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Griffiths page 124

  1. Feb 11, 2008 #1
    [SOLVED] Griffiths page 124

    1. The problem statement, all variables and given/known data
    Stop reading if you do not have Griffith's E and M book.

    Can someone explain why he says "But in the second case only the upper region contains a nonzero field, ... " ?

    I don't understand why the field is necessarily zero in the lower region?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 11, 2008 #2

    kdv

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    Because the lower region is inside a conductor, no?
     
  4. Feb 11, 2008 #3
    The conductor is just the x-y plane I think.
     
  5. Feb 11, 2008 #4

    Ben Niehoff

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    Take a spherical conductor. Remember that the potential is constant on the inside. Now simply take the limit as the radius goes to infinity. Now, you have an infinite conducting plane. One side of it (half-space) is "outside", the other side is "inside". And, the region inside a conductor has zero electric field (and therefore zero energy density).

    Thus, the conducting plane divides infinite space in half; one half is occupied by charges, but the other half is empty. The field within the latter half is zero.

    When you solve the problem via images, you are introducing a fictitious charge on the "other side" that replicates the boundary conditions. When you calculate the total energy, you have to ignore the energy due to this fictitious charge.

    Alternatively, you can say that the field on the "other side" of the conductor due to the charge q is exactly canceled by the field due to the surface charge distribution on the conductor. Again, since the field is zero, the energy density in that half of space is zero.
     
  6. Feb 11, 2008 #5
    Do you mean circular? And do you mean filled or hollow?
     
  7. Feb 11, 2008 #6

    Ben Niehoff

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    No, spherical. An infinite plane can be considered a sphere of infinite radius.
     
  8. Feb 11, 2008 #7
    What?!!!? In the problem, the x-y plane is the conductor. There is now way you can blow up a sphere large enough so that the it will coincide with the x-y plane since it will only be locally flat. You're example might make sense if it were a circular disk. I have no idea what you are saying and why you are using a sphere. Please clarify.
     
  9. Feb 11, 2008 #8

    Ben Niehoff

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    Yes, a sphere is only locally flat. But if you blow it up to infinite radius, while simultaneously taking the center infinitely far away, you will end up with a plane. Essentially, that "locally flat" part expands until it fills the entire xy plane. It's the same reason the Earth looks flat, even though it isn't. The difference is that the Earth has only a finite radius--in the case of an infinite radius, the "looks flat" part becomes all that matters.

    At any rate, you can just take a cube of infinite side length, if you want. The point is that one half-space is "inside", and the other half-space is "outside", where the infinite plane serves as the boundary between inside and outside.

    And as you well know, the electric field inside a conductor is zero.

    Here is a crude drawing:

    Code (Text):
     
     
           * Q (point charge)

          (outside)
    -------------------------------------- (infinite conducting plane)
          (inside)
     
     
     
     
    The electric field is zero on the "inside".
     
  10. Feb 11, 2008 #9
    But its only inside the "meat" of the conductor that the electric field is zero. See the first sentence on page 90. That's why I was asking whether your conductor was "hollow" or not. Apparently it is hollow, and thus I don't think we can say just a priori that the field inside is zero. But maybe I am mistaken.
     
  11. Feb 11, 2008 #10

    Ben Niehoff

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    It doesn't matter whether the conductor is hollow or not. Often, it doesn't even matter whether the shell of the conductor has holes in it or not! This is why a Faraday cage works.

    If a (solid) conductor has a net charge, then all of the charge resides on the surface. From this, it should be clear that the conductor can be hollowed out without changing anything.

    It is, perhaps, a curious mathematical fact that the field within a region R can always be canceled by some appropriate distribution of charges on the boundary of R. This follows from the Laplace equation.
     
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