Four metal plates are arranged as shown. Find | vedclass

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(C) Let the plates be numbered $1, 2, 3, 4$ from top to bottom.Plate $1$ is connected to $X$.Plates $2$ and $4$ are connected together.Plate $3$ is co

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$ \frac{2}{3} \frac{\varepsilon_{0} A}{d} $

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(C) Let the plates be numbered $1, 2, 3, 4$ from top to bottom.Plate $1$ is connected to $X$.Plates $2$ and $4$ are connected together.Plate $3$ is connected to $Y$.There are three capacitors formed between adjacent plates:$C_1$ between plate $1$ and $2$,$C_2$ between plate $2$ and $3$,and $C_3$ between plate $3$ and $4$.Each capacitor has capacitance $C = \frac{\varepsilon_0 A}{d}$.Plate $1$ is at potential $V_X$. Plate $3$ is at potential $V_Y$.Plates $2$ and $4$ are at the same potential,let's call it $V_P$.Capacitor $C_1$ is between $X$ and $P$. Capacitor $C_2$ is between $P$ and $Y$. Capacitor $C_3$ is between $Y$ and $P$.Capacitors $C_2$ and $C_3$ are in parallel between $P$ and $Y$,so their equivalent capacitance is $C_2 + C_3 = 2C$.This combination is in series with $C_1$.The equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{2C} = \frac{1}{C} + \frac{1}{2C} = \frac{3}{2C}$.Therefore,$C_{eq} = \frac{2}{3} C = \frac{2}{3} \frac{\varepsilon_0 A}{d}$.

Four metal plates are arranged as shown. Find the equivalent capacitance between $ X $ and $ Y $. ($ A $ = Area of each plate,$ d $ = distance between adjacent plates)

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