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(A) The electric field $E$ between the plates is given by $E = \frac{V}{d} = \frac{200}{0.01} = 2 	imes 10^4 \ V/m$.For the oil drop to float,the ele

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$6$

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(A) The electric field $E$ between the plates is given by $E = \frac{V}{d} = \frac{200}{0.01} = 2 	imes 10^4 \ V/m$.For the oil drop to float,the electric force must balance the gravitational force: $qE = mg$.Here,$q = ne$,where $n$ is the number of electrons and $e = 1.6 	imes 10^{-19} \ C$.The mass of the spherical oil drop is $m = ho V_{drop} = ho \left( \frac{4}{3} \pi r^3 ight)$.Substituting the values: $n 	imes (1.6 	imes 10^{-19}) 	imes (2 	imes 10^4) = 900 	imes \frac{4}{3} 	imes 3.14 	imes (8 	imes 10^{-7})^3 	imes 10$.$n 	imes 3.2 	imes 10^{-15} = 1200 	imes 3.14 	imes 512 	imes 10^{-21} 	imes 10$.$n 	imes 3.2 	imes 10^{-15} = 1.93 	imes 10^{-14}$.$n = \frac{1.93 	imes 10^{-14}}{3.2 	imes 10^{-15}} \approx 6.03$.Thus,the number of electrons is $6$.

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