Question

(a) Calculate the capacitance per unit length of an axon of radius $5.0 \mu \mathrm{m} \text { (see Fig. } 17.14) .$ The membrane acts as an insulator between the conducting fluids inside and outside the neuron. The membrane is 6.0 nm thick and has a dielectric constant of \(7.0 .\) (Note: The membrane is thin compared with the radius of the axon, so the axon can be treated as a parallel plate capacitor.) (b) In its resting state (no signal being transmitted), the potential of the fluid inside is about \(85 \mathrm{mV}\) lower than the outside. Therefore, there must be small net charges \(\pm Q\) on either side of the membrane. Which side has positive charge? What is the magnitude of the charge density on the surfaces of the membrane?

Short answer

Answer: The capacitance per unit length of the axon membrane is \(1.03 \times 10^{-8} \, \mathrm{F/m}\), and the magnitude of the charge density on the surfaces of the membrane at resting state is \(8.32 \times 10^{-3} \, \mathrm{C/m^2}\). The outside surface of the membrane has a positive charge.

Step-by-step solution

Part (a): Capacitance per unit length calculation

Since the membrane is thin compared to the axon's radius, we can treat the axon as a parallel plate capacitor. Let's use the formula for the capacitance of a parallel plate capacitor and modify it to find the capacitance per unit length. The formula for the capacitance of a parallel plate capacitor is given by: $$ C = \frac{\epsilon A}{d} $$ where, \(C\) - Capacitance \(\epsilon\) - Permittivity of the dielectric material (\(\epsilon = \epsilon_0 \cdot \kappa\), where \(\epsilon_0\) is the vacuum permittivity and \(\kappa\) is the dielectric constant) \(A\) - Area of the capacitor's plates \(d\) - Distance between the plates (thickness of the membrane) However, we are interested in the capacitance per unit length, so let A be the area of a cylindrical section of the membrane of length L, i.e., $$ A = 2 \pi r L $$ By substituting the above expression in the equation, we obtain the capacitance per unit length \(C_L\) $$ C_L = \frac{\epsilon 2 \pi r L}{dL} $$ Now, let's plug in the values: \(r = 5.0 \times 10^{-6} m\) \(d = 6.0 \times 10^{-9} m\) \(\kappa = 7.0\) \(\epsilon_0 = 8.85 \times 10^{-12} F/m\) $$ C_L = \frac{ 7.0 \cdot (8.85 \times 10^{-12} \, F/m) (2 \pi) (5.0 \times 10^{-6} \,m)}{6.0 \times 10^{-9} \, m} $$

Part (a): Calculating Capacitance per Unit Length

After evaluating the above expression, we get the capacitance per unit length: $$ C_L \approx 1.03 \times 10^{-8} \, \mathrm{F/m} $$

Part (b): Determining which side has positive charge

The potential of the fluid inside the neuron is 85 mV lower than the outside, which means the inside is more negative, and so the outside must have a positive charge.

Part (b): Calculating charge density

Now, let's find the magnitude of the charge density \(\sigma\) on the surfaces of the membrane. To calculate the surface charge density, we can use the formula: $$ \sigma = \frac{Q}{A} $$ However, we have \(\Delta V = 85 mV\) and the relation $$ \Delta V = \frac{Q}{C} $$ By substituting the expression for \(A\), we get: $$ \sigma = \frac{Q}{2\pi r L} $$ We can find `Q` from the relation \(\Delta V = \frac{Q}{C}\), then plug the value into the expression for \(\sigma\). After substituting the relevant values, we get: $$ \sigma = \frac{(85 \times 10^{-3} V) (6.0 \times 10^{-9} m)}{(1.03 \times 10^{-8} F/m) (2 \pi) (5.0 \times 10^{-6} m) L} $$

Part (b): Calculating Surface Charge Density

After evaluating the above expression, we get the surface charge density: $$ \sigma \approx 8.32 \times 10^{-3} \, \mathrm{C/m^2} $$ So, the capacitance per unit length of axon is \(1.03 \times 10^{-8} \, \mathrm{F/m}\), and the magnitude of the charge density on the surfaces of the membrane is \(8.32 \times 10^{-3} \, \mathrm{C/m^2}\). The outside surface of the membrane has a positive charge.