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(C) The given system consists of four parallel metallic plates. Let the outer plates be connected to point $P$ and the inner plates be connected to po

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$2 \varepsilon_{0} \frac{A}{d}$

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(C) The given system consists of four parallel metallic plates. Let the outer plates be connected to point $P$ and the inner plates be connected to point $Q$.Between each pair of adjacent plates,a capacitor is formed. Since there are four plates,there are three gaps between them. However,based on the connections shown:$1$. The top plate is connected to $P$.$2$. The second plate is connected to $Q$.$3$. The third plate is connected to $Q$.$4$. The bottom plate is connected to $P$.This configuration creates two capacitors in parallel,each with plate area $A$ and separation $d$. The first capacitor is formed by the top plate $(P)$ and the second plate $(Q)$. The second capacitor is formed by the third plate $(Q)$ and the bottom plate $(P)$.The capacitance of each individual capacitor is $C = \frac{\varepsilon_0 A}{d}$.Since these two capacitors are connected in parallel,the equivalent capacitance of the system is:$C_{eq} = C_1 + C_2 = \frac{\varepsilon_0 A}{d} + \frac{\varepsilon_0 A}{d} = \frac{2 \varepsilon_0 A}{d}$.

Four metallic plates,each of surface area (of one side) $A$,are placed at a distance $d$ apart from each other. The two outer plates are connected to a point $P$ and the two inner plates are connected to another point $Q$ as shown in the figure below. Then,the capacitance of the system is

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