Problem 37 - University Physics with Modern Physics

Question

Axons. Neurons are the basic units of the nervous system. They contain long tubular structures called axons that propagate electrical signals away from the ends of the neurons. The axon contains a solution of potassium \(\left(\mathrm{K}^{+}\right)\) ions and large negative organic ions. The axon membrane prevents the large ions from leaking out, but the smaller \(\mathrm{K}^{+}\) ions are able to penetrate the membrane to some degree (Fig. E23.37). This leaves an excess negative charge on the inner surface of the axon membrane and an excess positive charge on the outer surface, resulting in a potential difference across the membrane that prevents further \(K^{+}\) ions from leaking out. Measurements show that this potential difference is typically about 70 \(\mathrm{mV}\) . The thickness of the axon membrane itself varies from about 5 to \(10 \mathrm{nm},\) so we'll use an average of 7.5 \(\mathrm{nm}\) . We can model the membrane as a large sheet having equal and opposite charge densities on its faces. (a) Find the electric field inside the axon membrane, assuming (not too realistically) that it is filled with air. Which way does it point: into or out of the axon? (b) Which is at a higher potential: the inside surface or the outside surface of the axon membrane?

Step-by-Step Solution

Answer

The electric field is approximately \(9.33 \times 10^6\, \text{V/m}\) and points into the axon; the outside surface is at a higher potential.

Step 1: Understanding the Problem

We need to determine the electric field inside the axon membrane and the potential across it. We're given the potential difference as 70 mV and the membrane's average thickness as 7.5 nm. The axon membrane can be modeled as a parallel plate capacitor with equal and opposite charge densities on its faces.

Step 1: Calculate the Electric Field

The relationship between potential difference (V), electric field (E), and distance (d) in a uniform field is given by the formula: \[ V = E \cdot d \].The potential difference is 70 mV (or 0.070 V) and the membrane thickness is 7.5 nm (or 7.5 x 10^-9 m). Rearranging for the electric field, we have: \[ E = \frac{V}{d} = \frac{0.070}{7.5 \times 10^{-9}} \] which yields \[ E \approx 9.33 \times 10^6 \text{ V/m} \]. This is the magnitude of the electric field inside the axon membrane.

Step 2: Determine the Direction of the Electric Field

By convention, the electric field direction is from the positive to the negative charge. Since the outside of the axon has a positive charge and the inside has a negative charge, the electric field points from the outside to the inside of the axon.

Step 3: Determine Higher Potential Surface

The electric field direction indicates potential differences. The direction from positive (outside) to negative (inside) charges implies the outside is at a higher potential compared to the inside.

Key Concepts

Axon Membrane

The axon membrane plays a critical role in how neurons transmit signals in the nervous system. Think of it like a thin boundary that controls what goes in and out of the axon, specifically ions. This boundary is crucial for creating electrical signals crucial for our nervous system's functioning. Within the axon, you find a mixture of different ions, mainly potassium ions \( (\mathrm{K}^{+}) \), and large negative organic ions. These large ions remain inside in contrast to the \( \mathrm{K}^{+} \) ions, which can sneak through the membrane to some extent.

When these ions move, they create a situation where the inside of the membrane becomes negatively charged, while the outside becomes positively charged. This setup across the axon membrane is essential for the electrical activity in neurons. The result is a potential difference, especially evident when considering that neurons have a resting potential, usually around 70 \( \mathrm{mV} \). This potential difference is maintained by selectively allowing certain ions to pass through the membrane, creating a balance of charge across it.

Potential Difference

Potential difference, often called voltage, is a measure of electric potential energy between two points. In the context of our axon membrane, this describes the difference in charge from one side to the other. Imagine this like the water pressure difference between two ends of a pipe; higher pressure on one side pushes the flow toward the lower-pressure side. For axons, this means the inside is typically at a lower potential (negative) compared to the outside (positive).

The potential difference of 70 \( \mathrm{mV} \) across the membrane is a result of the ion distribution we've talked about. This voltage is critical because it helps neurons to create and propagate electrical impulses, known as action potentials, essential for communication within the nervous system. The making of this potential difference involves the movement of ions, primarily sodium, and potassium, across the membrane, which is precisely regulated by ion channels. This difference drives the electrical signals along the nerve fibers.

Parallel Plate Capacitor Model

Modeling the axon membrane as a parallel plate capacitor helps in simplifying how we understand the electric field and potential difference across it. A parallel plate capacitor consists of two conducting plates separated by a small distance – in our case, the axon membrane acts as this separator. This model means the inner and outer surfaces of the membrane can be thought of as two plates storing charge, one positive and one negative.

The formula connecting the electric field \( E \), potential difference \( V \), and distance \( d \) in this setup is \( V = E \cdot d \). By rearranging this formula, we can find the electric field as \( E = \frac{V}{d} \). With the given potential difference of 70 \( \mathrm{mV} \) and a membrane thickness of 7.5 \( \mathrm{nm} \), we calculated the electric field to be approximately \( 9.33 \times 10^6 \ \mathrm{V/m} \).

Understanding this model aids in visualizing the direction of the electric field, which goes from the positive outer surface to the negative inner surface. Thus, the higher charge (potential) is on the outside compared to the inside of the axon membrane.