JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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 \(..........\).

- A \(4\)
- B \(3\)
- C \(5\)
- D \(2\)
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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