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Gauss Law

  1. Jul 30, 2004 #1
    if u have an uncharged,perfectly conducting hollow sphere of thickness t and radius R, and a point charge Q is placed a distance D from the centre of the sphere so that R-t>D, what would the electric field, E, be?
    i just treated it the same as if Q was in the centre of the sphere and the electric field lines came perpendicular to the sphere and so

    [tex]E=\frac{Q}{4 \pi \epsilon_{0}D^2} [/tex]

    anyone got any better ideas?
     
  2. jcsd
  3. Jul 30, 2004 #2
    There is no reason to assume that you can do that. One the charge is offset from the center you've changed he symmetry of the situation.

    Pete
     
  4. Jul 30, 2004 #3

    Galileo

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    The point charge induces a charge (-Q) on the inner surface as to cancel the field inside the conductor. The remaining induced charge (Q) will distribute itself over the surface of the sphere.
    I'm pretty certain it will spread uniformly so that the electric field you got is correct.
     
  5. Jul 30, 2004 #4

    Doc Al

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    I assume you meant to write:
    [tex]E=\frac{Q}{4 \pi \epsilon_{0}r^2} [/tex]
    That would be correct for the field for r > R. (r = 0 is the center of the sphere.)

    See my answer here: https://www.physicsforums.com/showthread.php?t=37090
     
  6. Jul 30, 2004 #5

    turin

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    There is no electric field inside the conductor. The conductor accomplishes this by arranging a layer of charge in an appropriate distribution on the outer surface of the sphere with a total amount equal to -Q. The inner surface of the sphere therefore develops a negative charge deficit equal to this (+Q) and also distributed appropriately (since there is no field inside the conductor). This all results in a negative image charge somwhere below the surface of the sphere, not quite in the center, and not necessarily near the surface, and not necessarily equal to -Q. I don't remember what the expression is off the top of my head; sorry. Look up the "method of images."
     
  7. Jul 31, 2004 #6
    lol, thanks for the responses guys, very much appreciated
    but i'm still completely stumped by the question
    just to make it clear i want to find an expression for the electric field outside the sphere (after reading my first post i didnt think my question was very clear)

    thanks again
     
  8. Jul 31, 2004 #7

    Doc Al

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    What's stumping you? I gave the expression for the field outside the sphere in my first response. Now... to explain why that's the right answer requires a little more...
     
  9. Jul 31, 2004 #8

    Doc Al

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    Right.
    I believe you have it backwards. The -Q is induced on the inner surface. This surface charge is arranged to exactly cancel the field from the charge within the cavity, thus producing zero field everywhere inside the conductor.
     
  10. Jul 31, 2004 #9
    i just typed out my response and realised something. i couldnt see the difference between the formula i gave and the formula u gave but now i think i see it.
    so by the sounds of it, it doesnt matter where you place the charge Q inside the sphere, E outside the sphere depends on the position only from the centre of the sphere. is this right?
     
    Last edited: Aug 1, 2004
  11. Aug 1, 2004 #10

    Doc Al

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    That is correct.
     
  12. Aug 1, 2004 #11

    turin

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    Doc Al,
    Oops. I read the post too quickly. I was responding under the false impression that the point charge, Q, was placed outside the conductor. Sorry to all.
     
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