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Application of Gauss' law - point charge in hollow sphere

  1. Dec 11, 2016 #1
    Firstly I appologize, that I am not native english speaker and I dont study physics(but cybernetics we are getting just some general knowledge about physics), but hopeffuly I will write this right.
    1. The problem statement, all variables and given/known data

    We know that inside of a conductive object is protected from influence of outside electrical charges.


    2. Questions

    a. Analogically, is outside of the conductive object protected from influence of inside electrical charges? For example from point charge Q situated inside the hollow sphere at a random spot.
    b. The point charge is directly in the middle of the cavity of hollow sphere. Consider what is specific about this setup.
    fyzika.jpg


    3. The attempt at a solution

    a. If I place a point charge Q+ inside the sphere, then because the sphere is conductive( therfore has free electrons ), the electrons will move to the inside of the shell to "cancel" electric field from inside charge. This will leave the outside surface of the shell positively charged, therfore creating electromagnetic field on the outside. Meaning outside of the shell is not protected from the inside charge. That is what I came with, but I am supposed to proof this by some equation, I dont really know where to start here.....

    b.I guess - by the picture - I am supposed to express E. I can divide this problem into three parts: inside the cavity of the shell, outside the shell and in the shell object itself.

    1. Inside cavity: For inside cavity I used standard Gauss law therefore i got :
    [tex]E = \frac{1}{4\pi\epsilon}\cdot\frac{Q}{r^2}[/tex]
    2. In the shell: Here the E should be 0 after electrons cancel the point charge and stop moving. Again I should proof this but I dont really know where to start.

    3. Outside: I am not really sure here, but I guess it should be:
    [tex]E = \frac{1}{4\pi\epsilon}\cdot\frac{Q}{(r-R_2)^2}[/tex]

    So I would be very happy if you could help me with the two proofs and if you could check, if the rest of my thoughts are allright. Thank you in advance.​
     
    Last edited: Dec 11, 2016
  2. jcsd
  3. Dec 11, 2016 #2
    Gauss' law is true for any closed surface you care to consider. What does it tell you if you make a Gaussian sphere larger than your conductor?
     
  4. Dec 11, 2016 #3
    Ok, so outside the conductor it should be :
    [tex]E = \frac{1}{4\pi\epsilon}\cdot\frac{Q}{r^2}[/tex]
    Q will be same as inside the conductor.

    And in the shell I think it will look like this:
    [tex]E = \frac{1}{4\pi\epsilon}\cdot\frac{-Q_-+Q_+}{r^2} = 0[/tex]
     
  5. Dec 11, 2016 #4
    Yep.
     
  6. Dec 14, 2016 #5
    Thank you...
    I have got one supplementary:
    "Would the E in the shell be also 0 if the point charge wouldnt be exactly in the middle?"
    It should be, but there is a problem calculating
    [tex]\vec{E}\cdot\vec{n} =
    |\vec{E}|\cdot|\vec{n}|\cdot cos(\alpha)[/tex]
    because the angles are all different, so I have to come up with some generalization. Here, I dont really know where to start at all..
     
  7. Dec 14, 2016 #6
    The field inside an ideal conductor must always be zero. The charges are free to move, so they will always move until the field is balanced and they feel no more force.

    And yes the problem is harder to calculate when the symmetry is broken.
     
  8. Dec 14, 2016 #7
    Sorry, I didn't answer the rest of your question. With the charge off center I don't think there is a short cut. I think you have to solve the Poisson equation with Dirichlet boundary conditions. This procedure is one of the least favorite tasks for first year graduate students and I probably shouldn't try to describe it in this forum. To tell the truth I haven't thought about it since graduate school because I solve all such problems numerically. Differential equations all become simple when you have a computer!
     
  9. Dec 14, 2016 #8
    On the other hand you can pretty much guess what the equipotential lines will look like, and a lot of reasoning can be done without an exact solution.

    OOOHHH!!! It's hard to calculate what happens inside the conductor, but given that it must be an equipotential you should be able to use Gauss law to say something very interesting about the field outside.
     
  10. Dec 14, 2016 #9
    Oh yeah, I forgot to write, that I am supposed to do it for the conductor. Dont really know what am I actually supposed to do, because I dont think Its possible to calculate it for the inside...
     
  11. Dec 14, 2016 #10
    Good. Like I said, by definition it has to be an equipotential and Gauss law applied at the outside will tell you something very interesting.
     
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