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(B) Let the four plates be numbered $1, 2, 3, 4$ from top to bottom. The outer plates $1$ and $4$ are connected together. Points $A$ and $B$ are conne

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

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(B) Let the four plates be numbered $1, 2, 3, 4$ from top to bottom. The outer plates $1$ and $4$ are connected together. Points $A$ and $B$ are connected to plates $2$ and $3$ respectively.Between plates $1$ and $2$, there is a capacitor $C_1 = \frac{\varepsilon_0 S}{d}$.Between plates $2$ and $3$, there is a capacitor $C_2 = \frac{\varepsilon_0 S}{d}$.Between plates $3$ and $4$, there is a capacitor $C_3 = \frac{\varepsilon_0 S}{d}$.Since plates $1$ and $4$ are connected, they are at the same potential. Let this potential be $V_0$. Plate $2$ is at potential $V_A$ and plate $3$ is at potential $V_B$.The capacitor $C_1$ is connected between $V_A$ and $V_0$. The capacitor $C_3$ is connected between $V_B$ and $V_0$. The capacitor $C_2$ is connected between $V_A$ and $V_B$.This is equivalent to two capacitors $C_1$ and $C_3$ in series, which are then in parallel with $C_2$. Since $C_1 = C_2 = C_3 = C = \frac{\varepsilon_0 S}{d}$, the equivalent capacitance $C_{eq}$ is:$C_{eq} = C_2 + (C_1 	ext{ in series with } C_3)$$C_{eq} = C + \frac{C 	imes C}{C + C} = C + \frac{C}{2} = \frac{3C}{2}$Substituting $C = \frac{\varepsilon_0 S}{d}$, we get:$C_{eq} = \frac{3}{2} \frac{\varepsilon_0 S}{d}$

Four identical metal plates are located in air at equal distance $d$ from each other. The area of each plate is $S$. If the outer most plates are connected by a conducting wire as shown in the figure, the capacitance between points $A$ and $B$ will be

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