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(C) The given circuit consists of $4$ capacitors connected in parallel. Each capacitor is formed by two adjacent plates with area $A$ and separation $

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$\frac{\varepsilon_0 AV}{d}, \frac{-2\varepsilon_0 AV}{d}$

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(C) The given circuit consists of $4$ capacitors connected in parallel. Each capacitor is formed by two adjacent plates with area $A$ and separation $d$. The capacitance of each capacitor is $C = \frac{\varepsilon_0 A}{d}$.Plate $1$ is connected to the positive terminal of the battery. It forms one capacitor with plate $2$. The charge on plate $1$ is $q_1 = +CV = +\frac{\varepsilon_0 AV}{d}$.Plate $4$ is connected to the negative terminal of the battery. It acts as one plate for two capacitors: one with plate $3$ and one with plate $5$. Since plate $4$ is connected to the negative terminal,the charge on both sides of plate $4$ will be negative. The total charge on plate $4$ is $q_4 = -CV - CV = -2CV = -\frac{2\varepsilon_0 AV}{d}$.Thus,the charges on plates $1$ and $4$ are $\frac{\varepsilon_0 AV}{d}$ and $-\frac{2\varepsilon_0 AV}{d}$ respectively.

Five identical plates each of area $A$ are joined as shown in the figure. The distance between the plates is $d$. The plates are connected to a potential difference of $V \text{ volts}$. The charge on plates $1$ and $4$ will be

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