Extracellular potential via a point source

Question

You can stimulate neurons extracellularly - that is, you place an electrode at a certain distance.
Assuming a punctiform electrode and a homogeneous medium, the extracellular potential (in mV) reduces to
$$V = frac{rho_{ext} cdot I}{4 cdot pi cdot r}$$
where I is the applied current ($mu A$), $rho$ (given in $k Omega cm$) is the resisitivity of the extracellular medium and $r$ is the Euclidean distance to the cell.
Unfortunately, it is not given why this external potential simplifies like this....
How do I arrive at this? Is this a simplification from Maxwell theory?

verdelite · Accepted Answer

This formula can be derived. You do not use Maxwell theory, but use Ohm's law and the assumptions that the medium is uniform and the other electrode is far far away.
You insert current $I$  from this electrode. Current $I$ will travel through the medium and reach the other electrode. Since the other electrode is far far away, we assume the current is sphere symmetrical, like radiation. Thus the density of $I$ is proportional to $1/r^2$, thus the drop of $Delta V$ is porportional to $1/r^2$ (by Ohms law). You integrate $Delta V$ from $r$ to $infty$ to obtain potential at $r$, assuming potential at $infty$ is $0$. You reach $1/r$ law by integrating $1/r^2$.
(I have studied neurons for 3 years to feel familiar with your question...)

Extracellular potential via a point source

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