$\frac{\text{QV}}{\ell }=\frac{4}{3}\pi {\text{r}}^3\rho \text{g}$

When the oil drop is falling freely under the effect of gravity is a viscous medium with terminal speed $v,$ then

$m\mathrm{g}=6\mathrm{\pi }\mathrm{ }\mathrm{\eta } r v$                   ...(i)

To move the oil drop upward with terminal velocity $v$ if $E$ is the electric field intensity applied, the

$Eq=m\mathrm{g}+6\pi \mathrm{\eta }rv=m\mathrm{g}+m\mathrm{g}=2m\mathrm{g}$