# Millikan's oil drop experiment (charge of the electron)

Experimentalists: Nikolai Joseph and Bradley Knockel

## Goal

I want to measure the charge of an electron by measuring the charge on a bunch of oil droplets and seeing if I can find that my calculated charges are integer multiples of some fundamental charge. I don't really see why the charge of the electron gets all the attention with this experiment since the charge of a proton is also measured. The currently accepted value for fundamental charge is $e=1.60\times10^{-19} C$. The charge of the electron is e.

## Equipment

Some Equipment Needed: the Millikan device (including plate charging switch), power source, atomizer, mineral oil, multimeter, banana cords, and stopwatch.
• power source (should go up to 500 V direct current)
• atomizer (to spray oil droplets)
• 2 multimeters
• banana cords
• banana plug patch cords
• DC transformer for light
• micrometer
• mineral oil
• stopwatch
• THE MILLIKAN DEVICE! (scope, viewing chamber, light, level, plate charging switch, focusing wire, thermistor, etc.) (Model AP-8210 by PASCO scientific)

## Setup

1. plugged in power supply to wall and Millikan device (turned off)
2. hooked up multimeter using banana plug patch cords to check voltage from power supply
3. leveled the Millikan device
4. plugged in the light using DC transformer
5. focused viewing scope with focusing wire
6. aimed the lamp/light/filament
7. turned on power supply and checked it's voltage using first multimeter
8. attached another multimeter to the thermistor

## Values

### Known (given to as many significant figures as are reasonably certain):

• $d=7.59\times 10^{-3} m$ (plastic spacer width using micrometer)
• $\rho=8.86\times 10^2 \frac{kg}{m^3}$ (density of oil given on bottle)
• $g=9.8 \frac{m}{s^2}$ (gravitational acceleration)
• $p=8.5\times10^4 Pa$ (air pressure in Albuquerque)
• $b=8.20\times10^{-3} Pa\cdot m$ (some stupid constant)
• $l=1.0\times10^{-3} m$ (length droplet will be measured over)

### To be found when taking data:

• T (temperature from thermistor in °C)
• V (Voltage between plates in viewing chamber in volts)
• tf (time droplet takes to fall in no field in seconds)
• tr (time droplet takes to rise in field in seconds)

### To be calculated later:

• η (viscosity of air as a function of T found in a table in Pa*s)
• $v_f=\frac{l}{t_f}$ (average velocity of oil droplet falling in no field in m/s)
• $v_r=\frac{l}{t_r}$ (average velocity of oil droplet rising in a field in m/s)
• $a=\sqrt{\left(\frac{b}{2p}\right)^2+\frac{9\eta v_f}{2g\rho}}-\frac{b}{2p}$ (radius of droplet in meters)
• $q=\frac{4}{3}\pi\rho g d\frac{a^3}{V}\frac{\left(v_r+v_f\right)}{v_f}$ (charge of oil droplet in Coulombs)

## Comments on derivations of above equations

Using Stokes equation and Newton's 2nd law for a falling droplet in no field, one gets: $m\cdot g=9\pi\eta_{eff}a v_f$, where ηeff is a correction for small a. Substituting $m=\frac{4}{3}\pi a^3\rho$ and $\eta_{eff}=\eta\left(\frac{1}{1+\frac{b}{pa}}\right)$ into this equation and solving for a should get you the correct equation.

### Charge (q):

Newton's laws for a falling (in no field) and rising droplet create $m\cdot g=k v_f$ and $E\cdot q=mg+k v_r$, where k is how much the air effects the drag force and E is the electric field strength where up is positive. Eliminating k and then solving for q produces $q=\frac{mg\left(v_r+v_f\right)}{E v_f}$. If you substitute $m=\frac{4}{3}\pi a^3\rho$ and $E=\frac{V}{d}$ into this q equation, you should get the correct equation.

The sign V can get a little tricky when calculating q (all other values used to find q are positive). When the plate charging switch is set to negative, this means that the top plate is negative so the value for V should be positive. To get the droplet to rise, V will sometimes need to be positive and sometimes negative, which means the charge q will sometimes be positive or negative.

If you enjoy doing things differently, you can take velocity measurements with the field pushing the droplet down, in which case vr would be negative when finding q since the droplet is falling instead of rising. The equation for q is very flexible and can handle a negative vr. If you cannot get a power supply powerful enough to actually have the droplet of smallest mass and charge you can find rise, this is another instance where vr would need to be negative.

## Procedure

After turning off the external lights, we sprayed oil droplets into the viewing chamber using the atomizer by pumping droplet rich air into it. There is no science to this; we just kept trying over and over until droplets appeared in the center of the screen. We then selected drops that were barely falling through the viewing chamber in no electric field (we want drops that have little mass). From those drops, we selected one that moves slowly in a field (we want drops that have little charge). Perhaps, in hindsight, we should have selected some drops that were moving a little bit faster since most of our selected drops had plus or minus one electron, but we thought we were dealing with larger charges with our selected drops.

We measured the speed at which it falls, vf. Having a partner to hold the stopwatch and write data while the other person watches the droplet is very helpful. We then created an electric field that caused the droplet to rise and measured the speed, vr. We took many measurements of both of these speeds over and over on the same droplet. We then tried to introduce alpha particles using the thorium-232 source to change the charge of the oil droplet (to be either more positive or negative depending on how the collision between the oil and alpha particles occurred), but the droplet would often become lost in the viewing chamber before we could do this.

This process took practice, and it was hard to be sure that the droplet was not changing its charge unexpectedly. This happened a few times. Also, it probably would have been better if we would have waited for the power supply and thermistor to warm up to reduce fluctuations in voltage and temperature.

## Data

### Droplet 1, Charge A:

Our first observation for tr was very different and we suspect a change in charge, so we are discarding it, even though I am displaying it below.

• V=+503V
• T=23°C
 tf (s) tr (s) 41.3 47 49 51.3 45.5 43.9 10.9 4.5 4.6 4.8 4.9 4.8

### Droplet 2, Charge A:

• V=-503V
• T=26°C
 tf (s) tr (s) 59.2 60.1 69.9 62.6 9.6 9.3 9.3 9.1

### Droplet 2, Charge B:

Our first observation for tr was very different and we suspect a recording error, so we are discarding it, and I am displaying it below. We only took two falling times because these took much longer than the rising times and we were lazy.

• V=-504V
• T=26°C
 tf (s) tr (s) 85 87.1 2 1.43 1.53 1.43 1.52 1.51

### Droplet 3, Charge A:

• V=-504V
• T=27°C
 tf (s) tr (s) 42.3 47.2 50.8 47.1 12.1 12.1 12.9 13.5

### Droplet 6, Charge A:

Droplets 3B, 4 and 5 acquired either one or two data points before going out of focus and becoming lost.

• V=+505V
• T=27°C
 tf (s) tr (s) 57.5 63.5 63 10 9.7 9.1

## Calculations

Droplet/Charge $\eta\left(\times10^{-5} Pa\cdot s\right)$ $v_f\left(\times10^{-5} \frac{m}{s}\right)$ $v_r\left(\times10^{-4} \frac{m}{s}\right)$ $a\left(\times10^{-7} m\right)$ $\left| q\right|\left(\times10^{-19} C\right)$ Suspected Multiple of e
1A 1.83 2.17(17) 2.11(7) 4.08 4.00 2
2A 1.85 1.60(12) 1.07(2) 3.46 1.75 1
2B 1.85 1.16(2) 6.74(23) 2.89 7.81 4
3A 1.86 2.14(16) 0.79(4) 4.09 1.76 1
6A 1.86 1.63(9) 1.04(5) 3.52 1.75 1

## Analysis and Conclusion

My first observation is that making the charge more positive in 2B significantly decreases the size of the droplet. This is interesting because it shows that the collisions between the droplets and alpha particles are violent.

To find the charge of the electron, e, I first notice that we always chose the droplet with the smallest charge. Taking into account there being three of five droplets all having the same and low charge and that all five of the droplets are multiples of this charge, I conclude that the three droplets with smallest charge have charge e. To calculate e, I will set the sum of the charges equal to the sum of the suspected multiples of e.

$\sum \left|q\right|=\left(4.0044+1.7533+7.8087+1.7582+1.7539\right)\times10^{-19} C =\left(2+1+4+1+1\right)e$

I now can solve for e.

$e=1.90\times10^{-19} C$

Let's see how good of an experimentalist I am...

$Relative Error=\frac{\left|1.60\times10^{-19}C-e\right|}{\left|1.60\times10^{-19}C\right|}=0.19=19%$

The e I calculated is above the accepted value either because the accepted value is wrong (unlikely) or because of systematic error. A large part of this error may be due to not taking Albuquerque's high altitude into account when calculating viscosity, η.