In Millikan oil drop experiment, a charged drop of mass $1 .8 \times 10^{- 14}$ kg is 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

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answer is C.

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Detailed Solution

$QE = mg \to Q = \frac{mg}{E} \Rightarrow n = \frac{mgd}{Ve} \Rightarrow n = \frac{1 .8 \times 10^{- 14} \times 10 \times 0 .9 \times 10^{- 2}}{2 \times 10^{3} \times 1 .6 \times 10^{- 19}} = 5$

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