How to calculate the electric flux on something which has only one side like Klein bottle or a Möbius ring?

  • $\begingroup$ What is the source of the electric field? $\endgroup$ – harshit54 Mar 9 at 14:55
  • $\begingroup$ Well...I am talking about static field...As we all know,electric flux is defined as E multiples dS,and both of them should be vectors.The question is how to calculate it when dS is not applicable. $\endgroup$ – The Angel of Eliston Mar 9 at 15:21
  • $\begingroup$ @harshit54 Why make the question less general? $\endgroup$ – PiKindOfGuy Mar 9 at 15:32
  • $\begingroup$ You can look at physicsforums.com/threads/… $\endgroup$ – Vincent Fraticelli Mar 9 at 15:35

The flux would always be zero because for every $d\mathbf{A}$ pointing in one direction there's a corresponding $d\mathbf{A}$ pointing in the opposite direction.

From http://mathworld.wolfram.com/MoebiusStrip.html

This is a modified image from http://mathworld.wolfram.com/MoebiusStrip.html.

  • $\begingroup$ you are assuming that one can define flux through a non-orientable surface, ok but how do you that? $\endgroup$ – hyportnex Mar 9 at 16:52
  • $\begingroup$ @hyportnex Each surface element is orientable so orient one and the others follow. Then you end up with each surface element being doubly oriented. This is not a rigorous argument, but it seems to be correct. $\endgroup$ – PiKindOfGuy Mar 9 at 17:01
  • $\begingroup$ My knowledge of diffgeo is almost nonexistent to comment intelligently but I am skeptical about that. Maybe some expert here can opine, @ACuriousMind, probably_someone, QMechanic ? $\endgroup$ – hyportnex Mar 9 at 17:17

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