Question

The net charge difference across the membrane, just like the charge difference across the plates of a capacitor, is what leads to the voltage across the membrane.

How much excess charge (in picocoulombs, where 1 pc = 1x10^-12 C) must lie on either side of the membrane of an axon of length 2 cm to provide this potential difference (0.07 V) across the membrane? You may consider that a net positive charge with this value lies just outside the axon cell wall, and a negative charge with this value lies just inside cell wall, like the equal and opposite charges on capacitor plates.

Previous Info:

An average cylindrical axon has a cell wall that is d = 5 × 10 − 9  m thick and has a radius r = 5 × 10 − 6  m. The dielectric constant of the lipid bilayer κ = 7. Find the capacitance then of the axon cell wall membrane for an axon of length 4 cm. You may consider the cell membrane as having the geometry of a parallel plate capacitor by unrolling the cylinder, which yields parallel capacitor plates with an area A = 2 π r l where l = 1.0  cm is the length of the axon. Give your answer to the nearest nanofarad where 1 nanofarad = 1x10^-9 F.

Answer 1

Here, I have used simple concepts regarding potential difference, charge, capacitance and then using necessary formulae regarding them, I have calculated the answers.