- #1
karlhoffman_76
- 5
- 1
Hi guys, having a hard time figuring out where to start on a plasma physics related problem.
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##.
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)##
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?
Homework Statement
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##.
Homework 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)##
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?