This question can be challenging if you seriously attempt to
avoid formulae and explain in words. In fact, unless
you really understand what the formulae mean,
it's virtually impossible! Hence it is a perfect test of comprehension.
Those who simply write down the relevant formulae get no more than
half credit even if they are the "right" formulae.
On the other hand, there are many ways of explaining in words;
You get full credit for any that are not actually wrong.
First let's talk about a plasma, e.g. the ionosphere. Ions in a plasma are free to move, and so a constant electric field would produce a steady current. For "almost constant" electric fields (i.e. low frequency waves) the ionosphere might as well be a sheet of copper around the Earth, and like any good conductor it reflects electromagnetic waves - i.e. they do not propagate through it. However, as you turn up the frequency the ions can't move very far before the field has reversed, so the maximum amplitude of their motion gets smaller and smaller until all they do is jiggle a little as the wave propagates past them without difficulty. Also, for a given number of charged particles per unit volume, the distance between idividual charges is fixed. If the wavelength becomes short compared with that distance, one particle may go "left" while the next one goes "right" and they can no longer act "in concert" as a current density. So short wavelengths (high frequencies) "see" the plasma as a bunch of uncorrelated charges rather than a "conductor" and can propagate freely through it.
A hollow waveguide won't support TEM modes, so the "nice" propagation of a simple plane wave down the tube isn't allowed; we must have a wave reflecting off the sides of the tube and interfering with itself to produce standing waves with nodes at the surfaces. This imposes a constraint on the wavelength - it can't be longer than twice the width of the tube. And since the wave (viewed this way, as a plane wave "bouncing around") always propagates at , an upper limit on wavelength implies a lower limit on frequency.
The coaxial cable (or any other transmission line with two separate conductors that can be oppositely charged and carry opposite currents locally) will support TEM waves, which are basically localized plane waves propagating down the line without dispersion. They all travel at the same speed regardless of frequency and there are no restrictions on wavelength. "Anything goes."
ANSWER: Going back to the simple model of a plane wave reflecting off the walls of the waveguide to form standing waves, nothing has changed when the guide is filled with dielectric except that the original plane wave has a lower phase velocity instead of . The transverse wavevector components must still satisfy and , so where kz = k, or where . This differs from for an empty waveguide only by the factor .
We have (i)
Now for the B.C.: by inspection, since is strictly radial in direction. Similarly, since (strictly parallel to both surfaces), as well.
For the charge we do GAUSS' LAW (in integral form) on a short
cylindrical surface between the conductors, getting the familiar result
For the current we use AMPÈRE'S LAW (in integral form)
for a circular loop between the conductors to give the familiar result
, which we invert to get
Note that since outside the cable, the same two LAWs require that the charge and current on the outer conductor be equal and opposite to those on the inner conductor.
The magnetic field between the conductors is
and there is no magnetic field elsewhere.
The magnetic flux between the conductors
in a short section of length is thus
and so the inductance per unit length is
Similarly, a short length of the cable with equal and opposite charges on the inner and outer conductors has an electric field strength at radius s and thus a resultant voltage difference . Thus the capacitance per unit length is .
Putting the two together gives ,
in agreement with the previous result for Z.