3.4 Capacitance, Dielectrics & Axonal Cable Theory

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 ($C$) measures charge storage capacity per unit potential difference:

$$C = \frac{Q}{V}$$

• Units: Farads ($\text{F}$), where $1\text{ Farad} = 1\text{ Coulomb per Volt}$ ($1\text{ C}/\text{V}$). In biophysics and electronics, capacitances typically span microfarads ($\mu\text{F} = 10^{-6}\text{ F}$) to picofarads ($\text{pF} = 10^{-12}\text{ F}$).

Parallel-Plate Capacitors

For two parallel conducting plates of area $A$ separated by vacuum distance $d$:

$$C_0 = \frac{\epsilon_0 A}{d}$$

• Capacitance increases with larger plate surface area ($A \uparrow$).
• Capacitance increases with narrower separation distance ($d \downarrow$).

Dielectric Materials and Polarization

When an insulating material with dielectric constant $\kappa \ge 1.0$ (air $\kappa \approx 1.0$, lipid bilayer core $\kappa \approx 2-3$, pure water $\kappa \approx 80$) is inserted between the plates, electric field polarization enhances capacitance:

$$C' = \kappa C_0$$

Capacitor Combinations

Notice that capacitor network formulas are the mathematical inverse of resistor formulas:

Energy Stored in a Capacitor

The electrostatic potential energy ($U_C$) stored in a charged capacitor equals the work performed to charge it:

$$U_C = \frac{1}{2} C V^2 = \frac{1}{2} Q V = \frac{Q^2}{2 C}$$

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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:

1. Membrane Capacitance ($c_m$): The hydrophobic core of the phospholipid bilayer ($d \approx 3-5\text{ nm}$) forms a high-capacitance dielectric barrier separating intracellular and extracellular ionic solutions ($C_m \approx 1.0\ \mu\text{F}/\text{cm}^2$).
2. Membrane Resistance ($r_m$): Ion channels spanning the membrane provide leak conductances ($g_m = 1/r_m$).
3. Axoplasmic Longitudinal Resistance ($r_i$): The core cytoplasm presents internal resistance to axial current flow along the axon length ($r_i = \rho_{\text{axoplasm}} / (\pi r^2)$).

Cable Theory Mathematical Parameters

• Membrane Time Constant ($\tau$): $$\tau = r_m c_m$$ $\tau$ represents the time required for the transmembrane potential to reach $1 - 1/e \approx 63.2\%$ of its steady-state value during charging, or decay by $63.2\%$ toward resting potential. A smaller time constant permits more rapid membrane charging and faster action potential initiation.
• Membrane Length (Space) Constant ($\lambda$): $$\lambda = \sqrt{\frac{r_m}{r_i}}$$ $\lambda$ represents the distance along the neurite over which a subthreshold electrotonic voltage change decays to $1/e \approx 36.8\%$ of its initial amplitude ($\Delta V(x) = \Delta V_0 e^{-x/\lambda}$). To maximize the propagation distance of passive currents without signal loss, axons require high membrane resistance ($r_m \uparrow$) and low internal resistance ($r_i \downarrow$).

Myelin and Saltatory Conduction

Vertebrates achieve ultra-fast neural transmission without needing massive axon diameters (unlike the squid giant axon) through myelination:

1. 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 $d$ from $4\text{ nm}$ to over $1\ \mu\text{m}$ (a >250-fold increase). Because $C = \epsilon A / d$, membrane capacitance ($c_m$) drops precipitously. Less charge must be moved to depolarize the membrane, slashing the time required to charge the membrane.
2. Membrane Resistance Surges: Myelin covers leak channels, drastically increasing transverse membrane resistance ($r_m$). Because $\lambda = \sqrt{r_m / r_i}$, elevating $r_m$ dramatically lengthens the space constant ($\lambda$), allowing electrotonic currents to travel long distances internodally without dropping below the firing threshold.
3. Saltatory Conduction at Nodes of Ranvier: At periodic unmyelinated gaps called Nodes of Ranvier (spaced every $0.2-2.0\text{ mm}$), the axon membrane is exposed and packed with voltage-gated $\text{Na}^+$ channels at extraordinary density ($\sim 10,000-20,000/\mu\text{m}^2$, compared to $<25/\mu\text{m}^2$ 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 $120\text{ m}/\text{s}$) while saving cellular ATP by minimizing total ion exchange.
4. Pathology: Demyelinating Disorders: In Multiple Sclerosis (autoimmune attack on CNS myelin) and Guillain-Barré Syndrome (PNS demyelination), the destruction of myelin causes $c_m$ to spike and $r_m$ to plunge. Electrotonic currents leak out through unmyelinated internodal membrane, causing $\lambda$ to collapse below nodal spacing, producing conduction slowing or total conduction block.

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6. Worked Numerical Examples

Example 1: Multi-Branch Circuit Analysis

A $24\text{ V}$ battery with zero internal resistance is connected to a $6.0\ \Omega$ resistor ($R_1$) in series with a parallel network consisting of a $12\ \Omega$ resistor ($R_2$) and a $4.0\ \Omega$ resistor ($R_3$). Determine the total circuit current, the voltage drop across $R_3$, and the power dissipated by $R_1$.

Solution:

1. Calculate the equivalent resistance of the parallel combination ($R_{23}$): $$R_{23} = \frac{R_2 R_3}{R_2 + R_3} = \frac{12 \times 4.0}{12 + 4.0} = \frac{48}{16} = 3.0\ \Omega$$
2. Calculate total equivalent circuit resistance: $$R_{\text{total}} = R_1 + R_{23} = 6.0\ \Omega + 3.0\ \Omega = 9.0\ \Omega$$
3. Calculate total current using Ohm's law: $$I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{24\text{ V}}{9.0\ \Omega} = 2.67\text{ A}$$
4. Calculate the voltage drop across $R_1$: $$V_1 = I_{\text{total}} R_1 = (2.67\text{ A})(6.0\ \Omega) = 16.0\text{ V}$$
5. By Kirchhoff's loop rule, the remaining voltage drops across the parallel bank: $$V_{23} = V_{\text{total}} - V_1 = 24\text{ V} - 16\text{ V} = 8.0\text{ V}$$ Because $R_2$ and $R_3$ are in parallel, the voltage across $R_3$ is exactly $8.0\text{ V}$.
6. Calculate the power dissipated by $R_1$: $$P_1 = (I_{\text{total}})^2 R_1 = (2.67\text{ A})^2 \times 6.0\ \Omega \approx 42.7\text{ W}$$

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 ($d_B = 2 d_A$). By what factor is the length constant ($\lambda$) of Axon B larger than that of Axon A?

Solution:

1. Membrane resistance per unit length ($r_m$) is inversely proportional to membrane circumference: $$r_m \propto \frac{1}{\pi d} \implies r_m \propto \frac{1}{d}$$
2. Longitudinal internal resistance per unit length ($r_i$) is inversely proportional to cross-sectional area: $$r_i \propto \frac{1}{\pi (d/2)^2} \implies r_i \propto \frac{1}{d^2}$$
3. Substitute into the length constant formula: $$\lambda = \sqrt{\frac{r_m}{r_i}} \propto \sqrt{\frac{1/d}{1/d^2}} = \sqrt{d}$$
4. For a 2-fold increase in diameter ($d_B / d_A = 2$): $$\frac{\lambda_B}{\lambda_A} = \sqrt{2} \approx 1.41$$ Axon B has a length constant approximately $1.41\text{ times}$ greater than Axon A.