An oil drop of radius $\mathrm{r}$ and density $\rho$ is held stationary in a uniform vertically upwards…
An oil drop of radius $\mathrm{r}$ and density $\rho$ is held stationary in a uniform vertically upwards electric field 'E'. If $\rho_{0}( < \rho)$ is the density of air and e is quanta of charge, then the drop has-
deficiency of $\frac{4 \pi r^{3}\left(\rho-\rho_{0}\right) g}{3 \mathrm{eE}}$ electrons
deficiency of $\frac{4 \pi r^{2}\left(\rho-\rho_{0}\right) g}{\mathrm{eE}}$ electrons
Solution
Net downward force on the drop $=\frac{4}{3} \pi r^{3}\left(\rho-\rho_{0}\right) g$ For equilibrium, electric force must be upwards i.e. charge on the drop is positive. $\mathrm{neE}=\frac{4}{3} \pi \mathrm{r}^{3}\left(\rho-\rho_{0}\right) \mathrm{g}$ i.e. $\mathrm{n}=\frac{4 \pi \mathrm{r}^{3}\left(\rho-\rho_{0}\right) \mathrm{g}}{3 \mathrm{eE}}$
