An oil drop of radius r carrying a charge 'q' remains stationary in the presence of electric field of intensity E. If the density of oil is $ρ$ , then

E = \frac{4}{3} π r^{3} ρ g q

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E = \frac{4}{3} π r^{3} ρ g

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E = \frac{4}{3} π r^{3} ρ g / q

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E = \frac{4}{3} π r^{3} ρ / g^{3}

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Solution

The correct option is C $E = \frac{4}{3} π r^{3} ρ g / q$
As oil drop is stationery under presence of two fields, the electric and the gravitational, the forces must be in equilibrium.
$∴ F_{g} = F_{e}$
$\Rightarrow m g = q E$
$(\frac{4}{3} π r^{3}) p g = q E$
$∴ E = \frac{4}{3} π r^{3} p g / q$

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