= Solution
For a linear nondispersive medium with real absolute <permittivity> $\epsilon_a=\epsilon\epsilon_0$ and permeability $\mu_0$, instantaneous field <energy density> and <energy> flux are
$$
u=\frac12\epsilon_aE^2+\frac{B^2}{2\mu_0},\qquad \mathbf N=\mathbf E\times\mathbf H,
$$
with local balance $\partial_tu+\nabla\cdot\mathbf N=-\mathbf J\cdot\mathbf E$. <energy> stored locally and <energy> passing through a surface are distinct: integrating the balance relates their rates and the <power> given to charges.
Between the coaxial conductors, cylindrical Gauss and Ampère laws give
$$
\boxed{\mathbf E=\frac{V}{r\ln(b/a)}\widehat{\mathbf r},\qquad \mathbf B=\frac{\mu_0I}{2\pi r}\widehat{\boldsymbol\phi},\qquad a<r<b.}
$$
The inner line charge is $\lambda=2\pi\epsilon_a V/\ln(b/a)$, so the <capacitance> per <length> is $C=2\pi\epsilon_a/\ln(b/a)$. Integrating magnetic <energy> over the annulus gives $\int B^2(2\pi r\,dr)/(2\mu_0)=\mu_0I^2\ln(b/a)/(4\pi)=LI^2/2$, hence $L=\mu_0\ln(b/a)/(2\pi)$. Thus
$$
\boxed{Z_0=\sqrt{\frac LC}=\frac{\ln(b/a)}{2\pi}\sqrt{\frac{\mu_0}{\epsilon\epsilon_0}}.}
$$
These are per-length parameters for ideal conductors, excluding internal magnetic penetration.
Now let the <voltage> and current <amplitudes> be $V_0,I_0$ and multiply both transverse fields by $e^{i(kz-\omega t)}$. Their divergences vanish in the annular <dielectric>. The two nontrivial curl equations reduce to
$$
k\widetilde E_r=\omega\widetilde B_\phi,\qquad k\widetilde B_\phi=\omega\mu_0\epsilon_a\widetilde E_r.
$$
Consequently \b[both $k^2=\omega^2\mu_0\epsilon_a$ and $V_0=Z_0I_0$ are required for a forward <wave>]. Arbitrary independent static $V$ and $I$ cannot acquire the same forward-wave factor and still satisfy <Maxwell's equations>. This <TEM travelling-wave voltage-current relation> and the dispersion constraint identify the <transverse electromagnetic mode> and its <speed> $v_p=1/\sqrt{LC}$.
For the travelling <wave>, $\langle N_z\rangle=\operatorname{Re}(V_0I_0^*)/[4\pi r^2\ln(b/a)]$. Integrating over the <dielectric> annulus yields
$$
\boxed{\langle P\rangle=\frac12\operatorname{Re}(V_0I_0^*)=\frac{|V_0|^2}{2Z_0}=\frac{Z_0|I_0|^2}{2}.}
$$
This is the familiar circuit <power>, carried here through the <dielectric> rather than through the ideal metal. In <root mean square> notation it is $V_{\rm rms}I_{\rm rms}$ for the forward lossless <wave>; the corresponding static fields give $VI$ without a time-average factor. The average line <energy> is $(C|V_0|^2+L|I_0|^2)/4=C|V_0|^2/2$, whose product with <wave> <speed> agrees with the <power>.
Finite conductor resistance requires a small longitudinal electric field to drive current. Its <cross product> with the azimuthal <magnetic field> produces a radial Poynting component into each conductor: inward radially at the inner conductor and outward radially into the outer conductor. This <Poynting flux into lossy coaxial conductors> explains why forward <power> decreases along the cable as <energy> becomes <heat> in the metal. The mode is then only approximately TEM; strictly zero longitudinal electric field would be inconsistent with resistive current.
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