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The typical relationship equation $\mathbf{J}=\sigma \mathbf{E}$ implies that the electric field and the current density have the same direction. However, in a coaxial cable the Efield has a radial direction due to the opposite charge in the two conductors and the current density $\mathbf{J}$ has the direction along the wire, thus they are perpendicular. Why is this the case? Doesn't the Efield and $\mathbf{J}$ direction always coincide?

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However, in a coaxial cable the Efield has a radial direction due to the opposite charge in the two conductors and the current density J has the direction along the wire, thus they are perpendicular.

You're comparing the $\vec{J}$ in the conductive part of the coaxial structure with the $\vec{E}$ in the dielectric part of the structure.

If you used a crappy dielectric that allowed substantial leakage current, then leakage current would flow between the outer and inner conductors and the $\vec{J}$ in the dielectric would indeed be in the same direction as the $\vec{E}$.

When exciting the coax with an AC signal the $\vec{J}$ in the conductors doesn't necessarily correspond (either in magnitude or direction) to the $\vec{E}$ because not only electric field but also the magnetic field is significant in this system.

As others have pointed out, even in actual quasi-electrostatic scenarios in conductive materials, $\vec{J}$ and $\vec{E}$ don't necessarily point in the same direction and $\sigma$ might have to be considered as a tensor rather than a scalar. But that doesn't relate to the example you asked about.

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  • $\begingroup$ I think I start to understand. I think what confuses me the most is that since the efield has a radial direction instead of being along the wire, who is actually pushing the charge along the wire? Doesn't the electric field actually push the charge which is due to the potential ? Did you imply that the magnetic field is responsible for the current generation ? $\endgroup$
    – bigboss
    Mar 24 at 18:45
  • $\begingroup$ @bigboss coaxial cables are not isolated objects in space. They are used to connect devices. If they were not connected to anything, the current across them will be zero. Once you connect them to the circuit you want to measure you have a signal (voltage difference) along them that produces the current $\endgroup$ Mar 25 at 9:24
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$\vec J=\sigma\vec E$ holds (approximately) in materials. In the case of your example, the $\vec E$ field related to $\vec J$ would be the field resulting from the voltage difference that drives the current flowing in your wire: the size of the current density would be proportional to this external applied $\vec E$ resulting from the potential difference.

In a coaxial cable, one usually computes $\vec E$ for static (or quasistatic) charge distributions, and in this setup $\vec E$ is perpendicular (as it would in wire with constant static charge distribution). In fact, in a (perfect) conductor, there is no $\vec E$ inside a perfect conductor in the static limit, so the idea that you'd have an uniform current density $\vec J$ related to $\vec E$ inside a conductor shows this is not an electrostatics problem.

The $\vec E$ between the conductors in a coax cable has nothing to do with the current density and depends only on the (quasistatic) charge density on the surface of the conductor.

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  • $\begingroup$ However, when calculating the power flow with poynting vector, don't we use the radial electric field which is unrelated to current? If we assume that the cable is along z-axis, then the Poynting vector would result to be along z-axis direction due to the cross product of $\mathbf{E}\times \mathbf{H}$ resulting in a z-component. It kinda surprises me that the power flow relates to the radial E-field. $\endgroup$
    – bigboss
    Mar 24 at 14:26
  • $\begingroup$ Also, since the efield has a radial direction, how charge is pushed towards z-direction (assuming that the coaxial is along z). $\endgroup$
    – bigboss
    Mar 24 at 16:14
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    $\begingroup$ You want to be careful here: the power density can be written as $\vec E\cdot \vec J_c$ where $\vec J_c$ is the conduction current density. This is clearly along $\hat z$ so field in the dielectric (which is radial) will not contribute to this. $\endgroup$ Mar 24 at 18:46
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    $\begingroup$ The difficulty is that you are confusing the static field (radial) field with the electric along $\hat z$ that pushes the current. This field along $\hat z$ is generated by an outside source (say a voltage source), whereas the radial field would be generated by static charges (not currents). Basically the source of $\vec E$ along $\hat z$ is not part of your problem and is unrelated to the field between the two conductors of your coax cable. $\endgroup$ Mar 24 at 18:58
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    $\begingroup$ The power is dissipated through heat at the surface of the wire, which is radial to $\hat z$. $\endgroup$ Mar 24 at 19:16
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No $\boldsymbol{\mathrm{E}}$ and $\boldsymbol{\mathrm{J}}$ do not always point in the same direction. The $\boldsymbol{\sigma}$ is in general a tensor (the conductivity tensor), which you can think of as a matrix relating two vectors (with certain symmetry properties).

There's a good overview on the associated Wikipedia page.

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The relation \begin{equation} \mathbf{J} = \sigma \mathbf{E} \end{equation} is a simplification. More generally the conductivity $\sigma$ is a matrix (more precisely a tensor), hence the relation (using Einstein notation) is \begin{equation} \mathbf{J}_i = \sigma_{ij} \mathbf{E}_j \end{equation}

For the case of a coaxial cable, as you said, you have a component of the electric field in the radial direction, which, however, does not produce any current because of the insulator between the inner and outer conductor. On the other hand, you also have a component of the field along the cable, assuming that you are measuring a non-zero signal. The consequent voltage drop in turn produces the current along the cable.

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  • $\begingroup$ I'd add to the answer that the conductivity being a tensor, and the asked about case of the coax cable are different cases, and that the explanation to why the relation doesn't apply lies only in your 2nd paragraph. $\endgroup$ Mar 24 at 13:54
  • $\begingroup$ However, when calculating the power flow with poynting vector, don't we use the radial electric field which is unrelated to current? If we assume that the cable is along z-axis, then the Poynting vector would result to be along z-axis direction due to the cross product of $\mathbf{E}\times \mathbf{H}$ resulting in a z-component. It kinda surprises me that the power flow relates to the radial E-field. $\endgroup$
    – bigboss
    Mar 24 at 14:20
  • $\begingroup$ @bigboss Not sure if I understand correctly, but the magnetic field is caused by the current $\mathrm{J}$ flowing in the conductor, so yes the power depends on the radial field but also on the current. $\endgroup$ Mar 24 at 15:13
  • $\begingroup$ Hmm.. Let me rephrase my question... From what I understand, you said that apart from the E-field pointing radially outwards, there is also an E-field towards z-axis, i.e., along the cable $E_z$, which btw is created by whom? The radial E-field $E_r$ is due to the potential created by the differently charged inner and outer conductor. Who creates the E-field along the cable? And the initial question was about: usually power is associated with the current, which in this case has z-direction. However to calculate the Poynting vector and find the powerflow we use $E_r\times H_\phi$ and not $E_z$. $\endgroup$
    – bigboss
    Mar 24 at 15:23
  • $\begingroup$ 2/2 Continuing.. What I mean, is $E_z$ has the direction of $\mathbf{J}$, but to find a powerflow towards z-direction, we need the cross product of $E_r \times H_\phi$. Why the powerflow associates with the radial e-field and not the one following the current direction? $\endgroup$
    – bigboss
    Mar 24 at 15:25

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