 Solution

The fundamental mode of an ideal coaxial cable is a transverse electromagnetic mode: the electric field is radial between the conductors, the magnetic field is azimuthal, and energy flows along the cable through the dielectric. There is no cutoff for this mode; higher modes, conductor losses and dielectric dispersion can matter outside the ideal transmission-line regime.

Ignoring edge fields in the parallel-strip geometry, $E_x=V/a$ and $H_y=I/b$. Per unit length the charge is $ϵbV/a$ and the linked magnetic flux is $\mu aI/b$. Thus the parallel-plate transmission line hasFor a segment $dz$, charge conservation, equivalently the appropriate integral of the Ampère-Maxwell equation, gives $I(z)-I(z+dz)=Cdz{\partial }_{t}V$. Applying Faraday's law around a longitudinal rectangle between the plates gives $V(z)-V(z+dz)=Ldz{\partial }_{t}I$. These are the lossless telegrapher's equations,Differentiate the second with respect to $z$ and use the first, and conversely for $I$:The signal speed is $1/\sqrt{LC}=1/\sqrt{\mu ϵ}$, equal to $c$ for a vacuum-filled line. It is the speed of the coupled electromagnetic disturbance, rather than the electron drift velocity. Electrons locally adjust their motion as the field arrives; one electron need not travel the cable's length to transmit a signal. The Poynting vector in the dielectric integrates to power $VI$.