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I Explanation on the Electric Field in Griffiths' Textbook

  1. Jul 19, 2017 #1
    Hi. I'm going over the discontinuity aspects of the electric field as we cross the surface charge. A pillbox Gaussian surface was drawn and the electric field for "below" is throwing me off. Can someone explain why the electric field is pointing in the same direction from both sides of the surface? I supplied the image below. Thanks!
    C598XwYl61xVUe6uK65kRtgToayN3xKaeS9V4AFctuYMUuBESuUaoJZdjgozDKqg_b6bIs5tNeO3XAs04=w1496-h1058-no.png
     
  2. jcsd
  3. Jul 19, 2017 #2

    jtbell

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    Staff: Mentor

    I think he does it that way to indicate or emphasize that you need to use the same sign convention on both sides of the surface, which leads to eq. (2.31) having a minus sign, i.e. that you have to use the difference between the two fields. Also, it's actually possible for the field to be in the same direction on both sides of the surface, depending on the situation.

    Example #1: if the surface is the only charged object, then the field points upwards above the surface and downwards below it. Using the sign convention with +/- meaning up/down, this might give us something like $$(+5~\rm{N/C}) - (-5~\rm{N/C}) = \frac {\sigma} {\epsilon_0} \\ +10~\rm{N/C} = \frac {\sigma} {\epsilon_0}$$ Example #2: if we place the surface from example #1 between the plates of a large capactor that by itself produces a uniform field of 20 N/C upwards, the net fieid is now upwards on both sides of the surface, and we have $$(+25~\rm{N/C}) - (+15~\rm{N/C}) = \frac {\sigma} {\epsilon_0} \\ +10~\rm{N/C} = \frac {\sigma} {\epsilon_0}$$
     
  4. Jul 23, 2017 #3
    Thanks jtbell! I realized that I forgot to thank you and just put a thumbs up. In my attempt to find my post in the forums, I found a similar post but referencing Jackson book. I thought it was funny how I could have just searched for my question, since you answered that post, but I usually have no luck in searching. Thanks again!
     
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