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3.2 Electric Dipoles, Membrane Potentials & the Nernst Equation
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3.2 Electric Dipoles, Membrane Potentials & the Nernst Equation
Key Takeaways Explain This Topic AI
An electric dipole experiences external field torque tau = p E sin(theta) and possesses potential energy U = -p E cos(theta), with stable alignment at theta = 0 degrees.
The resting membrane potential (Vm approx -70 mV) is established by selective ion permeability (P_K >> P_Na) and modeled quantitatively via the Nernst and Goldman-Hodgkin-Katz equations.
The Nernst equation (Eion = (61.5/z) log([ion]out/[ion]in) mV at 37 °C) gives the equilibrium potential at which electrical and chemical driving forces on a single ion cancel; potassium's value near −90 mV anchors the resting potential, while sodium's near +60 mV supplies the driving force for depolarisation.
Last updated: September 2026
- 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 +q + q and − q -q − q ) separated by a fixed distance d d d .
Dipole Moment Vector
The electric dipole moment ( p ⃗ \vec{p} p ) is defined mathematically as: p ⃗ = q d ⃗ \vec{p} = q \vec{d} p = q d
Units : Coulomb-meters ( C ⋅ m \text{C}\cdot\text{m} C ⋅ m ) or Debyes ( 1 D = 3.336 × 10 − 30 C ⋅ m 1\text{ D} = 3.336 \times 10^{-30}\text{ C}\cdot\text{m} 1 D = 3.336 × 1 0 − 30 C ⋅ m ). Water has a substantial dipole moment of 1.85 D 1.85\text{ D} 1.85 D .
Vector Direction : In physics, p ⃗ \vec{p} p points from the negative charge ( − q -q − q ) toward the positive charge ( + q +q + q ). Note that in chemistry, dipole arrows are conventionally drawn from partial positive ( δ + \delta^+ δ + ) toward partial negative ( δ − \delta^- δ − ); 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 ( E ⃗ \vec{E} E ), the net translational force is zero because the forces on + q +q + q and − q -q − q cancel ( F ⃗ net = + q E ⃗ − q E ⃗ = 0 \vec{F}_{\text{net}} = +q\vec{E} - q\vec{E} = 0 F net = + q E − q E = 0 ). However, if the dipole is inclined at angle θ \theta θ relative to E ⃗ \vec{E} E , the separated forces produce a net torque ( τ ⃗ \vec{\tau} τ ): τ ⃗ = p ⃗ × E ⃗ ⟹ τ = p E sin θ \vec{\tau} = \vec{p} \times \vec{E} \implies \tau = p E \sin\theta τ = p × E ⟹ τ = pE sin θ
Torque is maximized when θ = 90 ∘ \theta = 90^\circ θ = 9 0 ∘ (dipole perpendicular to field: sin 90 ∘ = 1 \sin 90^\circ = 1 sin 9 0 ∘ = 1 ).
Torque is zero when θ = 0 ∘ \theta = 0^\circ θ = 0 ∘ (aligned) or θ = 180 ∘ \theta = 180^\circ θ = 18 0 ∘ (anti-aligned).
Potential Energy of a Dipole
The potential energy stored in an oriented dipole is: U = − p ⃗ ⋅ E ⃗ = − p E cos θ U = -\vec{p} \cdot \vec{E} = -p E \cos\theta U = − p ⋅ E = − pE cos θ
Stable Equilibrium ( θ = 0 ∘ \theta = 0^\circ θ = 0 ∘ ) : U = − p E U = -p E U = − pE . The dipole is aligned with E ⃗ \vec{E} E , yielding minimum potential energy.
Unstable Equilibrium ( θ = 180 ∘ \theta = 180^\circ θ = 18 0 ∘ ) : U = + p E U = +p E U = + pE . The dipole is anti-aligned with E ⃗ \vec{E} E , yielding maximum potential energy. Any infinitesimal perturbation generates a torque that rotates the dipole toward θ = 0 ∘ \theta = 0^\circ θ = 0 ∘ .
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 inside − V outside ≈ − 70 mV V_m = V_{\text{inside}} - V_{\text{outside}} \approx -70\text{ mV} V m = V inside − V outside ≈ − 70 mV ).
Transmembrane Ion Gradients
Active transport by the electrogenic Na + / K + \text{Na}^+/\text{K}^+ Na + / K + ATPase (which hydrolyzes 1 ATP to export 3 Na + 3\text{ Na}^+ 3 Na + ions and import 2 K + 2\text{ K}^+ 2 K + ions) establishes asymmetric ionic distributions:
Ion Intracellular [Conc] Extracellular [Conc] Chemical Gradient Direction Equilibrium Potential ( E ion E_{\text{ion}} E ion )
K + \text{K}^+ K + ∼ 140 − 150 mM \sim 140-150\text{ mM} ∼ 140 − 150 mM ∼ 4 − 5 mM \sim 4-5\text{ mM} ∼ 4 − 5 mM Outward (cytoplasm to ECF) ∼ − 90 to − 95 mV \sim -90\text{ to } -95\text{ mV} ∼ − 90 to − 95 mV
Na + \text{Na}^+ Na + ∼ 12 − 15 mM \sim 12-15\text{ mM} ∼ 12 − 15 mM ∼ 145 mM \sim 145\text{ mM} ∼ 145 mM Inward (ECF to cytoplasm) ∼ + 60 to + 65 mV \sim +60\text{ to } +65\text{ mV} ∼ + 60 to + 65 mV
Cl − \text{Cl}^- Cl − ∼ 4 − 10 mM \sim 4-10\text{ mM} ∼ 4 − 10 mM ∼ 110 − 120 mM \sim 110-120\text{ mM} ∼ 110 − 120 mM Inward (ECF to cytoplasm) ∼ − 65 to − 70 mV \sim -65\text{ to } -70\text{ mV} ∼ − 65 to − 70 mV
Ca 2 + \text{Ca}^{2+} Ca 2 + ∼ 100 nM \sim 100\text{ nM} ∼ 100 nM ( 10 − 4 mM 10^{-4}\text{ mM} 1 0 − 4 mM ) ∼ 1.5 − 2.0 mM \sim 1.5-2.0\text{ mM} ∼ 1.5 − 2.0 mM Inward (ECF to cytoplasm) ∼ + 120 to + 130 mV \sim +120\text{ to } +130\text{ mV} ∼ + 120 to + 130 mV
The Nernst Equation
The Nernst equation calculates the electrical potential ( E ion E_{\text{ion}} E ion ) required to balance an ion's chemical concentration gradient across a selectively permeable membrane, producing zero net thermodynamic flux (electrochemical equilibrium): E ion = R T z F ln ( [ ion ] out [ ion ] in ) E_{\text{ion}} = \frac{RT}{zF} \ln\left(\frac{[\text{ion}]_{\text{out}}}{[\text{ion}]_{\text{in}}}\right) E ion = z F RT ln ( [ ion ] in [ ion ] out )
At physiological body temperature ( 37 ∘ C = 310 K 37^\circ\text{C} = 310\text{ K} 3 7 ∘ C = 310 K ), converting the natural logarithm to base-10 yields the standard MCAT simplified form: E ion = 61.5 mV z log 10 ( [ ion ] out [ ion ] in ) E_{\text{ion}} = \frac{61.5\text{ mV}}{z} \log_{10}\left(\frac{[\text{ion}]_{\text{out}}}{[\text{ion}]_{\text{in}}}\right) E ion = z 61.5 mV lo g 10 ( [ ion ] in [ ion ] out )
Where:
R R R is the universal gas constant ( 8.314 J / ( mol ⋅ K ) 8.314\text{ J}/(\text{mol}\cdot\text{K}) 8.314 J / ( mol ⋅ K ) )
T T T is temperature in Kelvin ( 310 K 310\text{ K} 310 K )
z z z is the valence (charge) of the ion ( + 1 +1 + 1 for K + , Na + \text{K}^+, \text{Na}^+ K + , Na + ; + 2 +2 + 2 for Ca 2 + \text{Ca}^{2+} Ca 2 + ; − 1 -1 − 1 for Cl − \text{Cl}^- Cl − )
- F F F is Faraday's constant ( 96 , 485 C / mol e − 96,485\text{ C}/\text{mol } e^- 96 , 485 C / 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 V_m V m ) is a weighted average determined by the relative membrane permeabilities ( P ion P_{\text{ion}} P ion ) of all diffusing ions: V m = R T F ln ( P K [ K + ] out + P Na [ Na + ] out + P Cl [ Cl − ] in P K [ K + ] in + P Na [ Na + ] in + P Cl [ Cl − ] out ) 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) V m = F RT ln ( P K [ K + ] in + P Na [ Na + ] in + P Cl [ Cl − ] out P K [ K + ] out + P Na [ Na + ] out + P Cl [ Cl − ] in )
Important
Chloride Inversion in the GHK Equation : Notice that [ Cl − ] in [\text{Cl}^-]_{\text{in}} [ Cl − ] in appears in the numerator and [ Cl − ] out [\text{Cl}^-]_{\text{out}} [ Cl − ] out in the denominator. Because chloride carries a negative valence ( z = − 1 z = -1 z = − 1 ), its concentration ratio is inverted ( − ln ( x / y ) = ln ( y / x ) -\ln(x/y) = \ln(y/x) − ln ( x / y ) = ln ( y / x ) ) relative to monovalent cations.
At resting potential in typical neurons, resting potassium leak channels ( K ir K_{\text{ir}} K ir and K 2 P K_{2\text{P}} K 2 P ) remain open, whereas voltage-gated sodium channels are closed. The resting permeability ratio is approximately: P K : P Na : P Cl ≈ 1.0 : 0.04 : 0.45 P_{\text{K}} : P_{\text{Na}} : P_{\text{Cl}} \approx 1.0 : 0.04 : 0.45 P K : P Na : P Cl ≈ 1.0 : 0.04 : 0.45
Because P K ≫ P Na P_{\text{K}} \gg P_{\text{Na}} P K ≫ P Na , the membrane potential is drawn overwhelmingly toward E K E_{\text{K}} E K ( − 90 mV -90\text{ mV} − 90 mV ), stabilizing at roughly − 70 mV -70\text{ mV} − 70 mV .
6. Summary of Electrostatic Relationships
Quantity Symbol Formula Scalar or Vector SI Units Biological Manifestation
Coulomb Force F ⃗ e \vec{F}_e F e k ∣ q 1 q 2 ∣ / r 2 k\vert q_1 q_2\vert/r^2 k ∣ q 1 q 2 ∣/ r 2 Vector N \text{N} N Salt bridge attraction between Asp − ^{-} − and Lys + ^{+} +
Electric Field E ⃗ \vec{E} E k Q / r 2 = F ⃗ / q kQ/r^2 = \vec{F}/q k Q / r 2 = F / q Vector N / C \text{N}/\text{C} N / C or V / m \text{V}/\text{m} V / m Transmembrane electric field across lipid bilayer
Potential Energy U U U k q 1 q 2 / r = q V k q_1 q_2/r = qV k q 1 q 2 / r = q V Scalar J \text{J} J Free energy stored in separated ionic gradients
Electric Potential V V V k Q / r = U / q kQ/r = U/q k Q / r = U / q Scalar V \text{V} V ( J / C \text{J}/\text{C} J / C ) Resting membrane potential ( V m ≈ − 70 mV V_m \approx -70\text{ mV} V m ≈ − 70 mV )
Dipole Moment p ⃗ \vec{p} p q d ⃗ q\vec{d} q d Vector C ⋅ m \text{C}\cdot\text{m} C ⋅ m Water molecule polarity and peptide backbone dipole
- 7. Worked Numerical Examples
- Example 1: Transmembrane Electric Field Magnitude
A neuron has a resting membrane potential of − 70 mV -70\text{ mV} − 70 mV across a lipid bilayer with an effective dielectric thickness of 5.0 nm 5.0\text{ nm} 5.0 nm ( 5.0 × 10 − 9 m 5.0 \times 10^{-9}\text{ m} 5.0 × 1 0 − 9 m ). Calculate the magnitude of the electric field ( E E E ) within the membrane and the electrostatic force experienced by an extracellular divalent calcium ion ( Ca 2 + \text{Ca}^{2+} Ca 2 + ) entering the channel pore.
Solution :
Model the membrane as a uniform electric field between parallel plates: E = ∣ Δ V ∣ d = 70 × 10 − 3 V 5.0 × 10 − 9 m = 1.4 × 10 7 V / m 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} E = d ∣Δ V ∣ = 5.0 × 1 0 − 9 m 70 × 1 0 − 3 V = 1.4 × 1 0 7 V / m Notice the astronomical field strength : 14 million volts per meter 14\text{ million volts per meter} 14 million volts per meter ! Biological membranes sustain immense electric fields due to nanometer-scale dimensions.
The charge on Ca 2 + \text{Ca}^{2+} Ca 2 + is q = + 2 e = 2 ( 1.602 × 10 − 19 C ) = 3.204 × 10 − 19 C q = +2e = 2(1.602 \times 10^{-19}\text{ C}) = 3.204 \times 10^{-19}\text{ C} q = + 2 e = 2 ( 1.602 × 1 0 − 19 C ) = 3.204 × 1 0 − 19 C .
Calculate the force: F e = q E = ( 3.204 × 10 − 19 C ) ( 1.4 × 10 7 N / C ) = 4.49 × 10 − 12 N = 4.49 pN 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} F e = qE = ( 3.204 × 1 0 − 19 C ) ( 1.4 × 1 0 7 N / C ) = 4.49 × 1 0 − 12 N = 4.49 pN
Example 2: Calculating Nernst Potential for Sodium
During the peak of an action potential, [ Na + ] out = 145 mM [\text{Na}^+]_{\text{out}} = 145\text{ mM} [ Na + ] out = 145 mM and [ Na + ] in = 14.5 mM [\text{Na}^+]_{\text{in}} = 14.5\text{ mM} [ Na + ] in = 14.5 mM . Determine the sodium equilibrium potential ( E Na E_{\text{Na}} E Na ) at 37 ∘ C 37^\circ\text{C} 3 7 ∘ C .
Solution :
For sodium, z = + 1 z = +1 z = + 1 . The ratio is [ Na + ] out / [ Na + ] in = 145 / 14.5 = 10 [\text{Na}^+]_{\text{out}} / [\text{Na}^+]_{\text{in}} = 145 / 14.5 = 10 [ Na + ] out / [ Na + ] in = 145/14.5 = 10 .
Apply the simplified physiological Nernst formula: E Na = 61.5 mV + 1 log 10 ( 10 ) = 61.5 mV × 1 = + 61.5 mV E_{\text{Na}} = \frac{61.5\text{ mV}}{+1} \log_{10}(10) = 61.5\text{ mV} \times 1 = +61.5\text{ mV} E Na = + 1 61.5 mV lo g 10 ( 10 ) = 61.5 mV × 1 = + 61.5 mV This positive equilibrium potential drives massive inward Na + \text{Na}^+ Na + flux during the depolarization phase of an action potential.