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# An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55×104NC−1 in Millikans oil drop experiment. The density of the oil is 1.26 g cm−3. Estimate the radius of the drop. [g = 9.8 ms−2; e=1.60×10−19 C].

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#### Excess electrons on an oil drop, n=12Electric field intensity, E=2.55×104NC−1Density of oil, ρ=1.26gm/cm3=1.26×103kg/m3Acceleration due to gravity, g=9.81ms−2Charge on an electron, e=1.6×10−19CRadius of the oil drop =rForce (F) due to electric field E is equal to the weight of the oil drop (W).F=WEq=mgEne=43πr3×ρ×gWhere, q= Net charge on the oil drop = nem= Mass of the oil drop = Volume of the oil drop × Density of oil =43πr3×ρ∴r=3√3Ene4πρg=3√3×2.55×104×12×1.6×10−194×3.14×1.26×103×9.81=3√946.09×10−21=9.82×10−7m=9.82×10−4mmTherefore, the radius of the oil drop is 9.82×10−4mm.

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An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55×104NC1 in Millikans oil drop experiment. The density of the oil is 1.26 g cm3. Estimate the radius of the drop. [g = 9.8 ms2; e=1.60×1019 C].
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An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 104 N C−1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm−3. Estimate the radius of the drop. (g = 9.81 m s−2; e = 1.60 × 10−19 C).

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