Confused by the Question (oil drop charge)

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Homework Help Overview

The discussion revolves around a physics problem related to Millikan's Oil Drop Experiment, specifically focusing on determining the charge of an oil drop suspended in an electric field. Participants are exploring the relationship between mass, electric field, and charge, as well as how to interpret the results in the context of the experiment.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of the equation q=mg/E to find the charge and question how additional information, such as plate separation and potential difference, impacts the calculation. There is also inquiry into the interpretation of negative charge values and the calculation of excess or deficit electrons.

Discussion Status

Some participants have provided calculations and interpretations of the results, while others express uncertainty about the implications of negative charge values and the method for determining the number of excess electrons. There is an ongoing exploration of the problem without a clear consensus on the interpretation of results.

Contextual Notes

Participants are working with specific values for mass, voltage, and plate separation, but there is a noted lack of clarity on how these values interact within the equations used. The discussion includes considerations of the implications of negative charge in the context of electron count.

linuxux
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This is a short question, from my understanding, to determine the charge of an object we can use the equation q=\frac{mg}{E} , where m is mass of object and E is voltage of electrical field. I am asked to find the charge on a drop of oil suspended in the field. I was given both the mass and the applied voltage of the field, but i was also given distance between the field and also told that the lower plate is "at a lower potential". How do the other pieces of information contribute to the charge since they are not factors in the equation? i also have to find number of excess of deficit electrons, how do i do that? (perhaps by divide by charge of electron?)

the givens were oil drop mass: 2.6 x 10^15 kg, plate separation = .005 m ,potential difference between plate = 270 V , and it also says the lower plate is at a lower potential.

to determine E, i used E = v / d , where v is voltage, and d is plate separation, then i used the above equation to determine q.
 
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This is Millikan's Oil Drop Experiment - where the oil drop is 'suspended' between parallel plates. Normally the electric field is: E=N/C, ie force per coulomb of charge. In parallel plates the field is basically constant, and can instead be calculated by E=V/d. So if you have the distance and the voltage that gives you the electric field E. Stick that into your first equation and hopefully you'll get an integer multiple of electron charge :D
 
i calculated the E using the given voltage distance between plates, but the q i got was -4.7\cdot10^{-19} . i used -9.8 for my g value.

When i divide this number by the charge of an electron , e=1.60\cdot10^{-19} , i get 2.93 !

What went wrong?
 
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Well, we don't have the given values, so we can't tell you where you went wrong.
 
okay, i added the givens.
 
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the negative value of the charge would indicated an excess of electrons, so i thought dividing by the charge of an electron would give me the number of excess electrons, not sure if that's right however.
 
still need help one this one...
 
An electron is negatively charged so I don't think it's a problem getting a negative value for the charge. As for the answer, 2.93 is as close to 3 as makes no difference (considering how small a mass you're talking about) so I'd say that's all ok...You're supposed to get integer multiples of the electron charge so it's fine.

I don't know about the excess electrons sorry.
 

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