A capacitance of 2  μF is required in an electrical circuit across a potential difference of 1 . 0…

Let us assume that we connect $n$(whole number) capacitors in series such that the potential difference across the combination is $1000 \mathrm{V}$. This means the potential difference across each capacitor is

$V_{capacitor}=\frac{1000}{n} \mathrm{V}\le 300 \mathrm{V}$

$n⩾\frac{1000}{300}=3.333$.

$n_{min}=4$

This means we need to connect at least $4$ capacitors in series to make sure that the potential across each one is less than $300 \mathrm{V}$.

$C_{eq}=\frac{m}{n}C=2 \mathrm{\mu F}$

$$\begin{gathered} \frac{m}{4}=2 \\ \Rightarrow m=8 \end{gathered}$$

$\therefore$ Minimum no. of capacitors $=8\times 4=32$