# Physics307L F07:People/Gooden/Notebook/070924

## Contents

**Introduction**

- This lab is a reproduction of the Milikan Oil Drop experiment done by Robert Milikan in the early 20th century to discover the charge of an electron. The main idea of the experiment is that using oil droplets, which are very very small and having them fall through an electric field we measure the relative strength of the field on the drops (a force) to determine how much charge is on each drop. Since the drops are incredibly tiny there will only be several electrons on each drop contributing to its total charge.

**Set Up**

- 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)

(taken from Bradleys lab book)

**Procedure**

^{SJK 03:32, 7 November 2007 (CST)}

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.

(This was a great procedure, written by bradely explaining the experiment)

- Antonio came and helped out due to difficulties we had in seeing droplets of oil

**Data**

^{SJK 03:34, 7 November 2007 (CST)}

- Drop 1

**Fall Time** | **Rise Time**

7.18 2.97 6.22 3.01 6.43 2.72 5.98 2.44 6.63 2.97 6.27 2.57 6.51 2.55 7.09 2.73 7.18 2.78 6.93 2.98 7.2 3.74 6.49 3.47 6.61 3.87

- Drop 2

**Fall Time** | **Rise Time**

5.54 3.73 5.47 4.32 5.26 4.16 5.75 4.67 5.34 4.07 5.83 4.6 5.73 4.33 5.63 3.83 6.07 4.38 5.58 4.12 5.79 4.16 5.69 4.44 6.32 4.51

- Drop 3

**Fall Time** | **Rise Time**

7.81 2.08 7.78 1.87 7.49 2.03 8.45 1.96

- Drop 4

**Fall Time** | **Rise Time**

8.07 1.98 8.19 2.02 8.24 2.19 8.07 1.96

**Important Numbers**

d= 7.511 mm (.007511 meters) - width of spacer

Plate voltage: [math]V=500.6 Volts[/math]

Density of Oil [math]\rho = 886 kg/m^3[/math]

Resistance between plates: 2.101 Ohms

Temperature [math]T=23.5 C[/math]

- The temperature remained approximately constant

**Equations**

- These are equations relating the constants and other important parameters, that allow us to calculate the charge of each drop.

[math]\alpha = [/math]radius (m) [math]\eta = [/math]viscosity of air

[math]m = [/math]mass (kg) [math]V_f = [/math]fall velocity [math]V_r = [/math]Rise velocity

[math]Q = [/math]charge of drop (C) [math]p = [/math]barrometric pressur

- [math]\alpha=\sqrt{\frac{b^{2}}{{4p}^{2}}+\frac{9nV_f}{2g\rho}}-\frac{b}{2p}[/math]

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

- [math]E=\frac{V}{300d}[/math]

- [math]Q=mg\frac{\left( V_f+V_r\right)}{EV_f}[/math]

- [math]Q=\frac{4}{3}\pi\rho g\left(\sqrt{\frac{b^{2}}{{4p}^{2}}+\frac{9nV_f}{2g\rho}}-\frac{b}{2p}\right)^{3}\frac{\left( V_f+V_r\right)}{EV_f}[/math]

**Measurments**

^{SJK 03:34, 7 November 2007 (CST)}

- These are our results in calculating the charge of each drop for each measurment given the equations above. All are in units of Coulombs (C)

- From
**Drop 1**:

8.8183e-019 9.9109e-019 1.0321e-018 1.1862e-018 9.4522e-019 1.0979e-018 1.0692e-018 9.4674e-019 9.2446e-019 9.0721e-019 7.5149e-019 8.6793e-019 7.9566e-019

- From
**Drop 2**:

9.6276e-019 8.9552e-019 9.5124e-019 8.1552e-019 9.4861e-019 8.1101e-019 8.5381e-019 9.3304e-019 8.0116e-019 9.0186e-019 8.6478e-019 8.4769e-019 7.5696e-019

- From
**Drop 3**:

1.0749e-018 1.1736e-018 1.1323e-018 1.0576e-018

- From
**Drop 4**:

1.0897e-018 1.0602e-018 9.8943e-019 1.0987e-018

**Data Analysis**

- We must look at our data and I believe discard some of the measurments that seem to be very far off in their particular set

- for
**Drop 1**I would like to discard measurments 11 and 13 because they are fairly far off from the other 11 measurments are due to difficulty in seeing the drops those measurments could be tainted with error.^{SJK 03:36, 7 November 2007 (CST)}

- for
**Drop 2**I will keep all measurments.

- for
**Drop 3**and**Drop 4**I also want to keep all measurments.

- Taking the mean of each data set, i.e the mean of all measurments for Drop 1,Drop 2, etc., we find the the mean charge on the drops are:

Drop 1: 9.7981e-019 C +/- 1.01831e-019Drop 2: 8.7261e-019 C +/- 6.563e-019Drop 3: 1.1096e-018 C +/- 5.3289e-20Drop 4: 1.0595e-018 C +/- 4.9527e-20

**RESULTS**

Using the given mean values above, we must make some assumptions and guesses. We know (or have a good guess) that there is some basic unit of charge, off of which all other charges are multiples of. So lets take some guesses for these four values as to what multiple of e they are. Then we want to use the equation below to find e.

[math]\sum \left|q\right|=\left(9.7981+8.7261+11.096+10.0595\right)\times10^{-19} C =\left(6+5+7+6\right)e[/math]

I now can solve for [math]e[/math].

[math]e=1.653\times10^{-19} C[/math]

and the error that we find is

[math]Relative Error=\frac{\left|1.60\times10^{-19}C-e\right|}{\left|1.60\times10^{-19}C\right|}=0.032=3.2%[/math]