An oil drop of radius 2 , mm with a density 3 | vedclass

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(B) For the oil drop to be stationary,the electric force must balance the gravitational force: $qE = Mg$.Here,$q = ne$,where $n$ is the number of exce

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$1.73 	imes 10^{10}$

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(B) For the oil drop to be stationary,the electric force must balance the gravitational force: $qE = Mg$.Here,$q = ne$,where $n$ is the number of excess electrons and $e = 1.6 	imes 10^{-19} \, C$.The mass $M = 	ext{density} (ho) 	imes 	ext{volume} (V) = ho 	imes \frac{4}{3} \pi r^3$.Given: $ho = 3 \, g \, cm^{-3} = 3000 \, kg \, m^{-3}$,$r = 2 \, mm = 2 	imes 10^{-3} \, m$,$E = 3.55 	imes 10^{5} \, V \, m^{-1}$,$g = 9.81 \, m \, s^{-2}$.Substituting the values: $n 	imes (1.6 	imes 10^{-19}) 	imes (3.55 	imes 10^{5}) = 3000 	imes \frac{4}{3} 	imes \pi 	imes (2 	imes 10^{-3})^3 	imes 9.81$.$n 	imes 5.68 	imes 10^{-14} = 4000 	imes 3.14159 	imes 8 	imes 10^{-9} 	imes 9.81$.$n 	imes 5.68 	imes 10^{-14} = 9.833 	imes 10^{-4}$.$n = \frac{9.833 	imes 10^{-4}}{5.68 	imes 10^{-14}} \approx 1.73 	imes 10^{10}$.

An oil drop of radius $2 \, mm$ with a density $3 \, g \, cm^{-3}$ is held stationary under a constant electric field $3.55 \times 10^{5} \, V \, m^{-1}$ in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess? (Consider $g = 9.81 \, m \, s^{-2}$)

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