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Regarding linearly polarized electric fields that are produced by a dipole antenna and electric fields from a current carrying wire, are the equipotential surface the same as the magnetic fields? Because both are perpendicular to the electric fields.

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  • $\begingroup$ Since the magnetic field is not-zero and changes in time I would not be so sure the electric field is perpendicular to the equipotential surfaces, since you have to account for the contribution coming from the vector potential, i.e. $\mathbf{E} = - \nabla \Phi - \frac{\partial \mathbf{A}}{\partial t} $. $\endgroup$
    – secavara
    May 6, 2023 at 17:48
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    $\begingroup$ Magnetic fields aren't surfaces, so they aren't equipotential surfaces. In radiation problems, the (scalar) potential is also not uniquely defined, so I'm not sure if it makes sense to speak unambiguously of equipotential surfaces. Please clarify your question. Are you asking if E- and H-fields are always perpendicular? $\endgroup$
    – Puk
    May 6, 2023 at 17:57
  • $\begingroup$ @Puk unlike the $\mathbf H$ or $\mathbf E$ fields that are line creatures the magnetic flux density $\mathbf B$ is a surface creature although that has nothing to do with being equipotential or otherwise, see physics.stackexchange.com/questions/410714/… in a propagating wave of linear polarization the plane of the $B$ field in which it acts is the one spanned by the vectors $\mathbf {\hat k}$ and $\mathbf E$. $\endgroup$
    – hyportnex
    May 6, 2023 at 18:07
  • $\begingroup$ @secavara I'm bad at maths, but most drawings show the equipotential surface lines and electric field lines to be perpendicular. Not sure if that applies to every circumstances $\endgroup$
    – SnoopyKid
    May 6, 2023 at 18:18
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    $\begingroup$ @secavara You're right, but the situation is actually even worse than that. In a time-dependent situation, the equipotential conditions $\Phi =$ constant is not gauge invariant. So the equipotential surfaces are not in any way physically observable. Equipotentials really only make sense in electrostatics (or as approximations when any time dependence is a small correction on top of electrostatics). $\endgroup$
    – Buzz
    May 6, 2023 at 18:20

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I recently saw this. The equipotential surface(s) of an electric dipole do look like the locus of magnetic field lines of a perpendicular magnetic dipole. Kind of means you can make a velocity selector by sticking an electric and magnetic dipole perpendicularly.

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    $\begingroup$ I was also wondering whether all equipotential surface represent some kind of magnetic field. $\endgroup$ Sep 14, 2023 at 12:21
  • $\begingroup$ Definitely not. An isolated electrostatic charge is a counterexample. $\endgroup$
    – Dale
    Sep 14, 2023 at 12:30
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    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Sep 14, 2023 at 13:55

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