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(B) Let the capacitance of each parallel plate capacitor be $C = \frac{A\varepsilon_0}{d}$.From the circuit diagram,we can identify the arrangement of

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$\frac{3A\varepsilon _0}{5d}$

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(B) Let the capacitance of each parallel plate capacitor be $C = \frac{A\varepsilon_0}{d}$.From the circuit diagram,we can identify the arrangement of capacitors.There are four capacitors in total.Between $A$ and $C$,there is one capacitor of capacitance $C$.Between $C$ and $B$,there is one capacitor of capacitance $C$.Between $A$ and $C$ (via $D$),there are two capacitors in series,each of capacitance $C$,forming a branch with equivalent capacitance $\frac{C}{2}$.This branch is in parallel with the first capacitor between $A$ and $C$.So,the equivalent capacitance between $A$ and $C$ is $C_{AC} = C + \frac{C}{2} = \frac{3C}{2}$.Now,$C_{AC}$ is in series with the capacitor between $C$ and $B$.Therefore,the total equivalent capacitance $C_{eq}$ is given by:$\frac{1}{C_{eq}} = \frac{1}{C_{AC}} + \frac{1}{C} = \frac{1}{3C/2} + \frac{1}{C} = \frac{2}{3C} + \frac{1}{C} = \frac{2+3}{3C} = \frac{5}{3C}$.Thus,$C_{eq} = \frac{3C}{5} = \frac{3A\varepsilon_0}{5d}$.

Find the equivalent capacitance between $A$ and $B$.

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