Confused by the Question (oil drop charge)

In summary, the conversation discusses the use of an equation to determine the charge of an object in an electrical field. The equation is q = (mg)/E, where m is the mass of the object and E is the voltage of the field. It also mentions the use of Millikan's Oil Drop Experiment to determine the charge of a suspended oil drop. The conversation includes givens such as the mass of the oil drop, the distance between the plates, and the voltage of the field. It also explores the idea of excess electrons and how to calculate them. Ultimately, there is some confusion about the results obtained and the number of excess electrons.
  • #1
linuxux
133
0
This is a short question, from my understanding, to determine the charge of an object we can use the equation [tex]q=\frac{mg}{E}[/tex] , 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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  • #2
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
 
  • #3
i calculated the E using the given voltage distance between plates, but the q i got was [tex]-4.7\cdot10^{-19}[/tex] . i used -9.8 for my g value.

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

What went wrong?
 
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  • #4
Well, we don't have the given values, so we can't tell you where you went wrong.
 
  • #5
okay, i added the givens.
 
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  • #6
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.
 
  • #7
still need help one this one...
 
  • #8
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.
 

1. What is "Confused by the Question (oil drop charge)"?

"Confused by the Question (oil drop charge)" is a common phrase used in the scientific community to describe a state of confusion or uncertainty about a particular question or topic related to the charge on oil drops in an experiment.

2. What causes the confusion in understanding oil drop charge?

The confusion in understanding oil drop charge often arises from the complex and counterintuitive nature of the experiment itself. The experiment involves observing the movement of oil drops in an electric field and using this data to determine the charge on the drops. This can be a difficult concept to grasp, especially for those who are not familiar with the principles of electrostatics.

3. How can I better understand the concept of oil drop charge?

To better understand the concept of oil drop charge, it is important to have a strong understanding of electrostatics and the forces acting on charged particles. It may also be helpful to review the experimental setup and procedures, as well as any relevant equations and theories.

4. How does the charge on oil drops affect the results of the experiment?

The charge on oil drops plays a crucial role in the results of the experiment. The magnitude and sign of the charge will determine the direction and speed of the drops' movement in the electric field. This data is then used to calculate the charge on the drops and can also provide insights into the properties of the oil itself.

5. What are some common misconceptions about oil drop charge?

One common misconception about oil drop charge is that the drops have a fixed charge, when in reality, the charge on the drops can vary depending on factors such as the type of oil and the surrounding environment. Additionally, some may mistakenly believe that the charge on the oil drops is directly related to the charge on the plates in the experiment, when in fact, the two are independent variables.

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