Answer to Question #184084 in Optics for Noby

Question #184084

Given that the EM wave has a form of Ex=Eo cos(wt-kz+∅) and it can also be written as Ex=Eo[k(vt-z)+∅] where v=w/k is the velocity.

(a)show that the wave satisfy Maxwell equation

(b)show that v=cuberoot of permeability note,permetivity note permetivity


1
Expert's answer
2021-04-23T10:11:56-0400

Ex=Eocos(ωtkx+ϕo)d2Exdx2=0    d2Exdy2=0   d2Exdz2=k2Eocos(ωtkz+ϕo)......(i)E_x=E_ocos(\omega t-kx+\phi_o)\\\dfrac{d^2E_x }{dx^2}=0\ \ \ \ \dfrac{d^2E_x }{dy^2}=0\ \ \ \dfrac{d^2E_x }{dz^2}=-k^2E_ocos(\omega t-kz+\phi_o)......(i)


d2Exdt2=ω2Eocos(ωtky+ϕo)......(ii)\dfrac{d^2E_x }{dt^2}=-\omega^2E_ocos(\omega t-ky+\phi_o)......(ii)


Substituting (i) and (ii) in wave equation we get,

k2Eocos(ωtkz+ϕo)+ϵoϵrμo+ω2Eocos(ωtky+ϕo)=0ω2k2=1ϵoϵrμoωk=(ϵoϵrμo)12v=(ϵoϵrμo)1-k^2E_ocos(\omega t-kz+\phi_o) + \epsilon_o\epsilon_r\mu_o+\omega^2E_ocos(\omega t-ky+\phi_o)=0\\ \Rightarrow\dfrac{\omega^2}{k^2}=\dfrac{1}{ \epsilon_o\epsilon_r\mu_o}\\ \Rightarrow \dfrac{\omega}{k}=( \epsilon_o\epsilon_r\mu_o)^{-\frac{1}{2}}\\ \Rightarrow v=(\sqrt{ \epsilon_o\epsilon_r\mu_o})^{-1}




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