JEE MAIN - Physics (2023 - 6th April Evening Shift - No. 22)
As shown in the figure, two parallel plate capacitors having equal plate area of $$200 \mathrm{~cm}^{2}$$ are joined in such a way that $$a \neq b$$. The equivalent capacitance of the combination is $$x \in_{0} \mathrm{~F}$$. The value of $$x$$ is ____________.
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Answer
5
Explanation
$$
\begin{aligned}
& c=\frac{\varepsilon_0 A}{(d-c)} \\\\
& =\frac{\varepsilon_0 \times 200 \times 10^{-4}}{4 \times 10^{-3}} \\\\
& \therefore x=5
\end{aligned}
$$
The situation is equivalent to a conducting slab placed between the plates
The situation is equivalent to a conducting slab placed between the plates
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