Exercise 7.5.3: Where the power flows and how fast the energy moves

Half the power flows inside $r=\sqrt{ab}=\sqrt{0.45\times 1.571}=0.841$ mm, the geometric mean of the radii. The inner half of the gap, $\rho <(a+b)/2=1.010$ mm, carries $\ln (2.245)/1.25=64.7\%$.

With $L=\ln (b/a)=1.25$:

since ${\mu }_0/{\eta }^2=\epsilon$. The two energies are equal, as in a plane wave.

$P=V_0^2/2Z_0=100/100=1.00$ W, so

In symbols $v_E=\frac{\pi V_0^2/(\eta L)}{\pi \epsilon V_0^2/L}=\frac{1}{\eta \epsilon }=\frac{1}{\sqrt{{\mu }_0\epsilon }}$, independent of the radii.

Both energy densities fall as $1/{\rho }^2$: at $\rho =a$ they are $(b/a{)}^2={\mathrm{e}}^{2.5}=12.2$ times larger than at $\rho =b$.

$W_e^{\prime }=\frac{1}{4}{C}^{\prime }{V}_0^2=\frac{1}{4}\times 10^{-10}\times 100=2.5$ nJ/m and $W_m^{\prime }=\frac{1}{4}{L}^{\prime }{I}_0^2=\frac{1}{4}\times 2.5\times 10^{-7}\times 0.2^2=2.5$ nJ/m with $I_0=V_0/Z_0=0.2$ A: the circuit picture holds exactly for the TEM wave.