# Electrolytes and electric field

Let me assume, that I have an arbitrary electric field. Is there any way to determine what happens to this electric field if it is appolied to a charge in let me say water with natrium chloride in it? I guess there is some kind of response to the electric field by the ions, so that it will just "die out", as the electrolytes will screen it. Is there an equation that would confirm this idea?

-

To illustrate the effect, consider an immobile spherical charge $Q$ at the center of your coordinate system, surrounded by small mobile charge carriers of charge $\pm q$. Gauss's law will give you the potential $\phi(r)$ as a (spherically symmetric) function of $Q$ and the number of mobile charges contained within the radius $r$. The potential energy of a single mobile charge $q$ located at $r$ is $U = q\phi(r)$. Because these charges can easily move around, they will follow a Boltzmann distribution. Combining those two facts will give you a differential equation for the potential everywhere. It's a nasty equation, because it contains an $\exp(\phi(r)/kT)$ term.
If the potential is much less than $kT$, you can Taylor expand the expoential as $\exp(\phi/kT) = 1 + \phi(r)/kT + O(\phi^2)$. That gives you a linear differential equation for $\phi$. The result of that equation is that $\phi(r) \propto \frac{e^{-\kappa r}}{r}$. So in addition to the $1/r$ dependence you see from the immobile charge, you acquire an exponential fall-off due to the "cloud" of mobile charges. $\kappa^{-1}$ is called the Debye screening length. The most important characteristic is that the Debye length is proportional to the square root of the concentration of your mobile ions.
It should be possible to write out the Poisson-Boltzmann equation for an arbitrary fixed charge/potential. And if the ionic strength that you care about is sufficiently weak, it should be possible to also write down the corresponding Debye-Huckel equation. My guess is that the primary effect of adding the ions would still be the $e^{-\kappa r}$ fall-off.