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I have a finite surface charge (in yellow) which cuts a Gaussian surface (in green).

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Red points are the points of singularity on the Gaussian surface (i.e. where $\vec{E}$ is undefined).

Does this singularity (at red points) refrain us from computing flux over the Gaussian surface? Or is there a way to compute the flux? Any help appreciated.

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No, not really. The surface has an area of $0$ thus charge $0$, and mathematican's way is to introduce the convention $0\cdot \infty$.

If you really want to think about calculate it in a physical way, use symmetry.

Then it's clear that the flux from each single point at the boundary cancels out.

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  • $\begingroup$ Note that fields at the singularity points are not only infinite (this does not create ambiguity) but also points in all direction (this creates ambiguity). $\endgroup$ – N.G.Tyson Nov 5 '18 at 15:40
  • $\begingroup$ "Then it's clear that the flux from each single point at the boundary cancels out". How? Please elaborate. $\endgroup$ – N.G.Tyson Nov 5 '18 at 15:41
  • $\begingroup$ At the edge, up down canceled, at the "right angle", the cross angle canceled out. The math is different here so don't think about your daily Remiann integeral. $\endgroup$ – J C Nov 5 '18 at 15:57
  • $\begingroup$ If $\vec{E}$ is constant, then OK. But what if $\vec{E}$ is varying? $\endgroup$ – N.G.Tyson Nov 5 '18 at 16:13
  • $\begingroup$ @faheemahmed400 finite surface, the question is assuming symmetry. $\endgroup$ – J C Nov 5 '18 at 18:21

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