Exercise 7.3.5: Smaller with a dielectric, but lossier

$n=1.5$: $a^{\prime }=2/1.5=1.333\mathrm{cm}$, $b^{\prime }=0.667\mathrm{cm}$. In both guides $f_{co,10}=7.5\mathrm{GHz}$, $f/f_c=1.5$ and $s=0.745$.

${\alpha }_c^{\prime }=\frac{R_s}{({\eta }_0/n)(b/n)s}[\dots ]=n^2{\alpha }_c={\epsilon }_r{\alpha }_c$. With $R_s=27.7\mathrm{m\Omega }$ at $11.25\mathrm{GHz}$ and the bracket $1+\frac{2b}{a}\frac{4}{9}=1.444$:

$k=2\pi fn/c=353.4rad/m$:

The total is $0.690dB/m$, $5.6$ times the $0.124dB/m$ of the air guide.

$P=E_0^{2}ab/(4Z_{TE})$: $ab$ is divided by ${\epsilon }_r$ and $Z_{TE}=\eta /s$ by $n$, so $P^{\prime }=P/n$:

The scaled guide has ${\epsilon }_r$ times less area: $2/2.25=0.889\mathrm{c}{\mathrm{m}}^2$ against $2\mathrm{c}{\mathrm{m}}^2$. At the same field the power per unit area grows by $n$ (the filled guide has the lower impedance) and the area falls by $n^2$, net $1/n$.