Physics - Electrostatics Question with Solution | TestHub

Let us assume that we connect $\mathrm{n}$ (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

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

$n⩾\frac{1000}{300}=3.333...$

${\mathrm{n}}_{\min }=4$

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

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

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

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