Question

A parallel plate capacitor is constructed using three different dielectric materials, as shown in the figure. Find the effective capacitance across $A$ and $B$.

• A. $\frac{17}{9}\frac{{ϵ}_{0}A}{d}$
• B. $\frac{19}{7}\frac{{ϵ}_{0}A}{d}$
• C. $\frac{17}{12}\frac{{ϵ}_{0}A}{d}$
• D. $\frac{12}{7}\frac{{ϵ}_{0}A}{d}$

Answer 1

The correct option is B $\frac{19}{7}\frac{{ϵ}_{0}A}{d}$

Given:

$K_1=2;K_2=3$ and $K_3=4$

Now,

$C_1^{\prime }=K_1{C}_1=K_1\left(\frac{{ϵ}_0\left(A/2\right)}{d}\right)=K_1\left(\frac{{ϵ}_{0}A}{2d}\right)$

Similarly, $C_2^{\prime }$ and $C_3^{\prime }$ becomes

$C_2^{\prime }=K_2{C}_2=K_2\left[\frac{{ϵ}_0\left(A/2\right)}{\left(d/2\right)}\right]=K_2\left(\frac{{ϵ}_{0}A}{d}\right)$

$C_3^{\prime }=K_3{C}_3=K_3\left[\frac{{ϵ}_0\left(A/2\right)}{\left(d/2\right)}\right]=K_3\left(\frac{{ϵ}_{0}A}{d}\right)$

From the given figure, it is clear that $C_2^{\prime }$ and $C_3^{\prime }$ are in series, and together they are making parallel combination with $C_1^{\prime }$. Therefore, equivalent capacitance of the given combination will be

$C_{eq}=C_1^{\prime }||(C_2^{\prime },C_3^{\prime })=C_1^{\prime }+\frac{C_2^{\prime }{C}_3^{\prime }}{C_2^{\prime }+C_3^{\prime }}$

$\Rightarrow C_{eq}=K_1\left(\frac{{ϵ}_{0}A}{2d}\right)+\frac{K_2\left(\frac{{ϵ}_{0}A}{d}\right)\times K_3\left(\frac{{ϵ}_{0}A}{d}\right)}{K_2\left(\frac{{ϵ}_{0}A}{d}\right)+K_3\left(\frac{{ϵ}_{0}A}{d}\right)}$

$\Rightarrow C_{eq}=[\frac{K_1}{2}+\frac{K_2{K}_3}{K_2+K_3}]\left(\frac{{ϵ}_{0}A}{d}\right)$

After putting the values, we get

$C_{eq}=[\frac{2}{2}+\frac{3\times 4}{3+4}]\left(\frac{{ϵ}_{0}A}{d}\right)$


$\therefore C_{eq}=\frac{19}{7}\left(\frac{{ϵ}_{0}A}{d}\right)$

Hence, option $\left(B\right)$ is the correct answer.

Why this question ? To understand the difference between the series and parallel combination of capacitors.