{2k_3}\right]}^{-1}$