= 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 gives
$$
A=2\pi r\ell+2\pi r^2=6.3460\times10^{-10}\ \mathrm{m^2},\qquad
\mathcal V=\pi r^2\ell=3.1416\times10^{-16}\ \mathrm{m^3}.
$$
The <membrane capacitance> is $C=C_mA=6.3460$ pF. Thus $\Delta Q=C\Delta V=5.3941\times10^{-13}$ C, corresponding to $\Delta N=\Delta Q/e=3.3667\times10^6$ monovalent <sodium> <ions>. Converting charge to <amount of substance> using the <Faraday constant>,
$$
\boxed{\Delta c_{\rm Na}=\frac{C_m A\Delta V}{F\mathcal V}
=0.01780\ \mathrm{mol\,m^{-3}}=0.01780\ \mathrm{mM},\qquad
\frac{\Delta c_{\rm Na}}{15\ \mathrm{mM}}=1.186\times10^{-3}.}
$$
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 \b[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.
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