In a Millikan's oil drop experiment,the charg | vedclass

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(B) According to the quantization of electric charge,the total charge $Q$ on an object is given by $Q = ne$,where $n$ is an integer representing the n

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$4$

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(B) According to the quantization of electric charge,the total charge $Q$ on an object is given by $Q = ne$,where $n$ is an integer representing the number of excess electrons and $e$ is the elementary charge $(e \approx 1.6 	imes 10^{-19} \ C)$.Given $Q = 6.35 	imes 10^{-19} \ C$.We need to find $n = \frac{Q}{e}$.Substituting the values: $n = \frac{6.35 	imes 10^{-19}}{1.6 	imes 10^{-19}} = \frac{6.35}{1.6} \approx 3.968$.Since the number of electrons $n$ must be an integer,we round $3.968$ to the nearest integer,which is $4$.Therefore,the number of excess electrons on the drop is $4$.

In a Millikan's oil drop experiment,the charge on an oil drop is calculated to be $6.35 \times 10^{-19} \ C$. The number of excess electrons on the drop is:

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