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(D) Let the capacitance of each pair of adjacent plates be $C = \frac{{\varepsilon _0}A}{d}$.From the given circuit diagram,we can identify the arrang

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$\frac{3{\varepsilon _0}A}{2d}$

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(D) Let the capacitance of each pair of adjacent plates be $C = \frac{{\varepsilon _0}A}{d}$.From the given circuit diagram,we can identify the arrangement of capacitors.The top plate and the second plate form one capacitor $C_1$.The second plate and the third plate form another capacitor $C_2$.The third plate and the fourth plate form a third capacitor $C_3$.Based on the connections,the capacitor $C_1$ is in parallel with the series combination of $C_2$ and $C_3$.Thus,the equivalent capacitance is $C_{eq} = C_1 + \frac{C_2 \cdot C_3}{C_2 + C_3} = C + \frac{C \cdot C}{C + C} = C + \frac{C}{2} = \frac{3C}{2}$.Substituting $C = \frac{{\varepsilon _0}A}{d}$,we get $C_{eq} = \frac{3{\varepsilon _0}A}{2d}$.

Four metallic plates,each with a surface area of one side $A$,are placed at a distance $d$ from each other. The plates are connected as shown in the circuit diagram. Then the capacitance of the system between $a$ and $b$ is

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