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(B) To withstand a potential difference of $1000 \,V$ using capacitors that can only withstand $400 \,V$,we need to connect at least $n$ capacitors in

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$18$

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(B) To withstand a potential difference of $1000 \,V$ using capacitors that can only withstand $400 \,V$,we need to connect at least $n$ capacitors in series such that $n 	imes 400 \,V \geq 1000 \,V$. Thus,$n \geq 2.5$,so we must use at least $n = 3$ capacitors in series. The equivalent capacitance of $3$ capacitors of $2 \,\mu F$ each in series is $C_s = \frac{2 \,\mu F}{3}$. We need an equivalent capacitance of $4 \,\mu F$. Let $m$ be the number of such series branches connected in parallel. Then,$m 	imes C_s = 4 \,\mu F \Rightarrow m 	imes \frac{2}{3} \,\mu F = 4 \,\mu F \Rightarrow m = \frac{4 	imes 3}{2} = 6$. Total number of capacitors = (number of branches in parallel) $	imes$ (number of capacitors in series) = $6 	imes 3 = 18$.

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