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