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(D) Let the plates be numbered $1, 2, 3, 4, 5$ from top to bottom.Plate $1$ and $4$ are connected together. Plate $3$ and $5$ are connected to termina

Vedclass · Accepted Answer

$\frac{5}{3} \frac{\varepsilon_0 A}{d}$

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(D) Let the plates be numbered $1, 2, 3, 4, 5$ from top to bottom.Plate $1$ and $4$ are connected together. Plate $3$ and $5$ are connected to terminal $Q$. Plate $2$ is connected to terminal $P$.The capacitance of each pair of adjacent plates is $C = \frac{\varepsilon_0 A}{d}$.There are $4$ capacitors formed between the $5$ plates: $C_{12}, C_{23}, C_{34}, C_{45}$.- $C_{12}$ is between plate $1$ (connected to $4$) and plate $2$ $(P)$.- $C_{23}$ is between plate $2$ $(P)$ and plate $3$ $(Q)$.- $C_{34}$ is between plate $3$ $(Q)$ and plate $4$ (connected to $1$).- $C_{45}$ is between plate $4$ (connected to $1$) and plate $5$ $(Q)$.By analyzing the connections,we find that the equivalent circuit consists of capacitors in a combination of series and parallel. The net capacitance is calculated as $C_{	ext{net}} = \frac{5}{3} C = \frac{5 \varepsilon_0 A}{3d}$.Thus,the correct option is $(d)$.

The following arrangement consists of five identical metal plates parallel to each other. The area of each plate is $A$ and the separation between successive plates is $d$. The capacitance between $P$ and $Q$ is

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