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Charge on inner surface of conducting shell

  1. Jan 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Three thin (but finite thickness) concentric conducting shells have radii R1 < R2 < R3 and charges -2Q, +Q, and +3Q respectively, where Q is some positive number. What is the charge on the inner surface of the outmost shell?

    2. Relevant equations

    Charge on a conductor is always on outermost surface?

    3. The attempt at a solution

    So therefore the charge on the inner surface of the outermost shell should be zero? I do not know if this is right. So basically the charge of each shell is transferred to the outermost surface of the outermost shell? I am confused. The answer is 0, right? Just out of curiosity then what's the charge on the outer surface of the outermost shell?
     
  2. jcsd
  3. Jan 18, 2009 #2

    Doc Al

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    That's not true.

    But what is true is that the electric field within the conducting material must be zero. (In the electrostatic case.) Use that fact and Gauss's law to figure out what the charge must be on the inner surface.
     
  4. Jan 18, 2009 #3
    OK wait, so, if there are 3 concentric conducting shells,

    The electric field within the actual solid part of each shell is 0? Is that what you mean?
    What is the electric field in the empty space between two of the shells?

    If the inner surface of the outermost shell had excess charge, why would it stay there? Wouldn't it transfer immediately to the outside?
     
  5. Jan 18, 2009 #4

    Doc Al

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    Yes.
    It depends on the charge contained within the shells. Use Gauss's law to find out.

    No, why should it? What holds it there is the field from the charge within the hollow shell. Again, use Gauss's law.
     
  6. Jan 18, 2009 #5
    Hey thanks a lot for your help by the way
    Also here is what I've done so far:

    R3=radius of outermost shell (and R2 and R1 respectively for middle and innermost shell)
    -2Q = charge on innermost shell
    +Q = charge on middle shell
    +3Q = charge on outermost shell

    Gauss law, so electric field on inner surface of outermost shell due to charge enclosed by inner 2 shells is
    k(-2Q +Q)/R3^2

    Electric field on inner surface of outermost shell due to charge of outermost shell... is 0?

    I know the total electric field within the solid part of the outermost shell should be 0, but I cannot figure out how to set it that way. I am adding K(-Q)/R3^2 to what to get 0? The problem does not provide me with the actual thickness of the shell... but I feel that I should be working with 2 different radii to be able to get a solution, but besides R3 why would I use a different one?

    Also I'm sorry for asking so many questions :X Thanks for your help so much!
     
  7. Jan 18, 2009 #6

    Doc Al

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    OK. These are the total charges on each shell. Those charges may be distributed over the inner and outer surfaces of each shell.

    Do this: Imagine a Gaussian surface right inside the conducting material of the outer shell. (What's the field there?) The charge contained within that surface consists of: The charges from the two inner shells plus whatever charge is on the inner surface of the outer shell. What must the total charge be within that Gaussian surface? What can you deduce from that?
     
  8. Jan 18, 2009 #7
    OK so total charge on outer shell = 3Q = charge on inner surface + charge on outer surface

    Total electric field inside = 0
    let charge on inner surface = w
    then charge on outer surface = (3Q - w)
    charge from inner spheres = -Q

    (-kQ + kw)/r^2 + (3Q-w)/r^2 = 0

    (w-Q) = (3Q-w)
    2w = 4Q
    w = 2Q

    so the charge on the inner surface of the outermost shell is +2Q

    idk if that is even remotely right?
     
  9. Jan 18, 2009 #8

    Doc Al

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    Good.

    Good. So what's the total charge within that Gaussian surface (mentioned in my last post) in terms of Q and w? And what must that total charge equal? (That will tell you what w is.)

    Not sure what you're doing here. Forget about trying to add fields.
     
  10. Jan 18, 2009 #9
    OOH sorry,

    so Gauss law:

    integral (E . dA) = Qenc/epsiolon = (w + -Q)/epsilon

    0 = Qenc

    w - Q = 0
    w = +Q = charge on inner surface (and therefore 2Q = charge on outer surface)

    wait is that all i have to do? I'm very sorry for being so slow to get it.. lol
     
  11. Jan 18, 2009 #10

    Doc Al

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    Now you've got it! :wink:
     
  12. Jan 18, 2009 #11
    your help is really appreciated!
     
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