Answer to Question #155524 in Electricity and Magnetism for Riola Sadtoi

Question #155524

Question Link: https://ibb.co/xzWC2Ly


1
Expert's answer
2021-01-14T16:40:56-0500

As the displacement should satisfy the given equation, we will insert the expressions of "\\vec{x}, \\vec{E}" into this equation :

"m(\\vec{x_0}e^{-i\\omega t})''+m\\gamma (\\vec{x_0}e^{-i\\omega t})' =-e\\vec{E}_0 e^{-i\\omega t}"

"-\\omega^2 m\\vec{x_0}e^{-i\\omega t} -im\\omega\\gamma \\vec{x_0} e^{-i\\omega t} = -e \\vec{E}_0 e^{-i\\omega t}"

"\\vec{x_0}=\\frac{e\\vec{E_0}}{m(\\omega^2+ i\\omega \\gamma)}" , now multiply both by "e^{-i\\omega t}"

"\\vec{x}(\\omega, t) = \\frac{e \\vec{E}(t)}{m(\\omega^2+i\\omega \\gamma)}"

By definition, polarization vector is

"\\vec{P}(\\omega) = \\rho \\cdot \\vec{x}" , where "\\rho" is charge density. Thus, if "n" is electron concentration, then

"\\vec{P}(\\omega) = \\frac{-ne^2\\vec{E}(t)}{m(\\omega^2+i\\omega\\gamma)}"

By definition of permittivity :

"\\epsilon_0\\vec{E}+\\vec{P}=\\epsilon_0\\epsilon\\vec{E}"

"\\epsilon (\\omega)= 1 - \\frac{ne^2}{\\epsilon_0m(\\omega^2+i\\omega\\gamma)}"

"\\epsilon(\\omega)= 1-\\frac{\\omega_p^2}{\\omega^2+i\\omega\\gamma}, \\omega_p^2=\\frac{ne^2}{\\epsilon_0m}"


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