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(B) In the given arrangement, there are four plates. Let the plates be numbered $1, 2, 3,$ and $4$ from top to bottom.Plate $1$ and $4$ are connected

Vedclass · Accepted Answer

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

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(B) In the given arrangement, there are four plates. Let the plates be numbered $1, 2, 3,$ and $4$ from top to bottom.Plate $1$ and $4$ are connected to point $A$.Plate $2$ and $3$ are connected to point $B$.This forms three capacitors in parallel between the plates:$1$. Capacitor $C_1$ between plate $1$ and $2$.$2$. Capacitor $C_2$ between plate $2$ and $3$.$3$. Capacitor $C_3$ between plate $3$ and $4$.However, since plates $2$ and $3$ are both connected to point $B$, they are at the same potential. Thus, the potential difference across the capacitor formed by plates $2$ and $3$ is zero. This capacitor is short-circuited and does not contribute to the equivalent capacitance.The effective capacitors are $C_1$ (between $1$ and $2$) and $C_3$ (between $3$ and $4$).Both are in parallel, so $C_{eq} = C_1 + C_3 = \frac{\varepsilon_0 A}{d} + \frac{\varepsilon_0 A}{d} = \frac{2\varepsilon_0 A}{d}$.

Four metallic plates are arranged as shown in the figure. If the distance between each adjacent plate is $d$ and the area of each plate is $A$, then the equivalent capacitance of the system between points $A$ and $B$ is (Given $d << \sqrt{A}$):

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