www.neuron.yale.edu

I've implemented a simple simulation with a passive cylindrical section (400 um long, 1 um diam) and an external oscillating electric field at 50 Hz.
I don’t see any difference when I change the extracellular capacitance from its default value of 0 (the xc[0] or xc[1] in the extracellular mechanism). Any suggestion?

What did you do about xg?

Nothing.
I keep the xg default value.

And the default value of xg is what?

It's 1e+09 mho/cm2.

xc and xg are in parallel. From linear circuit theory we know that, for a resistor and capacitor in parallel, the frequency at which the current through the resistor equals the current through the capacitor is fc = 1/(2*PI*tau). Also, at frequencies smaller than fc/5, to a very good first approximation the current through the capacitor can be ignored--which means that the paralllel RC circuit acts like a resistor.

So let's imagine that you increased xc from 0 to 1 uF/cm2. The time constant of xc and xg in parallel is
tau = (1 (uF/cm2)) / (1e9 (S/cm2)) = 1e-9 uF/S = 1e-9 ohm*uF = 1e-15 ohm*F = 1e-15 second
so
fc is ~ 0.16e15 = 160,000 GHz
which means that for frequencies below 32,000 GHz the parallel combination of xc and xg is electrically equivalent to a resistor. Another way of putting this is: you can't see the effect of the capacitance because the capacitor is shorted out by a huge conductance.

Which means that, if you want to see the effect of xc, you either have to increase ____ or decrease ____ by a large amount.