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Q.

In a Millikan's oil drop experiment the charge on an oil drop is calculated to be 6.35 x 10-19 C. The number of excess electrons on the drop is

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a

4.2

b

6

c

3.9

d

4

answer is B.

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Detailed Solution

n=Qe=6.35×10−191.6×10−19≈4

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