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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
    [tex]-\sigma/2\epsilon_0[/tex].


    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]
    [tex]\epsilon_0(E_1+E_2)=\sigma[/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

    gabbagabbahey

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    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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