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 3 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.

Answer 1

Answer:

Assuming that the radius of the axon is 1 $\mu\text{m}$ and that the thickness of the membrane is $9\,\text{nm}$, $Q\approx13 \, \text{pC}$

Explanation:

So this is basically the problem of a cylindrical capacitor. You have a very long cylinder, compared to its radius, so the approximation of an infinite cylinder works quite well. For an infinite cylinder, electric field is perpendicular to the surface of the inner face of the membrane of the axon. Consider the gaussian surface of a solid cylinder with radius r and hight L. Then Gauss' law,

$\oint_{\partial V}\vec{E}\cdot d\vec{A}=\frac{Q}{\epsilon_0}$,

being $\partial V$ the surface of the gaussian surface and $Q$ the total charge inside it, Gauss' law becomes

$EA=\frac{Q}{\epsilon_0}\Rightarrow\vec{E}=\frac{Q}{\epsilon_0A}\hat{r}.$

The area A is that of the cylinder (without top and base), A=$2\pi r L$. Now, to obtain the potential difference we integrate in a path that is a straight line in the direction of the radial vector $\hat{r}$, and between the radius of the Axon, say $R=1 \mu\text{m}$ and the radius plus its thickness, $thick=9\text{nm}$

$V=-\oint\vec{E}\cdot d\hat{r}=-\int_{R}^{R+thick}\frac{Q}{\epsilon_0r2\pi L}\hat{r}\cdot \hat{r}=-\frac{Q}{\epsilon_02\pi L}\ln(\frac{R+thick}{R})$. So the net charge is given by $Q=\dfrac{V2\pi L\epsilon_0}{\ln(\frac{R+thick}{R})}$, ( the - sign is unimportant here as we are interested in the value of the charge). Computing all the numbers,

$Q=\dfrac{0.07 V 8.85\times10^{-12}F/m*0.03m}{\ln(\frac{1\times10^{-6}+9\times10^{-9}}{1\times10^{-6}})}\approx 13 \text{pC}$.