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. Again, make sure to include LaTeX: \kappa=7κ = 7 for the lipid bilayer.

a)  Q = 1.24 10⁻² pC,  b) Q = 8.68 10⁻² pC

Explanation:

a) the capacitance is defined

          C =

           Q = ε₀  

let's calculate

           Q = 8.85 10⁻¹²  0.07

           Q = 0.6195 10⁻¹²  

where a is the area of ​​the membrane

            A = d L

            Q = 0.6195 10⁻¹² Ll

            Q = 0.6195 10⁻¹² 0.02

             Q = 1.24 10⁻¹⁰ C

             Q = 1.24 10⁻² pC

B) the membrane is full of fat with k = 7

            C =

            Q = k ε₀  

             Q = k Q₀

            Q = 7   1.24 10⁻²

            Q = 8.68 10⁻² pC

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