Lab help (Millikan oil drop experiment)

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SUMMARY

The discussion focuses on the Millikan oil drop experiment simulation, where participants used graphing calculators to analyze 15 results. Key equations include q = (mgd)/V for charge calculation and m = (4/3)πr³P for mass determination, with P being the density of vegetable oil at 923 kg/m³. The conversation emphasizes the importance of plotting voltage against mass to derive a line of best fit, which aids in accurately calculating charge by determining the gradient value representing V/d.

PREREQUISITES
  • Understanding of basic physics concepts such as electric force and gravitational force
  • Familiarity with the Millikan oil drop experiment methodology
  • Proficiency in using graphing calculators for data analysis
  • Knowledge of density calculations and volume formulas
NEXT STEPS
  • Learn how to derive and interpret the line of best fit in experimental data analysis
  • Study the implications of varying droplet sizes on charge calculations
  • Explore advanced graphing techniques for better data visualization
  • Investigate the significance of error reduction in experimental physics
USEFUL FOR

Students conducting physics experiments, educators teaching the Millikan oil drop experiment, and anyone interested in understanding charge calculations and data analysis in scientific research.

Matt1234
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Hello,
We did a lab in class yesterday and we are expected to write a report and graph our results. The lab was a simulation of the Millikan oil drop experiment. We used graphing calculators to do the simulation and took down 15 results. One thing i don't understand is how to come up with the charge for any given value.

Homework Statement








Homework Equations



q= (m g r) / V
Where
m = mass
g = 9.8
r = distance OR plate separation (i am not sure, i have 2 values that could be R)
V = voltage


The other equation he gave us was m = (4/3) Pi r^3 P
pi = 3.14...
r = radius
P = density of oil

I looked up the density of oil to be:
vegetable oil
(0.923 grams) per (cubic centimeter) = 923 kilograms per (cubic meter)

The Attempt at a Solution



I tried some calculations i kept coming up with different values for M (which i think is the case since no two oil droplets would be the same mass. ) Yet the equation he gave us for calculating mass using the density confuses me since m is NOT going to be constant.

I am required to:
Show a sample calculation for determining the mass of the droplet and the charge on the droplet, along with the rest of my report.

Can someone please help me make sense of this?

Thank you.
 
Last edited:
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Well, we know that E= V/d, where V is the voltage applied and d is the distance between the plates, and that the electric force is given by F=Eq = Vq/d.

This must be equal to our weight force, F=mg, so that mg=Vq/d, giving q= mgd/V.

Your value R in the second column in the radius of each oil drop. Hence, the mass of the drop is just its volume times its density, or the formula you have above. The next thing you should be doing is drawing up another column, and write up the corresponding masses for each sample of different radius. Yes, you will get many different masses, each corresponding to a different Voltage.

You took 15 samples for a reason, you knew you would get different answers, all with some error. The point of taking 15 is to reduce your error, as some will lie above the right value, some less. To incorporate all your samples information, you have to now plot a graph of Voltage against Mass for your 15 samples. Your points should look at least approximately linear.

The next step is to take a good judgment and draw a line of best fit - a straight line that best approximates the data points. A good line of best fit should have approximately the same number of sample points above the line as there are below it. Once you have a good line of best fit, it is easy to determine that lines gradient. But remember what this gradient value represents, it is a "good" value for V/d, by good I mean that it sort of incorporates the information of all your data points. Now that we have a good value of V/d, call this value T, we know from before that q= mgd/V = gd/T. As we know the separation distance of the plates and gravitational acceleration, we can calculate q.
 
Thank you so much, that helped a lot. I used the formulas and compared to my results so i could find P to be 855 which is a reasonable number for oil.

Thank you sir!
 

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