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Gauss's Law is in free space is \[\nabla\cdot\vec{E}=\frac{\rho_{ext}}{\epsilon_0}\]. In a plasma $\epsilon_0$ becomes $\tensor{\epsilon}$ and the relation becomes \[\nabla\cdot(\epsilon\vec{E})=\rho_{ext}\] and noting that $\vec{E}=-\nabla\phi$ and tensor $\kappa=\tensor{\epsilon}\epsilon_0$, Gauss's Law can be rewritten as \[\vec{D}=\epsilon_0\tensor{\kappa}\cdot\vec{E}\] where $\tensor{\kappa}$ is the dielectric tensor.   For a cold plasma in an electric field, there will be no contribution to the to the dielectric tensor due to the ions having low thermal velocity in comparison to the electrons. Then the dielectric tensor has components \[\kappa_{\perp}=\kappa_{xx}=\kappa_{yy}=1-\frac{\omega_{pe}^2}{\omega^2-\omega_{ce}^2}\] [2]  \[\kappa_{xy}=\kappa_{yx}=\frac{i\omega_{ce}\omega_{ce}^2}{\omega^2-\omega_{ce}^2}\]  \[\kappa_{\parallel}=1-\frac{\omega_{pe}^2}{\omega^2}\] \[\kappa_{zz}=\kappa_{\parallel}=1-\frac{\omega_{pe}^2}{\omega^2}\]  It is needed to define a potential for an oscillating point charge in this system. Define $\rho$ from Gauss's law as \[\rho_{ext}=qe^{-i\omega t}\sigma(\vec{r})\]  where $\sigma(\vec{r})$ is the delta function at $\vec{r}$ at zero. Use fourier analysis on Gaus's law and note that $E=-\nabla\phi$ to solve for the potential.It is obtained that $\phi(r,z)=\frac{q}{4\pi\epsilon_{o}\sqrt{\rho^2+z^2}}$ where now $\rho$ is referring to radius. The resonance cone phenomena is described by electric fields so take the negative gradient of $\phi$ in cylindrical coordinates and the electric field is in the radial direction is \[E_r=-\frac{qe^{i\omega t}}{4\pi\epsilon_{0}\kappa_{\perp}\sqrt{\kappa_{\parallel}}}\left(\frac{\rho}{(\frac{z^2}{\kappa_{\parallel}}+\frac{\rho^2}{\kappa_{\perp}})^{3/2}}\right)\] where $\kappa_{\perp}$ is perpendicular to the background magnetic field. The resonance cones that are observed are described by the electric field in the radial direction. The resonance cone angle is found when looking for at what point does the electric field becomes infinite.This occurs when the condition $\frac{z^2}{\kappa_{\parallel}}+\frac{\rho^2}{\kappa_{\perp}}=0$. Define $\tan^2\theta=\frac{\rho}{z}$ then the condition when the radial electric field becomes infinite is when $\tan^2\theta=\frac{-\kappa_{\perp}}{\kappa{\parallel}}$ where $\theta$ is the resonance cone angle. This equation can be simplified further by taking the definintions of $\kappa_{\perp}$ and $\kappa_{\parallel}$ above and using the condition that $\omega<\omega_{ce},\omega_{pe}$, then \[\tan\theta\approx f\sqrt{\frac{1}{f_{pe}^2}+\frac{1}{f_{ce}^2}}\]