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(D) Let the capacitance of each pair of adjacent plates be $C = \frac{\varepsilon_0 A}{d}$.From the figure,plates $1$ and $4$ are connected together.

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$\frac{3}{2}\frac{\varepsilon_0 A}{d}$

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(D) Let the capacitance of each pair of adjacent plates be $C = \frac{\varepsilon_0 A}{d}$.From the figure,plates $1$ and $4$ are connected together. Plate $2$ is connected to terminal $A$,and plate $3$ is connected to terminal $B$.$1$. The capacitor formed by plates $2$ and $3$ is directly connected between $A$ and $B$. Its capacitance is $C$.$2$. The capacitor formed by plates $1$ and $2$ is in series with the capacitor formed by plates $3$ and $4$. Let these be $C_1$ and $C_2$ respectively,where $C_1 = C_2 = C$.$3$. The equivalent capacitance of this series combination is $\frac{1}{C_s} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C}$,so $C_s = \frac{C}{2}$.$4$. This series combination is in parallel with the capacitor formed by plates $2$ and $3$. Thus,the total capacitance $C_{AB} = C + C_s = C + \frac{C}{2} = \frac{3}{2}C$.Substituting $C = \frac{\varepsilon_0 A}{d}$,we get $C_{AB} = \frac{3}{2} \frac{\varepsilon_0 A}{d}$.

Four identical metal plates are arranged in air at equal distances from each other as shown in the figure. Each plate has an area $A$. Find the equivalent capacitance of the system between points $A$ and $B$.

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