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Homework Help: Why separation of variables? (electrostatics)

  1. Dec 14, 2008 #1
    "A sphere of homogeneous linear dielectric material is palcced in an otherwise uniform electric field E. Find the electric field inside the sphere."

    Griffiths uses separation of variables to solve laplace's equation in the interior of the sphere. I have two questions.

    (1) How can you try to solve laplace's equation? isn't there bound charge in the dielectric (so the charge isn't 0?) laplace's equation can only be solved in regions where the charge is zero, right? or is that only free charge?

    (2) Why can't we use gauss's law for dielectrics (integral of the displacement flux equals the enclosed free charge) to find the field? is it because there is no free charge?
  2. jcsd
  3. Dec 14, 2008 #2
    (1) it's true that there's bound charge in the dielectric. Would you agree however that there's no free charge in the dielectric. If so, then we can use equation (4.22) of Griffiths (3rd ed.) and obtain:

    [tex]\nabla \cdot \textbf{D}=0[/tex]
    because [tex] \rho_f=0 [/tex]

    Since the material is a linear dielectric, then equation (4.32) gives us

    [tex] \textbf{D} = \varepsilon \textbf{E} [/tex]

    Substitution into our original equation gives us:

    [tex] \varepsilon \nabla \cdot \textbf{E} = 0 [/tex]
    [tex] \nabla \cdot \textbf{E} = 0 [/tex]

    Then using the conventional [tex] -\nabla \phi = \textbf{E} [/tex]

    gives us:
    [tex] \nabla^2 \phi = 0 [/tex]

    (However, the [tex] \varepsilon [/tex] dependence comes into the boundary conditions)

    (2) You can, but it won't be very useful because now you don't have any nice symmetries (for example, the field won't be radial, or something. It's true that in the end the field inside just points uniformly in the z direction, but you don't know that to begin with).
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