Past exam of the biological physics course of the University of Oxford / 2016 / 1 / a / Solution

Treat the membrane as a capacitor, and count the additional positive charge on its inner surface. With radius $r=10^{-6}$ m and length $\ell =10^{-4}$ m, including both end caps givesThe membrane capacitance is $C=C_{m}A=6.3460$ pF. Thus $\Delta Q=C\Delta V=5.3941\times 10^{-13}$ C, corresponding to $\Delta N=\Delta Q/e=3.3667\times 10^6$ monovalent sodium ions. Converting charge to amount of substance using the Faraday constant,Omitting the ends changes the answer to $0.01762$ mM, a one-percent correction. A large membrane potential change therefore requires only about $0.12\%$ of the intracellular sodium population to enter. The bulk liquid remains nearly electrically neutral; excess charge resides chiefly near the cell membrane. This explains why an action potential need not exhaust the concentration difference that drives it, and why fixed reversal potentials are a useful approximation during one spike. This membrane capacitive ion budget assumes sodium supplies the minimum net charging requirement. Simultaneous opposing electric currents can require more sodium entry; repeated spikes eventually require the sodium-potassium ATPase to maintain the gradients.