Millikan Oil Drop: Atom Size & Ions

[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

 

[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.

 

[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.

 

[scientifico]

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

 

[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.

 

[jonlg_uk]

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

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.

 

[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?

 

[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.

 

[scientifico]

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

 

[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

 

[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?

 

[sophiecentaur]

There are pictures of this in textbooks 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.

 

[scientifico]

There are pictures of this in textbooks 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.

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

 

[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]

 

[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??

 

[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).

 

[utban]

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

 

[sophiecentaur]

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

What have you actually read about the Millikan experiment? Do you realize 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.

 

[utban]

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

 

[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.
Try re-stating your question.

 

[utban]

sir I've read this article twice in my course books...now one of my teacherz asked me that when we balance the electrical force and grav.force(weight) the drop stops moving bcoz two forces become equal in magnitude and opposite in direction similarly when we move on and take the concept of drag force,we make drag force and weight(grav.force) equal and opposite in direction.in accordance to above case the drop should stop moving but it does not happen instead drop moves with terminal velocity...why this happens...i repeat that in my point of view drop should stop moving due to cancellation of equal and opposite forces in latter case.kindly help me out.

 

[sophiecentaur]

sir I've read this article twice in my course books...now one of my teacherz asked me that when we balance the electrical force and grav.force(weight) the drop stops moving bcoz two forces become equal in magnitude and opposite in direction similarly when we move on and take the concept of drag force,we make drag force and weight(grav.force) equal and opposite in direction.in accordance to above case the drop should stop moving but it does not happen instead drop moves with terminal velocity...why this happens...i repeat that in my point of view drop should stop moving due to cancellation of equal and opposite forces in latter case.kindly help me out.

As I said in my earlier post: "Do you realize 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?"
When the voltage is applied, the drop is stationary, when the voltage is removed, the drop reaches terminal velocity. There are two halves to the experiment. The weight is established when the drop is falling. This weight force is the same one that is balanced by the field when it is stationary. The two measurements give you the value of the charge.

 

[xAxis]

sir I've read this article twice in my course books...now one of my teacherz asked me that when we balance the electrical force and grav.force(weight) the drop stops moving bcoz two forces become equal in magnitude and opposite in direction similarly when we move on and take the concept of drag force,we make drag force and weight(grav.force) equal and opposite in direction.in accordance to above case the drop should stop moving but it does not happen instead drop moves with terminal velocity...why this happens...i repeat that in my point of view drop should stop moving due to cancellation of equal and opposite forces in latter case.kindly help me out.

When the forces cancel each other, the body will continue to move at the current velocity. This is what terminal velocity is.
Generaly when the forces ballance, it doesn't mean that the object will stop. It just means that it will stop accelerating/decelerating. This is just the manifestation of Newtons first law.

 

[sophiecentaur]

Does no one realize that the experiment is in TWO PARTS? When it is stationary and when it is at terminal velocity, it is in equilibrium.
I suggest reading the Wiki pages on it, rather than arguing at cross purposes.

 

[sophiecentaur]

sir I've read this article twice in my course books...now one of my teacherz asked me that when we balance the electrical force and grav.force(weight) the drop stops moving bcoz two forces become equal in magnitude and opposite in direction similarly when we move on and take the concept of drag force,we make drag force and weight(grav.force) equal and opposite in direction.in accordance to above case the drop should stop moving but it does not happen instead drop moves with terminal velocity...why this happens...i repeat that in my point of view drop should stop moving due to cancellation of equal and opposite forces in latter case.kindly help me out.

Aha! I think I have spotted where your thinking is flawed. When a body falls and the resistance force equals the weight, it does not stop, but has reached its terminal velocity. (Newton's First Law of motion). The drop stops accelerating but doesn't stop moving.