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Infinitely Long Charged Cylinder in a Maxwellian Plasma

  1. May 16, 2015 #1
    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. jcsd
  3. May 21, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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