Question

Ignoring reflection at the air-water boundary, if the amplitude of a 10 GHz incident wave in air is 20 V/m at the water surface, at what depth will it be down to 1 µV/m? Water is characterized by er = 81, µr = 1, and σ = 0.1 S/m. Can water described above, at 10 GHz, be described as a low-loss dielectric, good conductor, or "in-between"?

Answer 1

0.80267 m

Step-by-step explanation:

E(z) = Electric field = 1 µV/m

`E_0` = 20 V/m

z = Depth

`\sigma` = Conductivity = 0.1 S/m

`\epsilon_r` = 81

`\mu` = Impedance of free space =
`120\pi\ \Omega`

Frequency is given by

`E(z)=E_0e^(-\alpha z)`

Parameter is given by

`\alpha=(\sigma)/(2)\sqrt{(\mu)/(\epsilon_r)}\\\Rightarrow \alpha=(1)/(2)\sqrt{((120\pi)^2)/(81)}\\\Rightarrow \alpha=20.94395\ N_p/m`

From the first equation

`1* 10^(-6)=20e^(-20.94395z)\\\Rightarrow ln(1* 10^(-6))/(20)=-20.94395z\\\Rightarrow z=(ln(1* 10^(-6))/(20))/(-20.94395)\\\Rightarrow z=0.80267\ m`

The depth is 0.80267 m