Question:
Four identical rectangular plates with length, l = 2 cm and breadth, b = \(\frac{3}{2}\) cm are arranged as shown in figure. The equivalent capacitance between A and C is \(\frac{x\varepsilon_0}{d}\). The value of x is ______________. (Round off to the Nearest Integer)
Four identical rectangular plates with length, l = 2 cm and breadth, b = \(\frac{3}{2}\) cm are arranged as shown in figure. The equivalent capacitance between A and C is \(\frac{x\varepsilon_0}{d}\). The value of x is ______________. (Round off to the Nearest Integer)
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Plates B and D are joined by a wire, so the BC and CD capacitors are in parallel, and this pair is in series with AB.
Updated On: Nov 4, 2026
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Correct Answer: 2
Solution and Explanation
Area of each plate \(=2\times\frac{3}{2}=3\ \mathrm{cm}^2\); each adjacent pair has capacitance \(C_0=\frac{\varepsilon_0 A}{d}=\frac{3\varepsilon_0}{d}\) (area taken in cm\(^2\)).
The three capacitors are AB, BC and CD. Plates B and D are connected by the wire, so BC and CD both lie between the common node (B, D) and C: they are in parallel, giving \(2C_0\).
This combination is in series with AB (between A and the node B): \(C_{eq}=\frac{C_0\cdot 2C_0}{C_0+2C_0}=\frac{2}{3}C_0=\frac{2}{3}\cdot\frac{3\varepsilon_0}{d}=\frac{2\varepsilon_0}{d}\).
Hence x = 2.
Answer: 2
The three capacitors are AB, BC and CD. Plates B and D are connected by the wire, so BC and CD both lie between the common node (B, D) and C: they are in parallel, giving \(2C_0\).
This combination is in series with AB (between A and the node B): \(C_{eq}=\frac{C_0\cdot 2C_0}{C_0+2C_0}=\frac{2}{3}C_0=\frac{2}{3}\cdot\frac{3\varepsilon_0}{d}=\frac{2\varepsilon_0}{d}\).
Hence x = 2.
Answer: 2
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