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Question

An oil drop of radius 2 mm with a density 3 g cm3 is held stationary under a constant electric field 3.55×105 V m1 in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess? (Consider g=9.81 m s2)

A
1.73×1010
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B
48.8×1011
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C
1.73×1012
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D
17.3×1010
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Solution

The correct option is A 1.73×1010
Suppose the number of excess electrons is n.

Force due to electric field, Fe=qE=(ne)E
Weight of the oil drop, W=mg

Since drop is stationary, Fe=W neE=mg
n=mgeE=ρ×43πR3×geE

n=(3×1000)×43×3.14×(2×103)3×9.81.6×1019×3.55×105

n=984704×1055.68=1.73×1010

n=1.73×1010

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