Question

A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants $K_1,K_2,K_3$ 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. $\frac{1}{K}$ = $\frac{K_1{K}_2}{K_1+K_2}+2K_3$
• 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 = $\frac{K_1{K}_3}{K_1+K_3}+\frac{K_2{K}_3}{K_2+K_3}$

$\text{Common mistake[Refer Figure 1]:}$

The common mistake students do in this problem is that they assume $C_1$ and $C_2$ in parallel and their combination in series with $C_3$. Which is completely wrong method. [Refer Fig. 1]

For two capacitors to be taken in parallel, their both ends should be at same potential.

Here, the upper ends of $C_1$ and $C_2$ are at same potential as they are connected to A through a metal plate.

But their lower ends being connected to a dielectric $K_3$ are not at same potential. As Electric field will be different in both dielectrics $K_1$ and $K_2$, since $K_1$ $\ne$ $K_2$.

$\text{Step 1: Divide}{K}_3\text{region into two equal parts and find capacitance of each parts [Refer Fig. 2]}$

Hence, To solve this problem, we divide dielectric $K_3$ in two equal parts as shown in figure 2. So,

$C_1=\frac{K_1{ϵ}_0\left(A/2\right)}{d/2}=\frac{K_1{ϵ}_{0}A}{d}$

$C_2=\frac{K_2\left({ϵ}_0\right)\left(A/2\right)}{d/2}=\frac{K_2{ϵ}_{0}A}{d}$

$C_3=\frac{K_3{ϵ}_0\left(A/2\right)}{d/2}=\frac{K_3\left({ϵ}_{0}A\right)}{d}$

$C_4=\frac{K_3{ϵ}_0\left(A/2\right)}{d/2}=\frac{K_3{ϵ}_{0}A}{d}$

Now, From figure, Clearly, we can assume $C_1$ and $C_3$ in series and $C_2$ and $C_4$, further their combinations will be in parallel.

$\text{Step 2: Equivalent capacitance [Refer Fig. 3]}$

Combining series combinations in figure 3, we get:

$C_{eq}=\frac{C_1{C}_3}{C_1+C_3}+\frac{C_2{C}_4}{C_2+C_4}$

Putting values of $C_1,C_2,C_3$ and $C_4$ from Step $\left(1\right)$

$C_{eq}=\frac{K_1{K}_3}{K_1+K_3}\frac{{ϵ}_{0}A}{d}+\frac{K_2{K}_3}{K_2+K_3}\frac{{ϵ}_{0}A}{d}$

$\text{Step 3: To find}$ $K_{eq}$, $\text{Compare}{C}_{eq}\text{with}\frac{K_{eq}{ϵ}_{0}A}{d}$ $\text{[Refer Fig. 4]}$

$C_{eq}=\frac{K_{eq}{ϵ}_{0}A}{d}$

$\Rightarrow \frac{K_{eq}{ϵ}_{0}A}{d}=\frac{K_1{K}_3}{K_1+K_3}\frac{{ϵ}_{0}A}{d}+\frac{K_2{K}_3}{K_2+K_3}\frac{{ϵ}_{0}A}{d}$

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

Hence equivalent dielectric constant will be as given in Option $D$