# Millikan Experience

1. Jun 13, 2012

### scientifico

Hello, in the Millikan experience have the oil drops vaporized similar dimensions to one atom of oil or they are still too big?
And do they need to be ionized to interact with the electric field of the plates?

Thanks

2. Jun 13, 2012

### sophiecentaur

They need to be small enough so that they fall slowly through the air and they need to have one or more extra electrons (or one or more too few). This charge occurs naturally when the atomiser forms the drops. Timing the rate at which they fall through the air tells you their mass (Stoke's Law) because you know the density of the oil. The value of the applied field needed to suspend one drop tells you the ratio of the charge to mass. It involves hundreds (/more) measurements on individual drops to extract good answers (averages) which will yield the smallest difference in charge between all the drops observed. This smallest difference is e. Brilliant bit of experimenting. Brilliant bloke.

3. Jun 13, 2012

### K^2

Some Millikan Oil Drop setups have an alpha source to "randomize" charges on drops. Makes the experiment significantly easier if you use it right.

4. Jun 14, 2012

### scientifico

If a drop wouldn't have extra or less electrons, won't it interact with the electric field?

5. Jun 14, 2012

### sophiecentaur

It wouldn't interact with a uniform field. Ionising drops with radiation would be better than with an aerosol but there can be safety issues with even pootling quantities of radioactive stuff. Particularly in Schools.

6. Jun 14, 2012

### jonlg_uk

Not quiet sure what you mean, but think of it this way, the more electrons the on the drop the faster it will travel towards a plate of the opposite polarity. In addition to the number of electrons dictating the speed of travel the mass of the oil drop will also have an effect. Milikan was able to calculate the mass of the drop because we know the forces acting on the oil drop when it is falling in air and has reached its terminal velocity vf can be equated to the following:

m*g =vf*k

m=mass of oil drop
g= gravitational force
vf= terminal velocity of the oil drop as it falls
k = frictional coefficient of air

Terminal velocity is reached when the frictional forces of the air k*vf equal the force due to gravity, m*g, Thus we can backwards calculate the number of electrons on a oil drop.

NOTE: That he had to first calculate the mass of the oil drop , before he could calculate the charge and hence the number of electrons on the oil drop. The mass of the oil droplet is calculated WITHOUT an Electric field applied.

7. Jun 14, 2012

### scientifico

For no extra or less electrons I meant a neutral drop (negative charge equilibrated by positive charge)... is the number of electron in a medium drop small or big?

8. Jun 14, 2012

### sophiecentaur

An uncharged object (dust or piece of paper) is attracted to a charged rod because the field is non-uniform and the object becomes polarised then attracted.

9. Jun 14, 2012

### scientifico

You mean when I charge a rod some line of the field have an higher intensity and can polarize a neutral object?

10. Jun 14, 2012

### sophiecentaur

That's how you can pick up dust on a comb. When the dust bits are polarised, there is slightly more attractive force on the near end than repulsive force on the far end. The field needs to be fanning out for it to work

11. Jun 15, 2012

### scientifico

but why an uniform field cannot polarize a piece of paper for example?
maybe because seen that every line have the same intensity, they nullify the work each others?

12. Jun 15, 2012

### sophiecentaur

There are pictures of this in text books IIRC but with a radial field, the density of the lines is higher at the point of contact.
I might ask you if you can think of a better explanation for this???
Remember, the effect is always to produce attraction, whatever the charge on the rod, so the dust must be neutral.

13. Jun 15, 2012

### scientifico

You mean why at contact point the lines are more dense?
I think because the electric field force increase when the distance decrease

14. Jun 15, 2012

### sophiecentaur

'Lines of force' are a way of describing the field. That's all. There's an inverse square law (approx) at work so field gets less with distance.
[Edit: The effect doesn't occur with a uniform field - for instance, between parallel plates because there are erqual and opposite forces on the ends of dust particles]

Last edited: Jun 15, 2012
15. Jul 13, 2012

### utban

hey me wants to know that in milikan experiment oil drop stops when elec.force and weight equals due to equality in magnitude and opposite in direction but drop moves with uniform velocity when drag force and weight becomes equal in magnitude still opposite in direction....why??

16. Jul 13, 2012

### sophiecentaur

What would you expect to happen? Do you know about 'terminal velocity'? Look it up and you can see how that will happen to an oil drop just the same as a free fall sky diver. (In fact it is the limit for all objects falling in a medium - given enough time).

17. Jul 15, 2012

### utban

i know abt terminal velocity but couldn't understand what ur ans. is

18. Jul 16, 2012

### sophiecentaur

What have you actually read about the Millikan experiment? Do you realise that it consists of two halves - one to establish the mass of the drop and one to find the voltage needed to cause the drop to be held stationary?
The terminal velocity calculation works using Stokes law and relies on knowing the density of the oil being used and the viscosity of the air (both of these are easy to find). If you can measure the terminal velocity of the drop then you can tell its weight. You start off by applying a particular voltage across the plates and wait for heavy and light drops to disperse, adjusting the volts to get just a small number of drops stationary. You then turn the volts off and time their rate of fall. Between the two, there is enough information to deduce the charge on the drop. This will be e, 2e or 3e. You have to measure hundreds of different drops to eliminate significant errors and to find the lowest common multiple - that is also a clever process.

19. Jul 16, 2012

### utban

now wht is the relevency of this ans.with the one i asked sir....

20. Jul 16, 2012

### sophiecentaur

I'm, afraid I can't see what your problem is. I have told you the basic principles of the Millikan experiments and pointed out how Stokes Law is used. What more can I do? Have you actually read about this anywhere else? Perhaps you could put a little more into this rather than relying on me for all of the help.