Electrical and Magnetic Properties


Bioelectric signal

Typical amplitude

Electrocardiogram (EKG/ECG, heart)

1 mV

Electroencephalogram (EEG, brain waves)

10–100 $$\upmu $$V

Electromyogram (EMG, muscle)

300 $$\upmu $$V

Transmembrane potential

100 mV

Electro-oculogram (EOG, eye)

500 $$\upmu $$V


Using data from [3, 16 ]



We have addressed the importance of electricity in the body only briefly in previous chapters. In Chap. 3 we examined the electromyograms (EMGs) of muscle activity (Fig. 3.​13), in Chap. 5 we saw that muscles are activated by electrical stimuli and the release of Ca$$^{2+}$$ ions, and in Chap. 8 we learned that the polarization and depolarization of cell membranes in the heart provide the signals for electrocardiograms (EKGs, ECGs). We now discuss such electrical interactions in more depth as we focus on the electrical properties of the body, the propagation of electrical signals in the axons of nerves, and electrical potentials in the body (Table 12.1).

It is impossible to overemphasize the importance of this human “bioelectricity.” The function of every cell depends on it. Every neuron in the brain, every neuron transmitting any information within the body, every neuron enabling skeletal, cardiac, and smooth muscles is yet another vital example. This chapter is largely a discussion of the physics of the motion of positive and negative ions in the blood and cells. We will be concerned with the motion of these ions across membrane boundaries, as in neurons, but not the underlying biology that controls these ion channels. Electric voltages measured at different places in the body describe electrical activity, as is seen in Table 12.1. We will emphasize the propagation of electrical signals in nerves and monitoring the EKG signals from the heart.

Electric and magnetic fields are closely coupled in many areas of physics; for example, electromagnetic waves (visible light, radio waves, X-rays, and so on) consist of electric and magnetic fields oscillating in phase. Magnetic fields appear when current flows. Although current flow is important in the body, the resulting magnetic fields appear to be relatively unimportant and we will address magnetism in the body only briefly in this chapter. For more details on the electrical and magnetic properties of the body see [2, 15, 17, 18, 23, 31].


12.1 Review of Electrical Properties


We first review the various elements of electrostatics and current flow needed to understand electricity in the body, including the flow of an electrical pulse along an axon.

The electric field at a distance r caused by a point charge q is given by Coulomb’s Law:


$$\begin{aligned} \mathbf {E}=\frac{kq\mathbf {r}}{r^{3}}=\frac{kq\mathsf {r}}{r^{2}}, \end{aligned}$$

(12.1)
where the vector from the charge to the point is $$\mathbf {r}=r\mathsf {r}$$ and $$\mathsf {r}$$ is a unit vector from that charge to the point of interest, as illustrated in Fig. 12.1a, b. The constant $$k = 8.99 \, \times \, 10^{9}$$ Nm$$^{2}/$$C$$^{2}$$ for a charge in vacuum, where C stands for coulombs, and can also be expressed as $$ 1/4\pi \epsilon _{0}$$. In a medium of dielectric constant $$\epsilon $$ (where $$ \epsilon = 1$$ in vacuum), $$k=1/4\pi \epsilon _{0}\epsilon $$.

A114622_2_En_12_Fig1_HTML.gif


Fig. 12.1
The electric field vectors for a positive and b negative charges are shown, along with those for c a dipole of two charges $$+q$$ and $$-q$$, separated by a distance d, so the magnitude of the dipole moment is $$P=qd$$. The direction of the dipole moment is seen by the arrow within the dipole in (d). d also shows the potential along the different radial directions shown for this dipole, and the $$1/r^{2}$$ decrease in each of these voltages

The potential of that charge is


$$\begin{aligned} V=\frac{kq}{r}. \end{aligned}$$

(12.2)
and, as here, the potential is usually defined to be zero as r approaches infinity. The potential difference (or voltage) between two points “b” and “a” caused by a field is


$$\begin{aligned} \Delta V=V_{\mathrm {b}}-V_{\mathrm {a}}=-\int _{\mathbf {r}_{a}}^{\mathbf {r}_{b}} \mathbf {E}\cdot \mathrm{d}\mathbf {r}. \end{aligned}$$

(12.3)
This can also be expressed as


$$\begin{aligned} \mathbf {E=-\nabla }V \end{aligned}$$

(12.4)
or in one dimension as


$$\begin{aligned} E\mathrm { }\mathbf {=-}\frac{\mathrm{d}V}{\mathrm{d}x}. \end{aligned}$$

(12.5)
If there are two charges q and $$-q$$ in vacuum separated, say a distance d in the $$\mathsf {d}$$ unit vector direction (so the vector between them is $$ \mathbf {d}=d\mathsf {d}$$), the electric field is the vector sum of the contributions from (12.1) and the electric potential is still obtained using (12.3) (Fig. 12.1c, d). For $$r\gg d$$, the expression for the potential can be simplified to give


$$\begin{aligned} V=\frac{k\mathbf {P}.\mathbf {r}}{r^{3}}, \end{aligned}$$

(12.6)
where $$\mathbf {P}=q\mathbf {d}$$ is the electric dipole moment vector, which has magnitude $$P=qd$$ and points in the $$\mathsf {d}$$ direction. If the angle between the dipole vector $$\mathbf {P}$$ and distance vector $$\mathbf {r}$$ is $$ \theta $$, then this equation becomes


$$\begin{aligned} V=\frac{kP\cos \theta }{r^{2}}. \end{aligned}$$

(12.7)
Similarly, we can calculate the dipole moment for many charges separated by various distances. Evaluation of the fields caused by such electric dipoles is of particular value when there is no net charge in the collection of charges, as is true most everywhere in the body.

Now let us consider a moving particle with charge q (in coulombs, C), the current, $$I=\mathrm{d}q/\mathrm{d}t$$, associated with such a charge or charges (which is the change in charge per unit time), and the associated current density, $$J=I/A$$ (which is the current flowing per unit area A). Charge is conserved, meaning that it is neither created nor lost. It also means that the vector sum of all currents entering a volume or a small volume element (such as a node) is zero in steady state (Fig. 12.2). This conservation of current (and charge) is known as Kirchhoff’s 1st Law


$$\begin{aligned} \sum _{n}I_{n}=0. \end{aligned}$$

(12.8)
(The direction of current flow is important here, even thought the current is being expressed as a scalar.)

A114622_2_En_12_Fig2_HTML.gif


Fig. 12.2
Kirchhoff’s 1st Law, showing that the algebraic sum of the current flows to a point must be zero (if charge is not accumulating or being depleted at that point) (Note that least one of the current flows must be negative, i.e., it must point outward)


A114622_2_En_12_Fig3_HTML.gif


Fig. 12.3
a Ohm’s Law and b evaluating resistance R from resistivity $$\rho $$

When a current flows along a material with resistance R (in ohms, $$ \varOmega $$) (which we called $$R_{\mathrm {elect}}$$ in other chapters), there is a voltage drop V (in volts V) (which we called $$V_{\mathrm {elect}}$$ in other chapters) across the material given by Ohm’s Law (Fig. 12.3a)


$$\begin{aligned} V=IR. \end{aligned}$$

(12.9)
The resistance is an extensive property that depends on the intensive property resistivity $$\rho $$ of the material, and the cross-sectional area A and length L of the structure (Fig. 12.3b)


$$\begin{aligned} R=\frac{\rho L}{A}. \end{aligned}$$

(12.10)
For a cylinder with radius a, we have $$A=\pi a^{2}$$ and $$R=\rho L/\pi a^{2}$$. More generally, for a structure with uniform cross-section, the resistance R is proportional to length and we can define a resistance per unit length


$$\begin{aligned} r=\frac{R}{L}=\frac{\rho }{A}, \end{aligned}$$

(12.11)
which equals $$\rho /\pi a^{2}$$ for conduction along a cylinder. The conductance G (units S (siemens), 1 S = 1 mho = 1/ohm $$=1/\varOmega $$) is 1 / R, the conductivity is $$\sigma =1/\rho $$, and the conductance per unit area $$g=G/A=1/ RA=1/\rho L$$. In the body, charged ions, such as Na$$^{+}$$, K$$^{+}$$ , Ca$$^{2+}$$, Cl$$^{-}$$, and negatively-charged proteins, are the important carriers of charge. Electrons are the charge carriers in most man-made electronic circuits.

A voltage or potential difference V can also develop between two structures, one with a charge $$+q$$ and the other with charge $$-q$$, because of the electric fields that run from one to the other. This voltage is


$$\begin{aligned} V=\frac{q}{C}, \end{aligned}$$

(12.12)
where C is the capacitance (in farads, F) of the system (called $$C_{\mathrm {elect}}$$ in other chapters). The capacitance C depends on the geometry of these two structures. For example, they could be two parallel plates or two concentric cylinders (Fig. 12.4), which is similar to the axon of a neuron.

A114622_2_En_12_Fig4_HTML.gif


Fig. 12.4
Capacitance for a parallel plates and b cylindrical shells

For two parallel plates with area A separated a distance b by an insulator with dielectric constant $$\kappa $$, we see


$$\begin{aligned} C_{\mathrm {parallel \;plates}}=\frac{\kappa \epsilon _{0}A}{b}. \end{aligned}$$

(12.13)
The charge density on each plate is $$\sigma =q/A$$.

The algebraic sum of all voltages along a closed loop circuit equals zero (Fig. 12.5). This is known as Kirchhoff’s 2nd Law


$$\begin{aligned} \sum _{n}V_{n}=0. \end{aligned}$$

(12.14)


A114622_2_En_12_Fig5_HTML.gif


Fig. 12.5
Kirchhoff’s 2nd Law, showing that the algebraic sum of the potential drops (voltages) along a closed loop is zero



Table 12.2
The molar conductance at infinite dilution $$\varLambda _{0,i}$$ for different ions






















Ion
$$\varLambda _{0,i}$$ (1/ohm-m-M)

H$$^{+}$$

34.9

OH$$^{-}$$

19.8

Na$$^{+}$$

5.0

Cl$$^{-}$$

7.6


Using data from [33]


12.2 Electrical Properties of Body Tissues



12.2.1 Electrical Conduction Through Blood and Tissues


When voltage is applied across a metal, a current flows because electrons move under the influence of an electric field. When a voltage is applied across a solution containing positive and negative ions, current flows because both ions move under the influence of the electric field. The conductivity $$\sigma $$ of a solution is the sum of the contributions to the current flow for each ion. For low concentrations of these ions, this contribution is proportional to the concentration $$n_{i}$$ for that ion, with a proportionality constant $$\varLambda _{0,i}$$, so


$$\begin{aligned} \sigma =\sum _{i}n_{i}\varLambda _{0,i}. \end{aligned}$$

(12.15)
Table 12.2 gives $$\varLambda _{0,i}$$, the molar conductance at infinite dilution for several common ions, while Table 12.3 gives typical concentrations of common ions in the blood and in cells. The resistance of a path can be determined using $$\rho =1/\sigma $$ and $$R=\rho L/A$$ (12.10).

As with many materials, body tissues have dielectric properties, but still have some conductivity, and therefore can be considered as leaky dielectrics. The resistivity of body tissues is shown in Table 12.4 and Fig. 12.6.


Table 12.3
Ionic concentrations in blood and cell cytoplasm of unbound ions






















































Ion
Blood
Cytoplasm
Ratio
 
concentration
concentration
  

Na$$^{+}$$

145 mM
12 mM
12:1

K$$^{+}$$

4 mM
140 mM
1:35

H$$^{+}$$

40 nM
100 nM
1:2.5

Mg$$^{2+}$$

1.5 mM
0.8 mM
1.9:1

Ca$$^{2+}$$

1.8 mM
100 nM
18:1

Cl$$^{-}$$

115 mM
4 mM
29:1

HCO$$_{3}^{-}$$

25 mM
10 mM
2.5:1


Using data from [34]



Table 12.4
Low frequency resistivity of some body tissues, in ohm-m ($$\varOmega $$-m)


























































Tissue
Resistivity

Cerebrospinal fluid
0.650

Blood plasma
0.7

Whole blood
1.6 (Hct = 45%)

Skeletal muscle
  

– Longitudinal
1.25–3.45

– Transverse
6.75–18.0

Liver
7

Lung
  

– Inspired
17.0

– Expired
8.0

Neural tissue (as in brain)
  

– Gray matter
2.8

– White matter
6.8

Fat
20

Bone
$$>$$” src=”http://basicmedicalkey.com/wp-content/uploads/2017/06/A114622_2_En_12_Chapter_IEq70.gif”></SPAN>40</DIV></TD></TR><br /> <TR class=noclass><br /> <TD align=left><br /> <DIV class=SimplePara>Skin</DIV></TD><br /> <TD align=left> </TD></TR><br /> <TR class=noclass><br /> <TD align=left><br /> <DIV class=SimplePara>– Wet</DIV></TD><br /> <TD align=left><br /> <DIV class=SimplePara>10<SPAN id=IEq71 class=InlineEquation><IMG alt=$$^{5}$$ src=

– Dry
10$$^{7}$$


Using data from [3, 9, 23]



A114622_2_En_12_Fig6_HTML.gif


Fig. 12.6
Cross-section of the thorax, with the electrical resistivity of six types of tissues (From [23]. Used with permission)


A114622_2_En_12_Fig7_HTML.gif


Fig. 12.7
Structure of a neuron (From [29])


12.3 Nerve Conduction


Figure 12.7 shows the structure of nerve cells or neurons with a nucleus, dendrites that receive information across synapses, an axon, and the axon terminals and synapses for signal transmission to other neurons. There are many such neuron axons in a nerve. Unmyelinated axons have no sheath surrounding them. Myelinated axons have myelin sheaths in some regions, which are separated by nodes of Ranvier (ron-vee-ay’). These sheaths are formed by Schwann cells that are wrapped around the axon (Fig. 12.8), with successive wrapped cells separated at a node of Ranvier (Fig. 12.7). We will concentrate on how an electrical impulse travels along such axons.

A114622_2_En_12_Fig8_HTML.gif


Fig. 12.8
The successive wrapping of Schwann cells about the axon of a neuron to form the myelin sheath of a myelinated nerve (From [29])

Approximately 2/3 of the axon fibers in the body are unmyelinated. They have radii of 0.05–$$0.6 \;\upmu $$m and a conduction speed of u (in m/s) $$\approx 1.8 \sqrt{a}$$, where a is the radius of the axon (in $$\upmu $$m). Myelinated fibers have outer radii of 0.5–$$10 \;\upmu $$m and a conduction speed of u (in m/s) $$\approx 12(a+b)\approx 17a$$, where b is the myelin sheath thickness (in $$\upmu $$m) (and $$a+b$$ is the total axon radius). The spacing between the nodes of Ranvier is $$\approx $$ $$280a$$.

Neurons whose axons travel from sensing areas to the spinal cord are called afferent neurons or input or sensory neurons. (They are “affected” by conditions that are sensed.) Neurons whose axons leave the ventral surface of the brain stem and the spinal cord to convey signals away from the central nervous system are efferent neurons or motor neurons, and these neurons exercise motor control. (They “effect” a change.) There are approximately 10 million afferent neurons, 100 billion neurons in the brain with 100 trillion synapses, and a half a million efferent neurons, so there are roughly 20 sensory neurons for every motor neuron and several thousand central processing neurons for every input or output neuron for processing. Bundles of these neuron axons are called nerves outside of the brain and tracts inside the brain. Details about the nervous system are given in [25].

There are approximately $$1-2 \, \times \, 10^{6}$$ optical nerves from the $$1-2 \, \times \, 10^{8}$$ rods and cones in our eyes, 20,000 nerves from the 30,000 hair cells in our ears, 2,000 nerves from the $$10^{7}$$ smell cells in our noses, 2,000 nerves from the $$10^{8}$$ taste sensing cells in our tongues, 10,000 nerves from the 500,000 touch-sensitive cells throughout our body, and many (but an uncertain number of) nerves from the $$3 \, \times \, 10^{6}$$ pain cells throughout our body.


12.3.1 Cell Membranes and Ion Distributions


The cell membrane divides the intracellular and extracellular regions, in neurons and other cells. There are Na$$^{+}$$, K $$^{+}$$, Cl $$^{-}$$, negatively-charged proteins, and other charged species both in the neurons (intracellular) and in the extracellular medium. The concentrations of these ions are such that there is charge neutrality (i.e., an equal number of positive and negative charges) in both the intracellular and extracellular fluids. However, there are negative charges on the inside of the cell membrane and positive charges on the outside of this membrane that produce a resting potential of $$-70$$ mV (Fig. 12.9). This means that the intracellular medium is at $$-70$$ mV, when the extracellular potential is arbitrarily defined to be 0 V, as is the custom. Only potential differences are significant, so we are not limiting the analysis by fixing the extracellular potential. This resting potential is the usual potential difference when there is no unusual neural activity. This is known as the polarized state . (The propagation of an electrical signal would constitute unusual activity.)

A114622_2_En_12_Fig9_HTML.gif


Fig. 12.9
Ion concentrations (in mmol/L) in a typical mammalian axon nerve cell ($$n_{\mathrm {i}}$$) and in the extracellular fluid surrounding it ($$n_{\mathrm {o}}$$), and their ratios ($$n_{\mathrm {i}}/n_{\mathrm {o}}$$) (Based on [17, 18])

While there is charge neutrality both inside and outside the membrane, the concentrations of each ion are not equal inside and outside the cell, as we will see. The differences in ion concentrations inside and outside the cell membrane are due to a dynamic balance. When there are changes in the permeability of the cell membrane to different charged species, there are transient net charge imbalances that change the potential across the cell membrane. An increase in the membrane potential from $$-70$$ mV, such as to the $$ -60$$ mV seen in Fig. 12.10, is known as depolarization, while a decrease from $$-70$$ mV to say $$-80$$ mV is called hyperpolarization . Depolarization is due to the net flow of positive charges into the cell or negative charges to regions outside the cell. Hyperpolarization is due to the net flow of negative charges into the cell or positive charges to outside the cell. Such changes in ion permeability are often termed as changes in the ion channel.

A114622_2_En_12_Fig10_HTML.gif


Fig. 12.10
The membrane resting potential of $$-70$$ mV (inside the membrane relative to the always fixed 0 mV outside)—the polarized state, along with potential disturbances showing depolarization (voltage increases from the resting potential value), repolarization (returns to the resting potential), and hyperpolarization (decreases from the resting potential)

Figure 12.9 also shows the concentrations of some of the important charged species inside and outside the cell under resting (i.e., polarized) conditions. We see that there are many more Na$$^{+}$$ outside (145 mmol/L) than inside (15 mmol/L) the cell, but many more K$$^{+}$$ inside (150 mmol/L) than outside (5 mmol/L). Including miscellaneous positive ions outside the cell, there are 165 mmol/L of positive ions both inside and outside the cell. Similarly, there are many more Cl$$^{-}$$ outside (125 mmol/L) than inside (9 mmol/L) the cell, but many more miscellaneous negative ions (including proteins) inside (156 mmol/L) than outside (30 mmol/L). There are also 165 mmol/L of negative ions both inside and outside the cell.

There are several driving forces that determine the ionic concentrations, in general, and these intracellular and extracellular concentrations, in particular:

1.

There is the natural tendency for concentrations to be uniform everywhere, so when there are concentration gradients across the cell membrane there are flows of these species from the regions of higher concentration to regions of lower concentrations, to equalize the intracellular and extracellular concentrations. This is described by Fick’s First Law of Diffusion (7.​64), $$J_{\mathrm {diff}}=-D_{\mathrm {diff}}$$ $$\mathrm{d}n/\mathrm{d}x$$ , where $$J_{\mathrm {diff}}$$ is the flux of ions in the x direction (the number of ions flowing across a unit area in a unit time ), $$D_{\mathrm {diff}}$$ is the diffusion constant, n is the local concentration of ions, and $$\mathrm{d}n/\mathrm{d}x$$ is the local concentration gradient.

 

2.

Because the potential is negative inside the cell, we would expect positive ions to enter the cell and be more dominant in the intracellular fluid than the extracellular fluid and for there to be such concentration gradients; this is true for K$$^{+}$$ but not for Na$$^{+}$$. Similarly, we expect negative ions to leave the cell because of the resting potential and be more dominant outside the cell than inside—and again for there to be concentration gradients; this is true for Cl$$^{-}$$ but not for the negatively-charged proteins, which form the bulk of the miscellaneous negative ions.

When charged species are in an electric field, they get accelerated and eventually attain a steady-state drift velocity, $$v_{\mathrm {drift}}$$, because of collisions that act as a drag force. As shown in Problem 12.7, the drift velocity of a given ion is


$$\begin{aligned} v_{\mathrm {drift}}=\mu E, \end{aligned}$$

(12.16)
where $$\mu $$ is called the mobility and E is the electric field. The flux of ions due to this electric field is


$$\begin{aligned} J_{\mathrm {elect}}=nv_{\mathrm {drift}}=n\mu E. \end{aligned}$$

(12.17)

 

3.

The cell membrane permeability and active processes cause the ion concentrations on either side of the membrane to deviate from the values expected from diffusion and the motion of charges in electric fields. The cell membranes are permeable to K$$^{+}$$ and Cl$$^{-}$$, which explains why they behave as expected. Proteins are never permeable to the cell membrane, which is why the concentration of negative-protein ions is unexpectedly high inside. The chemical mechanism called the Na$$^{+}$$ pump (or the Na$$^{+}$$–K$$^{+}$$ pump) actively transports 3Na$$^{+}$$ from inside to outside the cell for every 2K$$^{+}$$ it transports from outside to inside the cell; this keeps Na$$^{+}$$ outside the cell and K$$ ^{+}$$ inside.

 
The high Na$$^{+}$$ concentration outside the cell is the result of the Na$$ ^{+} $$ pump fighting against the driving electrical forces and the tendency to equalize concentrations (Fig. 12.11). The high K$$^{+}$$ concentration inside the cell is the result of the electric forces and the Na$$^{+}$$ pump fighting against the tendency to equalize concentrations. The high Cl$$^{-}$$ concentration outside the cell is the result of the electrical forces fighting against the tendency to equalize concentrations. The concentration of negative protein ions is unexpectedly high inside because they are large and not permeable to the cell membrane.

A114622_2_En_12_Fig11_HTML.gif


Fig. 12.11
Mechanisms for ion flow across a polarized cell membrane that determine the resting membrane potential

Figure 12.12 depicts the directions of motion for charged and neutral molecules for either the random thermal motion in diffusion or the directed effect due to an electric field. Figure 12.13 shows how a concentrated band of charged and neutral molecules changes due to either diffusion or an electric field.

A114622_2_En_12_Fig12_HTML.gif


Fig. 12.12
The direction of motion for charged and neutral molecules due to a diffusion only (at a given instant) and b an electric field only

Ionic Distributions (Advanced Topic)

What are the expected ionic distributions due to the membrane potential? First, let us consider the expected distributions for several steady state

A114622_2_En_12_Fig13_HTML.gif


Fig. 12.13
An initial band of charged and neutral molecules (in a) changes very differently by the uniform thermal spreading in diffusion (in b) and the separation caused by an electric field (in c). a Before, b After, diffusion only, c After, electric field only

conditions for a given ion. In steady state, the net flow of ions into any region is zero, so $$J_{\mathrm {diff}}+J_{\mathrm {elect}}=0$$ and using (7.​64) and (12.17) we see that


$$\begin{aligned} D_{\mathrm {diff}}\frac{\mathrm{d}n}{\mathrm{d}x}=n\mu E. \end{aligned}$$

(12.18)
(In steady state n does not depend on time, so the partial derivative in (7.​64) is not needed here.)

The diffusion coefficient, $$D_{\mathrm {diff}}$$, and mobility, $$\mu $$, are actually closely related. Consider a cylinder of cross-sectional area A and length $$\mathrm{d}x$$ along the x direction than contains a density n of ions of charge q. When an electric field E is applied along the x direction, the ions in the cylinder feel a force $$(nq)(A \mathrm{d}x)E$$, where nq is the total charge per unit volume and $$A\mathrm{d}x$$ is the volume. The mechanical force on this cylinder is due to the difference between the pressure $$\times $$ area on one side wall, AP(x), and that on the other side wall, $$ AP(x+\mathrm{d}x)=A[P(x)+(\mathrm{d}P/\mathrm{d}x)\mathrm{d}x]$$, or $$-A(\mathrm{d}P/\mathrm{d}x)\mathrm{d}x$$. The sum of these forces is zero in steady state, so $$\mathrm{d}P/\mathrm{d}x=nqE$$. Using the ideal gas law (7.​2) $$P=nk_{\mathrm {B}}T$$ (which is an approximation here and where n is now the number of molecules per unit volume because $$k_{\mathrm {B}}$$ is used instead of the gas constant R), we see that $$\mathrm{d}P/\mathrm{d}x=k_{\mathrm {B}}T(\mathrm{d}n/\mathrm{d}x)$$ or $$ k_{\mathrm {B}}T(\mathrm{d}n/\mathrm{d}x)=nqE$$. Comparing this to (12.18), gives the Einstein equation


$$\begin{aligned} \mu =\frac{qD_{\mathrm {diff}}}{k_{\mathrm {B}}T}, \end{aligned}$$

(12.19)
a result we will use soon.

Now let us consider the charge current due to two ions, one of charge q (which we will say is $$>$$” src=”http://basicmedicalkey.com/wp-content/uploads/2017/06/A114622_2_En_12_Chapter_IEq141.gif”></SPAN>0), with density <SPAN id=IEq142 class=InlineEquation><IMG alt=$$n_{+}$$ src= and mobility $$ \mu _{+}$$, and the other of charge $$-q$$, with density $$n_{-}$$ and mobility $$ \mu _{-}$$. (We will now define the mobilities as being positive, so for this negative ion $$v_{\mathrm {drift}}=-\mu _{-}E$$.) If these are the only two ions, charge neutrality gives $$n_{+}=n_{-}=n$$. The particle flux of each is determined by the concentration gradient of each and the motion of each in an electric field. The $$j_{\mathrm {flux}}$$ charge flux (or current density) is the ion charge $$\times $$ the ion flux. For the positive ion: $$j_{ \mathrm {flux,+}}= q(J_{\mathrm {diff,+}}+J_{\mathrm {elect,+}})=-qD_{\mathrm {diff,+}}(\mathrm{d}n/\mathrm{d}x)\,+\, qn\mu _{+}E$$ and for the negative ion it is: $$j_{\mathrm {flux,-}}=-q(J_{\mathrm {diff,-} }+J_{\mathrm {elect,-}})=qD_{\mathrm {diff,-}}(\mathrm{d}n/\mathrm{d}x)+qn\mu _{-}E$$, so the total current density is


$$\begin{aligned} j_{\mathrm {flux}}=-q(D_{\mathrm {diff,+}}-D_{\mathrm {diff,-}})\frac{\mathrm{d}n}{\mathrm{d}x}+qn(\mu _{+}+\mu _{-})E, \end{aligned}$$

(12.20)
which is known as the Nernst–Planck equation . This can also be written as


$$\begin{aligned} j_{\mathrm {flux}}=qn(\mu _{+}+\mu _{-})\left( E-\frac{D_{\mathrm {diff,+}}-D_{\mathrm {diff,-}}}{\mu _{+}+\mu _{-}} \; \frac{\mathrm{d}\ln n}{\mathrm{d}x}\right) , \end{aligned}$$

(12.21)
where we have expressed $$(\mathrm{d}n/\mathrm{d}x)/n$$ as $$\mathrm{d}\ln n/\mathrm{d}x$$. The prefactor on the right-hand side is the conductivity, $$\sigma =qn(\mu _{+}+\mu _{-})$$. (Also, the factor $$\varLambda _{0,i}$$ in (12.15) clearly equals $$q_{i}\mu _{i}$$.)

When there is no net current flow, we see


$$\begin{aligned} E=\frac{D_{\mathrm {diff,+}}-D_{\mathrm {diff,-}}}{\mu _{+}+\mu _{-}}\;\frac{\mathrm{d}\ln n}{\mathrm{d}x}. \end{aligned}$$

(12.22)
So, the voltage between two points, such as from the inside (with subscript i) of the membrane to the outside (with subscript o) is


$$\begin{aligned} \Delta V= & {} V_{\mathrm {i}}-V_{\mathrm {o}}=-\int _{\mathrm {outside}}^{ \mathrm {inside}}E\mathrm { }\;\mathrm{d}x \end{aligned}$$

(12.23)



$$\begin{aligned}= & {} -\frac{D_{\mathrm {diff,+}}-D_{\mathrm {diff,-}}}{\mu _{+}+\mu _{-}}\int _{\mathrm {outside}}^{\mathrm {inside} }\frac{\mathrm{d}\ln n}{\mathrm{d}x}\mathrm { }\;\mathrm{d}x \end{aligned}$$

(12.24)



$$\begin{aligned}= & {} -\frac{D_{\mathrm {diff,+}}-D_{\mathrm {diff,-}}}{\mu _{+}+\mu _{-}}\;\ln (n_{ \mathrm {i}}/n_{\mathrm {o}}). \end{aligned}$$

(12.25)
Using the Einstein relation, (12.19), we know that $$D_{\mathrm {diff}}=\mu k_{\mathrm {B}}T/q$$, and so


$$\begin{aligned} \Delta V=-\frac{k_{\mathrm {B}}T}{q}\frac{\mu _{+}-\mu _{-}}{\mu _{+}+\mu _{-}} \;\ln (n_{\mathrm {i}}/n_{\mathrm {o}}). \end{aligned}$$

(12.26)
This is the Nernst equation .

Let us apply this to a membrane that is impermeable to negative ions, so $$\mu _{-}=0$$ and


$$\begin{aligned} \Delta V=-\frac{k_{\mathrm {B}}T}{q}\;\ln (n_{\mathrm {i}}/n_{\mathrm {o}}). \end{aligned}$$

(12.27)
Calling the charge $$q=Ze$$, where e is the magnitude of an elementary charge (electron or proton), we see that


$$\begin{aligned} \frac{n_{\mathrm {i}}}{n_{\mathrm {o}}}=\exp \left( -Ze(V_{\mathrm {i} }-V_{\mathrm {o}})/k_{\mathrm {B}}T\right) . \end{aligned}$$

(12.28)
This ratio is known as the Donnan ratio and this is known as Donnan equilibrium. The ion densities are considered constant within both the inside and outside regions.

This expression can also be derived by using the Maxwell–Boltzmann distribution, which gives the probability of a state being occupied, P(E, T), if it has an energy E and is in thermal equilibrium with the environment at temperature T


$$\begin{aligned} P(E,T)=A\exp (-E/k_{\mathrm {B}}T). \end{aligned}$$

(12.29)
The potential energy of the charge is $$E=ZeV$$. If a given species were in thermal equilibrium we would expect that its concentration n would be proportional to $$\exp (-ZeV_{\mathrm {local}}/k_{\mathrm {B}}T)$$, where $$V_{ \mathrm {local}}$$ is the local potential, or more exactly


$$\begin{aligned} n=n_{\infty }\exp (-ZeV_{\mathrm {local}}/k_{\mathrm {B}}T), \end{aligned}$$

(12.30)
where $$n_{\infty }$$ is the concentration very far away, where the potential is zero. In particular, we would expect the ratio of the concentrations for each ion inside and outside the cell gives (12.28).

This Donnan ratio includes the physics of the first two driving forces explained earlier, as well as the physics of thermal equilibration. For the resting potential $$V_{\mathrm {i}}-V_{\mathrm {o}} = -70$$ mV at $$T = 310$$ K (core body temperature) and $$Z = +1$$, we expect $$n_{\mathrm {i} }/n_{\mathrm {o}} = 13.7$$ for this “Donnan” equilibrium; for Na$$^{+}$$ this ratio is 15/145 = 0.103 and for K$$^{+}$$ it is 150/5 = 30. For $$Z = -1$$ we expect $$n_{\mathrm { i}}/n_{\mathrm {o}} = 1/13.7 = 0.073$$; for Cl$$^{-}$$ this ratio is 9/125 = 0.072 and for miscellaneous singly negative charge ions it is 156/30 = 5.2. There is relatively good agreement for K$$^{+}$$ and Cl$$^{-}$$, and great disagreement for Na$$^{+}$$ and the Misc.$$^{-}$$ for the reasons given earlier, such as the Na$$^{+}$$ pump for Na$$^{+}$$. (The agreement is not perfect for K$$^{+}$$ because the Na$$^{+}$$ pump brings K$$^{+}$$ into the cell.)

The theoretical Nernst potential $$V_{\mathrm {Nernst}}$$ is the potential that would lead to the observed concentration ratios


$$\begin{aligned} \left( \frac{n_{\mathrm {i}}}{n_{\mathrm {o}}}\right) _{\mathrm {observed} }=\exp (-ZeV_{\mathrm {Nernst}}/k_{\mathrm {B}}T). \end{aligned}$$

(12.31)
For Na$$^{+}$$ it is 61 mV, for K$$^{+}$$ it is $$-91$$ mV, and for Cl$$^{-}$$ it is $$-70$$ mV.

When the Nernst equation (12.27) is generalized to include the effects of many ions, such as Na$$^{+}$$, K$$^{+}$$, and Cl$$^{-}$$, and membrane permeability, the Goldman Voltage equation is obtained


$$\begin{aligned} \Delta V=-\frac{k_{\mathrm {B}}T}{q}\ln \frac{p_{\mathrm {Na}}n_{\mathrm {Na,i}}+p_{ \mathrm {K}}n_{\mathrm {K,i}}+p_{\mathrm {Cl}}n_{\mathrm {Cl,i}}}{p_{\mathrm {Na}}n_{\mathrm { Na,o}}+p_{\mathrm {K}}n_{\mathrm {K,o}}+p_{\mathrm {Cl}}n_{\mathrm {Cl,o}}}, \end{aligned}$$

(12.32)
with membrane permeabilities p (and with the subscripts i for inside and o for outside). For neurons and sensory cells the permeability for Cl$$^{-}$$ is so small that it can often be neglected, and we find:


$$\begin{aligned} \Delta V=-\frac{k_{\mathrm {B}}T}{q}\ln \frac{p_{\mathrm {Na}}n_{\mathrm {Na,i}}+p_{ \mathrm {K}}n_{\mathrm {K,i}}}{p_{\mathrm {Na}}n_{\mathrm {Na,o}}+p_{\mathrm {K}}n_{\mathrm { K,o}}} \end{aligned}$$

(12.33)
or


$$\begin{aligned} \Delta V=-\frac{k_{\mathrm {B}}T}{q}\ln \frac{\alpha n_{\mathrm {Na,i}}+n_{\mathrm { K,i}}}{\alpha n_{\mathrm {Na,o}}+n_{\mathrm {K,o}}}, \end{aligned}$$

(12.34)
with $$\alpha =p_{\mathrm {Na}}/p_{\mathrm {K}}$$. Using the earlier concentrations and $$\alpha = 0.02$$, this resting potential difference is $$-75$$ mV, which is closer to the real resting potential than $$V_{\mathrm {Nernst}}=-91$$  mV for K$$^{+}$$.

Poisson–Boltzmann Equation (Advanced Topic)

So far we have determined the concentration ratios for a given potential. A more general problem, and one that is a bit beyond our scope, is to determine the potential V by using (12.1) and (12.3) and the densities of charges in the region. In other words, we also need to couple the potential with the distributions of ions.

By integrating the field over a surface $$\mathbf {a}$$, such as a sphere, around the charge q, Coulomb’s Law (12.1) becomes Gauss’ Law


$$\begin{aligned} \int \mathbf {E}\cdot \mathrm{d}\mathbf {a}=\frac{1}{\epsilon _{0}\epsilon }\;q. \end{aligned}$$

(12.35)
This can be converted into the differential form


$$\begin{aligned} \mathbf {\nabla }\cdot \mathbf {E}=\frac{1}{\epsilon _{0}\epsilon }\;\rho , \end{aligned}$$

(12.36)
where $$\rho $$ is the charge density


$$\begin{aligned} \rho =\sum _{i}Z_{i}en_{i}. \end{aligned}$$

(12.37)
In one dimension, this form of Gauss’ Law becomes


$$\begin{aligned} \frac{\mathrm{d}E}{\mathrm{d}x}=\frac{1}{\epsilon _{0}\epsilon }\;\rho . \end{aligned}$$

(12.38)
Using the relation between electric field and potential (12.4) and (12.5) these become Poisson’s equation
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Jun 11, 2017 | Posted by in GENERAL & FAMILY MEDICINE | Comments Off on Electrical and Magnetic Properties

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