Q 12-02-101JEE MainJEE Main 2023 (6 Apr, Shift 2)Medium
As shown in the figure, two parallel plate capacitors having equal plate area of $200\ \text{cm}^2$ are joined in such a way that $a\ne b$. The equivalent capacitance of the combination is $x\epsilon_0$ F. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 5
The inner two plates are joined, so the combination is two capacitors (gaps $a$ and $b$) in series:
$$\frac1C=\frac{a}{\epsilon_0A}+\frac{b}{\epsilon_0A}\ \Rightarrow\ C=\frac{\epsilon_0A}{a+b}$$
$a+b=d-c=4\ \text{mm}$: $C=\dfrac{\epsilon_0\times200\times10^{-4}}{4\times10^{-3}}=5\epsilon_0$.
Solution by Sreeraj P, M.Sc Physics