Question

A parallel plate capacitor of area $ A$, plate separation $ d$ and capacitance $ C$ is filled with three different dielectric materials having dielectric constant $ {K}_{1}, {K}_{2}, {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?

\( A = \) Area of plates

• 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. $\frac{1}{K}=\frac{K_1{K}_2}{K_1+K_2}+2K_3$
• D. $K=\frac{K_1{K}_3}{K_1+K_3}+\frac{K_2{K}_3}{K_2+K_3}$

Answer 1

The correct option is D

$K=\frac{K_1{K}_3}{K_1+K_3}+\frac{K_2{K}_3}{K_2+K_3}$

Step 1: Given

A parallel plate capacitor of area $A$, plate separation $d$ and capacitance $C$ is filled with three different dielectric materials having dielectric constant $K_1,K_2,K_3$ as shown in figure.

Step 2: To find

If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor, then its dielectric constant $K=?$

Step 3: Formula used

Capacitance,

$C=\frac{K{\epsilon }_{0}A}{d}$

where $K$ is the dielectric constant, $d$ is plate separation, ${\epsilon }_0$ is the permittivity of free space and $A$ is the area.

Step 4: Calculation

We consider the given dielectrics as four dielectrics of same area $\frac{A}{2}$ and same plate separation $\frac{d}{2}$as shown in the figure.

Capacitance at the part with dielectric constant $K_1$,

$C_1=\frac{K_1{\epsilon }_0\left({\displaystyle \frac{A}{2}}\right)}{\left({\displaystyle \frac{d}{2}}\right)}$

Capacitance at the part with dielectric constant $K_2$,

$C_2=\frac{K_2{\epsilon }_0\left({\displaystyle \frac{A}{2}}\right)}{\left({\displaystyle \frac{d}{2}}\right)}$

Capacitance at the part with dielectric constant $K_3$,

$$\begin{gathered} C_3=\frac{K_3{\epsilon }_0\left({\displaystyle \frac{A}{2}}\right)}{\left({\displaystyle \frac{d}{2}}\right)} \\ {C}_4=\frac{K_3{\epsilon }_0\left({\displaystyle \frac{A}{2}}\right)}{\left({\displaystyle \frac{d}{2}}\right)} \end{gathered}$$

Equivalent capacitance in series,

$$\begin{gathered} \frac{1}{C_{1,3}}=\frac{1}{C_1}+\frac{1}{C_3} \\ \frac{1}{C_{1,3}}=\frac{C_3+C_1}{C_1{C}_3} \\ {C}_{1,3}=\frac{C_1{C}_3}{C_3+C_1} \\ \frac{1}{C_{2,4}}=\frac{1}{C_2}+\frac{1}{C_4} \\ \frac{1}{C_{2,4}}=\frac{C_4+C_2}{C_2{C}_4} \\ {C}_{2,4}=\frac{C_2{C}_4}{C_4+C_2} \end{gathered}$$

Since $C_{1,3}and{C}_{2,4}$ are in parallel equivalent resistance,

$$\begin{gathered} C=C_{1,3}+C_{2,4} \\ \Rightarrow C=\frac{C_1{C}_3}{C_1+C_3}+\frac{C_2{C}_4}{C_2+C_4} \\ \Rightarrow C=\frac{\left(\frac{K_1{\epsilon }_{0}A}{d}\right)\left(\frac{K_3{\epsilon }_{0}A}{d}\right)}{\left(\frac{K_1{\epsilon }_{0}A}{d}\right)+\left(\frac{K_3{\epsilon }_{0}A}{d}\right)}+\frac{\left(\frac{K_2{\epsilon }_{0}A}{d}\right)\left(\frac{K_3{\epsilon }_{0}A}{d}\right)}{\left(\frac{K_2{\epsilon }_{0}A}{d}\right)+\left(\frac{K_3{\epsilon }_{0}A}{d}\right)} \\ \Rightarrow C=\frac{{\left(\frac{{\epsilon }_{0}A}{d}\right)}^2{K}_1{K}_3}{\left(\frac{{\epsilon }_{0}A}{d}\right)\left(K_1+K_3\right)}+\frac{{\left(\frac{{\epsilon }_{0}A}{d}\right)}^2{K}_2{K}_3}{\left(\frac{{\epsilon }_{0}A}{d}\right)\left(K_2+K_3\right)} \\ \Rightarrow C=\left(\frac{{\epsilon }_{0}A}{d}\right)\left(\frac{K_1{K}_3}{\left(K_1+K_3\right)}+\frac{K_2{K}_3}{\left(K_2+K_3\right)}\right) \\ \Rightarrow \frac{K{\epsilon }_{0}A}{d}=\left(\frac{{\epsilon }_{0}A}{d}\right)\left(\frac{K_1{K}_3}{\left(K_1+K_3\right)}+\frac{K_2{K}_3}{\left(K_2+K_3\right)}\right)\left(\because C=\frac{K{\epsilon }_{0}A}{d}\right) \\ \Rightarrow K=\frac{K_1{K}_3}{K_1+K_3}+\frac{K_2{K}_3}{K_2+K_3} \end{gathered}$$

Therefore, 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 $K=\frac{K_1{K}_3}{K_1+K_3}+\frac{K_2{K}_3}{K_2+K_3}$

Hence, option D is correct.