Question

Given that the skin depth of graphite at 100 (MHz) is 0.16 (mm), determine (a) the conductivity of graphite, and (b) the distance that a 1 (GHz) wave travels in graphite such that its field intensity is reduced by 30 (dB).

Answer 1

Answer:

the answer is below

Explanation:

a) The conductivity of graphite (σ) is calculated using the formula:

$\delta=\frac{1}{\sqrt{\pi f \mu \sigma} }\\\\\sigma =\frac{1}{\pi f \mu \delta^2}$

where f = frequency = 100 MHz, δ = skin depth = 0.16 mm = 0.00016 m, μ = 0.0000012

Substituting:

$\sigma =\frac{1}{\pi *10^6* 0.0000012*0.00016^2}=0.99*10^4\ S/m$

b) f = 1 GHz = 10⁹ Hz.

$\alpha=\sqrt{\pi f \mu \sigma} = \sqrt{0.0000012*10^9*\pi*0.99*10^5}=1.98*10^4\ Np/m\\\\20log_{10} e^{-\alpha z}=-30\ dB\\\\(-\alpha z)log_{10} e=-1.5 \\\\z=\frac{-1.5}{log_{10} e*-\alpha} =1.75*10^{-4}\ m=0.175\ mm$