Why does the wave velocity of EM signals in a wire depend on the insulator's permittivity?

Question

I'm in an electromagnetism class and struggling with a concept. The textbook derived several equations which state that the wave velocity of an electrical signal in a transmission line depends on the permittivity and magnetic permeability of the insulator. Why is it the insulator's permittivity that determines wave velocity and not the wire's, the medium the wave is traveling through?

Our textbook used the following steps:

L' and C', the combined inductance per unit length and capacitance per unit length of the transmission lines, are related to the electric permittivity and magnetic permeability by:

$$L'C'=\mu \epsilon$$

(If I'm reading the text correctly, epsilon and mu belong to the surrounding medium. This, I think, is the source of my confusion. Why is this relation true?)

If we assume the transmission lines are lossless, then under the lumped-element model, the wave number and angular velocity are related by:

$$\beta=\omega \sqrt{L'C'} = \omega \sqrt{\mu \epsilon}$$

Giving the phase velocity:

$$u_p = \frac{\omega}{\beta} = \frac{1}{\sqrt{L'C'}} = \frac{1}{\sqrt{\mu \epsilon}}$$

Answer 1

As it turns out, the currents and charge density oscillations in the metal wire are not the only parts of the wave. There are also the fields! And where are the fields? Well, if the wire is a good conductor, then the fields will barely penetrate it at all! The fields reside almost entirely in the insulator. They swirl around the metal wire as the wave propagates, all the while interacting with the insulator’s molecules. So it’s no surprise that the properties of the wave (speed, wavelength, attenuation) will depend on the properties of the insulator.

Answer 2

In a transmission line (and in fact in all circuits) electric energy is actually transmitted as electromagnetic waves whose energy is concentrated in the space between the wires.

In a transmission line the conductive conductors serve as boundary conditions which “anchor” the EM field in space between the conductors. This EM wave travels in the space between the conductors which is filled with dielectric. For this reason the speed of the wave is the speed of light in the dielectric.

Answer 3

There are different ways to consider wires:

• The lumped-element model, where all the properties of the circuit are reduced to capacitances, inductances and resistances (if necessary, taken per unit length)
• Full electrodynamic description, where any transmission line can be viewed as a waveguide

If we adopt the latter description, a typical waveguide is a cylindrical wire with a dielectric inside. One can solve the Maxwell equations for the field inside and find that the field is a standing wave in the cross-sectional directions and a propagating wave along the wire. The metallic part of the waveguide is conducting and mainly induces the boundary conditions on the field (as there is almost no field inside a conductor). Thus, all the field is concentrated within the dielectric and adjusted according to its permittivity and permeability.

The classical description of waveguides using the lumped-element model is given by the telegrapher’s equations. In this case we do not need to consider fields, but only the potential differences and currents. Although one or other description may appear more natural in certain contexts, they are largely equivalent (admitting that the lumped-element model is rather contrived when applied to optical fiber), hence the equivalence between the $LC$ and $\mu\epsilon$.