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(B) Let the potential of the positive terminal of the battery be $V$ and the negative terminal be $0$.Plates $1, 3,$ and $5$ are connected to the posi

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$\frac{{{\varepsilon _0}AV}}{d},\frac{{ - 2{\varepsilon _0}AV}}{d}$

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(B) Let the potential of the positive terminal of the battery be $V$ and the negative terminal be $0$.Plates $1, 3,$ and $5$ are connected to the positive terminal,so their potential is $V$.Plates $2$ and $4$ are connected to the negative terminal,so their potential is $0$.This arrangement forms capacitors between adjacent plates.Capacitor $C_1$ is formed by plates $1$ and $2$,$C_2$ by plates $2$ and $3$,$C_3$ by plates $3$ and $4$,and $C_4$ by plates $4$ and $5$.Each capacitor has capacitance $C = \frac{{{\varepsilon _0}A}}{d}$.Plate $1$ has only one face forming a capacitor with plate $2$. The charge on plate $1$ is $Q_1 = C 	imes (V - 0) = \frac{{{\varepsilon _0}AV}}{d}$.Plate $4$ has two faces: one forming a capacitor with plate $3$ and another with plate $5$. Both plates $3$ and $5$ are at potential $V$,while plate $4$ is at potential $0$.The charge on plate $4$ is $Q_4 = C 	imes (0 - V) + C 	imes (0 - V) = -\frac{{{\varepsilon _0}AV}}{d} - \frac{{{\varepsilon _0}AV}}{d} = -\frac{{2{\varepsilon _0}AV}}{d}$.

In the given system,the area of each plate is $A$ and the distance between two consecutive plates is $d$. What is the charge on plates $1$ and $4$?

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