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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 \(..........\).

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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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