An oil drop carrying a charge of 2 electrons has a mass of $3.2\times {10}^{-17}\text{}kg.$  If is falling freely in air with terminal speed. The electric field required to make the drop move upwards with the same speed is

Without electric field when oil drop is moving downwards,

$mg=6\pi \eta r\upsilon$

Under el. Field of strength E, oil drop is to move upwards

$\therefore$     $F_e-mg=6\pi \eta r\upsilon =mg$                            …using$\left(i\right)$

$\left(2e\right)E=2mg$

        $E=\frac{mg}{e}=\frac{3.2\times {10}^{-17}\times 10}{1.6\times {10}^{-19}}$

            $=2\times {10}^{3}V{m}^{-1}$