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

Answer 1

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

Step-by-step explanation:

a) the capacitance is defined

C =
`(Q)/(\Delta V) = \epsilon_o (A)/(d)`

Q = ε₀
`(A)/(d) \ \Delta V`

let's calculate

Q = 8.85 10⁻¹² 0.07
`(A)/(d)`

Q = 0.6195 10⁻¹²
`(A)/(d)`

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)/(\Delta V) = k \epsilon_o \ (A)/(d)`

Q = k ε₀
`(A)/(d) \ \Delta V`

Q = k Q₀

Q = 7 1.24 10⁻²

Q = 8.68 10⁻² pC