A charged oil drop falls with terminal velocity $v_{0}$ in the absence of electric field. An electric field $E$ keeps it stationary. The drop acquires charge $3q$, it starts moving upwards with velocity $v_{0}$. The initial charge on the drop is

Question

Dr. Abubacker · Accepted Answer

(c) When drop is stationary, then
$q_{1}E = 6\pi\ \eta r\ v_{0}$ or $q_{1} = 6\pi\ \eta r\ v_{0}/E$
When drop moves upwards, then
$3q = \frac{6\pi\ \eta r\left( v_{0} + v_{0} ight)}{E} = 2 	imes \left( \frac{6\pi\ \eta rv_{0}}{E} ight) = 2q_{1}$
$	herefore\ \ q_{1} = \frac{3}{2}q$

Q. A charged oil drop falls with terminal velocity $v_{0}$ in the absence of electric field. An electric field $E$ keeps it stationary. The drop acquires charge $3q$, it starts moving upwards with velocity $v_{0}$. The initial charge on the drop is

Explanation

Reviewed by Dr. Abubacker

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