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3.4 Capacitance, Dielectrics & Axonal Cable Theory
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3.4 Capacitance, Dielectrics & Axonal Cable Theory
Key Takeaways Explain This Topic AI
Capacitors store charge and potential energy (U = 1/2 C V^2); inserting a dielectric material of constant kappa amplifies capacitance to C' = kappa C.
Myelin sheaths increase axonal membrane resistance r_m and drastically decrease capacitance C_m, increasing the length constant lambda and reducing the time constant tau to drive rapid saltatory conduction.
Capacitors combine opposite to resistors: capacitance adds directly in parallel (Ceq = C1 + C2) and reciprocally in series (1/Ceq = 1/C1 + 1/C2), while an isolated charged capacitor holds Q constant so inserting a dielectric lowers V and stored energy rather than raising charge.
Last updated: September 2026
- 3.4 Capacitance, Dielectrics & Axonal Cable Theory
- 4. Capacitance, Dielectrics, and Capacitor Networks
A capacitor stores separated electric charges and potential energy across an electric field established between two conducting surfaces.
Definition of Capacitance
Capacitance ( C C C ) measures charge storage capacity per unit potential difference: C = Q V C = \frac{Q}{V} C = V Q
Units : Farads ( F \text{F} F ), where 1 Farad = 1 Coulomb per Volt 1\text{ Farad} = 1\text{ Coulomb per Volt} 1 Farad = 1 Coulomb per Volt ( 1 C / V 1\text{ C}/\text{V} 1 C / V ). In biophysics and electronics, capacitances typically span microfarads ( μ F = 10 − 6 F \mu\text{F} = 10^{-6}\text{ F} μ F = 1 0 − 6 F ) to picofarads ( pF = 10 − 12 F \text{pF} = 10^{-12}\text{ F} pF = 1 0 − 12 F ).
Parallel-Plate Capacitors
For two parallel conducting plates of area A A A separated by vacuum distance d d d : C 0 = ϵ 0 A d C_0 = \frac{\epsilon_0 A}{d} C 0 = d ϵ 0 A
Capacitance increases with larger plate surface area ( A ↑ A \uparrow A ↑ ).
Capacitance increases with narrower separation distance ( d ↓ d \downarrow d ↓ ).
Dielectric Materials and Polarization
When an insulating material with dielectric constant κ ≥ 1.0 \kappa \ge 1.0 κ ≥ 1.0 (air κ ≈ 1.0 \kappa \approx 1.0 κ ≈ 1.0 , lipid bilayer core κ ≈ 2 − 3 \kappa \approx 2-3 κ ≈ 2 − 3 , pure water κ ≈ 80 \kappa \approx 80 κ ≈ 80 ) is inserted between the plates, electric field polarization enhances capacitance: C ′ = κ C 0 C' = \kappa C_0 C ′ = κ C 0
Caution
The Disconnected vs. Connected Capacitor Distinction : On MCAT passage questions, verify whether the battery remains connected when the dielectric is inserted:
Disconnected (Isolated) Capacitor : Charge Q Q Q is trapped and remains constant ( Q ′ = Q 0 Q' = Q_0 Q ′ = Q 0 ). Therefore, as C C C increases by factor κ \kappa κ , voltage drops ( V ′ = V 0 / κ V' = V_0 / \kappa V ′ = V 0 / κ ), electric field decreases ( E ′ = E 0 / κ E' = E_0 / \kappa E ′ = E 0 / κ ), and stored energy decreases ( U ′ = U 0 / κ U' = U_0 / \kappa U ′ = U 0 / κ ).
Connected to Battery : Voltage V V V is fixed by the power source ( V ′ = V 0 V' = V_0 V ′ = V 0 ). Therefore, as C C C increases by factor κ \kappa κ , charge increases ( Q ′ = κ Q 0 Q' = \kappa Q_0 Q ′ = κ Q 0 ), electric field is unchanged ( E ′ = V / d = E 0 E' = V/d = E_0 E ′ = V / d = E 0 ), and stored energy increases ( U ′ = κ U 0 U' = \kappa U_0 U ′ = κ U 0 ).
Capacitor Combinations
Notice that capacitor network formulas are the mathematical inverse of resistor formulas:
Topology Equivalent Capacitance Formula Invariant Parameter Physical Rationale
Parallel C eq = C 1 + C 2 + C 3 + … C_{\text{eq}} = C_1 + C_2 + C_3 + \dots C eq = C 1 + C 2 + C 3 + … Voltage ( V total = V 1 = V 2 V_{\text{total}} = V_1 = V_2 V total = V 1 = V 2 ) Combining plate surface areas ( A eff = A 1 + A 2 A_{\text{eff}} = A_1 + A_2 A eff = A 1 + A 2 )
Series 1 C eq = 1 C 1 + 1 C 2 + … \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots C eq 1 = C 1 1 + C 2 1 + … Charge ( Q total = Q 1 = Q 2 Q_{\text{total}} = Q_1 = Q_2 Q total = Q 1 = Q 2 ) Adding plate separation distances ( d eff = d 1 + d 2 d_{\text{eff}} = d_1 + d_2 d eff = d 1 + d 2 )
Energy Stored in a Capacitor
The electrostatic potential energy ( U C U_C U C ) stored in a charged capacitor equals the work performed to charge it: U C = 1 2 C V 2 = 1 2 Q V = Q 2 2 C U_C = \frac{1}{2} C V^2 = \frac{1}{2} Q V = \frac{Q^2}{2 C} U C = 2 1 C V 2 = 2 1 Q V = 2 C Q 2
5. Neurobiological Application: Axon Cable Theory and Saltatory Conduction
In cellular neurobiology, an unmyelinated axon functions as a leaky cylindrical cable containing three primary electrical elements:
Membrane Capacitance ( c m c_m c m ) : The hydrophobic core of the phospholipid bilayer ( d ≈ 3 − 5 nm d \approx 3-5\text{ nm} d ≈ 3 − 5 nm ) forms a high-capacitance dielectric barrier separating intracellular and extracellular ionic solutions ( C m ≈ 1.0 μ F / cm 2 C_m \approx 1.0\ \mu\text{F}/\text{cm}^2 C m ≈ 1.0 μ F / cm 2 ).
Membrane Resistance ( r m r_m r m ) : Ion channels spanning the membrane provide leak conductances ( g m = 1 / r m g_m = 1/r_m g m = 1/ r m ).
Axoplasmic Longitudinal Resistance ( r i r_i r i ) : The core cytoplasm presents internal resistance to axial current flow along the axon length ( r i = ρ axoplasm / ( π r 2 ) r_i = \rho_{\text{axoplasm}} / (\pi r^2) r i = ρ axoplasm / ( π r 2 ) ).
Cable Theory Mathematical Parameters
Membrane Time Constant ( τ \tau τ ) : τ = r m c m \tau = r_m c_m τ = r m c m τ \tau τ represents the time required for the transmembrane potential to reach 1 − 1 / e ≈ 63.2 % 1 - 1/e \approx 63.2\% 1 − 1/ e ≈ 63.2% of its steady-state value during charging, or decay by 63.2 % 63.2\% 63.2% toward resting potential. A smaller time constant permits more rapid membrane charging and faster action potential initiation.
Membrane Length (Space) Constant ( λ \lambda λ ) : λ = r m r i \lambda = \sqrt{\frac{r_m}{r_i}} λ = r i r m λ \lambda λ represents the distance along the neurite over which a subthreshold electrotonic voltage change decays to 1 / e ≈ 36.8 % 1/e \approx 36.8\% 1/ e ≈ 36.8% of its initial amplitude ( Δ V ( x ) = Δ V 0 e − x / λ \Delta V(x) = \Delta V_0 e^{-x/\lambda} Δ V ( x ) = Δ V 0 e − x / λ ). To maximize the propagation distance of passive currents without signal loss, axons require high membrane resistance ( r m ↑ r_m \uparrow r m ↑ ) and low internal resistance ( r i ↓ r_i \downarrow r i ↓ ) .
Myelin and Saltatory Conduction
Vertebrates achieve ultra-fast neural transmission without needing massive axon diameters (unlike the squid giant axon) through myelination:
Capacitance Plummets : Specialized glial cells (Schwann cells in the PNS, oligodendrocytes in the CNS) wrap up to 100-300 concentric layers of lipid membrane around the axon. This increases effective dielectric thickness d d d from 4 nm 4\text{ nm} 4 nm to over 1 μ m 1\ \mu\text{m} 1 μ m (a >250-fold increase). Because C = ϵ A / d C = \epsilon A / d C = ϵ A / d , membrane capacitance ( c m c_m c m ) drops precipitously . Less charge must be moved to depolarize the membrane, slashing the time required to charge the membrane.
Membrane Resistance Surges : Myelin covers leak channels, drastically increasing transverse membrane resistance ( r m r_m r m ). Because λ = r m / r i \lambda = \sqrt{r_m / r_i} λ = r m / r i , elevating r m r_m r m dramatically lengthens the space constant ( λ \lambda λ ) , allowing electrotonic currents to travel long distances internodally without dropping below the firing threshold.
Saltatory Conduction at Nodes of Ranvier : At periodic unmyelinated gaps called Nodes of Ranvier (spaced every 0.2 − 2.0 mm 0.2-2.0\text{ mm} 0.2 − 2.0 mm ), the axon membrane is exposed and packed with voltage-gated Na + \text{Na}^+ Na + channels at extraordinary density ( ∼ 10 , 000 − 20 , 000 / μ m 2 \sim 10,000-20,000/\mu\text{m}^2 ∼ 10 , 000 − 20 , 000/ μ m 2 , compared to < 25 / μ m 2 <25/\mu\text{m}^2 < 25/ μ m 2 under myelin). The action potential "jumps" from node to node via rapid passive internodal electrotonic conduction, boosted actively at each node, increasing conduction speed up to 100-fold (reaching 120 m / s 120\text{ m}/\text{s} 120 m / s ) while saving cellular ATP by minimizing total ion exchange.
Pathology: Demyelinating Disorders : In Multiple Sclerosis (autoimmune attack on CNS myelin) and Guillain-Barré Syndrome (PNS demyelination), the destruction of myelin causes c m c_m c m to spike and r m r_m r m to plunge. Electrotonic currents leak out through unmyelinated internodal membrane, causing λ \lambda λ to collapse below nodal spacing, producing conduction slowing or total conduction block.
- 6. Worked Numerical Examples
- Example 1: Multi-Branch Circuit Analysis
A 24 V 24\text{ V} 24 V battery with zero internal resistance is connected to a 6.0 Ω 6.0\ \Omega 6.0 Ω resistor ( R 1 R_1 R 1 ) in series with a parallel network consisting of a 12 Ω 12\ \Omega 12 Ω resistor ( R 2 R_2 R 2 ) and a 4.0 Ω 4.0\ \Omega 4.0 Ω resistor ( R 3 R_3 R 3 ). Determine the total circuit current, the voltage drop across R 3 R_3 R 3 , and the power dissipated by R 1 R_1 R 1 .
Solution :
Calculate the equivalent resistance of the parallel combination ( R 23 R_{23} R 23 ): R 23 = R 2 R 3 R 2 + R 3 = 12 × 4.0 12 + 4.0 = 48 16 = 3.0 Ω R_{23} = \frac{R_2 R_3}{R_2 + R_3} = \frac{12 \times 4.0}{12 + 4.0} = \frac{48}{16} = 3.0\ \Omega R 23 = R 2 + R 3 R 2 R 3 = 12 + 4.0 12 × 4.0 = 16 48 = 3.0 Ω
Calculate total equivalent circuit resistance: R total = R 1 + R 23 = 6.0 Ω + 3.0 Ω = 9.0 Ω R_{\text{total}} = R_1 + R_{23} = 6.0\ \Omega + 3.0\ \Omega = 9.0\ \Omega R total = R 1 + R 23 = 6.0 Ω + 3.0 Ω = 9.0 Ω
Calculate total current using Ohm's law: I total = V R total = 24 V 9.0 Ω = 2.67 A I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{24\text{ V}}{9.0\ \Omega} = 2.67\text{ A} I total = R total V = 9.0 Ω 24 V = 2.67 A
Calculate the voltage drop across R 1 R_1 R 1 : V 1 = I total R 1 = ( 2.67 A ) ( 6.0 Ω ) = 16.0 V V_1 = I_{\text{total}} R_1 = (2.67\text{ A})(6.0\ \Omega) = 16.0\text{ V} V 1 = I total R 1 = ( 2.67 A ) ( 6.0 Ω ) = 16.0 V
By Kirchhoff's loop rule, the remaining voltage drops across the parallel bank: V 23 = V total − V 1 = 24 V − 16 V = 8.0 V V_{23} = V_{\text{total}} - V_1 = 24\text{ V} - 16\text{ V} = 8.0\text{ V} V 23 = V total − V 1 = 24 V − 16 V = 8.0 V Because R 2 R_2 R 2 and R 3 R_3 R 3 are in parallel, the voltage across R 3 R_3 R 3 is exactly 8.0 V 8.0\text{ V} 8.0 V .
Calculate the power dissipated by R 1 R_1 R 1 : P 1 = ( I total ) 2 R 1 = ( 2.67 A ) 2 × 6.0 Ω ≈ 42.7 W P_1 = (I_{\text{total}})^2 R_1 = (2.67\text{ A})^2 \times 6.0\ \Omega \approx 42.7\text{ W} P 1 = ( I total ) 2 R 1 = ( 2.67 A ) 2 × 6.0 Ω ≈ 42.7 W
Example 2: Axon Diameter and Length Constant
A neuroscientist compares two unmyelinated axons with identical membrane resistivity and axoplasm composition. Axon B has twice the diameter of Axon A ( d B = 2 d A d_B = 2 d_A d B = 2 d A ). By what factor is the length constant ( λ \lambda λ ) of Axon B larger than that of Axon A?
Solution :
Membrane resistance per unit length ( r m r_m r m ) is inversely proportional to membrane circumference: r m ∝ 1 π d ⟹ r m ∝ 1 d r_m \propto \frac{1}{\pi d} \implies r_m \propto \frac{1}{d} r m ∝ π d 1 ⟹ r m ∝ d 1
Longitudinal internal resistance per unit length ( r i r_i r i ) is inversely proportional to cross-sectional area: r i ∝ 1 π ( d / 2 ) 2 ⟹ r i ∝ 1 d 2 r_i \propto \frac{1}{\pi (d/2)^2} \implies r_i \propto \frac{1}{d^2} r i ∝ π ( d /2 ) 2 1 ⟹ r i ∝ d 2 1
Substitute into the length constant formula: λ = r m r i ∝ 1 / d 1 / d 2 = d \lambda = \sqrt{\frac{r_m}{r_i}} \propto \sqrt{\frac{1/d}{1/d^2}} = \sqrt{d} λ = r i r m ∝ 1/ d 2 1/ d = d
For a 2-fold increase in diameter ( d B / d A = 2 d_B / d_A = 2 d B / d A = 2 ): λ B λ A = 2 ≈ 1.41 \frac{\lambda_B}{\lambda_A} = \sqrt{2} \approx 1.41 λ A λ B = 2 ≈ 1.41 Axon B has a length constant approximately 1.41 times 1.41\text{ times} 1.41 times greater than Axon A.