3.2 Electric Dipoles, Membrane Potentials & the Nernst Equation

3.2 Electric Dipoles, Membrane Potentials & the Nernst Equation

4. Electric Dipoles and Torque in External Fields

An electric dipole consists of two equal and opposite charges ($+q$ and $-q$) separated by a fixed distance $d$.

Dipole Moment Vector

The electric dipole moment ($\vec{p}$) is defined mathematically as:

$$\vec{p} = q \vec{d}$$

• Units: Coulomb-meters ($\text{C}\cdot\text{m}$) or Debyes ($1\text{ D} = 3.336 \times 10^{-30}\text{ C}\cdot\text{m}$). Water has a substantial dipole moment of $1.85\text{ D}$.
• Vector Direction: In physics, $\vec{p}$ points from the negative charge ($-q$) toward the positive charge ($+q$). Note that in chemistry, dipole arrows are conventionally drawn from partial positive ($\delta^+$) toward partial negative ($\delta^-$); MCAT physics items use the standard negative-to-positive physics convention.

Torque on a Dipole in a Uniform Field

When an electric dipole is placed in a uniform electric field ($\vec{E}$), the net translational force is zero because the forces on $+q$ and $-q$ cancel ($\vec{F}_{\text{net}} = +q\vec{E} - q\vec{E} = 0$). However, if the dipole is inclined at angle $\theta$ relative to $\vec{E}$, the separated forces produce a net torque ($\vec{\tau}$):

$$\vec{\tau} = \vec{p} \times \vec{E} \implies \tau = p E \sin\theta$$

• Torque is maximized when $\theta = 90^\circ$ (dipole perpendicular to field: $\sin 90^\circ = 1$).
• Torque is zero when $\theta = 0^\circ$ (aligned) or $\theta = 180^\circ$ (anti-aligned).

Potential Energy of a Dipole

The potential energy stored in an oriented dipole is:

$$U = -\vec{p} \cdot \vec{E} = -p E \cos\theta$$

• Stable Equilibrium ($\theta = 0^\circ$): $U = -p E$. The dipole is aligned with $\vec{E}$, yielding minimum potential energy.
• Unstable Equilibrium ($\theta = 180^\circ$): $U = +p E$. The dipole is anti-aligned with $\vec{E}$, yielding maximum potential energy. Any infinitesimal perturbation generates a torque that rotates the dipole toward $\theta = 0^\circ$.

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5. Biological Application: Resting Membrane Potential and Electrophysiology

In living cells, particularly neurons and cardiomyocytes, the hydrophobic core of the plasma membrane acts as an electrical insulator separating aqueous electrolytic solutions. Selective ion transport establishes an electric potential difference called the resting membrane potential ($V_m = V_{\text{inside}} - V_{\text{outside}} \approx -70\text{ mV}$).

Transmembrane Ion Gradients

Active transport by the electrogenic $\text{Na}^+/\text{K}^+$ ATPase (which hydrolyzes 1 ATP to export $3\text{ Na}^+$ ions and import $2\text{ K}^+$ ions) establishes asymmetric ionic distributions:

The Nernst Equation

The Nernst equation calculates the electrical potential ($E_{\text{ion}}$) required to balance an ion's chemical concentration gradient across a selectively permeable membrane, producing zero net thermodynamic flux (electrochemical equilibrium):

$$E_{\text{ion}} = \frac{RT}{zF} \ln\left(\frac{[\text{ion}]_{\text{out}}}{[\text{ion}]_{\text{in}}}\right)$$

At physiological body temperature ($37^\circ\text{C} = 310\text{ K}$), converting the natural logarithm to base-10 yields the standard MCAT simplified form:

$$E_{\text{ion}} = \frac{61.5\text{ mV}}{z} \log_{10}\left(\frac{[\text{ion}]_{\text{out}}}{[\text{ion}]_{\text{in}}}\right)$$

Where:

• $R$ is the universal gas constant ($8.314\text{ J}/(\text{mol}\cdot\text{K})$)
• $T$ is temperature in Kelvin ($310\text{ K}$)
• $z$ is the valence (charge) of the ion ($+1$ for $\text{K}^+, \text{Na}^+$; $+2$ for $\text{Ca}^{2+}$; $-1$ for $\text{Cl}^-$)
• $F$ is Faraday's constant ($96,485\text{ C}/\text{mol } e^-$)

Goldman-Hodgkin-Katz (GHK) Voltage Equation

Because real membranes are permeable to multiple ion species simultaneously, the overall resting membrane potential ($V_m$) is a weighted average determined by the relative membrane permeabilities ($P_{\text{ion}}$) of all diffusing ions:

$$V_m = \frac{RT}{F} \ln\left(\frac{P_{\text{K}}[\text{K}^+]_{\text{out}} + P_{\text{Na}}[\text{Na}^+]_{\text{out}} + P_{\text{Cl}}[\text{Cl}^-]_{\text{in}}}{P_{\text{K}}[\text{K}^+]_{\text{in}} + P_{\text{Na}}[\text{Na}^+]_{\text{in}} + P_{\text{Cl}}[\text{Cl}^-]_{\text{out}}}\right)$$

At resting potential in typical neurons, resting potassium leak channels ($K_{\text{ir}}$ and $K_{2\text{P}}$) remain open, whereas voltage-gated sodium channels are closed. The resting permeability ratio is approximately:

$$P_{\text{K}} : P_{\text{Na}} : P_{\text{Cl}} \approx 1.0 : 0.04 : 0.45$$

Because $P_{\text{K}} \gg P_{\text{Na}}$, the membrane potential is drawn overwhelmingly toward $E_{\text{K}}$ ($-90\text{ mV}$), stabilizing at roughly $-70\text{ mV}$.

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6. Summary of Electrostatic Relationships

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7. Worked Numerical Examples

Example 1: Transmembrane Electric Field Magnitude

A neuron has a resting membrane potential of $-70\text{ mV}$ across a lipid bilayer with an effective dielectric thickness of $5.0\text{ nm}$ ($5.0 \times 10^{-9}\text{ m}$). Calculate the magnitude of the electric field ($E$) within the membrane and the electrostatic force experienced by an extracellular divalent calcium ion ($\text{Ca}^{2+}$) entering the channel pore.

Solution:

1. Model the membrane as a uniform electric field between parallel plates: $$E = \frac{|\Delta V|}{d} = \frac{70 \times 10^{-3}\text{ V}}{5.0 \times 10^{-9}\text{ m}} = 1.4 \times 10^7\text{ V}/\text{m}$$ Notice the astronomical field strength: $14\text{ million volts per meter}$! Biological membranes sustain immense electric fields due to nanometer-scale dimensions.
2. The charge on $\text{Ca}^{2+}$ is $q = +2e = 2(1.602 \times 10^{-19}\text{ C}) = 3.204 \times 10^{-19}\text{ C}$.
3. Calculate the force: $$F_e = q E = (3.204 \times 10^{-19}\text{ C})(1.4 \times 10^7\text{ N}/\text{C}) = 4.49 \times 10^{-12}\text{ N} = 4.49\text{ pN}$$

Example 2: Calculating Nernst Potential for Sodium

During the peak of an action potential, $[\text{Na}^+]_{\text{out}} = 145\text{ mM}$ and $[\text{Na}^+]_{\text{in}} = 14.5\text{ mM}$. Determine the sodium equilibrium potential ($E_{\text{Na}}$) at $37^\circ\text{C}$.

Solution:

1. For sodium, $z = +1$. The ratio is $[\text{Na}^+]_{\text{out}} / [\text{Na}^+]_{\text{in}} = 145 / 14.5 = 10$.
2. Apply the simplified physiological Nernst formula: $$E_{\text{Na}} = \frac{61.5\text{ mV}}{+1} \log_{10}(10) = 61.5\text{ mV} \times 1 = +61.5\text{ mV}$$ This positive equilibrium potential drives massive inward $\text{Na}^+$ flux during the depolarization phase of an action potential.