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I know that in a conductor body, in an electrostatic situation (Where $\vec E=0$ in the interior), the E field must be perpendicular to the surface outside because it is solely generated by electric charges and therefore it is conservative. Due to this, the line integral $\int \vec E \cdot d\vec l$ must be zero for any closed path partially in and partially out of the body. This prohibits any tangential component.

My question is: Does this still stand for E fields produced by changing magnetic fields? ie. A neutral body being hit by a EM wave, or the same body being near a solenoid with changing current. My intuition tells me it should not stand, since those E fields are non-conservative according to Faraday's law, so the argument that the line integral must be zero is not enough. Moreover, it is not even an electrostatic situation, so could this be generalized (The title of the post)?

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  • $\begingroup$ You've answered your question yourself. What more do you need? It's indeed that the perpendicularity relies on a static version of Faraday's law... $\endgroup$
    – kricheli
    Commented Dec 1, 2022 at 17:17
  • $\begingroup$ Now I'm studying stationary EM waves with the book by Sears & Zemansky. To produce them, it's stated that an incident wave must hit a conductor surface to produce a reflected wave to superpose with the incident one. The wave acts perpendicular to the surface. They argue that since "the E field must always be perpendicular to a conductor surface" there can't be any E field in the surface and therfore this is a boundary condition. I've read in other posts that indeed this happens, but I find the argument used by Sears & Zemansky awkward and I do not fully grasp what they mean with that. $\endgroup$
    – user336281
    Commented Dec 1, 2022 at 17:47
  • $\begingroup$ I was hoping for a rigurous proof to the title of the post, or a counterproof, rather than just my own unfounded thoughts. $\endgroup$
    – user336281
    Commented Dec 1, 2022 at 17:49

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It is indeed possible for the E field lines to be non-orthogonal to the surface of a conductor with finite conductance.

Remember, if the E field is derived from a scalar potential, $\phi$, then the E field being perpendicular to the surface of the conductor implies that the surface is equipotential. So there are two ways to make the E field non-parallel to the conductor:

  1. have $-\frac{\partial \mathbf A}{\partial t}\ne 0$ so that the E field is not derived only from a scalar potential

  2. have $ \phi$ be spatially non-constant on the conductor surface so that even the static E field is non-perpendicular

Case 1) can be achieved using an external time-varying magnetic field, as you suggested. Case 2) can be achieved simply by having a current in a conductor with finite conductivity.

Here is a paper that describes a semi-quantitative approach for determining the surface charge on a conductor based exactly on how much the E-field “bends” at the surface.

A semiquantitative treatment of surface charges in DC circuits. Rainer Mueller. Am. J. Phys. 80 (9), September 2012. http://dx.doi.org/10.1119/1.4731722

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