Five identical metal plates, each of area A, | vedclass

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(C) Let $C = \frac{{\varepsilon _0}A}{d}$ be the capacitance of a single pair of adjacent plates.From the diagram, plate $2$ is connected to terminal

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

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(C) Let $C = \frac{{\varepsilon _0}A}{d}$ be the capacitance of a single pair of adjacent plates.From the diagram, plate $2$ is connected to terminal $A$. Plates $3$ and $5$ are connected to terminal $B$. Since plates $1$ and $4$ are connected together, they form an isolated node.Let the potential of $A$ be $V_A$ and $B$ be $V_B$. Let the potential of the node $(1, 4)$ be $V_x$.The capacitors are formed between adjacent plates: $C_{12}, C_{23}, C_{34}, C_{45}$.$C_{12}$ is between $V_x$ and $V_A$.$C_{23}$ is between $V_A$ and $V_B$.$C_{34}$ is between $V_B$ and $V_x$.$C_{45}$ is between $V_x$ and $V_B$.Node $V_x$ has capacitors $C_{12}, C_{34}, C_{45}$ connected to it. Since $C_{34}$ and $C_{45}$ are in parallel (both between $V_x$ and $V_B$), their equivalent is $2C$. This $2C$ is in series with $C_{12}$ (between $V_x$ and $V_A$).The equivalent capacitance of the branch containing $V_x$ is $C' = \frac{C \cdot 2C}{C + 2C} = \frac{2}{3}C$.This branch is in parallel with $C_{23}$ (between $V_A$ and $V_B$).$C_{eq} = C + \frac{2}{3}C = \frac{5}{3}C = \frac{5{\varepsilon _0}A}{3d}$.

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