JEE Main 2023 — Capacitance Question with Solution

The given data is 

$$\begin{gathered} A=200\times {10}^{-4} {\mathrm{m}}^2 \\ d=5 \times {10}^{-3} \mathrm{m} \end{gathered}$$

The value of $a+b=4\times {10}^{-3} \mathrm{m}$

The formula for a parallel plate capacitor is 

$C=\frac{{\epsilon }_{0}A}{D}$

Let the capacitor with gap $a$ be $C_1$ and capacitor with gap $b$ be $C_2$

$C_1=\frac{{\epsilon }_{0}A}{a}, C_2=\frac{{\epsilon }_{0}A}{b}$

The equivalent capacitance is 

$$\begin{gathered} \frac{1}{C_{eq}}=\frac{a}{{\epsilon }_{0}A}+\frac{b}{{\epsilon }_{0}A} \\ \Rightarrow C_{eq}=\frac{{\epsilon }_{0}A}{a+b} \end{gathered}$$

It is given that $C_{eq}=x{\epsilon }_0 \mathrm{F}$

So, 

$$\begin{gathered} \frac{{\epsilon }_{0}A}{a+b}=x{\epsilon }_0 \\ \Rightarrow x=\frac{A}{a+b} \\ \Rightarrow x=\frac{200\times {10}^{-4}}{4\times {10}^{-3}}=5 \end{gathered}$$