# Stokes' law experiment

1. Oct 2, 2010

### ananthu

I am not able to understand the following situations.

In stokes' experiment a tiny lead shot falls freely under gravity in a highly viscous column of liquid. When the viscous force becomes equal to the net weight of the lead shot it is said that the shot moves down with a constant velocity known as terminal velocity. The explanation given is that the net force on the shot becomes zero at this point as the upward acting viscous force on the shot cancels the net down ward force due to the weight of the shot. That is the acceleration of the lead shot becomes zero which means the increase in velocity of the shot becomes zero. So the lead shot, instead of hanging at rest at that point itself, actually moves down with a constant terminal velocity.

Now think of this situation. In Millikan's oil drop method,an oil drop is initially made to fall under gravity in between two plates through air.Here also the oil drop attains the constant terminal velocity under gravity when the net force on the oil drop is zero.

But when you apply a suitable upward acting electric field on the charged oil drop, you can alter the field in such a way that the drop either moves up or stand still in air. When the drop is made to stand still without moving either up or down, again we take that the net force acting on the drop is zero.

Here, we take the net weight of the drop is equal to the force due to the electric field,ie. (mg = Eq) while solving problems.

My doubt is this:

In the previous case also the net force on the lead shot is zero but the shot did not stop and stay still but continued to move downwards with a constant velocity where as in the second case how is that the charged oil drop could be made to stand still by making the net force acting on it zero?

Then which statement is true, ie. when the net force on a freely falling body through a viscous medium becomes zero, it continues to move down but with a constant velocity (first case),
or when the net force on a freely falling body through a viscous medium becomes zero,
it just stops and stands still there itself (second case)?

2. Oct 2, 2010

### Born2bwire

Drag. As an object moves through a fluid, the force exerted by the fluid on the object (and obviously vice-versa) is drag and this is dependent upon the speed of the object. So as the shot falls through the oil, it accelerates and increases its speed. But as its speed increases the force of drag also increases. Eventually they come to a point where they cancel each other out and reach terminal velocity. Since there is no net force, the shot keeps descending on its merry way at the speed necessary to create the requisite amount of drag force.

In the Millikan oil-drop experiment, terminal velocity is probably not considered. This is because the viscosity of air is very low compared to that of the oil and the oil drops would not have been able to fall long enough to achieve terminal velocity. The oil drops are basically sprayed into the electric field, so they do not have an appreciable amount of time to be acted upon by gravity before being taken up by the electric field. So with judicious positioning of the spray, you can ensure that a reasonable number of the oil drops will enter the electric field with charges and without any vertical velocity. Once in the field, they will experience zero net force in the vertical direction and thus will not gain any vertical velocity and remain at the same vertical position (we can assume that we can make the field's area large enough that the drag on the oil drop will eventually stop it from drifting out of the field in the horizontal direction).

3. Oct 4, 2010

### ananthu

But, we are using actually using the terminal velocities in the Millikan's oil drop method while calculating the charge of the drop.

charge on the oil drop is given by Q= 1/E n3/2(v=v1{9v/2(p-s)g} where n= coefficient of viscosity of air, v and v1 are the terminal velocities in the absence and presence of electric field respectively, p = the density of the oil and the s = density of air.

4. Oct 4, 2010