User:Nathan Giannini/Notebook/Physics 307L/101101

SJK 20:13, 21 December 2010 (EST)
20:13, 21 December 2010 (EST)
Good primary notebook.

Set-Up

• First we leveled the plateform for the Oil Drop Apparatus.
• We measure the width of the plastic spacer with a micrometer. This is the plate separation.
• Then we focused the viewing scope by using the focusing wire and adjusted the light.
• We connected the power supply to the plate voltage connectors using bannana plug patch plugs and applied 500V. We also connected a multimeter in parralel with the apparatus and power supply to better monitor the voltage.
• We connected another multimeter to the thermistor connectors to measure the resistance in the thermistor. This gave us the temperature within the viewing chamber.
• We used the atomizer to introduce oil drops into the chamber. This was done with the ionization source lever in the Spray Droplet Position, which allowed air to escape the chamber.
• Once there were droplets in the chamber we moved the ionization source lever to the Off Position. We tried to find a droplet with a fall velocity between 7 and 15 seconds. This was one of the hard parts of the experiment.
• When we chose a drop we measured as many fall and rise times as we could. The fall times are the times it takes the drops to fall from one major reticule line to the next. The major reticule lines are 0.5 mm apart. The rise times are the same, but with the voltage applied between the plates so that the drops rise against the force of gravity.
• For the last oil drop we measured several fall and rise times and then moved the ionization lever to the ON position and measured several more fall and rise times.

Equipment

• Millikan Oil Drop Apparatus
• Tel-Atomic (50V and 500V power supply)
• Oil of density 886 Kg/m^3

Data

Particle 3

• Velocity measurements are average values
• Standard Deviations obtained by STDEV in google docs
• Standard Deviation in Rise Time: $0.28seconds\,\!$
• Average Rise Time: $t_r=5.76seconds\,\!$
• Average Rise Velocity: $v_r=8.67 \cdot 10^{-5}+/-1.7 \cdot 10^{-6}m/s\,\!$
• Standard Deviation in Fall Time: $0.56seconds\,\!$
• Average Fall Time: $t_f=12.80seconds\,\!$
• Average Fall Velocity: $v_f=3.91 \cdot 10^{-5}+/-4.4 \cdot 10^{-6}m/s\,\!$

Calculating Charge of Particle 3

Errors here are such that they incur the greatest amount of error in the charge.

• I used a three step approach to the charge calculation.
• First I calculated the radius a of the oil drop
$a=\sqrt{(b / 2 p)^2+9 n v_f/2 g \rho}{-b/2p}\,\!$
• $a=5.667 \cdot 10^{-7}+/-3.362 \cdot 10^{-8}m\,\!$
• Then I calculated the mass
$m={4/3 \pi a^3 \rho}\,\!$
• $m=6.754 \cdot 10^{-16}+/-1.249 \cdot 10^{-16}Kg\,\!$
• Finally I calculated the total charge
$q=\frac{mg(v_f+v_r)}{(V/d)v_f}\,\!$
$q=3.3448 \cdot 10^{-19}+/-3.986 \cdot 10^{-20}C\,\!$

Elementry charge

• Knowing that the accepted value for e is $e=1.6 \cdot 10^{-19}C\,\!$, it appears that particle 3 has 2e.
• Dividing q by 2 I get:
$e=1.6724 \cdot 10^{-19}+/-1.993 \cdot 10^{-20}C\,\!$

Particle 4

• Velocity measurements are average values
• Standard Deviations obtained by STDEV in google docs
• Standard Deviation in Rise Time: $1.25seconds\,\!$
• Average Rise Time: $t_r=4.41seconds\,\!$
• Average Rise Velocity: $v_r=1.13 \cdot 10^{-4}+/-2.5 \cdot 10^{-5} m/s\,\!$
• Standard Deviation in Fall Time: $1.02seconds\,\!$
• Average Fall Time: $t_f=16.47seconds\,\!$
• Average Fall Velocity: $v_f=3.04 \cdot 10^{-5}+/-1.9 \cdot 10^{-6} m/s\,\!$

Calculating Charge of Particle 4

• Errors here are such that they incur max error in charge.
• I used a three step approach to the charge calculation.
• First I calculated the radius a of the oil drop
$a=\sqrt{(b/2p)^2+9nv_f/2g\rho}{-b/2p}\,\!$
• $a=4.952 \cdot 10^{-7}+/-1.66 \cdot 10^{-8}m\,\!$
• Then I calculated the mass
$m={4/3 \pi a^3 \rho}\,\!$
• $m=4.507 \cdot 10^{-16}+/-4.68 \cdot 10^{-17}Kg\,\!$
• Finally I calculated the total charge
$q=\frac{mg(v_f+v_r)}{(V/d)v_f}\,\!$
$q=3.269 \cdot 10^{-19}+/-7.643 \cdot 10^{-20}C\,\!$

Elementry charge

• Knowing that the accepted value for e is $e=1.6 \cdot 10^{-19}C\,\!$, it appears that particle 4 has 2e.
• Dividing q by 2 I get:
$e=1.6345 \cdot 10^{-19}+/-3.822 \cdot 10^{-20}C\,\!$

Particle 5

• I have eliminated bad data from this part of my calculations.
• Velocity measurements are average values
• Standard Deviations obtained by STDEV in google docs
• Standard Deviation in Rise Time: $0.43seconds\,\!$
• Average Rise Time: $t_r=4.88seconds\,\!$
• Average Rise Velocity: $v_r=1.03 \cdot 10^{-4}+/-8.6 \cdot 10^{-6} m/s\,\!$
• Standard Deviation in Fall Time: $0.99seconds\,\!$
• Average Fall Time: $t_f=16.26seconds\,\!$
• Average Fall Velocity: $v_f=3.29 \cdot 10^{-5}+/-2.1 \cdot 10^{-6} m/s\,\!$

Calculating Charge of Particle 5

• I used a three step approach to the charge calculation.
• First I calculated the radius a of the oil drop
$a=\sqrt{(b/2p)^2+9nv_f/2g\rho}{-b/2p}\,\!$
• $a=5.178 \cdot 10^{-7}+/-1.76 \cdot 10^{-8}m\,\!$
• Then I calculated the mass
$m={4/3 \pi a^3 \rho}\,\!$
• $m=5.152 \cdot 10^{-16}+/-5.44 \cdot 10^{-17}Kg\,\!$
• Finally I calculated the total charge
$q=\frac{mg(v_f+v_r)}{(V/d)v_f}\,\!$
$q=3.272 \cdot 10^{-19}+/-3.97 \cdot 10^{-20}C\,\!$

Elementry charge

• Knowing that the accepted value for e is $e=1.6 \cdot 10^{-19}C\,\!$, it appears that particle 5 has 2e.
• Dividing q by 2 I get:
$e=1.636 \cdot 10^{-19}+/-1.94 \cdot 10^{-20}C\,\!$

Particle 7

• Velocity measurements are average values
• Standard Deviations obtained by STDEV in google docs
• Standard Deviation in Rise Time: $0.92seconds\,\!$
• Average Rise Time: $t_r=3.93seconds\,\!$
• Average Rise Velocity: $v_r=1.27 \cdot 10^{-4}+/-3.0 \cdot 10^{-5} m/s\,\!$
• Standard Deviation in Fall Time: $1.34seconds\,\!$
• Average Fall Time: $t_f=13.91seconds\,\!$
• Average Fall Velocity: $v_f=3.59 \cdot 10^{-5}+/-3.6 \cdot 10^{-6} m/s\,\!$

Calculating Charge of Particle 7

• I used a three step approach to the charge calculation.
• First I calculated the radius a of the oil drop
$a=\sqrt{(b/2p)^2+9nv_f/2g\rho}{-b/2p}\,\!$
• $a=5.428 \cdot 10^{-7}+/-2.86 \cdot 10^{-8}m\,\!$
• Then I calculated the mass
$m={4/3 \pi a^3 \rho}\,\!$
• $m=5.935 \cdot 10^{-16}+/-9.89 \cdot 10^{-17}Kg\,\!$
• Finally I calculated the total charge
$q=\frac{mg(v_f+v_r)}{(V/d)v_f}\,\!$
$q=4.141 \cdot 10^{-19}+/-1.16 \cdot 10^{-19}C\,\!$

Elementry charge

• Knowing that the accepted value for e is $e=1.6 \cdot 10^{-19}C\,\!$, it appears that particle 7 has 2e.
• Dividing q by 2 I get:
$e=2.067 \cdot 10^{-19}+/-5.78 \cdot 10^{-20}C\,\!$
• also we notice that this one has a possibility for a 3e charge on it, so for this we get:
$e=1.765 \cdot 10^{-19}+/-3.85 \cdot 10^{-20}C\,\!$

Particle 8

• Velocity measurements are average values
• Standard Deviations obtained by STDEV in google docs
• Standard Deviation in Rise Time: $1.16seconds\,\!$
• Average Rise Time: $t_r=12.997seconds\,\!$
• Average Rise Velocity: $v_r=3.85 \cdot 10^{-5}+/-3.2 \cdot 10^{-6} m/s\,\!$
• Standard Deviation in Fall Time: $1.05seconds\,\!$
• Average Fall Time: $t_f=15.871seconds\,\!$
• Average Fall Velocity: $v_f=3.15 \cdot 10^{-5}+/-2.2 \cdot 10^{-6} m/s\,\!$

Calculating Charge of Particle 8

• I used a three step approach to the charge calculation.
• First I calculated the radius a of the oil drop
$a=\sqrt{(b/2p)^2+9nv_f/2g\rho}{-b/2p}\,\!$
• $a=5.060 \cdot 10^{-7}+/-1.89 \cdot 10^{-8}m\,\!$
• Then I calculated the mass
$m={4/3 \pi a^3 \rho}\,\!$
• $m=4.808 \cdot 10^{-16}+/-5.59 \cdot 10^{-17}Kg\,\!$
• Finally I calculated the total charge
$q=\frac{mg(v_f+v_r)}{(V/d)v_f}\,\!$
$q=1.635 \cdot 10^{-19}+/-2.02 \cdot 10^{-20}C\,\!$

Elementry charge

• Knowing that the accepted value for e is $e=1.6 \cdot 10^{-19}C\,\!$, it appears that particle 8 has 1e.
• Therefore, I get:
$e=1.635 \cdot 10^{-19}+/-2.02 \cdot 10^{-20}C\,\!$

Acknowledgements

• Randy Llafler - for helping me figure out how my calculations were off, for the set-up and the equations seen on this page.