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Griffiths (electrodynamics) problem 4.19, part b

  1. Dec 6, 2008 #1
    When finding the polarization of the dielectric portion between the two plates using P=[tex]\epsilon_{0}[/tex] [tex]X_{e}[/tex]E, why would we use the E-field in the air portion (as is done in the solutions manual and here: http://www.getofftheinternet.net/s_10077.pdf ) instead of solving for the E-field in the dielectric portion and using that to find P?

    Also, why is the surface charge density over the air segment equal to the TOTAL surface charge density? Isn't it the free charge density in this problem? I'm pretty confused, so any help will be massively appreciated.
     
    Last edited: Dec 6, 2008
  2. jcsd
  3. Dec 6, 2008 #2

    turin

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    P and E0 contribute to E in the material by vector summing. P is the part of E that is due to the charge separation, and E0 is the reason for the charge separation.
     
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