Gauss' Law for nonconducting sphere

  • Thread starter armolinasf
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Homework Statement



There is a nonconducting solid sphere with uniform charge density rho and radius R=15 but it has a smaller hollow spherical cavity of radius a = 5, it is not centered: the distance between the center of the larger sphere is a distance b = 6

I'm trying to find the electric field at a point (1) a/2 to the left of the cavity's center and (2) a distance a/2 above the cavity's center.


The Attempt at a Solution



I figured that since gauss' law states that E=q(enc)/(eps*A), the electric field would be zero in all three vector directions since there is no charge enclosed in the region, however there is an electric field in the i direction for both (1) and (2) - according to my solutions guide - and I can't see why since there seems to be no charge enclosed.

Thanks for the help.
 

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Answers and Replies

  • #2
BruceW
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the charge distribution outside the Gaussian surface is not symmetrical, so you cannot use that argument. Its going to be slightly more difficult than that. What principles do you think you can use to make this problem easier?
 
  • #3
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So for my argument to work the outer charge distribution has to be symmetrical? Why is that exactly?

But taking that into consideration I can see that it is symmetrical with respect to the y and z axes which would explain why the electric field has an x component only.
 
  • #4
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I know that there is a superposition of forces involved but I'm having difficulty seeing exactly how/where it applies
 
  • #5
BruceW
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So for my argument to work the outer charge distribution has to be symmetrical? Why is that exactly?

If you use a Gaussian surface inside the cavity, and don't use symmetry arguments, then what exactly does Gauss' law tell you?
 
  • #6
BruceW
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I know that there is a superposition of forces involved but I'm having difficulty seeing exactly how/where it applies

Yes, superposition is the key to making this problem easier. you need to think outside the box a little bit, but once you realise the trick, you won't forget it again :)
 
  • #7
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Gauss' law relates flux and charge and flux measures the electric field over a given area. Symmetry arguments are used to define the area through which the electric field "flows" correct? so if there is not enough symmetry to define a solvable integral then it dosen't allow you to define flux and so gauss' law doesn't tell you anything.

Am I on the right track there?

As far as the problem goes...

Since there is a symmetry about the y and z axis the flux entering and exiting must be equal and hence the electric field in those directions must cancel and equal zero.

I tried finding the field of a sphere that included the point I was interested in using E = (rho*b)/(3 * eps) and got the correct answer, but I still don't feel like I understand this problem.....
 
  • #8
BruceW
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Gauss' law relates flux and charge and flux measures the electric field over a given area. Symmetry arguments are used to define the area through which the electric field "flows" correct? so if there is not enough symmetry to define a solvable integral then it dosen't allow you to define flux and so gauss' law doesn't tell you anything.
Yes, pretty much. Gauss' law for charge distributions without symmetry still does tell us something, but it doesn't give us a simple integral.

armolinasf said:
Since there is a symmetry about the y and z axis the flux entering and exiting must be equal and hence the electric field in those directions must cancel and equal zero.
hmm. I'm not sure what you mean here. Do you mean that if we imagine a small Gaussian cube inside the cavity, then the flux over the faces perpendicular to the x axis will cancel due to symmetry? But the question asks us to calculate the electric field at a point which is not on the x axis. So I am not convinced by this argument.

armolinasf said:
I tried finding the field of a sphere that included the point I was interested in using E = (rho*b)/(3 * eps) and got the correct answer, but I still don't feel like I understand this problem.....
You used E = (rho*b)/(3 * eps) to get the answer? But that is the answer already (without the direction).

A hint: In this question, we want a symmetric charge distribution. Or at least a superposition of two symmetric charge distributions, right? So for this problem, we could view it either as one non-symmetric charge distribution, or...
 

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