In the figure,the area of each plate is A and | vedclass

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(A) The given system consists of four plates. Let the capacitance of each pair of adjacent plates be $C = \frac{A \varepsilon_0}{d}$.Based on the circ

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

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(A) The given system consists of four plates. Let the capacitance of each pair of adjacent plates be $C = \frac{A \varepsilon_0}{d}$.Based on the circuit diagram,plates $1$ and $3$ are connected together,and plates $2$ and $4$ are connected to terminals $A$ and $B$ respectively.This forms two capacitors in parallel (between plates $1-2$ and $3-2$) which are then in series with a third capacitor (between plates $3-4$).Let $C_1$ be the capacitance between plates $1$ and $2$,$C_2$ between $3$ and $2$,and $C_3$ between $3$ and $4$. Thus,$C_1 = C_2 = C_3 = C$.The parallel combination of $C_1$ and $C_2$ gives $C' = C_1 + C_2 = C + C = 2C$.This $C'$ is in series with $C_3$. The equivalent capacitance $C_{eq}$ is given by:$C_{eq} = \frac{C' \cdot C_3}{C' + C_3} = \frac{(2C) \cdot C}{2C + C} = \frac{2C^2}{3C} = \frac{2}{3}C$.Substituting $C = \frac{A \varepsilon_0}{d}$,we get $C_{eq} = \frac{2}{3} \frac{A \varepsilon_0}{d}$.

In the figure,the area of each plate is $A$ and the distance between consecutive plates is $d$. What is the effective capacitance between points $A$ and $B$?

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