Exercise 6.2.9: Is half an ounce of copper enough at 28 GHz?

$\delta =17.5\mu \mathrm{m}$ when $f=1/(\pi {\mu }_0\sigma {\delta }^2)=14.3\mathrm{MHz}$. At $28\mathrm{GHz}$ the foil is $17.5/0.395=44.3$ skin depths thick, and the field at the back is down by $8.686\times 44.3=385\mathrm{dB}$: the back face carries nothing.

The loss per unit area between the surface and the depth $t$, divided by the total,

At $28\mathrm{GHz}$, $t=1\mu \mathrm{m}$ gives $1-e^{-5.06}=99.4\%$ and $t=0.3\mu \mathrm{m}$ gives $1-e^{-1.52}=78.1\%$. At $1\mathrm{GHz}$ ($\delta =2.09\mu \mathrm{m}$), $t=1\mu \mathrm{m}$ gives $61.6\%$.

$1-e^{-2t/\delta }=0.99$ gives $t=\frac{\delta }{2}\ln 100=0.91\mu \mathrm{m}$.

$1-e^{-2t/\delta }$ at $t=\delta$ is $86.5\%$, the familiar statement that most of the current flows within one skin depth, and the $99\%$ depth $2.3\delta$ is the $20\mathrm{dB}$ depth of the field, as it must be for a power fraction of $10^{-2}$. At $28\mathrm{GHz}$ practically all the current flows inside the micro-roughness of the copper, which lengthens its path: measured losses of standard foil exceed the smooth-surface $R_s$ by up to a factor of two. The thickness of the foil, by contrast, could be a hundred times smaller without any electrical effect.