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(D) Let the capacitance of each pair of plates be $C = \frac{\varepsilon_0 A}{d}$.Based on the arrangement,the system can be modeled as two capacitors

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

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(D) Let the capacitance of each pair of plates be $C = \frac{\varepsilon_0 A}{d}$.Based on the arrangement,the system can be modeled as two capacitors in series ($C_1$ and $C_2$) which are then in parallel with a third capacitor $(C_3)$.Here,$C_1 = C_2 = C_3 = C = \frac{\varepsilon_0 A}{d}$.The equivalent capacitance of the series combination of $C_1$ and $C_2$ is $C_{12} = \frac{C_1 C_2}{C_1 + C_2} = \frac{C^2}{2C} = \frac{C}{2}$.Now,$C_{12}$ and $C_3$ are in parallel,so the total capacitance $C_T$ is:$C_T = C_{12} + C_3 = \frac{C}{2} + C = \frac{3C}{2}$.Substituting $C = \frac{\varepsilon_0 A}{d}$,we get:$C_T = \frac{3 \varepsilon_0 A}{2 d}$.

Four metal plates,each with surface area $A$ on one side,are placed with separation '$d$' as shown in the figure. The capacitance between $a$ and $b$ is ($\varepsilon_0$ = permittivity of free space).

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