Millikan oil drop experiment lab help

1. Oct 18, 2011

superiority

1. The problem statement, all variables and given/known data
I'm doing Millikan's oil drop experiment in a lab. I've got a set of measurements and all that, but in analysis of the data I can't seem to get values for the charges on individual drops that seem reasonable. I find that each drop is carrying between 10 and 300 elementary charges. I've gone through my work over and over, and I'd appreciate if anybody here could point out if I'm doing something wrong.

2. Relevant equations
The equations given for an oil drop falling through air and rising in an electric field are, respectively:
$$6 \pi r \eta u_g = \frac{4}{3}\pi r^3 (\sigma - \rho ) g$$
$$6 \pi r \eta u_E = qE - \frac{4}{3} \pi r^3 (\sigma - \rho ) g$$
where r is the radius of the drop, η is viscosity of air, ug is the speed at which the drop falls under the influence of gravity, uE is the speed at which the drop rises in an electric field, σ is the density of the oil, ρ is the density of air, q is the charge on the drop, and E is the electric field strength. From these two equations are derived expressions for the radius and for the charge on a drop:
$$r = \sqrt{\frac{9 \eta u_g}{2 (\sigma - \rho ) g}}$$
$$q = 6 \pi \frac{d}{V} \eta (u_E + u_g ) r$$
Where d is the distance between capacitor plates and V is the voltage applied.
Also used is Millikan's correction to Stokes' law for particles at small velocities, which involves multiplying the speed by $(1 + \frac{k}{r P})^{-1}$, where P is air pressure and k is a (supplied) constant. This gives a value slightly smaller than one, and the final calculation for q looks like:
$$q = 6 \pi \frac{d}{V} \eta (u_E + u_g ) \sqrt{\frac{9 \eta u_g}{2 (\sigma -\rho ) g}}\left( 1 + \frac{k}{r P} \right) ^{-\frac{3}{2}}$$
The procedure, having made various measurements, is to first find the radius r, then use that to calculate the correction and from that the charge.
These equations are all directly out of the instruction sheet for the experiment. I didn't have to derive anything myself.

3. The attempt at a solution
I have a whole bunch of data points for different drops, so I'll use two here and go through the working. If anything looks like it's the wrong order of magnitude or something, please point it out!

The distance the droplets were measured falling/rising over was 5x10-3 m. Viscosity of air η was found (using an empirical formula based on room temperature) to be 1.808x10-5 N m s-2. Density of the oil σ was 860 kg m-3. Density of air ρ was 1.2101 kg m-3 The distance d between capacitor plates was measured as 7.53x10-3 m. Applied voltage V was 500 V. Air pressure P was 103,058 Pa. k was given as equal to 8.226x10-3 Pa m.

I measured one drop (D1) falling freely over the 5 mm distance in 3.99 seconds, and rising in the electric field in 3.09 seconds (D1 was the quickest drop both falling and rising). Another (D2) fell in 14.82 seconds and rose in 9.37 seconds (D2 was more typical).
Dividing distance by time, I found that D1 fell at ug1 = 1250x10-6 m s-1 and rose at uE1 = 1620x10-6 m s-1, while D2 fell at ug2 = 338x10-6 m s-1 and rose at uE2 = 530x10-6 m s-1.
I plugged the ug values into the equation for r above, and found that D1 had radius r1 = 34.8x10-7 m and D2 had radius r2 = 18.1x10-7 m. Then the Stokes correction values were calculated to be c1 = 1.0229 and c2 = 1.0442.
Then using the formula for charge, I found the charges on the drops q1 = 495x10-19 C and q2 = 75.7x10-19 C. Letting q0 = 1.6x10-19 C, the elementary charge, then q1 ≈ 309q0 and q2 ≈ 47q0. This is much larger than I expected, and furthermore, they aren't all that close to being integer multiples of q0 (it's more like 309.4 and 47.3 than 309 and 47). Note that q1 is the largest value for the charge I found, and the next-largest in my data set is only ≈77q0 (with the smallest being ≈12q0). Can anybody see if there's any mistake here?

2. Oct 18, 2011

phyzguy

I don't see anything that you've done wrong. My recollection from doing this experiment many years ago was that there were many drops with large charges that moved very quickly. We needed to wait until the electric field swept these out of the way so we could find drops that had smaller charges and moved more slowly in the electric field, then we turned up the voltage to hold these stationary. Maybe your answers are correct and you were just focusing on the wrong drops?

3. Oct 18, 2011

Liquidxlax

I'm currently performing the same experiment and i found a great way to do it. you need to make an equation such that you can related all your different fields and velocities of a single drop to evaluate for you charge q which would be actually some integer multiplied by q. When you do this for all your drops and plot a histogram you should be able to see a relationship between each of the drops that they are all multiplicities of the charge of an electron