An oil drop of 12 excess electrons is held st | vedclass

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Question

(B) For the oil drop to be stationary,the upward electric force must balance the downward gravitational force.$F_e = W$$qE = mg$Given $n = 12$ excess

Vedclass · Accepted Answer

$9.82 	imes 10^{-4} \; mm$.

Vedclass · Answer

(B) For the oil drop to be stationary,the upward electric force must balance the downward gravitational force.$F_e = W$$qE = mg$Given $n = 12$ excess electrons,the total charge $q = ne = 12 	imes 1.60 	imes 10^{-19} \; C = 1.92 	imes 10^{-18} \; C$.The mass $m$ of the drop is given by $m = V ho = \frac{4}{3} \pi r^3 ho$.Substituting these into the equilibrium equation:$neE = \frac{4}{3} \pi r^3 ho g$$r^3 = \frac{3neE}{4 \pi ho g}$Given $ho = 1.26 \; g \, cm^{-3} = 1260 \; kg \, m^{-3}$,$E = 2.55 	imes 10^4 \; N \, C^{-1}$,and $g = 9.81 \; m \, s^{-2}$:$r^3 = \frac{3 	imes (1.92 	imes 10^{-18}) 	imes (2.55 	imes 10^4)}{4 	imes 3.14159 	imes 1260 	imes 9.81}$$r^3 = \frac{1.4688 	imes 10^{-13}}{155167.6} \approx 9.466 	imes 10^{-19} \; m^3$$r = (9.466 	imes 10^{-19})^{1/3} \approx 9.82 	imes 10^{-7} \; m$Converting to $mm$: $r = 9.82 	imes 10^{-7} 	imes 10^3 \; mm = 9.82 	imes 10^{-4} \; mm$.

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