Does the Debye Length Change with Applied Potential Strength in a Plasma?

[Sharkey4123]

The debye length is the effective length over which electrostatic potential disturbances are "screened out" in a plasma.

So if I drop a charged point particle in a plasma, then I expect after some Debye Length, D, from this point charge, I can see no difference between any other point in the plasma.

Now, if I were to increase the charge of that particle and drop it in the plasma again, would the Debye length still be the same as before? Even if I considered this new charged particle to have any arbitrarily large charge?

Surely, the electrostatic potential created by this new charge would be much larger than the previous case, and thus "extend" further, taking "more" of the plasma to screen it out, hence a larger Debye Length. Is this correct?

 

[Sharkey4123]

I've probably figured this out myself, but should this be of use to anyone I'll leave it here for reference, or someone can correct me.

I believe that if I were to add a larger charge, there would simply be more ions/electrons (depending on the sign of the charge) attracted to within the Debye length.

So instead of an "increased" Debye length, I'd have more of the appropriate particle within the Debye length.

 

[the_wolfman]

Roughly speaking if you apply a potential to a plasma then the potential in a plasma will decay as [itex] \phi_0 e^{-\frac{\left|x\right|}{\lambda_D}} [/itex] where [itex] \lambda_D[/itex] is the Debye length and [itex] \phi_0[/itex] is the applied potential.

The Debye length is a measure of how fast the applied potential decays. It is independent of the strength of the applied potential. It is the same if I apply a 1V potential or a 10V potential. However a larger potential will penetrate farther into a plasma. This isn't due to a change in the Debye length. You're just applying a stronger field so it takes more Debye lengths for it to decay away.