Physics - Electrostatics Question with Solution | TestHub

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