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. $\frac{2}{k}=\frac{3}{k_1+k_2+k_3}+\frac{1}{k_4}$

Answer 1

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

Here the capacitance, $C=\frac{Ak{ϵ}_0}{d}$

The above three capacitors $C_1,C_2,C_3$ are in parallel and then it is in series with $C_4$

Here, $C_1=\frac{\left(A/3\right)k_1{ϵ}_0}{d/2}=\frac{2k_1}{3k}C$,

$C_2=\frac{\left(A/3\right)k_2{ϵ}_0}{d/2}=\frac{2k_2}{3k}C$,

$C_3=\frac{\left(A/3\right)k_3{ϵ}_0}{d/2}=\frac{2k_1}{3k}C$,

$C_4=\frac{\left(A\right)k_4{ϵ}_0}{d/2}=\frac{2k_4}{k}C$

Now the equivalent capacitance for the combination of four capacitors is

$1/C_{eq}=1/(C_1+C_2+C_3)+1/C_4$

or $1/C=\frac{3k}{2C}\left[\frac{1}{k_1+k_2+k_3}\right]+\frac{k}{2k_{4}C}$ (as $C_{eq}=C$)

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