Physics

In Milikan's oid 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 $$45^{o}$$ with the vertical. Neglecting buoyant forc due dto the air, what is the viscous force acting on the drop in first case? 


SOLUTION
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 Fv then due to terminal velocity v then we can write

$$mg−qE=Fv$$.....[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

$$2qE=Fv$$.....[2] [Neglecting buoyant force due to the air]

Combining [1] and [2] we get

$$mg−12Fv=Fv$$

$$\Rightarrow Fv=\dfrac{2}{3}mg$$
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