# User:Boleszek/Notebook/Physics 307l, Junior Lab, Boleszek/2008/11/03

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## Procedure

Equipment setup

1. Measuring the thickness of the plastic between the plates. D for capacitor: 7.75mm, 7.76mm on other side, so 7.755mm on average.
2. Cleaned capacitor with dry paper towel, took care to remove any particles
3. Placed the focusing wire in place, connected light and focused on the wire
4. Adjusted the light to best illuminate the wire exactly as explained in manual. Removed the focusing wire.
5. There are 5 small divisions per large division in the viewer grid
6. Attached 500 volt power supply.

The procedure can be found online at Pasco's website

density of Squib#5597 Mineral Oil given by manual is 886kg/m^3

Measurements
Time measurements are between two major reticle lines (.5mm)
Initial Trials – We took a bit of time to simply play with the tools to be sure we understood what we were doing.
Barometric pressure: 30.41 in / 1029.7 hPa (from http://www.wunderground.com/US/NM/Albuquerque.html)
Measured 500.8 V on the capacitor
Temperature within the chamber 26C
Added droplets with a single squeeze of the atomizer. Immediately saw drops within the viewer. Picked a small, slow moving one and timed it.
Picked a drop, Setting top plate positive, it went up

## Data Acquisition

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
Data is also recorded sequentially instead of in a table so that precise time of the temperature readings can be seen

Procedure is for one person to watch the drop and control the apparatus, calling out to the next person who has the stop watch. The person with the stop watch records the time in the "lab book". There is some inherent human error in this method, but it seems to be the best we can do with the tools on hand. A good stop watch that records consecutive times would be very helpful to improve accuracy and to speed up acquisition process. SJK 15:46, 16 November 2008 (EST)
15:46, 16 November 2008 (EST)
people found java applets that could do this and seemed to work well (even allowing a single individual to record times w/o looking at the computer)

It is very tiring watching these drops. Setting the apparatus to a comfortable height is essential as this is a lengthy process and discomfort inhibits good, patient data acquisition. A better shield around the eye piece would also help. Or better yet, one of those CCD adapters that can project the image to a computer screen.

### Day One

Drop 1

• Up under positive charge on top plate: 4.33, 4.10, 4.27
• Down with both plates grounded: 17.56, 19.24
• Down with charge on bottom plate:

Lost drop to view

Drop 2

• Up under positive charge on top plate: 13.12 17.46 11.27 12.55 11.58
• Down with both plates grounded: 18.17 19.46 19.86
• Down with charge on bottom plate:

Drop 3

• Up under positive charge on top plate: 8.62, 7.36, 7.55
• Down with plates grounded: 23.08, 26.5, 26.98
• Down with negative charge on top plate:

Measured temperature: 27C

Drop 4

• Up under positive charge on top plate: 2.24s, 2.56s, 2.37s
• Down with plates grounded: 7.17s, 6.46s, 6.81s
• Down with negative charge on top plate:

Tried for about 10 minutes to change the charge on this drop. No joy.

### Day Two

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.

• Barometric Pressure: 29.82 in / 1009.7 hPa (Falling)
• initial temp.: 24C
• plate voltage: 502V

Drop 5

• Up under positive charge on top plate: 1.87, 17.42 (behavior has changed)
• Down with both plates grounded: 18.23, 15.27, 21.40

Drop 6

• Up under positive charge on top plate:4.90,4.68, 4.12, 4.42, 5.11
• Down with both plates grounded: 28.72

Plate Voltage: 506V Temperature of chamber: 25.5C

Drop 7

• Up under positive charge on top plate: 4.81, 4.65, 5.09, 4.73, 5.06
• Down with both plates grounded: 48.93, 54.83, 55.55

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.

Drop 8

• Up under positive charge on top plate: 3.09, 3.09, 3.21
• Down with both plates grounded: 93.81

Lost it

Chamber temp 26.5 Charge on plates: 507V

Drop 9

• Up under positive charge on top plate: 10.33, 9.05, 9.17
• Down with both plates grounded: 10.14, 10.18, 10.05

Exposure begins (under + charge no motion occurs, very stable) lost him

Drop 10

• Up under positive charge on top plate: 19.02, 18.11, 16.71
• Down with both plates grounded: 6.09, 5.88, 5.56,

Exposure begins (after a few sec looses charge)

• Up under positive charge on top plate:
• Down with both plates grounded: 5.52

lost him

Drop 11

• Up under positive charge on top plate: 22.67, 5.34 (change in charge noticed), 5.70, 5.36
• Down with both plates grounded: 5.31, 5.90, 5.93, 5.83

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

Drop 12

• Up under positive charge on top plate: 8.93, 8.84, 4.12, 3.48
• Down with both plates grounded: 22.00, 22.34, 25.62

Exposure begins (charge is quickly reversed)

• Up under negative charge on top plate: 3.68, 3.50, 3.65, 3.43, 1.15, 1.15, 1.30, 1.21, 1.36
• Down with both plates grounded: 22.70, 20.45

chamber temperaturte: 28C

Voltage: 506.5V

Drop 13

• Up under positive charge on top plate: 2.58, 2.45, 2.98
• Down with both plates grounded: 17.34,

premature end to data acquisition

Drop 14

• Up under positive charge on top plate: 1.81, 2.00, 1.58
• Down with both plates grounded: 7.08, 7.63, 7.75

Exposure begins (no change for a while, then noticeable slow down)

• Up under positive charge on top plate: 15.53, 20.33, 15.95, 17.81, 19.33
• Down with both plates grounded: 7.46, 7.87

Second Exposure

• Up under positive charge on top plate: 13.27, 1.15, 1.18, 1.21
• Down with both plates grounded: 6.59, 7.43

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

1. not so small that the drop floats on air, but not so massive that weight pulls down despite an electric force pushing upwards.
2. smallest amount of charge possible, which corresponds to slow upwards movement

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:

$q = \frac{4}{3}\pi \rho g \left[ \sqrt{\left( \frac{b}{2p}\right)^2 + \frac{9 \eta v_f}{2 g \rho}}-\frac{b}{2p}\right]^3 \frac{v_f + v_r}{Ev_f}$

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

$m g = k v_f \,$

and

$E q = m g + k v_r \,$

eliminating k and solving for q we obtain

$q = \frac{mg(v_f + v_r)}{E v_f}$

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):

$m= \frac{4}{3} \pi a^3 \rho$

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

$a = \sqrt{\left( \frac{b}{2p}\right)^2 + \frac{9 \eta v_f}{2 g \rho}}-\frac{b}{2p}$

Which can be plugged back into the expression for m such that

$m= \frac{4}{3} \pi \left[ \sqrt{\left( \frac{b}{2p}\right)^2 + \frac{9 \eta v_f}{2 g \rho}}-\frac{b}{2p} \right]^3 \rho$

Then we plug this back into the small expression for q and obtain the big one.
However, one complication remains...There is an E in the expression for q, but how do we know E without knowing the capacitance or at least the surface area of the parallel plates? The manual provides the relation

$E(e.s.u) = \frac{V(volts)}{300d(cm)}$

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.

It is obvious that there are many variables and constants I have to deal with here, so I find it helpful to make a table of the constants that are fixed and those that depend on the measurements.

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)
16:11, 16 November 2008 (EST)
I remember last year when looking over a student's millikan lab, I started reading about atomospheric pressure, then started reading about hurricanes, and naming conventions of hurricanes and who knows what else. Well, I remember some of that, but not the answer to the question I had: "is air pressure "corrected" for altitude?" I don't remember the answer, but I'm putting a note here now, so we'll remember to think about that later. It's possible that the air pressure at 5,000 feet is actually less than the weather report, because it's possible they correct for the altitude above sea level...

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.

The Calculations

• Drop Charge

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).
I should note that as a check to the validity of my calculations I compare my radius with that provided in an example calculation in the Teachers Guide at the end of the manual:
a = 4.9*10^-7m ± 0.2*10^-7m

All values are presented with ± their stdms when relevant.

• Drop 2 (only 1 charge)

rise time (s) : 13.12 17.46 11.27 12.55 11.58
fall time (s) : 18.17 19.46 19.86
Temp = 27°C
p = 102970 Pa
η = 1.858*10^-5 Ns/m^2
V = 500.8V
E = 6.46*10^4V/m
v_r = 3.88*10^-5m/s ± 3.28*10-6m/s
v_f = 2.61*10^-5m/s ± 6.95*10-7m/s
a = 4.63*10^-7m ± 1.33*10^-8m
m = 3.69*10^-16kg ± 3.18*10^-17kg

$q=1.39*10^-19 C \pm 1.41*10^-20 C \,$

Wow, that's actually close to the expected value!

• Drop 9 (only 1 charge)

rise time (s) : 10.33, 9.05, 9.17
fall time (s) : 10.14, 10.18, 10.05
Temp = 26.5°C
p = 100955 Pa
η = 1.855*10^-5 Ns/m^2
V = 507V
E = 6.54*10^4V/m
v_r = 5.27*10^-5m/s ± 2.26*10^-6m/s
v_f = 4.94*10^-5m/s ± 1.87*10^-7m/s
a = 6.50*10^-7m ± 2.61*10^-9m
m = 1.02*10^-15kg ± 1.23*10^-17kg

$q=3.15*10^-19 C \pm 7.97*10^-21 C \,$

This is probably the charge due to two electrons.

• Drop 11 (2 charges)

--1st charge
rise time (s) : 22.67
fall time (s) : 5.31, 5.90, 5.93, 5.83
Temp = 27.8°C
p = 100955 Pa
η = 1.861*10^-5 Ns/m^2
V = 507V
E = 6.54*10^4V/m
v_r = 2.21*10^-5m/s ± 0m/s (only one data point)
v_f = 8.72*10^-5m/s ± 2.21*10^-6m/s
a = 8.78*10^-7m ± 2.32*10^-8m
m = 2.51*10^-15kg ± 1.20*10^-16kg

$q=4.71*10^-19 C \pm 3.75*10^-20 C \,$

--2nd charge
rise time (s) : 5.34, 5.70, 5.36
fall time (s) : 5.31, 5.90, 5.93, 5.83
Temp = 27.8°C
p = 100955 Pa
η = 1.861*10^-5 Ns/m^2
V = 507V
E = 6.54*10^4V/m
v_r = 9.15*10^-5m/s ± 1.96*10^-6m/s
v_f = 8.72*10^-5m/s ± 2.21*10^-6m/s
a = 8.78*10^-7m ± 2.32*10^-8m
m = 2.51*10^-15kg ± 1.20*10^-16kg

$q=7.70*10^-19 C \pm 6.26*10^-20 C \,$

• Drop 12 (3 charges)

--1st charge
rise time (s) : 8.93, 8.84
fall time (s) : 22.00, 22.34, 25.62, 22.70, 20.45
Temp = 28°C
p = 100955 Pa
η = 1.862*10^-5 Ns/m^2
V = 506.5V
E = 6.54*10^4V/m
v_r = 5.63*10^-5m/s ± 2.85*10^-7m/s
v_f = 2.22*10^-5m/s ± 8.27*10^-7m/s
a = 4.24*10^-7m ± 1.72*10^-8m
m = 2.83*10^-16kg ± 3.44*10^-17kg

$q=1.50*10^-19 C \pm 1.87*10^-20 C\,$

--2nd charge
rise time (s) : 3.48 3.68, 3.50, 3.65, 3.43
fall time (s) : 22.00, 22.34, 25.62, 22.70, 20.45
Temp = 28°C
p = 100955 Pa
η = 1.862*10^-5 Ns/m^2
V = 506.5V
E = 6.54*10^4V/m
v_r = 1.41*10^-4m/s ± 1.96*10^-6m/s
v_f = 2.22*10^-5m/s ± 8.27*10^-7m/s
a = 4.24*10^-7m ± 1.72*10^-8m
m = 2.83*10^-16kg ± 3.44*10^-17kg

$q=3.92*10^-19 C \pm 3.94*10^-20 C \,$

--3rd charge
rise time (s) : 1.15, 1.15, 1.30, 1.21, 1.36
fall time (s) : 22.00, 22.34, 25.62, 22.70, 20.45
Temp = 28°C
p = 100955 Pa
η = 1.862*10^-5 Ns/m^2
V = 506.5V
E = 6.54*10^4V/m
v_r = 4.07*10^-4m/s ± 1.38*10^-5m/s
v_f = 2.22*10^-5m/s ± 8.27*10^-7m/s
a = 4.24*10^-7m ± 1.72*10^-8m
m = 2.83*10^-16kg ± 3.44*10^-17kg

$q=8.20*10^-19 C \pm 1.07*10^-19 C \,$

• Drop 14 (2 charges)

--1st charge
rise time (s) : 1.81, 2.00, 1.58
fall time (s) : 7.08, 7.63, 7.75, 7.46, 7.87
Temp = 28°C
p = 100955 Pa
η = 1.862*10^-5 Ns/m^2
V = 506.5V
E = 6.54*10^4V/m
v_r = 2.81*10^-4m/s ± 1.90*10^-5m/s
v_f = 6.62*10^-5m/s ± 1.20*10^-6m/s
a = 7.60*10^-7m ± 1.45*10^-8m
m = 1.63*10^-15kg ± 9.33*10^-17kg

$q=3.81*10^-18 C \pm 1.03*10^-19 C\,$

--2nd charge
rise time (s) : 15.53, 20.33, 15.95, 17.81, 19.33
fall time (s) : 7.08, 7.63, 7.75, 7.46, 7.87
Temp = 28°C
p = 100955 Pa
η = 1.862*10^-5 Ns/m^2
V = 506.5V
E = 6.54*10^4V/m
v_r = 2.84*10^-5m/s ± 1.49*10^-6m/s
v_f = 6.62*10^-5m/s ± 1.20*10^-6m/s
a = 7.60*10^-7m ± 1.45*10^-8m
m = 1.63*10^-15kg ± 9.33*10^-17kg

$q=1.04*10^-18 C \pm 2.08*10^-20 C \,$

• Electron Charge

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).
accepted value of e = 1.60*10^-19C

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
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

$measured\quad e = \frac{1.76 + 1.65 + 1.56 + 1.495}{4}*10^-19=1.61625*10^-19 C$

Since I don't actually have precision up to the third decimal I can drop the 625 and I get the wonderful result:

$e=1.61*10^-19 C \,$

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):
Δe=1/6[(1.41*10^-20)+(7.97*10^-21)+(3.75*10^-20)+(6.26*10^-20)+(1.87*10^-20)+(3.94*10^-20\)]

$\Delta e= 3.0045*10^-20 C \,$
SJK 16:52, 16 November 2008 (EST)
16:52, 16 November 2008 (EST)
I'm not sure I quite follow and / or agree with the "difference" method...e.g., if the charges were all 0.5e, 1.5e, etc. then you'd be getting rid of a 1/2 charge effect. It does look, however, like you spent a lot of time looking over this data and in particular the error analysis, and it looks like you're applying many of the concepts very well. one very important issue is exclusion of data points. My opinion is probably that you're throwing too many data points away...however the very good practice you are employing is that you're talking about all of the data you're keeping or excluding, and that is most important. It's very good scientific practice to discuss the exclusion of data and to make objective arguments. I'm not an expert, but supposedly Millikan excluded fractional charges himself, and many people criticize him for this...but if people know that he did this, then it's at least commendable that he took notes about what he was doing so that others could figure this out.

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.

distributions of four time measurements are obviously not always normal

Though I expect that the measured time intervals should be normally distributed, I have not made enough measurements to actually have a consistent normal distribution as can be seen in the figure to the right. Therefore a confident statement about confidence intervals here would not be based on the actual data. However, since the stdm of my measured electron charge is known I can calculate the 95% confidence interval that I would get if I took enough data points to get a true normal distribution (of course my stdm would probably shrink as well, but since it's already only 18% of the mean I think this calculation will not be totally unsound). I found a nice website which provides an interactive calculation tool for normal distributions and from which I gathered that "95% of the area of a normal distribution is within 1.96 standard deviations of the mean". Then the upper and lower limits of the 95% confidence interval are just

$Lower\, Limit=1.61*10^-19 -\left(1.96\right)\left(3.00*10^-20\right)=1.02*10^-19 C$
$Upper\, Limit=1.61*10^-19 +\left(1.96\right)\left(3.00*10^-20\right)=2.20*10^-19 C$

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.
I feel it is important to mention that, at least as far as I understand, it is possible to calculate the 99%, 95%, 68%, and any other percent confidence interval of a normal distribution if one simply knows the standard deviation (though the std is not really necessary in some cases) and how many stds away from the mean the boundary of a certain percentage of area under the normal distribution will reach. So the confidence error is not an absolutely fundamental and unique error quantifier such as the variance, std, and stdm. Rather, by using these quantities it allows us to quantify how often a value will be measured at any given distance away from the mean.

## Error Propagation

SJK 16:27, 16 November 2008 (EST)
16:27, 16 November 2008 (EST)
I'll admit that I'm not pouring over your algebra here...but does look like you are implementing things properly (and thus learning valuable skills).

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.
Since none of the variables are correlated I can apply the general formula for calculating the total variance (error) from the sum in quadrature of errors of individual "perpendicular" variables:

$\sigma_X^2=\left (\frac{\partial f}{\partial A}\sigma_A\right )^2+\left (\frac{\partial f}{\partial B}\sigma_B\right )^2+\left (\frac{\partial f}{\partial C}\sigma_C\right )^2+\cdots$

For my calculations I will replace σ with the standard deviation of the mean, represented by Δ.

$\Delta x= \sqrt{\frac{1}{N} \frac{\sum_{i=1}^N (x_i - \overline{x})^2}{N-1}}\,$
• given that v=d/t, velocity errors are calculated according to the formula:
$\Delta v = \left( \left(\frac{d}{dt} \frac{d}{t} \Delta t \right)^2\right)^{1/2} = \frac{d}{t^2}\Delta t = v\left(\frac{\Delta t}{t}\right)$

So I really am dealing with the relative error

$\frac{\Delta v}{v} = \frac{\Delta t}{t}$

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.

• The radius errors are calculated from the v_f error according to the formula:
$\Delta a = \left( \left(\frac{d}{dv_f} \left(\sqrt{\left( \frac{b}{2p}\right)^2 + \frac{9 \eta v_f}{2 g \rho}}-\frac{b}{2p}\right)\Delta v_f\right)^2\right)^{1/2} = \frac{9 \eta}{2 g \rho} \left[\frac{\Delta v_f}{\sqrt{\left( \frac{b}{2p}\right)^2 + \frac{9 \eta v_f}{2 g \rho}}} \right]$
• The mass is just some constant multiplied by a^3 so its error is easily calculated to be:
$\Delta m= 4 \pi \rho a^2 \Delta a \,$
• Finally, the charge errors are calculated from the velocity and mass errors:

since $q = \frac{mg(v_f + v_r)}{Ev_f}$ it follows that

$\Delta q = \sqrt{\left (\frac{\partial}{\partial m}\frac{mg(v_f + v_r)}{Ev_f}\Delta_m\right )^2+\left (\frac{\partial}{\partial v_r}\frac{mg(v_f + v_r)}{Ev_f}\Delta v_r\right )^2+\left (\frac{\partial}{\partial v_f}\frac{mg(v_f + v_r)}{Ev_f}\Delta v_f\right )^2}$
$= \sqrt{\left(\frac{g(v_f + v_r)}{Ev_f}\Delta_m\right)^2+ \left(\frac{mg}{Ev_f}\Delta v_r\right)^2+ \left(-\frac{mg v_r}{E v_f^2}\Delta v_f\right)^2}$

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 Remarks

1.) 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)
16:20, 16 November 2008 (EST)
though then you wouldn't have the convenience of terminal velocity (stokes drag) and you'd instead have free fall or constant acceleration...that might be even more annoying?

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)
16:21, 16 November 2008 (EST)
This would be fun to try