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\section{Theory}  \newcommand{\tensor}[1]{\stackrel{\leftrightarrow}{#1}}  Gaus's Gauss's  Law is in free space is \[\nabla\cdot\vec{E}=\frac{\rho}{\epsilon_0}\]. In a plasma $\epsilon_0$ becomes $\tensor{\epsilon}$ and the relation becomes \[\nabla\cdot(\epsilon\vec{E})=\rho\] and noting that $\vec{E}=-\nabla\phi$ and tensor $\kappa=\tensor{\epsilon}\epsilon_0$, Gaus's 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 \[\kappa_{xx}=\kappa_{yy}=1-\frac{\omega_{pe}^2}{\omega^2-\omega_{ce}^2}\]  \[\kappa_{xy}=\kappa_{yx}=\frac{i\omega_{ce}\omega_{ce}^2}{\omega^2-\omega_{ce}^2}\]  \[\kappa_{\parallel}=1-\frac{\omega_{pe}^2}{\omega^2}\]  It is needed to define a potential for an oscillating point charge in this system. Define $\rho$ from Gaus's Gauss's  law as \[\rho=qe^{-i\omega t}\sigma(\vec{r})\] 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 $\phi(r,z)=\frac{q}{4\pi\epsilon_{o}\sqrt{\rho^2+z^2}}$. The resonance cone phenomena is described by electric fields so take the negative gradient of $phi$ $\phi$  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.