Question

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 ??

Answer 1

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`