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Question: In Millikan oil drop experiment, a charged drop of mass \(1.8 \times 10^{- 14}kg\)is stationary betw...

In Millikan oil drop experiment, a charged drop of mass 1.8×10−14kg1.8 \times 10^{- 14}kgis stationary between its plates. The distance between its plates is 0.90 cm and potential difference is 2.0 kilo volts. The number of electrons on the drop is.

A

500

B

50

C

5

D

0

Answer

5

Explanation

Solution

QE=mg⇒Q=mgE⇒n=mgdVeQE = mg \Rightarrow Q = \frac{mg}{E} \Rightarrow n = \frac{mgd}{Ve}

⇒n=1.8×10−14×10×0.9×10−22×103×1.6×10−19=5\Rightarrow n = \frac{1.8 \times 10^{- 14} \times 10 \times 0.9 \times 10^{- 2}}{2 \times 10^{3} \times 1.6 \times 10^{- 19}} = 5