Five conducting parallel plates having area A | vedclass

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(C) Let the plates be numbered $1, 2, 3, 4, 5$. Point $A$ is connected to plates $1, 3, 5$ and point $B$ is connected to plates $2, 4$. The capacitors

Vedclass · Accepted Answer

$\frac{4\varepsilon_0 AE}{d}$

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(C) Let the plates be numbered $1, 2, 3, 4, 5$. Point $A$ is connected to plates $1, 3, 5$ and point $B$ is connected to plates $2, 4$. The capacitors are formed between adjacent plates: $C_{12}$ is between $1$ and $2$,$C_{23}$ is between $2$ and $3$,$C_{34}$ is between $3$ and $4$,and $C_{45}$ is between $4$ and $5$. Each capacitor has capacitance $C = \frac{\varepsilon_0 A}{d}$. Connecting the plates: Plates $1, 3, 5$ are at potential $V_A$. Plates $2, 4$ are at potential $V_B$. Capacitors are: $C_{12}$ (between $V_A$ and $V_B$),$C_{23}$ (between $V_B$ and $V_A$),$C_{34}$ (between $V_A$ and $V_B$),$C_{45}$ (between $V_B$ and $V_A$). All four capacitors are in parallel between points $A$ and $B$. Equivalent capacitance $C_{eq} = C_{12} + C_{23} + C_{34} + C_{45} = 4 	imes \frac{\varepsilon_0 A}{d} = \frac{4\varepsilon_0 A}{d}$. Charge flown $Q = C_{eq} E = \frac{4\varepsilon_0 AE}{d}$.

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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