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Gauss's Law Charged Hollow Conductor

  1. Jan 20, 2007 #1
    if there is not a uniform distribution of charge in the middle how is it possible that there's a uniform field outside. I know why it's true according to gauss's law but it doesn't make sense to me. if theres some positive charge near the interior wall somewhere then there will be an accumulation of negative charge near there in the conductor and therefore an accumulation of positive charge on the outside of the conductor near that spot. Please don't answer this using gauss's law. i would like a qualitative analysis

    this is not a hw question!
     
    Last edited: Jan 20, 2007
  2. jcsd
  3. Jan 20, 2007 #2
    Got a couple "shift" buttons on your keyboard? Spare some.
     
  4. Jan 20, 2007 #3
    :tongue: ..
     
  5. Jan 20, 2007 #4
    Do you mean the integral version of gauss's law? This version cannot be applied unless the charge is uniform according to certain special symmetries.

    Otherwise you mean the differential form of Gauss' law, in which case I can't understand the boundary conditions you are describing.
     
  6. Jan 20, 2007 #5
    Last edited: Jan 20, 2007
  7. Jan 20, 2007 #6
    I believe the positive charge that causes the accumulation of negative charge will cancel out the effect outside of the conductor.

    after all, if you only consider the negative charges when applying guess's law, you will get a non-zero electric field as well inside a conductor.
     
  8. Jan 20, 2007 #7
    the question i have pertain to the permutation of gauss's law that yield's a faraday cage. where whatever is inside is obfuscated by the conductor
     
  9. Jan 20, 2007 #8
  10. Jan 20, 2007 #9

    Gokul43201

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    The field outside a charged cylinder is not uniform - it falls away with distance from the cylinger.
     
  11. Jan 20, 2007 #10
    that's not what i mean. firstly i mean spherical conductor , secondly im questioning why it is the same at every point on the surface
     
  12. Jan 20, 2007 #11

    Gokul43201

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    Symmetry.

    Is there any reason to expect the charge density to be different at points [itex](r_1, \theta _1, \phi _1) [/itex] and [itex](r_1, \theta _2, \phi _2)~ [/itex] ?
     
  13. Jan 20, 2007 #12
    yes? because i said so. i obviously understand what symmetry means and what it implies? pictures in my book and according to my professor it doesn't matter where the charge is inside the hollow conductor the efield will be the same at every point on the surface
     
  14. Jan 22, 2007 #13

    Gokul43201

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    Just because you say so, doesn't mean it's physical. How do you make an asymmetric charge density in a conducting spherical shell (without applying an external field)? If it were non-conducting, I'd understand ...but in that case, the surface field would not be the same everywhere.
     
  15. Jan 22, 2007 #14
    not inside the conductor, inside the cavity
     
    Last edited: Jan 22, 2007
  16. Jan 22, 2007 #15

    Gokul43201

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    There are charges inside the cavity? What's holding these charges in place, preventing them from flying apart towards the shell?
     
  17. Jan 23, 2007 #16
    do i really have to do this? fine? suppose there is a spherical insulator with a charge place closed to it's surface. around this spherical insulator is wrapped a conductor with some charge. happy now?
     
  18. Jan 23, 2007 #17

    Gokul43201

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    Yes, the situation makes more sense now.

    You are picturing the induced charge on the inside surface of the conductor creating an oppositely induced charge on the outside surface near the location of the excess charge. The flaw is in assuming that that the charge "induced" on the outer surface must, in some manner, "reflect" the charge induced on the inner surface. This is not true, and it turns out that a uniform charge is induced all over the outer surface.

    To get a better intuitive grasp of this, ask yourself what it means to say that the charge built up on the inner surface induces charge on the outer surface. Really, all that happens, is that free electrons redistribute themselves in response to say, a positive charge placed near the inside surface. In other words, free electrons from all parts of the conductor approach the inner surface, near the point where the positive charge lives. In doing so, they leave behind positively charged pockets where they came from. These pockets can also be treated as though they are mobile charges. So, these pockets, or holes, will want to get as far away from each other as possible, and they do this by:
    (i) migrating to the outer surface, and
    (ii) maintaining a uniform density there.

    These holes, moreover, do not feel any effect from the electrons that are crowded near one spot on the inside, because, in the steady state, no electric field can propagate through the body of the conductor itself.
     
  19. Jan 23, 2007 #18
    i understand everything else except for this. why doesn't the field of the induced negative charge inside the conductor on the interior side propagate through the conductor? in fact i see a paradoxical thing in the way you explained it. the left over holes or positive repel each because of what else but their efield.

    edit just so you know im not trying to disprove guass's law as i have faith in my forefathers, i just want a clear understanding
     
  20. Jan 23, 2007 #19

    siddharth

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    Because, the field due to the induced charges exactly cancels with the field due to the charge in the cavity at all points in the conductor (Note that this is not so inside the cavity). Therefore, the asymmetrical effect of the charges present in the cavity and on the inner surface is not felt and the remaining charges distribute uniformly on the outer surface, as Gokul pointed out.
     
    Last edited: Jan 23, 2007
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