A typical cell membrane in an animal maintains a potential difference across the membrane of ΔV = 70 mV and the membrane has a thickness of about 8 nm. The capacitance of the membrane is about 1 microFarad per square cm.
(B) If we model the membrane by a simple "two thin sheets of charge" model separated by 8 nm with nothing between them, what would be the electric field be between the sheets and what would the charge density on the sheets of the membrane? Explain your reasoning.
Now, notice that:
[tex]Q=C\Delta V\\
\Delta V=E.L[/tex]
Where L is the thickness of the membrane.
We then can get the electric field by just using:
[tex]E=\frac{\Delta V}{L}=\frac{70\times10^{-3}}{8\times10^{-9}}=8.75\times10^{6}N.C^{-1}[/tex]
Now using the charge expression and plugging the the expression for [tex]\Delta V[/tex] into it and then solving for [tex]E[/tex] we get:
[tex]E=\frac{Q}{CL}[/tex]
Now let's multiply both denominator and numerator by [tex]\frac{1}{A}[/tex] this yields:
[tex]E=\frac{\sigma}{C_{\sigma}L}[/tex]
Where
[tex]\sigma=\frac{Q}{A}\\ C_{\sigma}=\frac{C}{A}[/tex]
Notice that the value in the wording of the problem is no actually the capacitance, but the capacitance per area [tex]C_{\sigma}[/tex], also [tex]\sigma[/tex] is the charge density.
Let's transform [tex]C_{\sigma}[/tex] to [tex]\frac{farad}{m^2}[/tex]:
[tex]\frac{1\mu F}{cm^2}=\frac{1\times10^{-2}F}{m^2}[/tex]
Solving for [tex]\sigma[/tex] we get:
[tex]sigma=E.C_{\sigma}.L=7\times10^{-4}\frac{Coulomb}{m^2}=7\times10^{-8}\frac{Coulomb}{cm^2}[/tex]
