An oil drop of radius 1 μ m is held stationary under a constant...

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