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A point charge inside infinite medium of dielectric

  1. Nov 2, 2009 #1
    A point charge inside infinite dielectric material

    1. The problem statement, all variables and given known data
    A point charge is an infinite medium of dielectric material having a relative permittivity [tex]\epsilonr[/tex]. <--- epsilon(sub r). Find the electric field vector and the potential function at any point in space, assuming that the potential is zero volts at infinity.
    2. Relevant equations
    D = [tex]\epsilon[/tex] * E + P
    P = [tex]\epsilon0[/tex] * [tex]\chie[/tex] vector E
    [tex]\epsilon[\tex] = [tex]\epsilon0[/tex] * [tex]\epsilonr[/tex]
    [tex]\chie[/tex] = [tex]\epsilonr[/tex] - 1


    well its not letting me put it in right so im gonna enter them in with () next to them
    D(vector) = epsilon*E(vector) + P(vector)
    P(vector) = epsilon(sub 0)*chi(sub e)*E(vector)
    epsilon = epsilon(sub 0)*epsilon(sub r)
    chi(sub e) = epsilon(sub r) - 1
    Electric Field of Point Charge = k*q/r^2
    Electric Field of Sphere = q/(4*pi*epsilon(sub0)*r^2)

    3. The attempt at a solution
    Im not quiet sure what my first step would be. I would think that a point charge inside a dielectric medium would have a reduced electric field, but it is infinite so wouldnt its electric field be nothing?
     
    Last edited: Nov 2, 2009
  2. jcsd
  3. Nov 2, 2009 #2

    gabbagabbahey

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    Just use Gauss' Law for [itex]\textbf{D}[/itex]....can you think of a Gaussian surface that will exploit the symmetry present?
     
  4. Nov 2, 2009 #3
    what do you mean by exploit?
     
  5. Nov 2, 2009 #4

    gabbagabbahey

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    "Exploit" is just another word for "use"
     
  6. Nov 2, 2009 #5
    I feel that a sphere represents a good shape.
     
  7. Nov 2, 2009 #6

    gabbagabbahey

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    Okay, but why? And where will the center of the sphere be?
     
  8. Nov 2, 2009 #7
    The center of the sphere will be located on the point charge
     
  9. Nov 2, 2009 #8
    so would my answer just be q/(4*pi*epsilon(sub0)*r^2)
    ?
     
  10. Nov 6, 2009 #9

    gabbagabbahey

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    Not quite...What is Gauss' Law in terms of [itex]\textbf{D}[/itex]? Do you see why Gauss' Law in terms of [itex]\textbf{E}[/itex] isn't helpful here?
     
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