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(B) Let $C = \frac{{\varepsilon _0}A}{d}$ be the capacitance of a single pair of adjacent plates.From the circuit diagram,we identify the capacitors f

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$\frac{4{\varepsilon _0}A}{d}$

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(B) Let $C = \frac{{\varepsilon _0}A}{d}$ be the capacitance of a single pair of adjacent plates.From the circuit diagram,we identify the capacitors formed between the plates:$1$. Capacitor between plate $2$ and $1$ (connected to $A$ and $B$ respectively).$2$. Capacitor between plate $2$ and $3$ (connected to $A$ and $B$ respectively).$3$. Capacitor between plate $4$ and $3$ (connected to $A$ and $B$ respectively).$4$. Capacitor between plate $4$ and $5$ (connected to $A$ and $B$ respectively).Analyzing the connections: Plate $2$ is at potential $A$. Plates $1, 3, 5$ are connected to $B$. Plate $4$ is connected to $1$,so it is also at potential $B$.Thus,we have capacitors between $(2,1)$,$(2,3)$,$(4,3)$,and $(4,5)$.Since $1, 3, 5$ are all at potential $B$ and $2$ is at $A$,we have capacitors $C_{21}, C_{23}$ in parallel. Also,$4$ is at $B$,so $C_{43}$ and $C_{45}$ are in parallel.Wait,looking at the diagram: $2$ is connected to $A$. $1, 3, 4, 5$ are connected to $B$. Actually,the plates form capacitors: $C_{21}$ (between $2$ and $1$),$C_{23}$ (between $2$ and $3$),$C_{43}$ (between $4$ and $3$),$C_{45}$ (between $4$ and $5$).Since $1, 3, 5$ are connected to $B$ and $4$ is connected to $1$,all plates $1, 3, 4, 5$ are at potential $B$. Plate $2$ is at $A$.This results in $4$ capacitors in parallel: $C_{eq} = C_{21} + C_{23} + C_{43} + C_{45} = 4C = \frac{4{\varepsilon _0}A}{d}$.

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