There are n electrons of charge e on a drop of oil of density $ρ .$ It is in equilibrium in an electric field E. Then radius of drop is

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${[\frac{2 n e E}{π ρ g}]}^{\frac{1}{3}}$

${[\frac{2 n e E}{4 π g}]}^{\frac{1}{2}}$

${[\frac{neE}{ρ g}]}^{\frac{1}{2}}$

${[\frac{3 n e E}{4 π ρ g}]}^{\frac{1}{3}}$

answer is C.

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Detailed Solution

Given data: n electrons e on a drop of oil of density $ρ .$

Concept used: electric field and electric fields lines

Detailed solution:

Due to the cancelling out of the gravity force and the electric operating on it, the drop is in equilibrium. Electrical power = $E \times n e$

Gravitational force =

$\frac{4}{3} π r^{3} ρ g = \frac{4}{3} π r^{3} ρ g = Ene \Rightarrow r = \sqrt[3]{\frac{3 E n e}{4 π ρ g}}$

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