# [Neuroscience] Re: Explaination of the direction of field potentials

jonesmat jonesmat at physiology.wisc.edu
Tue Jul 11 12:24:20 EST 2006

```Matthew Kirkcaldie wrote:

> I woulda thought the amount of charge involved in charging up the
> membrane would be minuscule, and would be drawn most from nearby
> solution, and would drop off with the square of distance at least,
> meaning that charging the membrane would be nigh-impossible to detect as
> a movement of ions away / toward an electrode sitting off the surface.

> It doesn't sound like it's going to add up to much current at the
> recording electrode ... !
>
> I would reckon minis involve a much greater migration of ions than
> charging up the membrane, and yet you have to be patched on with a good
> strong seal to see them.

Hi all,

Just for the sake of doing a back-of-the-envelope calculation, let's
assume 1 neuron rests at -60 mV, has a leak resistance of 250 MOhm, and
a capacitance of 50 pF (these aren't toooo far off of what you would
measure with a patch pipette in a pyramidal cell), but also assume that
it's a sphere to make the calculation tractable. Let's also, for a
moment, assume that the EPSC is a square-shaped current pulse injected
by an invisible  constant current source (i.e., the "dendrite"), and
that it is sufficient to depolarize the somatic membrane by 10 mV if it
were to reach steady state. The time constant of this cell is Rm*Cm =
12.5 ms, so reaching steady state will take about 5 times that long, or
62.5 ms.

Consider the resistive component of charge transfer: At rest, no net
current is flowing by definition. At the plateau of the EPSP, the
voltage has changed 10 mV from rest so the current required to get
there was Isyn = deltaV/Rm = 40 pA, and this needs to be maintained
over about 60 ms, yielding a total *resistive* charge transfer of 2.4
pC.

Meanwhile, at steady-state, the capacitance will have absorbed Q =
deltaV*C = 0.5 pC.

So under steady-state conditions, the capacitive charge transfer is
about 20% of the resistive charge transfer. This is small, but not
negligible. Somebody correct me if I make a mathematical blunder. I'm
prone to do that.

However, in real life, the EPSP is not a steady-state thing. The
voltage, charge, current etc are related by an equation something like
this:

CdV/dt = [Isyn - Gleak*(V-Eleak)]

where the quantity  Gleak*(V-Eleak) is the resistive component of
current (Eleak is the resting potential), and CdV/dt is the capacitive
component.

Looking at this equation, the entire capacitive component is by itself
on the left,  the difference of injected and resistive components are
on the right, and the two sides have to be equal. At rest, Isyn = 0,
and V = Eleak, so the right side is all zero, thus the left side must
be zero too. But if we inject an Isyn, this will change. In the first
jiffy after turning on the synapse, Isyn is positive, but V hasn't had
time to change yet (i.e., the capacitance hasn't charged), so the
resistive component through Gleak is still zero. To balance, the
*capacitive* component must become positive.

In other words, because of the time dependence of the equation, there
are points in time where one component (capacitive in this first
instance, where V is changing rapidly) will carry (essentially) *all*
the current, and the resistive component will carry (essentially) none
of it. Likewise, there are other points in time (i.e., steady-state)
where the resistive component will carry all the current and the
capacitive component will carry none of it.

In the steady-state case where we allowed the resistive current to stay
on for 60 ms, the total charge breakdown was 5/6 resistive and 1/6
capacitive. At early points in time, or any time when the voltage is
changing quickly, the relation is the opposite.

So, my inkling that the capacitive component dominates the charge that
produces the somatic fEPSP is consistent with the calculation above, as
long as one thinks of the EPSP as a *changing* voltage, rather than a