L1 · Chapter 1

The membrane stores charge like a capacitor

Sandbox Physics · 2026-10-31

Every electrical signal of a neuron is a voltage across its membrane. The membrane itself barely conducts, but the salt water on both sides does, so the membrane is a capacitor: put a little charge on one side and the voltage rises at once. The ion channels in the membrane are leaky pores, and each kind of channel has a battery behind it, set by a concentration difference. This chapter draws that circuit; every later chapter adds to it.

Charge on a thin insulating film

A cell membrane is a double layer of lipid molecules a few nanometres thick, and ions barely cross it. Inside is cytoplasm, outside is tissue fluid, and both are conducting salt solutions. Two conductors separated by a thin insulator make a capacitor.

For the squid giant axon, Hodgkin and Huxley used a membrane capacitance $C_M=1\ \mu\mathrm{F\,cm^{-2}}$ ; their measured mean was 0.91, with a range of 0.8 to 1.5 (Table 3 of the 1952 paper). So 1 nC of charge on each square centimetre changes the voltage by 1 mV.

This course writes $u$ for the membrane potential measured from rest, with $u>0$ when the inside becomes more positive (depolarisation). Charge conservation gives the membrane equation: an applied current partly charges the capacitor and partly crosses the channels. Gerstner and co-workers write it this way in their Eqs. (2.3) and (2.4); Eq. (1) of Hodgkin and Huxley says the same thing with the opposite sign convention.

Q=C_M\,u,\qquad C_M\frac{du}{dt}=-\bigl[I_{\mathrm{Na}}+I_{\mathrm{K}}+I_{L}\bigr]+I_{\mathrm{ext}}

Left: charge is proportional to voltage. Right: the membrane equation, with

I_{\mathrm{ext}}

the applied current and the three ionic currents counted positive outward. A very brief shock that delivers charge

q

moves the voltage from 0 to

q/C_M

The membrane as a circuit. The lipid layer stores charge like a capacitor

C_M

. Each kind of channel is a conductance

g_x

in series with a battery at its reversal potential

E_x

; the arrows mark the sodium and potassium conductances that change with voltage. An applied current

I_{\mathrm{ext}}

either charges the capacitor or leaves through the channels. Original schematic.

Source: A. L. Hodgkin and A. F. Huxley · W. Gerstner

A concentration difference is a battery

Ion pumps move sodium out and potassium in, so the inside of the cell is rich in potassium and poor in sodium, and the outside the reverse. Open a channel that passes only potassium and potassium flows out down its concentration gradient, carrying positive charge, so the inside turns negative. At some voltage the electric pull back balances the push of the concentration difference and the net current stops. That voltage is the equilibrium potential of potassium.

At equilibrium the ion density follows a Boltzmann distribution, which gives the Nernst formula (Eq. (2.2) of Gerstner and co-workers). For a mammalian neuron with 140 mM potassium inside and 5 mM outside, $k_BT/e=25.3\ \mathrm{mV}$ at 20 °C gives a potassium equilibrium potential of $-84\ \mathrm{mV}$ . Sodium at 145 mM outside and 10 mM inside gives $+68\ \mathrm{mV}$ .

When the voltage crosses the equilibrium potential the current reverses, so it is also called the reversal potential. In the circuit each kind of channel is a conductance in series with a battery, and its current is the conductance times the distance of the voltage from the reversal potential. Hodgkin and Huxley fitted three reversal potentials for the squid axon. In this course’s sign convention, measured from rest, they are $E_{\mathrm{Na}}=+115\ \mathrm{mV}$ for sodium, $E_{\mathrm{K}}=-12\ \mathrm{mV}$ for potassium and $E_L=+10.6\ \mathrm{mV}$ for the leak.

E=\frac{k_BT}{q}\,\ln\frac{n_{\mathrm{out}}}{n_{\mathrm{in}}},\qquad I_x=g_x\,(u-E_x)

Here

q

is the charge of the ion and

n

its concentration. Right: the current through one kind of channel with conductance

g_x

Source: W. Gerstner · A. L. Hodgkin and A. F. Huxley

At rest the currents cancel

Rest does not mean no current. In the Hodgkin–Huxley model at rest, the sodium conductance is $0.0106\ \mathrm{mS\,cm^{-2}}$ and sodium leaks in at $1.22\ \mu\mathrm{A\,cm^{-2}}$ . The potassium conductance is $0.367\ \mathrm{mS\,cm^{-2}}$ and potassium flows out at $4.40\ \mu\mathrm{A\,cm^{-2}}$ . The leak carries $3.18\ \mu\mathrm{A\,cm^{-2}}$ inward. The three add to zero and the voltage does not move.

That zero is built in. A footnote to Table 3 says the leak potential, $-10.613\ \mathrm{mV}$ in their signs, was chosen to make the total ionic current zero at rest. The steady inflow of sodium and outflow of potassium slowly run down the concentration differences, and the cell restores them with ion pumps. Their equations have no pump, because a pump does almost nothing on the millisecond scale of one spike.

g_{\mathrm{Na}}(u-E_{\mathrm{Na}})+g_{\mathrm{K}}(u-E_{\mathrm{K}})+g_L(u-E_L)=0\quad\text{at}\quad u=0

The resting condition. No single term is zero, but the sum is. With the numbers above the residual is below

0.005\ \mu\mathrm{A\,cm^{-2}}

, from rounding the leak potential to three decimals.

Source: A. L. Hodgkin and A. F. Huxley · W. Gerstner

Worked example: how many ions make a 15 mV shock

Problem. Hodgkin and Huxley used instantaneous shocks to set an initial depolarisation and then let the membrane go. How much charge per square centimetre depolarises the membrane by 15 mV at once? How many singly charged ions per square micrometre is that? If the channel conductances then stayed frozen at their resting values, how fast would the voltage fall back?

Model. The charge is $q=C_Mu_0$ . The number of ions is the charge divided by the elementary charge $e=1.602\times10^{-19}\ \mathrm{C}$ . With frozen conductances the membrane is a resistor in parallel with a capacitor, with time constant equal to the capacitance divided by the sum of the three resting conductances.

Numbers. $q=1\ \mu\mathrm{F\,cm^{-2}}\times15\ \mathrm{mV}=15\ \mathrm{nC\,cm^{-2}}$ . That is $15\times10^{-9}/1.602\times10^{-19}=9.36\times10^{10}\ \mathrm{cm^{-2}}$ ions, about 936 per square micrometre, or $0.155\ \mathrm{pmol\,cm^{-2}}$ . The resting conductances add to $0.0106+0.367+0.3=0.677\ \mathrm{mS\,cm^{-2}}$ , so $\tau=C_M/g=1.48\ \mathrm{ms}$ .

Explanation. A few hundred ions lift a square micrometre of membrane by 15 mV, while the cytoplasm behind it holds hundreds of millions of potassium ions. The voltage changes fast and the concentrations hardly at all, which is why the batteries can be treated as constant during a spike.

Check. The channels do not stay frozen. A 15 mV start is above threshold, and the sodium channels open; the next chapter shows how they respond to voltage. Lower the shock to 6 nC/cm² and the model’s highest voltage is 5.96 mV, after which it falls back. Only then does the membrane behave much like a plain resistor and capacitor.

Source: A. L. Hodgkin and A. F. Huxley · W. Gerstner

Try it in the Lab

The shock charge sets the starting point

Predict first: what voltage does a 6 nC/cm² shock produce? Will the voltage then keep rising or fall back?
Choose the Just below threshold preset and read the peak depolarization. Then raise the shock charge to 7 and see what changes.
Explain: charge divided by capacitance gives the starting point, and the channels decide where the voltage goes next. The next chapter explains the difference between 6 and 7.

Exercises with explained answers

A shock delivers 40 nC/cm². What is the starting depolarisation, and how many singly charged ions per square micrometre does it take?
Show the answer
The start is $40\ \mathrm{mV}$ . The ion count is $40\times10^{-9}/(1.602\times10^{-19}\times10^{8})\approx2.5\times10^{3}$ per square micrometre.
At 6.3 °C, what is the equilibrium potential of potassium with 140 mM inside and 5 mM outside? How does it compare with 20 °C?
Show the answer
$k_BT/e=24.1\ \mathrm{mV}$ , so $E_{\mathrm{K}}=24.1\times\ln(5/140)=-80\ \mathrm{mV}$ . Colder ions move less, so the same concentration ratio holds a smaller voltage, about 4 mV less than the −84 mV at 20 °C.
Table 5 of Hodgkin and Huxley gives a net sodium entry of 4.33 pmol per square centimetre for one propagated action potential at 18.5 °C. The axon radius is 238 µm. How much does the internal sodium concentration rise?
Show the answer
For a cylinder of radius $a$ , the volume behind each unit of membrane area is $a/2$ . So $\Delta c=2\times4.33\times10^{-12}\ \mathrm{mol\,cm^{-2}}/0.0238\ \mathrm{cm}=3.6\times10^{-10}\ \mathrm{mol\,cm^{-3}}$ , about 0.36 µM. Internal sodium is tens of millimolar, so one spike changes it by less than one part in a hundred thousand.
Sodium flows in all the time at rest. Why does the voltage not rise?
Show the answer
The outward potassium current and the leak current together cancel the sodium inflow, so the total current is zero and the charge on the capacitor does not change. The price is a slow loss of the concentration differences, which the cell pays back with pumps that consume energy.

Sources and what we recomputed

A. L. Hodgkin and A. F. Huxley, A quantitative description of membrane current and its application to conduction and excitation in nerve, Journal of Physiology 117, 500–544 (1952), PMC1392413
We read every page, pp. 500 to 544, from the scanned pages in PubMed Central. Sign convention on pp. 505 and 506: V is the displacement from rest with depolarisation negative, and inward current is positive. Rate constants: Eqs. (12), (13), (20), (21), (23) and (24). Eq. (26) and Table 3 on pp. 518 to 520: the full membrane equation and its constants at 6.3 °C, with the leak potential chosen so that the current is zero at rest. Numerical method on p. 523. Figs. 12, 13 and 21: threshold between 6 and 7 mV. Table 4 on p. 527: spike shape. Table 5 on pp. 532 and 533: sodium and potassium movements. Fig. 20: refractory period. Fig. 22: anode break. P. 541: their warning that the equations are an empirical description. We recomputed Tables 4 and 5 for the membrane action potentials, not the propagated one.
W. Gerstner, W. M. Kistler, R. Naud and L. Paninski, Neuronal Dynamics, Cambridge University Press (2014), online edition, Sections 2.1 and 2.2
Section 2.1: the Boltzmann factor, Eq. (2.1), and the Nernst potential, Eq. (2.2), with the sodium example near +67 mV and the potassium example near −83 mV. Section 2.2 (Ch2.S2.html): charge conservation on the membrane, Eqs. (2.3) and (2.4), the three ionic currents, Eq. (2.5), first-order gating, Eq. (2.6), and the response to a voltage step, Eq. (2.7). This book writes u as the membrane potential with depolarisation positive, the convention our course uses.