Question

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

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

Answer 1

The correct option is C

$k=\frac{2}{3}(k_1+k_2+k_3)+2k_4$
Here the capacitance, $C=\frac{Ak{ϵ}_0}{d}$
$C_{K_1},C_{K_2},C_{K_3}$ are in parallel combination and they are connected in series with $C_{K_4}$


Here, $C_{K_1}=\frac{\left(A/3\right)k_1{ϵ}_0}{d/2}=\frac{2k_1}{3}C$,
$C_{K_2}=\frac{\left(A/3\right)k_2{ϵ}_0}{d/2}=\frac{2k_2}{3}C$,

$C_{K_3}=\frac{\left(A/3\right)k_3{ϵ}_0}{d/2}=\frac{2k_3}{3}C$,

$C_{K_4}=\frac{\left(A\right)k_4{ϵ}_0}{d/2}=2k_{4}C$
Now the equivalent capacitance for the combination of four capacitors is
$\frac{1}{C_{eq}}=\frac{1}{(C_{k_1}+C_{k_2}+C_{k_3})}+\frac{1}{C_{K_4}}$
$C_{eq}=kC$

$\frac{1}{kC}=\frac{3}{2C}\left[\frac{1}{k_1+k_2+k_3}\right]+\frac{1}{2k_{4}C}$

or $\frac{1}{k}=\frac{3}{2(k_1+k_2+k_3)}+\frac{1}{2\left(k_4\right)}$
or
$k=\frac{2(k_1+k_2+k_3)}{3}+2\left(k_4\right)$