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

Question

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

Answer 1

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 :

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

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

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