The discussion revolves around the surface integral of the product of electric potential and electric field, specifically questioning why the integral ∫(φE).dS is considered to be zero in general electrostatic cases. Participants explore the implications of this integral in the context of electrostatics and energy derivations.
The discussion is active, with participants providing insights and clarifications regarding the assumptions behind the integral. There is an exploration of different interpretations, particularly concerning the conditions under which the integral may be considered zero.
Participants reference notes from a lecturer that involve the derivation of total electrostatic energy, indicating that assumptions about the surface being at infinity play a critical role in the evaluation of the integral.
the_wolfman said: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?
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.
the_wolfman said: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.