The intensity of the electric field required to keep a water drop of radius 10−5 cm just suspended in air when charged with one electron is approximately g=10 newton /kg,e=1.6×10−19 coulomb

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260 volt /cm

260 newton/coulomb

130 volt /cm

130 newton/coulomb

answer is B.

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Detailed Solution

For balance mg=eE⇒E=mgeAlso m=43πr3d=43×227×10−73×1000kg⇒E=4/3×22/7×10−73×1000×101.6×10−19=260 N/C

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