Question

Capacitance of the capacitor shown in the figure is

• A. $24\text{nF}$
• B. $20\text{nF}$
• C. $15\text{nF}$
• D. $10\text{nF}$

Answer 1

The correct option is A $24\text{nF}$

From the diagram, we can see that $K_2$ and $K_3$ are in series and together make parallel combination with $K_1$.

Let,
$C_{LHS}=$ capacitance of $K_1$
$C_{RHS}=$ equivalent capacitance of $K_2$ and $K_3$

So, equivalent capacitance of the system will be

$C_{eq}=C_{LHS}+C_{RHS}$

$\Rightarrow C_{eq}=\frac{K_1{ϵ}_0\left(A/2\right)}{2d}+\frac{\frac{K_2{ϵ}_0\left(A/2\right)}{d}\frac{K_3{ϵ}_0\left(A/2\right)}{d}}{\frac{K_2{ϵ}_0\left(A/2\right)}{d}+\frac{K_3{ϵ}_0\left(A/2\right)}{d}}$

$\Rightarrow C_{eq}=\frac{K_1{ϵ}_{0}A}{4d}+\frac{{ϵ}_{0}A}{2d}(\frac{K_2{K}_3}{K_2+K_3})$

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

$\Rightarrow C_{eq}=\frac{8.85\times {10}^{-12}\times 2}{2\times 0.001}(\frac{2}{2}+\frac{3\times 4}{3+4})$

$\Rightarrow C_{eq}=24\times {10}^{-9}=24\text{nF}$

Hence, option $\left(a\right)$ is correct.

$$\begin{array}{c}\text{Why this question ?}\\ \text{Concept - In this process right half of the}\\ \text{capacitor has two capacitors of dielectrics}\\ K_2\text{and}{K}_3\text{in series. Also, both}\\ \text{halves (right and left) are in parallel.}\end{array}$$