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(B) Let the potential of plate $1$ be $V_A$ and plate $3$ be $V_B$. The plates $2$ and $4$ are connected together,let their potential be $V_x$. Plate

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

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(B) Let the potential of plate $1$ be $V_A$ and plate $3$ be $V_B$. The plates $2$ and $4$ are connected together,let their potential be $V_x$. Plate $5$ is isolated.The capacitors are formed between adjacent plates:$C_1$ between $1$ and $2$ with capacitance $C = \frac{\varepsilon_0 A}{d}$.$C_2$ between $2$ and $3$ with capacitance $C = \frac{\varepsilon_0 A}{d}$.$C_3$ between $3$ and $4$ with capacitance $C = \frac{\varepsilon_0 A}{d}$.$C_4$ between $4$ and $5$ with capacitance $C = \frac{\varepsilon_0 A}{d}$.Since plate $5$ is isolated,it does not participate in the circuit. The circuit consists of $C_1$ in series with the parallel combination of $C_2$ and $C_3$.Equivalent capacitance $C_{eq} = \frac{C_1 	imes (C_2 + C_3)}{C_1 + C_2 + C_3} = \frac{C 	imes (2C)}{C + 2C} = \frac{2}{3}C = \frac{2\varepsilon_0 A}{3d}$.The charge flown through the cell is $Q = C_{eq} E = \frac{2\varepsilon_0 AE}{3d}$.

Five conducting parallel plates having area $A$ and separation between them $d$ are placed as shown in the figure. Plate number $2$ and $4$ are connected by a wire,and between point $A$ and $B$,a cell of emf $E$ is connected. The charge flown through the cell is

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