Let $m$ rows of $n$ series capacitor be taken then minimum number of capacitors required is




Also effective voltage is

$V^{\prime \prime }=1000=n\times 250$

$\Rightarrow n =\frac{1000}{250}=4$

Also these four capacitor are connected in series then effective capacitance is

$\frac{1}{C\prime \prime }=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{4}{8}$

$\Rightarrow C^{\prime \prime }=2\mathrm{\mu }\mathrm{F}$

$\therefore C^{\prime \prime \prime \prime }= 16=2\times m$

$\Rightarrow m=\frac{16}{2}=8$

Hence $N=m\times n=8\times 4=32$