As shown in the figure, two parallel plate capacitors having equal plate area of $200\,cm ^2$ are joined in such a way that $a \neq b$. The equivalent capacitance of the combination is $x \varepsilon_0 F$. The value of $x$ is $..........$.
JEE MAIN 2023, Diffcult
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$c=\frac{\varepsilon_0 A}{(d-c)}$
$=\frac{\varepsilon_0 \times 200 \times 10^{-4}}{4 \times 10^{-3}}$
$\therefore x=5$
The situation is equivalent to a conducting slab placed between the plates
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