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(B) In the given arrangement,the plates are connected such that they form capacitors in parallel.There are $n = 7$ plates,which form $n - 1 = 6$ capac

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$\frac{6 \varepsilon_0 A}{d}$

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(B) In the given arrangement,the plates are connected such that they form capacitors in parallel.There are $n = 7$ plates,which form $n - 1 = 6$ capacitors.Each capacitor has an area $A$ and separation $d$,so the capacitance of each is $C = \frac{\varepsilon_0 A}{d}$.Since all $6$ capacitors are connected in parallel between points $P$ and $Q$,the equivalent capacitance is:$C_{	ext{net}} = C_1 + C_2 + C_3 + C_4 + C_5 + C_6 = 6C$$C_{	ext{net}} = \frac{6 \varepsilon_0 A}{d}$

Seven identical plates each of area $A$ and successive separation $d$ are arranged as shown in the figure. The effective capacitance of the system between $P$ and $Q$ is

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