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A tiny spherical oil drop of radius r carrying a net charge q is balanced in still air with a vertical uniform electric field of strength E. When the field is switched off, the drop is observed to fall with terminal velocity v , viscosity of the air $= η$ and the density of oil = d. The magnitude of q is

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$q = \frac{6 πηrv}{E}$

$q = \frac{πηrv}{6 E}$

$q = \frac{6 ηrv}{πE}$

$q = \frac{6 πηr}{Ev}$

answer is D.

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Detailed Solution

In equilibrium, mg = qE ----(1)
In the absence of electric field, $mg = 6 πηrv$ ----(2)
$\begin{array}{l} qE = 6 πηrv \\ \Rightarrow q = \frac{6 πηrv}{E} \end{array}$

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