qE=Mg⇒neE=ρ(34πr3)×g
⇒n×1.6×10−19×3.55×105=3×103×34×π×(2×10−3)3×9.81
⇒n=173×10(3−9−5+19)⇒n=1.73×1010.
JEE Main 2021 — Physics Modern Physics
An oil drop of the radius 2mm with a density 3g cm−3 is held stationary under a constant electric field 3.55×105Vm−1 in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess? (considerg=9.81ms−2).
Held on 18 Mar 2021 · Verified 6 Jul 2026.
48.8×1011
1.73×1010
17.3×1010
1.73×1012
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