I have a coaxial cable with internal conductor of radius r1 and external conductor of radii r2 and r3. The material of the conductors has a conductivity \$\sigma_1\$. Between the conductors there is a imperfect dielectric of conductivity \$\sigma_2\$. I am asked to determine the conductance of the dielectric medium.

I have no problems doing the computation when I assume that the current density J has a radial direction. I come up with

$$G=\frac{2\pi \sigma_2}{\ln{\frac{r_2}{r_1}}}$$

My problem is understanding why the current is radial. I think I'm not correctly understanding the phenomena of stationary currents on dielectrics. I know that in this domain the electric field has 2 components: a normal one and a tangential one. The normal one is calculated the same way as it was in electrostatics, by applying Gauss Law. The tangential one, because of the continuity of the electric field is the same as inside the conductor. Now what I don't understand is how there is a current inside the dielectric. Is it because it's not a perfect one? But how is that current radial? Why that direction if the electric field has 2 components? Why don't we have a current in the same direction (longitudinal) as inside the conductors? Can anyone clarify me please? My questions are about the concept of electric currents on dielectric media, and not about the calculation of conductance.


3 Answers 3


Let's say we apply a DC voltage between the internal and the external conductors on the left end of a coaxial cable and leave the right end of the cable open.

The applied voltage will create radial field between the internal and external conductors and some leakage current will flow, along the radial lines, based on the conductance of the imperfect dielectric. Due to the leakage current, there will be a voltage drop along the length of the internal and external conductors, but, since their resistance is much much smaller than the resistance of the dielectric, this voltage drop and, therefore, the longitudinal electric field will be negligible and won't noticeably affect the radial electric field lines and the radial leakage current.

If we connect a load (say, a resistor) between the internal and external conductors on the right end of the cable, an additional, much more significant, current will flow in the internal and external conductors, which will create an additional small voltage drop along their length and, as a result, the strength of the longitudinal electric field will slightly increase.

This longitudinal electric field components, will slightly bend the radial electric field, but, assuming that most of the applied voltage will drop on the load, the radial lines still won't be significantly distorted and, therefore, the leakage current through the dielectric will still be predominantly radial.

The diagram below shows how the electric field lines (long red arrows) maintain their radial direction in the presence of the leakage current (short blue arrows) and slightly bend in the presence of the load current (long blue arrows) due to the added longitudinal electric field component (short red arrows).

enter image description here

In summary, the longitudinal electric field in the internal and external conductors will have some effect on the shape of the electric field between them and, as a result, the path of the leakage current through the dielectric may by slightly deviate from radial.


A coax cable has two conductors. The inner one is a normal wire, and the outer one is a hollow shell. Between these two has to be some sort of insulator, else the two conductors of the cable would be shorted to each other, and the cable would have little use.

This problem seems to be about the leakage current between the two conductors thru this insulation. For a perfect insulator, that current would be 0. Since the problem explicitly mentions "imperfect dielectric", it is addressing this leakage current.

That current necessarily flows between the two conductors. Think of a thin slice of a long cable. The current between the two conductors can only be radial. It has no reason to spiral in or out, for example. A long cable is simply a lot of these thin slices put together. The current is radial in each slice, and continues to be radial as you abut many slices.

You may be confusing the direction of the leakage current thru the insulator with the magnetic field around the inner conductor. The latter is tangential, but has nothing to do with the leakage current.

It appears this question is strictly about the DC leakage current, or resistance per length of cable. Dynamic effects therefore don't matter, and static magnetic fields are irrelevant.

When current is flowing in either conductor of the coax cable, there will be a axial voltage gradient. Maybe that's what you are getting hung up on. However, the conductivity of any reasonable insulator is so vastly lower than the conductors (they were chosen for those properties, after all), that any axial current flow thru the insulator is vanishingly small compared to the other currents.

  • \$\begingroup\$ Ok I think I'm understanding. Thank you for being clear! I have other question about the normal and tangent components of the electric field. Does the tangential component only exists on the boundaries (interface) of the dielectric? And then on the rest of the dielectric the E field is only normal? Or are we just ignoring the tangential field because, as you said, it's small compared to the normal one? \$\endgroup\$ Commented Oct 7, 2018 at 15:29
  • \$\begingroup\$ @Gran: There should be no tangential component at all, since current flow is perpendicular to that direction. There is no tangential voltage gradient. The radial voltage gradient is proportional to the voltage difference between the two conductor, and the axial proportional to the voltage drop along the conductors due to current flowing thru them. \$\endgroup\$ Commented Oct 7, 2018 at 16:23
  • \$\begingroup\$ But my books says and I'll quote: Although electric circuits are practically perfect tubes for J-field lines, you must be aware that the same is not true for the electric field E. In fact, E exists not only inside the circuit conductors but outside of them as well. At the conductor/insulator interface (conductor side) you have a purely tangential component for the electric field. At the conductor/insulator interface (insulator side), although J=0, you will not get E=0. \$\endgroup\$ Commented Oct 7, 2018 at 16:30
  • \$\begingroup\$ The electric field vector on the insulator side can be obtained by adding two orthogonal components. The magnitude of the normal component, which is usually the dominant term, depends on the local surface charge density. As for the tangential component, it is a simple matter to show (using curl E = 0) that it coincides with the one observed inside the conductor. \$\endgroup\$ Commented Oct 7, 2018 at 16:30
  • \$\begingroup\$ I'm having trouble relating this information... \$\endgroup\$ Commented Oct 7, 2018 at 16:30

Now what I don't understand is how there is a current inside the dielectric.

If \$\sigma_2 > 0\$, then if there is a non-zero electric field in the dielectric region, there will be a current as well.

You've calculated \$G\$, the conductance. As with any conductor, there won't actually be any current until you apply a voltage between its two terminals (the inner and outer conductors of the coax).

But how is that current radial?

Your source has probably assumed that \$\sigma_1 \gg \sigma_2\$, so that the outer conductor is essentially an equipotential surface.

Then, by symmetry, there's no reason for there to be a tangential component of \$\vec{E}\$ in the dielectric region (or anywhere else). And thus no tangential component of \$\vec{J}\$ either.


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