= 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 $\epsilon bV/a$ and the linked magnetic flux is $\mu aI/b$. Thus the <parallel-plate transmission line> has
$$
\boxed{C=\epsilon\frac ba,\qquad L=\mu\frac ab,\qquad LC=\mu\epsilon.}
$$
For a segment $dz$, <charge conservation>, equivalently the appropriate integral of the <Ampère-Maxwell equation>, gives $I(z)-I(z+dz)=C\,dz\,\partial_tV$. Applying <Faraday's law> around a longitudinal rectangle between the plates gives $V(z)-V(z+dz)=L\,dz\,\partial_tI$. These are the lossless <telegrapher's equations>,
$$
\partial_zI=-C\partial_tV,\qquad\partial_zV=-L\partial_tI.
$$
Differentiate the second with respect to $z$ and use the first, and conversely for $I$:
$$
\boxed{\partial_z^2V=LC\partial_t^2V,\qquad\partial_z^2I=LC\partial_t^2I.}
$$
The signal speed is $1/\sqrt{LC}=1/\sqrt{\mu\epsilon}$, 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 $VI$.
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