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(B) In the given arrangement,there are $6$ metallic plates. Plates $1, 3, 5$ are connected to point $P$,and plates $2, 4, 6$ are connected to point $Q

Vedclass · Accepted Answer

$5\varepsilon_0 A/d$

Vedclass · Answer

(B) In the given arrangement,there are $6$ metallic plates. Plates $1, 3, 5$ are connected to point $P$,and plates $2, 4, 6$ are connected to point $Q$. This configuration creates $5$ capacitors connected in parallel,where each capacitor is formed between two adjacent plates. The capacitance of each individual capacitor is $C = \frac{\varepsilon_0 A}{d}$. Since there are $5$ such capacitors in parallel,the equivalent capacitance $C_{eq}$ is given by: $C_{eq} = C + C + C + C + C = 5C = \frac{5\varepsilon_0 A}{d}$.

Six metallic plates,each with a surface area $A$ on one side,are placed at a distance $d$ from each other. The alternate plates are connected to points $P$ and $Q$ as shown in the figure. The capacitance of the system is:

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