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Homework Help: Gauss's law in Dielectrics

  1. Oct 4, 2004 #1
    Hey everyone, this is my first question on here, but I've seen you all be very helpful to others to I was hoping someone could clarify something for me.




    1. How does Gauss law modify in dielectric materials?
     
  2. jcsd
  3. Oct 4, 2004 #2

    Claude Bile

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    Gauss' law in matter is given by;

    [tex] \nabla.\vec D = \rho_{f} [/tex]

    Where D is the electric displacement vector and rho is the free charge density. D is realted to the Electric field thus;

    [tex] \vec D = \epsilon \vec E [/tex]

    Where epsilon is the permittivity of the medium.

    Claude.
     
  4. Oct 5, 2004 #3
    Hmm okay not sure I follow you.

    So what exactly is different about it compared to in any other material?
     
  5. Oct 5, 2004 #4

    Claude Bile

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    Sorry, I misread your earlier post.

    In a dielectric, the free charge density will be zero.

    Claude.
     
  6. Oct 5, 2004 #5
    Ahhh nice one,

    I think I get it now,

    Cheers matey
     
  7. Oct 6, 2004 #6

    ehild

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    Claude, this is wrong. There can be free charges in a dielectric. You thought of metals, I think.

    Gauss's law is the same for all substances: [tex]Div \B{D}=\rho_{free}[/tex].

    [tex] \B{D}=\epsilon\B{E}[/tex]. The difference is in the permittivity [tex]\epsilon[/tex].

    ehild
     
  8. Oct 6, 2004 #7

    Claude Bile

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    Yes, you are quite right.

    In optical media , Gauss' Law is frequently approximated to be divD = 0.

    Thanks ehild for pointing out by error.

    Claude.
     
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