JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2023
- Hard
- Numerical
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 \(..........\).

Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(5\)
Step-by-step Solution
Detailed explanation
\(c=\frac{\varepsilon_0 A}{(d-c)}\)
\(=\frac{\varepsilon_0 \times 200 \times 10^{-4}}{4 \times 10^{-3}}\)
\(\therefore x=5\)
The situation is equivalent to a conducting slab placed between the plates
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