Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linear dielectrics

  1. Jun 17, 2012 #1
    Say we have a linear dielectric with electric susceptibility χe a certain free charge density ρ. By using the fact that ∫D[itex]\cdot[/itex]da = Qfree-enclosed you can find the resulting electric field, because D = εE , where ε = ε0(1+χe).

    The above is very weird for me. It seems to me that you are getting out too much information compared to how much you have at the start.

    Let's look at it.

    We know that the free charges will induce a certain polarization. These will in turn induce and electric field which induces a polarization and so on. This is not easy to break up to infinity but we find for linear dielectrics that the total electric field due to polarization such that:

    E = ε0χeP

    But how do we have any information that allows us to compute either the total field in the end or the total polarization? All we know is the field due to the free charges and the constant of proportionality between the total resulting field and P. It just seems magical to me that you are actually able to calculate E or P with just this information.
     
  2. jcsd
  3. Jun 17, 2012 #2
    [itex]\vec{P}=\epsilon_{0}\chi_{e}\vec{E}[/itex]
    [itex]\vec{D}=\epsilon_{0}\vec{E}+\vec{P}=\epsilon_{0} \chi_{e}\vec{E}+\epsilon_{0} \vec{E}=\epsilon \vec{E}=\epsilon_{r} \epsilon_{0}\vec{E}[/itex]

    Basically, this is simply a way to account for the apparent reduced E field in a material of different dielectric constant. At least that's how I think of it.
     
    Last edited: Jun 17, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook