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Uniform Charge on Sphere

  1. Jan 7, 2008 #1
    Charge is distributed on the surface of a spherical conducting shell. A point particle with charge q is inside. If polarization effects are negligible the electrical force on the particle q is greatest when:

    a. it is near the inside surface of the balloon

    b. it is at the center of the balloon

    c. it is halfway between the balloon center and the inside surface

    d. it is anywhere inside (the force is same everywhere and is not zero)

    e. it is anywhere inside (the force is zero everywhere)

    I think e is the right answer, but apparently this is incorrect. If polarization effects are negligible, then that means the uniform charge distribution will be constant and the point particle should experience no net electrostatic force inside the shell.
     
  2. jcsd
  3. Jan 7, 2008 #2
    << solution deleted by berkeman >>
     
    Last edited by a moderator: Jan 8, 2008
  4. Jan 7, 2008 #3

    berkeman

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    Sorry, pixel. I've moved this to homework help, and deleted your solution. This is really a homework/coursework question, so the homework rules apply (no answers given out).
     
    Last edited: Jan 8, 2008
  5. Jan 7, 2008 #4

    berkeman

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    breez, start by thinking of two separated charges, and you move the test charge from directly between them, over closer to one of the two. Write the equation for the net E field and force on your test charge. What does it do as the test charge gets closer and closer to one of the two charges?
     
  6. Jan 8, 2008 #5
    Can you explain why my solution is incorrect?
     
  7. Jan 8, 2008 #6
    My reasoning is based on this:

    Theorem:
    If a charged particle is located inside a shell of uniform charge, there is no net electrostatic force on the particle from the shell.
     
  8. Jan 8, 2008 #7

    berkeman

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    You might be right, but that's not intuitive for me. What theorem exactly are you referring to? Can you provide a pointer?
     
  9. Jan 8, 2008 #8

    berkeman

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    Hmmm. I think I misread the question before. The charges would be on the outside surface of the conducting shell, no? It does seem then that the field inside would be zero. Like the 2nd example at this page, but with the inside of the solid conducting sphere hollowed out:

    http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/elesph.html
     
  10. Jan 8, 2008 #9

    mda

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    Yes... we can get to this from Gauss' law. Since we ignore polarization due to the test charge we know that all fields will have radial symmetry, and the result follows. (Actually for this question it doesn't matter which surface holds the charge, but Gauss' law and the property of metals indicates the outside).
     
  11. Jan 8, 2008 #10
    Yeah, this question seems really trivial with this theorem, but my physics class is online, and solution E was marked incorrect on the website that my course is based on. Perhaps it was an error? Is there a nuance to this question I am missing?
     
  12. Jan 8, 2008 #11

    berkeman

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    The only clue that I see at this point is that the word "balloon" is used in the body of the question. A balloon does not generally have a conducting surface. Have you posted the exact, total question with all of its text? Is there a figure or other information that you can add?
     
  13. Jan 8, 2008 #12

    mda

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    I was thinking the same thing, although being a insulator doesn't change anything if we are still ignoring polarization. Spherical charge distribution means zero interior field.
    The only thing I can see that would result in a non-zero force for an interior charge is asymmetric charge distribution, either due to an asymmetric geometry or allowing polarization effects.
     
  14. Jan 8, 2008 #13

    Doc Al

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    A conducting shell will have charge on its inner surface as well as its outer surface; the distribution of that inner surface charge depends on where the point charge is.

    Of course, if this is a non-conducting balloon...
     
  15. Jan 8, 2008 #14

    mda

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    True, but I would call this polarization, which we are told to ignore.
     
  16. Jan 8, 2008 #15

    Doc Al

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    I would not call charge rearrangement on a conductor polarization. If it's a conducting shell and you ignore the inner surface charge--what's the point of the problem?

    Of course if it's a non-conducting shell--like a balloon--then the answer is clear.
     
  17. Jan 8, 2008 #16
    True, this question seems pointless without the shell being a conductor, but the question also states to ignore polarization, the rearrangement of electrons in a conductor. Thus imho, this questions is very poor in that first we are told the shell is a conductor, then in the same sentence told to treat it as if an insulator.

    Yeah, so if charge rearrangement is present, I think the answer should be a, near the inside surface of the balloon.

    My reasoning: As the particle moves near the inside surface, the charge on the shell moves away from that point the particle is nearing, and hence a difference in charges builds up and the electrostatic force hence increases.

    Also, my textbook uses the terminology "polarization" to describe charge rearrangement.
     
  18. Jan 8, 2008 #17
    If we are talking about what is happening on the inside of the shell then we need to know where the charge is on the inside. I don't see that anywhere. (Edit: Looking at all the posts, Doc Al already said this, sorry for repeating.)

    Also, your reasoning seems incorrect. As the proton, or whatever it is, moves near the surface then the conductor will build up a collection of negative charges so as to make the field inside the conductor be zero. The charge will not move away, it will move towards. On the outside of the shell the positive charge will be uniformly distributed with total charge +q.
     
    Last edited: Jan 8, 2008
  19. Jan 9, 2008 #18

    Doc Al

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    I suspect it was a typo. Typical balloons are not conductors. (Edit: On the other hand, breez already said that answer E was marked wrong.)

    Right.

    I suspect you mean that if the particle is positively charged, that the positive charges on the shell move away. Right idea, but in most conductors (like metals) it's only the electrons that can move--so they move towards the particle. (Mindscrape pointed this out.) I'm sure you also realize that if within the conducting shell is a positively charged particle, then the inner surface of the shell will have an equal negative charge.

    You're right. "Polarization" is used to refer to any charge rearrangement. (My bad before.)
     
    Last edited: Jan 9, 2008
  20. Jan 9, 2008 #19

    Shooting Star

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    One thing just struck me, and it has got nothing to do with the Physics problem at hand. When berkeman deleted pixel's post so that answer couldn't be seen by the OP, already a mail has gone to the OP with pixel's post, and he would have already seen pixel's solution. (But of course, others could not.)

    Now, after a prolonged discussion after which no clearcut solution has emerged, due to various reasons, it would be all right to read pixel's post, I feel.
     
  21. Jan 9, 2008 #20

    berkeman

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    You are correct that since pixel's post was directly after the OP, the OP would have received e-mail notification of the post with the contents (if the OP has that option turned on in their profile). My deleting the answer was more in keeping with the PF policy of no direct answers.

    Pixel's thought was the same as I was trying to hint at with my post following his (strongest near the surface of the sphere), but by Gauss' law, apparently that shouldn't be correct either.
     
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