An oil drop of radius 2 mm with density 3 g $c m^{- 3}$ is held stationary under a constant electric field $3.55 \times 10^{5} V m^{- 1}$ in the Millikan’s oil 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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$1.73 \times 10^{10}$

$4.8 \times 10^{11}$

$17.3 \times 10^{10}$

$17.3 \times 10^{12}$

answer is A.

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Detailed Solution

$E l e c t r i c f o r c e i s b a l a n c e d b y g r a v i t a t i o n a l f o r c e . F_{E} = F_{G} \Rightarrow n e E = m g \Rightarrow n = \frac{m g}{e E} = \frac{ρ V \times g}{e E} = \frac{3000 \times \frac{4}{3} π {(2 \times 10^{- 3})}^{3} \times 9.81}{1.6 \times 10^{- 19} \times 3.55 \times 10^{5}} \Rightarrow n = 1.73 \times 10^{10}$