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Griffiths Problem 4.18

  1. Mar 16, 2008 #1
    [SOLVED] Griffiths Problem 4.18

    1. The problem statement, all variables and given/known data
    This question refers to Griffiths E and M book.

    Obviously we are supposed to use equation 4.23 here. But for that to be of any use, we need to assume that the electric displacement is orthogonal to the plates. We are told that the medium is linear, but that only means that [itex]\vec{D} = \epsilon \vec{E} [/itex] where [itex]\vec{E}[/itex] is the total electric field. What we need to assume is that [itex]\vec{D} [/itex] is parallel to [itex] \vec{E_0} [/itex] where [itex] \vec{E_0} [/itex] is the field that would be present if both slabs were removed. You cannot use the analysis below Figure 4.21 to show that because this is definitely a case that Griffiths warns us about in the paragraph above Figure 4.21. Therefore I do not see why [itex]\vec{D} [/itex] and [itex] \vec{E_0} [/itex] are parallel.


    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Mar 16, 2008
  2. jcsd
  3. Mar 17, 2008 #2
    anyone?
     
  4. Mar 17, 2008 #3

    G01

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    In that problem, the electric displacement is assumed to be orthogonal to the plates. Remember that the analysis of a parallel plate capacitor assumes E is parallel to the the plates, by assuming the plates are infinite. Similar reasoning is done with the displacement here as well.

    Check out the paragraph at the bottom of p182 in Griffith's, ending with Eq. 4.35 and 4.36. This explains the reasoning being used in solving this problem.
     
  5. Mar 17, 2008 #4
    But that reasoning only applies in the case that Figure 4.21 (or something similar) does not apply. But here we have two slabs of linear dielectric, so it is pretty clear that something similar to Figure 4.21 does apply.

    Also, why can we assume that D = 0 inside the metal plate? I do not see why that follows from E = 0. Is it true that there is never a "frozen-in" polarization inside of a conductor?
     
    Last edited: Mar 17, 2008
  6. Mar 17, 2008 #5

    G01

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    Sorry. I left my Griffith's book in the lab today so I don't know what figure you are referring too.

    As for your second question, I think it's safe to assume that for a perfect conductor, which we are talking about here will not have any frozen-in polarization.
     
  7. Mar 17, 2008 #6

    Ben Niehoff

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    You can establish the directions of E, D, and P from symmetry arguments alone:

    1. Consider the slabs to be infinite planes (this is the assumption you are making when you ignore fringing effects).

    2. A stack of parallel, infinite planes has both rotational and translational symmetry. It is entirely uniform and isotropic in the xy directions.

    3. Therefore, any solutions to the problem (be they scalar or vector fields) must possess the SAME symmetry.

    4. The only vector fields possessing this symmetry are those directed along a line perpendicular to the planes.

    Therefore, each of E, D, and P is perpendicular to the conducting planes.
     
  8. Mar 17, 2008 #7
    Why does 3 follow from 2?
     
  9. Mar 17, 2008 #8

    Ben Niehoff

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    Think what the consequences would be if it were not so. I.e., if you could set up two infinite parallel plates, how could you expect that the symmetry would be spontaneously violated in the D field?
     
  10. Mar 17, 2008 #9
    That's certainly the intuition, but there should be a mathematically rigorous way to prove that. This is the heart of my question and I guess I should have asked this when Griffiths said "by symmetry, E points away from the plane" in Example 2.4. But maybe that is something that is too advanced for Griffiths and that I will learn in Jackson...

    http://en.wikipedia.org/wiki/Spontaneous_symmetry_breaking
     
    Last edited: Mar 17, 2008
  11. Mar 17, 2008 #10

    Ben Niehoff

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    For this particular problem, it is actually enough to know that the two exterior plates are conducting. The E field is always perpendicular to a conducting surface in the vicinity of that surface. Since the dielectrics in this problem are presumed to be homogeneous and linear, then we have

    [tex]\vec D = \epsilon \vec E[/tex]

    and therefore D is perpendicular to the plates also. That D is vertical all throughout the region of interest then follows from the Laplace equation.

    For more general problems, I'm sure there is a proof that the fields must share the symmetries of the charge configuration, but I'm a bit too busy to try to prove it right now.
     
  12. Mar 18, 2008 #11
    But
    [tex]\vec E_{vac} \neq \vec E[/tex]

    All we know is that [tex]\vec E_{vac} [/tex] is perpendicular to the conducting surface. We want to show that [tex]\vec E [/tex] is as well. Anyway, it does seem intuitive and I understand that you are busy. I'll just mark this as solved and leave the mathematical proof as a black box for now.
     
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