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

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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 circuit diagram, we identify the arrangement of capacitors. The plates are numbered $1, 2, 3, 4$ from top to bottom. Plate $1$ and $4$ are connected together, and plate $2$ and $3$ are connected together.Capacitor $C_1$ is formed by plates $1$ and $2$. Capacitor $C_2$ is formed by plates $2$ and $3$. Capacitor $C_3$ is formed by plates $3$ and $4$.Based on the connections, $C_1$ and $C_3$ are in series, and this combination is in parallel with $C_2$.Thus, the equivalent capacitance is $C_{eq} = C_2 + \frac{C_1 \cdot C_3}{C_1 + C_3} = C + \frac{C \cdot C}{C + C} = C + \frac{C}{2} = \frac{3C}{2}$.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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