Physics - Electrostatics Question with Solution | TestHub

$E=3.65\times 10^4\mathrm{N}/\mathrm{C},r=1\mu \mathrm{m}=10^{-6}\mathrm{m}$ $\begin{array}{rl}{\rho }_{\text{oil }}& =1.26\mathrm{g}/{\mathrm{cm}}^3\\ & =1.26\times 10^3\mathrm{kg}/{\mathrm{m}}^3\end{array}$ Since, droplet is stationary hence weight of droplet $=$ force due to electric field $\Rightarrow \frac{4}{3}\pi r^3\cdot {\rho }_{\text{oil }}\cdot g=qE...(i)$ If $n$ be the number of excess electrons in the oil drop, then $q=ne$ Hence, from Eq. (i), $\frac{4}{3}\pi r^3{\rho }_{\text{oil }}\cdot g=neE$ $\begin{array}{rl}\Rightarrow n& =\frac{4\pi r^3{\rho }_{\text{oill }}g}{3eE}\\ & =\frac{4\times 3.14\times {\left(10^{-6}\right)}^3\times 1.26\times 10^3\times 10}{3\times 1.6\times 10^{-19}\times 3.65\times 10^4}\\ & =9.03\simeq 9\end{array}$