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

the answer is below

Step-by-step explanation:

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

`\delta=(1)/(√(\pi f \mu \sigma) )\\\\\sigma =(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 =(1)/(\pi *10^6* 0.0000012*0.00016^2)=0.99*10^4\ S/m`

b) f = 1 GHz = 10⁹ Hz.

`\alpha=√(\pi f \mu \sigma) = √(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=(-1.5)/(log_(10) e*-\alpha) =1.75*10^(-4)\ m=0.175\ mm`