Question

In the figure, a capacitor is filled with dielectrics. The resultant capacitance is,

• A. $\frac{2{ϵ}_{0}A}{d}[\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}]$
• B. $\frac{{ϵ}_{0}A}{d}[\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}]$
• C. $\frac{2{ϵ}_{0}A}{d}[k_1+k_2+k_3]$
• D. $\frac{{ϵ}_{0}A}{d}(\frac{k_1{k}_2}{k_1+k_2}+\frac{k_3}{2})$

Answer 1

The correct option is D $\frac{{ϵ}_{0}A}{d}(\frac{k_1{k}_2}{k_1+k_2}+\frac{k_3}{2})$

From the figure, $C_1$ and $C_2$ are in series and this combination is in parallel with $C_3$.
So, the circuit can be reduced as follows.


By using capacitance formula, we get,

$C_1$=$\frac{k_1{ϵ}_0\left(A/2\right)}{d/2}=\frac{k_1{ϵ}_{0}A}{d}$

and $C_2=\frac{k_2{ϵ}_0\left(A/2\right)}{d/2}=\frac{k_2{ϵ}_{0}A}{d}$

Equivalent of these two capacitors connected in series will be

$C_{series}=\frac{C_1{C}_2}{C_1+C_2}=\frac{k_1{k}_2}{k_1+k_2}(\frac{{ϵ}_{0}A}{d})$

Now, $C_3=\frac{k_3{ϵ}_0\left(A/2\right)}{d}=\frac{k_3{ϵ}_{0}A}{2d}$

This capacitor is in parallel to $C_{series}$

Hence, $C_{eq}=C_3+C_{series}=\frac{k_3{ϵ}_{0}A}{2d}+\frac{k_1{k}_2}{k_1+k_2}(\frac{{ϵ}_{0}A}{d})$

$\Rightarrow C_{eq}=\frac{{ϵ}_{0}A}{d}(\frac{k_1{k}_2}{k_1+k_2}+\frac{k_3}{2})$

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