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Gauss' Law in Dielectrics Part II

  1. Nov 17, 2004 #1
    OK I just want to start from the beginning and try to get the first part of this problem so I can get what is going on in my head and understand it. Here's the problem:

    A point charge q is imbedded in a solid material of dielectric constant K.

    A) Use Gauss's law as stated in equation [tex]\oint{K \vec{E} \cdot \vec{A}} \;=\; \frac{Q_{free}}{\epsilon_{0}}[/tex] to find the magnitude of the electric field due to the point charge q at a distance d from the charge.

    So how does it effect what is going on? I mean the dielectric? I don't get it!
  2. jcsd
  3. Nov 17, 2004 #2
    E = q/(4*pi*K*(epsilon_0)*(d^2))
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