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Boundary conditions of electric field?

  1. Apr 13, 2015 #1
    I'm reading griffiths electrodynamics and I am confused about a concept. Mainly because I might be interpreting it in different ways. Why does the equation contain an E with a negative in front? Namely, E_below. Isn't the Electric field pointing away from the surface with the surface charge density and isn't the surface differential also pointing away from the surface? So when using gaus's law, shouldn't the equation be EA +EA =(1/εo)σA or 2E =(1/εo)σ rather than E_above — E_below =(1/εo) σ. This doesn't make sense unless we are talking about an external electric field that is propagating towards and across the surface. Because in that case the dot product of the external electric field and surface differential, one will be positive and one will be negative. But if this is the case, what about the electric field from the surface charge itself? After all, guas's law is relating the electric field coming from the surface and the surface charge density. Can someone clarify what is happening?

    PS. I've included the pages. One of the pictures doesn't want to change orientation even after rotating it and uploading it again. image.jpg image.jpg
  2. jcsd
  3. Apr 13, 2015 #2


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    Think of it as ##\vec E_{above}\cdot (dA\hat z) + \vec E_{below}\cdot (dA(-\hat z))=
    ((\vec E_{above})_z - (\vec E_{below})_z)dA##
    Last edited: Apr 13, 2015
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