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Surface integral of potential*electric field

  1. Apr 10, 2013 #1
    could someone please explain why ∫(φE).dS (where φ is the potential and E is the electric field) is equal to zero in a general electrostatic case?

    Thank you
     
  2. jcsd
  3. Apr 10, 2013 #2
    Its not true in general.

    Consider the case with a point charge q at the r=0 and let the surface be r=1.

    We know that the potential on this surface will be constant so

    [itex]\oint \phi \vec E \cdot dS = \phi \oint \vec E \cdot dS \sim \phi Q_{enclosed} \neq 0 [/itex]


    If I may ask, what led you to the conclusion that it equals zero?
     
  4. Apr 10, 2013 #3
    Maybe they mean the scalar magnetic potential, which is -of course- zero in an electrostatic field.
     
  5. Apr 10, 2013 #4
    Thanks for the reply :)

    This is from notes written by my lecturer, I was just trying to understand his reasoning...
    It was in the derivation of total electrostatic energy due to a finite charge distribution, where he went from this expression U=1/2∫ρ(r)φ(r)dV to this one U=1/2∫εE2dV by means of substituting: ∇.E=ρ/ε and E=-∇φ where at some point he mentioned ∫∇.(φE)dV = ∫(φE).dS=0 ....
    I can write up the derivation if you'd like.
     
  6. Apr 10, 2013 #5
    No need. The trick is that he is assuming that the surface is out at infinity. The integral scales as Q^2/r and thus it goes to zero as r goes to infinity.
     
  7. Apr 11, 2013 #6
    got it. thanks so much for your help :)
     
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