# Infinitely Long Charged Cylinder in a Maxwellian Plasma

1. May 16, 2015

### karlhoffman_76

Hi guys, having a hard time figuring out where to start on a plasma physics related problem.

1. The problem statement, all variables and given/known data

An inﬁnitely long metallic cylinder with a radius of $r_0$ is immersed in a Maxwellian plasma. An electric potential, $\phi_0$, is applied to the cylinder. Assuming that the electrons are mobile but the ions are stationary, derive an expression for the potential as a function of radial distance, $r$, also including $r_0$, $\phi_0$, and the Debye length, $\lambda_D$.

2. Relevant equations

Poisson's equation in cylindrical coordinates for the radial component only,

$\nabla^2\phi=\frac{1}{r}\frac{\partial}{\partial r}(r\frac{\partial\phi}{\partial r})=\frac{-e}{\epsilon_0}(n_p-n_e)$

Boltzmann relation for electrons,

$n_e=n_0exp(e\phi/k_B T_e)$

3. The attempt at a solution

Inside the cylinder the electric fields will cancel, hence there is no potential there. For $0\leq r\leq r_0$,

$\phi(r)=0$

For $r>r_0$ first assume that far away from the cylinder $n_0=n_p=n_e$. Thus the expression for the divergence of the electric field becomes,

$\nabla^2\phi=\frac{1}{r}\frac{\partial}{\partial r}(r\frac{\partial\phi}{\partial r})=\frac{-e}{\epsilon_0}n_0[1-exp(e\phi/k_B T_e)]$

For large distances away from the cylinder $|e\phi|\ll k_B T_e$ and so expanding the exponential into a Taylor series and keeping only the first order term,

$\nabla^2\phi=\frac{1}{r}\frac{\partial}{\partial r}(r\frac{\partial\phi}{\partial r})=\frac{n_0 e^2}{k_B T_e \epsilon_0}\phi$

Expanding out the LHS,

$\nabla^2\phi=\frac{d^2\phi}{dr^2}+\frac{1}{r}\frac{d\phi}{dr}=\frac{n_0 e^2}{k_B T_e \epsilon_0}\phi$

A second order, non-linear ODE in $r$. I have not been able to figure out how to solve this equation. Although I only JUST now had a thought; at large distances from the cylinder the first order term will disappear. This is because as $r\to \infty$,

$\frac{1}{r}\frac{d\phi}{dr}\to 0$

The resulting ODE is easy to solve. Am I on the right track?

2. May 21, 2015