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(D) Let the capacitance of each pair of adjacent plates be $C_{0} = \frac{\varepsilon_{0} A}{d}$,where $A = l 	imes b = 2 	imes \frac{3}{2} = 3 \, c

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$2$

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(D) Let the capacitance of each pair of adjacent plates be $C_{0} = \frac{\varepsilon_{0} A}{d}$,where $A = l 	imes b = 2 	imes \frac{3}{2} = 3 \, cm^{2}$.From the figure,plates $B$ and $D$ are connected together.The capacitor is formed between plates $A$ and $B$,$B$ and $C$,and $C$ and $D$.Since $B$ and $D$ are connected,the plates $B$ and $D$ are at the same potential.The circuit consists of a capacitor between $A$ and $B$ (let this be $C_{1}$),and two capacitors in parallel between $B$ and $C$ (one between $B$ and $C$,and one between $C$ and $D$).Thus,$C_{eq} = C_{AB} 	ext{ in series with } (C_{BC} \parallel C_{CD})$.$C_{eq} = \frac{C_{0} 	imes (C_{0} + C_{0})}{C_{0} + (C_{0} + C_{0})} = \frac{C_{0} 	imes 2C_{0}}{3C_{0}} = \frac{2}{3} C_{0}$.Substituting $C_{0} = \frac{\varepsilon_{0} A}{d} = \frac{3 \varepsilon_{0}}{d}$,we get $C_{eq} = \frac{2}{3} 	imes \frac{3 \varepsilon_{0}}{d} = \frac{2 \varepsilon_{0}}{d}$.Comparing this with $\frac{x \varepsilon_{0}}{d}$,we find $x = 2$.

Four identical rectangular plates with length,$l = 2 \, cm$ and breadth,$b = \frac{3}{2} \, cm$ are arranged as shown in the 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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