An oil drop of radius r and density ρ is held stationary in a uniform vertically upwards electric field ‘E’. If ${\rho }_o (< \rho )$ is the density of air and e is quanta of charge, then the drop has–

The net downward force on the drop other than the electric force $=\frac{4}{3}{\mathrm{\pi r}}^3\left(\mathrm{\rho }\right)\mathrm{g} -\frac{4}{3}{\mathrm{\pi r}}^3\left({\mathrm{\rho }}_{\mathrm{o}}\right)\mathrm{g} =\frac{4}{3}{\mathrm{\pi r}}^3\left(\mathrm{\rho }-{\mathrm{\rho }}_{\mathrm{o}}\right)\mathrm{g}$ 

For equilibrium, electric force must be upwards i.e. charge on the drop is positive.

$neE=\frac{4}{3}{\mathrm{\pi r}}^3\left(\mathrm{\rho }-{\mathrm{\rho }}_{\mathrm{o}}\right)\mathrm{g}$   i.e.  $n=\frac{4{\mathrm{\pi r}}^3\left(\mathrm{\rho }-{\mathrm{\rho }}_{\mathrm{o}}\right)\mathrm{g}}{3eE}$