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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.

Qmechanic
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SnoopyKid
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  • 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} $. – secavara May 06 '23 at 17:48
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    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? – Puk May 06 '23 at 17:57
  • @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 https://physics.stackexchange.com/questions/410714/why-does-a-magnetic-field-curl-around-a-current-carrying-element/410735#410735 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 $\mathbf E$. – hyportnex May 06 '23 at 18:07
  • @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 – SnoopyKid May 06 '23 at 18:18
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    @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). – Buzz May 06 '23 at 18:20
  • @Puk I think E- and H-fields are not always perpendicular as they depend on the structure, geometry amd design of the source, correct? – SnoopyKid May 06 '23 at 18:20
  • @hyportnex do the equipotential surface represents electric field? I don't think electric fields themselves are two-dimensional because I think only imaginary lines are two-dimensional but two-dimensional things do not exist in real life – SnoopyKid May 06 '23 at 18:23
  • @Buzz so in the moving charge context, there is no such thing as equipotential surface? – SnoopyKid May 06 '23 at 18:24

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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.

TheTheoMess
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    I was also wondering whether all equipotential surface represent some kind of magnetic field. – TheTheoMess Sep 14 '23 at 12:21
  • Definitely not. An isolated electrostatic charge is a counterexample. – Dale Sep 14 '23 at 12:30
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