A dust particle of mass 4 × 10^ -12 mg in suspended in air under...

Mass of dust particle, $m=4 \times 10^{-12} \mathrm{mg}$$\begin{aligned} & =4 \times 10^{-12} \times 10^{-3} \mathrm{ g} \\ & =4 \times 10^{-12} \times 10^{-3} \times 10^{-3} \mathrm{ kg} \\ & =4 \times 10^{-18} \mathrm{ kg} \end{aligned}$Electric field, $E=50 \mathrm{ N} / \mathrm{C}$ Weight of dust particle, $W=m g=4 \times 10^{-18} \times 10=4 \times 10^{-17} \mathrm{ N}$ Electric force experienced by dust particle,\begin{aligned} & F_e=q E \\ & F_e=n e \cdot E=n \times 1.6 \times 10^{-19} \times 50 \end{aligned}where, $n$ is the number of electrons removed from neutral dust particle. $\therefore$ At balance condition,\begin{aligned} & \text{ Electric force }=\text{ Weight of dust particle } \\ & n \times 1.6 \times 10^{-19} \times 50=4 \times 10^{-17} \\ & n=\frac{4 \times 10^{-17}}{1.6 \times 10^{-19} \times 50} \\ & =\frac{400}{80}=5 \end{aligned}$$