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konstantin123 [22]
3 years ago
15

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"?
Physics
1 answer:
dimaraw [331]3 years ago
7 0

Answer:

0.80267 m

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

From the first equation

1\times 10^{-6}=20e^{-20.94395z}\\\Rightarrow ln\dfrac{1\times 10^{-6}}{20}=-20.94395z\\\Rightarrow z=\dfrac{ln\dfrac{1\times 10^{-6}}{20}}{-20.94395}\\\Rightarrow z=0.80267\ m

The depth is 0.80267 m

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