How Does Poisson's Equation Apply to Plasma Physics?

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SUMMARY

This discussion focuses on the application of Poisson's equation in plasma physics, specifically for a plasma composed of electrons and protons. The equation presented is \frac{1}{r^{2}}\frac{d}{dr}\left(r^{2}\frac{d\varphi\left(r\right)}{dr}\right) = \frac{2}{\lambda_{D}}\varphi\left(r\right), with the solution given as \varphi\left(r\right) = \frac{e}{4\pi\epsilon_{0}r}exp\left(-\frac{\sqrt{2}r}{\lambda_{D}}\right) as r approaches infinity. The user, Peter, initially struggled with integrating the equation but later found a method by substituting R(r) = r \varphi (r), leading to a homogeneous second-order differential equation. This approach clarified the path to the solution.

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peterjaybee
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Hello,

In my plasma physics book, poissons equation in relation to a plasma made up of electrons and protons is given as

\frac{1}{r^{2}}\frac{d}{dr}\left(r^{2}\frac{d\varphi\left(r\right)}{dr}\right) = \frac{2}{\lambda_{D}}\varphi\left(r\right)

The solution of this equation when phi(r) tends to 0 for r tending to infinity is

\varphi\left(r\right) = \frac{e}{4\pi\epsilon_{0}r}exp\left(-\frac{\sqrt{2}r}{\lambda_{D}}\right)

How do you get to this solution? I have tried multiplyin both sides by r^2, then integrating w.r.t. r, by parts, but I cannot get any further than that.

Regards,

Peter
 
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The simplest thing to do would be to substitute using:

R(r) = r \varphi (r)

Then you can rewrite it as:

\frac{1}{r}\frac{d}{dr} \left( r^2 \frac{d \varphi}{dr} \right) = \frac{2}{\lambda_d}R

You can continue from here by performing the chain rule to place the left hand side in terms of R.

EDIT: Nevermind, I made a mistake earlier on, fixed now. Sorry for any confusion.
 
Last edited:
I have now got a homogeneous 2nd order differential equation and from that an auxillary equation, which is beginning to take the form of the answer. Thanks
 

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