User:Boleszek/Notebook/Physics 307l, Junior Lab, Boleszek/2008/11/03
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The Millikan Oil Drop Experiment
The procedure can be found online at Pasco's website
density of Squib#5597 Mineral Oil given by manual is 886kg/m^3
Time data for drop movement is recorded in seconds, no units are put in below to make data easier to copy and paste into other applications
Lost drop to view
Once we find a candidate and get a few measurements, we will open the ionization source in hopes of getting at least a few of these to change charge.
Plate Voltage: 506V Temperature of chamber: 25.5C
Switched charge, stopped alpha exposure. Up with opposite charge (negative on top plate): 2.58, 2.21, 4.33, 4.70 Appeared to loose charge.
Chamber temp 26.5 Charge on plates: 507V
Exposure begins (under + charge no motion occurs, very stable) lost him
Exposure begins (after a few sec looses charge)
Exposure begins (won't go up, but can increase speed downwards, thus we determine the oil particle must have been too massive to rise under the force of one e-charge)
Chamber temperature: 27.8c Plate voltage: 507V
Exposure begins (charge is quickly reversed)
chamber temperaturte: 28C
premature end to data acquisition
Exposure begins (no change for a while, then noticeable slow down)
Temperature remains at 28°C. at the end of the day the pressure was 29.81 in / 1009.4 hPa (Rising)
Documentation of Charge Calculation Process
Choosing the Best Drops
I will calculate the electron charge from a chosen group of the "best candidates", but first I must clarify our rules for choosing a drop in the first place. Our criteria for choosing a candidate for observation are
From our batch of observed drops we can estimate a few that seem to have the lowest number of excess electrons and the highest number of consistent measurements. How do we do this? Well, thankfully we were not alone in our search for the singly charged drop as Millikan was. In the lab manual we are told that if a particle in free fall covers the distance between the major reticle lines (.5mm) in about 15s it will rise the same distance under the influence of an electric potential in about 15s if it has 1 excess electron, 7s if it has 2 excess electrons, and 3s if it has 3 excess electrons. If we have a drop that takes longer than 15s to fall the .5mm distance under gravity, then it is lighter and thus will probably rise faster, whereas a heavier particle will fall faster, but rise slower than the small one given that both particles had the same charge. Especially useful in calculating charge accurately is the observation of a change in the charge of drops due to alpha particle exposure, because the constant mass of the drop will allow a more accurate determination of the ration of the two charges, so whenever this happened we made sure to make as many measurements as possible.
The Charge Expression
The calculation of charge is algebraically intensive, but otherwise straightforward. The manual provides procedures to calculate the charge in e.s.u and in coulombs, so I choose the more familiar coulombs calculation.
The charge in coulombs is given by the formula:
Now this complicated - looking thing is really not that bad when we realize what each term represents. We know that there are 3 significant forces acting on a drop under an electric potential and only two when in free fall. Since the droplets are so small they reach their terminal velocity very quickly (on the order of milliseconds) so we can set opposing forces equal to each other (indeed when I observed the drops they seemed to move at constant velocity). Then we have two equations with k as the coefficient of air resistance
eliminating k and solving for q we obtain
Now this is a much nicer expression than the big one above, but we do not know the mass of our drop, so we take the fact that it is nearly spherical to multiply its volume by its density (ρ):
But unfortunately we do not know the radius either, so we must use Stoke's Law, relating the radius of a free falling spherical body to its velocity in a medium with coefficient of viscosity η. Since the velocities of the drops are so low a correction factor must be introduced to the original formula. In the end the radius is given by
Which can be plugged back into the expression for m such that
Then we plug this back into the small expression for q and obtain the big one.
So, unfortunately, I will still have to deal with e.s.u! I will have to convert my answer for E into V/m by the conversion factor 1V/m≈3.33×10^−5e.s.u.
FIXED d = 7.755*10^-3m - separation of the plates in the condenser ρ = 886kg/m^3 - density of oil g = 9.8m/s^2 acceleration of gravity b = constant, equal to 8.20 x 10-3 Pa • m p = 30.41 in / 1029.7 hPa for Day 1, 29.815 in / 1009.55 hPa average pressure for Day 2 - barometric pressure in pascals (does vary, but independently of experiment and only slightly over the course of a single day's measurements) EXPERIMENTALLY DEPENDENT a – radius of the drop in m vf – velocity of fall in m/s vr – velocity of rise in m/s V – potential difference across the plates in volts q – charge, in coulombs, carried by the droplet η– viscosity of air in poise ( Ns/m2)It should be noted that I used Wikipedia's page on Hectopascals to understand what hPa meant. I learned that "in everyday life, the pascal is perhaps best known from meteorological barometric pressure reports, where it occurs in the form of hectopascals (1 hPa = 100 Pa)". Also, because the value of η is temperature dependent, the manual provides us with a linear graph of η vs Temp in Appendix A so that determination of η is a matter of simple extrapolation.SJK 16:11, 16 November 2008 (EST)
By understanding which variables and constants depend on what I can make the VERY IMPORTANT CONCLUSION that the first term of the charge expression needs to be calculated only once, the second term must be calculated for each drop, and the third term must be calculated for each observed change in charge. This is how I will employ the charge expression.
I choose 5 of the 14 drops that I feel will be most promising and have the largest number of data points. These are Drops 2, 9, 11, 12, and 14. Some of them change their charge and some don't so the calculations will differ respectively. I do the actual calculations in the MATLAB code page(where calculations are divided into sections for each drop indicated by %% Drop#) but I document each step below, providing values of each relevant parameter along with standard deviations of the mean for values to which error applies (I explain why some values have errors and others don't later on in the Error Propagation section).
All values are presented with ± their stdms when relevant.
rise time (s) : 13.12 17.46 11.27 12.55 11.58
Wow, that's actually close to the expected value!
rise time (s) : 10.33, 9.05, 9.17
This is probably the charge due to two electrons.
After these charge calculations are complete, I order them from least to greatest and find the difference between each consecutive value. If I am lucky, the differences will all be similar numbers, or at least obvious multiples of an underlying value. The average of these differences should yield an approximate value for e (I do not use MATLAB for these simple calculations).
The charges have the following order and differences:
Charge (x10^-19) || Difference || Assumed charge 1) 1.39 || 3-1 = 1.76 || 1e 2) 1.50 || 3-2 = 1.65 || 1e 3) 3.15 || 4-3 = 1.56 || 1e X 3.92 (bad point) 4) 4.71 || 5-4 = 2.99 || 2e so 1e = 2.99/2 = 1.495 5) 7.70 || 6-5 = 0.50 || ? These values might have persuaded the first searchers 6) 8.20 || 7-6 = 2.20 || ? of electron charge to believe in half-charges, but we 7) 10.4 || know better these days, so I must conclude that these X 38.1 (I ignore this point too) values are the results of faulty data.
The 4 values that I choose from the table above are fairly close to each other. I average them and obtain
Since I don't actually have precision up to the third decimal I can drop the 625 and I get the wonderful result:
The stdm of my final value is obtained by averaging the stdms of the charge values that contributed to the calculation (which turn out to be every charge except the second of Drop12 and all of Drop14):
My value is 1% off from the accepted value of 1.60*10^-19 C and the accepted value is well within my margin of experimental error, so I am pleased with my results.
Since each limit differs from the mean by 5.88*10^-20 C (which is 36% of the mean) I can conclude that if I had made enough measurements I might expect that only 5% of them fall above 36% of the mean electron charge.
Error PropagationSJK 16:27, 16 November 2008 (EST)
I must admit that I will be ignoring a large number of error producing factors (such as voltage readings on voltmeter, slight variations in temperature and pressure, measurement of the plastic plate thickness), but I do so under the opinion that these errors contribute little to the total error which is mostly due to our inability to accurately time the travel of the droplets. Since the time measurement errors translate directly to the velocities, the errors of any terms containing velocities will have to be propagated. This results in the fact that, even if I choose to ignore so many possible causes of error, I must still calculate the error of a (radius depends on v_f), m (mass depends on radius), and q (depends on m, v_f, and v_r). Therefore a good deal of error propagation is still performed.
For my calculations I will replace σ with the standard deviation of the mean, represented by Δ.
So I really am dealing with the relative error
At first this conclusion bothered me because I thought of the extra factor of v as a set of values corresponding to the time data points, so I wondered "is there a different Δv for each data point?". But after some thought I realized that this error calculation is done with respect to the mean value of the time, and thus velocity, so the multiplicative factor "v" really is just a number, namely the mean of all the velocities calculated for a given charged drop.
since it follows that
I realize that since the only real "variable" that has errors is the time I could have done just one calculation for q error if I had rewritten the charge expression entirely in terms of t. But that would have resulted in a terribly complicated derivative, so I opted to calculate the error of each variable in q and treat those as the basis from which to propagate towards q. I make one code for these horrible calculations and apply it to each drop by tediously replacing variables on the MATLAB code page.
Retrospective Remarks1.) The manual claims that we can ignore the buoyant force that air has on the droplet since air density is only about one thousandth that of the oil, and though I did not take it into account myself I must say that a few of the drops seemed so light that they would randomly travel upwards when in free fall, so collisions with air molecules or micro air-currents in the chamber may actually be a noticeable source of error. I am tempted to think that this experiment should ideally be conducted in a vacuum, though it would undoubtedly be a great bother to get the micro droplets in for every measurement and retain the vacuum.SJK 16:20, 16 November 2008 (EST)
2.) In order to be more accurate I would have monitored the viscosity of air more consistently throughout the measurement process.
3.) Ideally, we should have made charge calculations throughout the measurement process so that we could begin to get a better idea of what physical effects correspond to which values of charge.
4.) Though the manual tells us to use a 500V potential, I notice that in the example calculation in the back a potential of only 386V is applied. A lower potential corresponds to slower drop speed, which is desirable since a slower object is easier to accurately time. Next time I would have made sure to lower the potential significantly.5.) Having only now read the very interesting historical notes at the end of the manual, I wish I had done so earlier. They explain that Millikan adjusted the potential such that a single moving drop could be suspended by a balance of upward electric force and downward gravitational pull. This method has the effect of greatly simplifying the calculation of q since the air resistance term in the balance of forces equation is gone. Then the only velocity that would have to be found would be v_f for the calculation of radius and the effect of the potential would be completely expressed by the E term in the expression without need for v_r.SJK 16:21, 16 November 2008 (EST)