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.
Best Answer
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).
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.