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Homework Help: A problem in Griffith's textbook of electrodynamics, P189.

  1. Sep 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Exercise 1:
    Suppose a infinte plane z=0 divides the 3-dimensional space into two parts. The region below z=0 is filled with linear dielectric material of susceptibility [itex]\chi[/itex]. The bound charge density at z=0 is [itex]\sigma[/itex]. Find the electric field of the two regions.

    Exercise 2:
    The original problem in Griffith's book is to calculate the force on a point charge q situated a distance d above the the origion (x=0,y=0,z=d). The first step is to find the electric field [itex]E_z[/itex] just inside the dielectric at z=0, which is due in part to q and in part to the bound charge itself. And he states that the latter contribution is

    In my opinion, we should write two equations(from Guass's law in terms of E and D respectively):
    [tex]\epsilon_0 E_1+\epsilon_0(1+\chi)E_2=0[/tex]
    Then [itex]E_2[/itex] is my answer, which is quite different from the expression in Griffith's book [itex]-\sigma/2\epsilon_0[/itex].

    I need a detailed solution to the electric field of this configuration.
    Last edited: Sep 10, 2012
  2. jcsd
  3. Sep 10, 2012 #2
    What's the meaning of those two equations you write?
  4. Sep 10, 2012 #3


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    Gold Member

    Why, in your opinion, should we write these two equations?

    Keep in mind that the boundary condition Eq 4.40 applies to the normal component of the total electric field when crossing a surface from one linear dielectric to another, not just the part of the (normal component) field due only to the surface charge there (which is what Griffiths' claims is [itex]\frac{\sigma}{2\epsilon_0}[/itex]).
  5. Sep 10, 2012 #4
    These two equations are two expressions of Gauss's law in terms of E and D respectively.
  6. Sep 10, 2012 #5
    I modified my question as you can see. Now as to Exercise 1, do these two equtions hold?
    Then the total electric field [itex]E_z[/itex] should be the summation of the two parts.
  7. Sep 10, 2012 #6
    You're right! However, The "E2" you write is not the contribution of the induced charge. The eletric field contribution of the induced charge is - sigma/(2*epsilon)

    The E2 which you write is the "total" electric field just below the plate due to charge q and those induced charge.
  8. Sep 10, 2012 #7
    I changed the way I put my question. What I don't understand is why the contribution from the bound charge is [itex]-\sigma/(2\epsilon_0)[/itex]. That is why I put forward Exercise 1, to consider the part of electric field from the bound charge. In this case, I think the equations are OK. Then [itex]E_2[/itex] is the result we are looking for, which can be added directly to the total electric field because of superposition principle.
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