# Physics307L F08:People/Le/Notebook/070917

## Contents

### Day 2

--Linh N Le 18:40, 17 September 2007 (EDT)

We got the experiment working, finally! The one noticeable difference was that the oil sprayer was full, which may have given a better spray. We also made the chamber really really bright to try to see the oil droplets better.

#### Procedure

Once we got the oil droplets into the capacitor, we turned the ionizer on to allow the oil droplets to gain excess electrons. (It was interesting to note that some of the droplets were ionized positively, seen as they were attracted to the positive plate).

After we let them settle, we tried to find and isolate a droplet that fell across the the major grid lines (.5mm)in about 15s. The droplets also had to rise up when the upper plate was charged positively and fall faster when the bottom plate was charged positively.

After that setup, we measure the fall time across the .5mm with the plates on neutral and the rise time when we turned the top plate positive.

We also made sure the keep track of the voltage and the resistance in the thermistor.

Ideally, we were to make measurements with some droplet, then increase its excess electrons and remeasure the data.

#### Data

Spacing of plates (as computed by the average of the data above): 8.10mm

Density of Squibb's Mineral Oil: 886kg/m^3

Atmospheric Pressure in Abq (as looked up on the internet): 8.33X10^4 Pa

##### Set 1

Voltage: 501V Resistance: 1.9982 M(ohms) Distance: .5mm

Fall Time (s) Rise Time (s)
18.38 3.31
14.38 3.15
14.11 5.14
13.7 5.7
13.8 5.19
14.72 5.52
14.5 5.23
13.13 5.27
15.37 5.37
Avg Fall Time Avg Rise Time
14.67s 4.877s

In this set, we were able to change the charge on the droplet and remeasure

Voltage=501.9V Resistance=1.976M(ohms)

Fall Time (s) Rise Time (s)
13.28
13.86 2.19
14.58 2.21
13.09 2.70
14.38 2.46
13.94 2.24
Avg Fall Time(s) Avg Rise Time(s)
13.855 2.36
##### Set 2

Voltage: 500V Resistance: 1.923M(ohms)

Fall Time(s) Rise Time(s)
15.97 2.99
14.50 3.13
13.80 2.95
15.50 2.69
15.17 2.75
15.36 3.36
Avg Fall Time(s) Avg Rise Time(s)
15.05 2.97

We changed the charge of the oil droplet in this set as well, but the droplet moved so fast, that it was very hard to sync Cary's observations and oral commands with my data collecting. As that is the case, we only took 2 data points.

Fall Time (s) Rise Time(s)
9.70 .72
10.32 .73
##### Set 3

Starting Voltage: 501.0 V Resistance: 2.07 M(ohms) Ending Voltage:501.8 V 2.041M(Ohms)

FALL TIME (s) RISE TIME (s)
14.87 4.81
17.53 4.87
17.17 4.85
18.67 4.57
17.89 4.53
17.77 4.37
15.93 4.56
16.49 4.90
18.97 4.61
18.41 5.11
19.24 2.50
Avg Fall Time(s) Avg Rise Time(s)
17.54 4.52
##### Set 4

Starting Voltage:501.4 V Resistance:1.945 M(Ohms)

Fall Time (s) Rise Time (s)
16.13 2.23
14.29 2.14
14.85 2.47
15.96 1.93
16.49 1.99
14.79 2.25
Avg Fall Time(s) Avg Rise Time(s)
15.42 2.17

#### Comment

The lab asked us to try to use the same droplet as many times as possible, and remeasure it after changing the charge. It worked once, but either the droplet moved too fast (to where we felt that the measurements would be very err'd), the droplet itself moved out of view, or was destroyed

#### Calculations

The apparatus' manual has a nice derivation for an equation to calculate the charge of the electrons found inside the oil drops.

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

q- the charge of the electron

p-barometric pressure-8.334Pa

d-distance between capacitor plates

g- acceleration due to gravity- $9.8 \frac{m}{s^2}$

b- constant 8.20E( − 3)Pa * m

a- radius in drop measured in meters

ρ-density of the oil which is $886 \frac{kg}{m^3}$

η- viscostity of air (found by the comparing temp inside the capacitor with chart in manual appendix A)

V- potential difference across the plates in Volts

vr- rise velocity (divding .5mm by rise time)

vf- falling velocity (dividing .5mm by rise time)

E electric field (found by $\frac{V}{d})$

##### Set 1A
see comment
Steven J. Koch 01:58, 2 October 2007 (EDT):Your data look really great! I have been waiting for a note that you are done with the summary, but now I suppose I should assume you are. I will look at it a few more times, but I'm not quite sure what you're doing at the end with the 1+1+1+2+1.5 thingy...

Voltage: 501V

Resistance: 1.9982 M(ohms)

Temp=25C

η = 1.8480NsmE − 5

$v_favg=3.408E-5 \frac{m}{s}$

$v_ravg=1.025E-4 \frac{m}{s}$

E= V/d= 61851.85 V/m

Plug everything into the formula: q=3.39E-19 C (yay for being in the same order of mag!)

If i divide this by the known value of e, i get about 2.12, telling me this droplet has 2 excess electrons.

##### Set 1B

Voltage=501.9

Resistance=1.976M(ohms)

Temp=25.5C

η = 1.8480NsmE − 5

$v_favg=3.608E-5 \frac{m}{s}$

$v_ravg=2.118E-4 \frac{m}{s}$

E= V/d= 61886.56 V/m

Painstakenly plug and chug into the formula and get q= 6.38E-19C

Divide by e and get 3.988, really close to 4

That means that we did get the oil drop to get a quantized amount more charge!(if thats not good data, then I dont know what is)

##### Set 2

Voltage: 500V

Resistance: 1.923M(ohms)

Temp= 26C

η = 1.8520NsmE − 5

$v_favg=3.225E-5 \frac{m}{s}$

$v_ravg=1.683E-4 \frac{m}{s}$

E= V/d= 61652.28 V/m

Another round of math and I get q=4.84E-19

Divide by e and i get 3.03, so about 3 excess electrons.

Looking at rise vel for sets 1A and 1B, this one was faster than 1A but slower than 1B, making our measurements pretty consistent with this notion of quantized energies.

##### Set 3

Voltage: 501.0 V

Resistance: 2.07 M(ohms)

Temp: 24C

η = 1.8440NsmE − 5

$v_favg=2.855E-5 \frac{m}{s}$

$v_ravg=1.106E-4 \frac{m}{s}$

E= V/d= 61775.59 V/m

Insert values into the formula and get q=3.08E-19C

Divide by e and get 1.92, which is very close to 2

##### Set 4

Voltage:501.4 V

Resistance:1.945 M(Ohms)

Temp:26C

η = 1.8520NsmE − 5

$v_favg=3.243E-5 \frac{m}{s}$

$v_ravg=2.304E-4 \frac{m}{s}$

E= V/d= 61901.23 V/m

Last but not least, calculations yield q as 6.34E-19C

which is 3.96 e, close to 4

##### Conclusion

If we look at set 1A and set 3, their charges are about even. I will call that charge "Q". Sets 1B and 4 are about twice as big, and set 2 is about 1.5times as big.

So now, we can take the avg of them and find some value for this "Q"

$Q= \frac{1+1+2+2+1.5}{5} x10^{-19}$

Q = 1.5x10 − 19

Now for a quick error check

$%error= \frac{|Actual-Experimental|}{|Actual|}x100$

$%error= \frac{|1.60217646 x 10^{-19}-1.5 x 10^{-19}|}{|1.60217646 x 10^{-19}|}x100$

%error = 6.37

#### Errors

This lab has its fair share of spots where error can be introduced:

• I went on the internet to find the barometric pressure in Albuquerque, but that is probably an estimated quanity and changes with specific altitude of your area and the weather at the time
• The viscosity of the air inside the capacitor is measured by the temperature inside the capacitor. That value is determined by measuring the resistance in a thermistor and then finding values on a chart.
• The temperature is estimated off a chart, but when you get a value for the resistance that falls between values, you have to round
• Once the temperature is found, you look at a graph to find the value, and as above, you have to round
• While measuring the fall and rise times of the droplets, it is uncertain how close or far the droplets are, so adjusting the microscope might change the distance that they travel (although it is assumed the apparatus is calibrated to prevent this)
• There is also "lag time" between measurements as one of us stared into the scope and the other was running the stopwatch
• In the calculations, I rounded the values we found as 2 times greater than or 3 times greater than ect. instead of using an acutal ratio
see comment
Steven J. Koch 01:56, 2 October 2007 (EDT):It is great that you reported the uncertainty here...but what are the units??? If this were real, new research, the risk is that you wouldn't be able to decipher years from now what the units on this measurement were...so, it's always necessary to record units with your answer to save from future ambiguity.

With these possible errors, I would say that my results are 1.5 +/- .3 E-19