neuroscience - What is the mechanism by which myelination reduces the capacitance of the axon membrane?

There are two mechanisms that have been proposed to me.

1) Layering of Schwann cell membrane with conducting fluid between the layers is analogous to several capacitors in series. Since capacitance in series add by the reciprocal rule (as resistors do in parallel), this reduces the total capacitance.

2) The myelin increases the distance between the 'plates' of the capacitor. For parallel plate capacitors $ C = \epsilon A/d$ where d = distance between the plates. Thus increasing the distance reduces the capacitance.

Which of these explanations best applies to myelin, or is it in fact a mixture of both?

Answer

Circuit analogies don't 100% apply to myelin because membranes have complex electrical properties, but both of those explanations work and they are in fact essentially interchangeable: Take a membrane with distance d across the membrane and capacitance c. Then we add some myelin to get a new capacitance C at a new distance D.

If you 4X the distance between plates (D = d * 4), C=c/4 (from the formula you posted as (2) ); if you add 3 extra plates (so now you have a total of 4 plates), C=1/(1/c + 1/c + 1/c + 1/c)=c/4.

Importantly, myelin also increases the membrane resistance, and because myelin is typically very thick compared to a normal membrane (~10nm for one layer vs. 500-2500nm for myelin), you can almost consider myelination to increase resistance to infinity (compared to the axial resistance of cytoplasm) and the capacitance to zero.

See this page for some more info.

Note that the reason these explanations are interchangeable is that there is effectively no distance between the added plates in series and no difference in capacitance for each individual capacitor/piece of membrane (for example, see this page).