Q: An oil drop carrying a charge of 2 electrons has a mass of 3.2 × 10-17 kg. It is falling freely in air with terminal speed. The electric field required to make the drop move upwards with the same speed is
(a) 2 × 103 V/m
(b) 4 × 103 V/m
(c) 3 × 103 V/m
(d) 8 × 103 V/m
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Ans: (a)
Sol: Without electric field when oil drop is moving downwards,
mg = 6 π η r v …(i)
Under electric field of strength E , oil drop is to move upwards
F – mg = 6 π η r v
F – mg = mg …using (i)
(2 e)E = 2 mg
E = mg/e
$\displaystyle E = \frac{3.2 \times 10^{-17}\times 10}{1.6 \times 10^{-19}} $
= 2 × 103 V /m