Physics
Electricity and Potential
The Millikan Oil Drop (Please refer to text section 17-1 )
Ninety-five years ago, Robert A. Millikan performed a famous experiment that resulted in the following two conclusions:
1. That charge is quantized. (You have to define this from class-time.)
2. In addition, the experiment determined that the electron's charge is 1.6 x 10-19 C.
You haven’t completed the assignment below until you have shown why the above two conclusions are proven by the experiment. You must do this in your own words as the conclusion.
An oversimplified experiment a lot like the one performed by Millikan is described in the dramatization below. You can solve the problem easily and reach the same conclusions.
CHARGED PLATE
Suspended oil drop: mass m; charge q
OPPOSITELY CHARGE PLATE
A tiny droplet of oil is suspended in the space between two equal, but oppositely charged plates. The oil drop floats motionless even though gravity is acting. It can do this because the drop itself carries a tiny amount of charge. The plates have been charged so that there is a potential difference of 5000 V between them. The plates are only spaced 5.0 cm apart from each other. The tiny oil drop’s mass is 8.2 x 10-12 g.
Answer each of the following, with the correct number of significant digits.
Also g must be 9.8 m/s2 for this, as if this were an experiment. The two significant figures in the value of g is crucial for the conclusion. Rounding to 10 m/s2 yields no conclusion.
Why do students think it’s tolerable to write non-thorough, sketchy math in the margins of this paper, and pass that off as defended solution? Won’t earn A’s.
I. Assume that the excess charges on the oil drop are electrons. If that is true, and the electric force is the kind that can act opposite to gravity, explain which plate is positive and which plate is negative.
II. What is the direction of the electric field between the plates?
III. What is the magnitude of the electric field between the plates? (You can’t use F=qE to do this, since you don’t know the charge on the oil. Use other information given.)
IV. Use the fact that the oil drop is in equilibrium to find the charge on the drop in C. (Hint: Now that you have the E field, you can use F=qE to solve for charge, provided you are a physicist and can figure out what the force value is equal to.
At this point, if you have gotten this far and never did an FBD, you have messed up.
V. Convert the charge's value to elementary charge units, e. You can easily look up the conversion from C’s to e’s in the book.
VI. "Millikan" found that some other oil drops were held in equilibrium when the potential difference was set to exactly 25,000 V. This field voltage is ____ times what it used to be. If these other oil drops were held motionless in this new field what does that tell you about their charge? Solve for this charge, and express the answer in units of e. (Here assume all oil drops here have the same mass. (Yeah, right!))
VII. Here is the amazing result. No higher voltage values above 25,000 V would suspend any particle whatsoever. It was absolutely impossible to suspend a particle of mass 8.2 x 10-12 g, if the voltage was greater than 25,000 V. What does this tell you about the charge value that it is impossible to have? (Think: when the voltage is greater, the charge is needs to be ______to keep the force the same. (Why do I know the force has to be the same? ______)
VIII. "Millikan" tried setting the potential to 10,000 V and looked for oil drops that would float motionless with this potential value. What would the charge need to be under these circumstances? Again convert to units of e.
If you’ve done the above problems without rounding your answers to two and only two digits, then you’ve violated the rules of significant figures and more importantly, you will be unable to find any significant conclusions that result from your answers. By thinking about the meaning of your three charge answers that you solved for in V, VI, and VIII, you can arrive at the exciting conclusion that Millikan discovered here in the great state of California.
IX. Another amazing result: "Millikan" could find absolutely no drops with the given mass that would float motionless when the potential was 10,000 V. This means that the charge value that corresponds to 10,000 V was a charge value that Millikan was unable to find for any of the oil drops. Use the answer VIII to explain why. (Your explanation was one of his conclusions and the reason he won the Nobel Prize.)
Don’t forget the two-part conclusion.
Hints: According the calculations you did, what was the smallest answer possible for charge? Answer this in e units for it to make sense. Additionally, what was significant about the charge value that was impossible to detect?
Guess what special particle’s charge value this experiment essentially discovered.