How to solve the following ?

In Millikan's oil drop experiment on applying a vertically upward electric field an oil drop (of mass m) moves vertically downward with certain terminal speed. On applying double the electric field in horizontal direction the drop moves making #45^@ # with vertical. Neglecting buoyant force due to the air, what is the viscous force acting on the drop in first case ??

1 Answer
Jan 1, 2018

In first case the droplet of mass #m# and charge #q# moves with a constant terminal velocity (say v) under the influence downward gravitational force #mg# and the upward electrical force #q# for the upward electric field #E# acted on it.

If the viscous force exerted by air be #F_v# then due to terminal velocity #v# then we can write

#mg-qE=F_v.....[1]# [Neglecting buoyant force due to the air]

At this state of downard motion of the droplet with terminal velocity #v# if an electric field double in magnitude is applied in horizontal direction on the droplet it is found to move in the direction #45^@# with the vertical. This is possible only if the droplet gains same terminal velocity #v# in horizontal direction. In this case horizontal electrical force should be balanced by the viscous force acting on the droplet in the horizontal direction. So we can write

#2qE=F_v.....[2]# [Neglecting buoyant force due to the air]
Combining [1] and #[2] we get

#mg-1/2F_v=F_v#

#=>F_v=2/3mg#