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Chapter 19: Problem 67

Nerve cells are long, thin cylinders along which electrical disturbances (nerve impulses) travel. The cell membrane of a typical nerve cell consists of an inner and an outer wall separated by a distance of \(0.10 \mu \mathrm{m}\). The electric field within the cell membrane is \(7.0 \times 10^{5} \mathrm{N} / \mathrm{C}\). Approximating the cell membrane as a parallel-plate capacitor, determine the magnitude of the charge density on the inner and outer cell walls.

Short Answer

Expert verified
The charge density is approximately \(6.195 \times 10^{-6} \text{ C/m}^2\).

Step by step solution

01

Understanding the problem

We are given the electric field within the cell membrane and the distance between the inner and outer walls. We need to determine the charge density on the membrane, assuming it is a parallel-plate capacitor.
02

Formula for electric field in capacitors

The electric field \( E \) in a parallel-plate capacitor is related to the surface charge density \( \sigma \) and the permittivity of the medium \( \varepsilon_0 \) by the formula: \[ E = \frac{\sigma}{\varepsilon_0} \] Where \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/(\text{N} \cdot \text{m}^2) \) is the permittivity of free space.
03

Solve for charge density (\( \sigma \))

Rearrange the formula to solve for the surface charge density \( \sigma \): \[ \sigma = E \cdot \varepsilon_0 \] Substitute the given electric field \( E = 7.0 \times 10^5 \text{ N/C} \) and the permittivity of free space \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/(\text{N} \cdot \text{m}^2) \) into the equation.
04

Perform the calculation

Substitute the values into the equation: \[ \sigma = 7.0 \times 10^5 \text{ N/C} \times 8.85 \times 10^{-12} \text{ C}^2/(\text{N} \cdot \text{m}^2) \] Calculate to find \( \sigma \).
05

Final result

Upon calculating, we find: \( \sigma = 6.195 \times 10^{-6} \text{ C/m}^2 \). This is the charge density on the inner and outer walls of the nerve cell membrane.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parallel-Plate Capacitor
Imagine two flat, conductive surfaces facing each other, separated by a small gap. This setup is called a parallel-plate capacitor. Parallel-plate capacitors are essential in electronics because they store electrical energy. They are used in various applications, from small circuits to complex systems like computer motherboards.
  • The two plates in a parallel-plate capacitor have opposite charges, which creates an electric field perpendicular to the plates.
  • Inside, the electric field is uniform, meaning it has the same strength and direction everywhere between the plates.
  • The potential difference (voltage) across the plates causes them to store electric energy.
Membranes of nerve cells can be compared to parallel-plate capacitors. Just like in capacitors, the cell membranes regulate ion flow and store energy in the form of electrical potential. Understanding this similarity helps in determining characteristics like charge density.
Charge Density
Charge density is a way to describe how much electric charge is concentrated in an area. In a parallel-plate capacitor, we often talk about surface charge density, which refers to the amount of charge per unit area on the plates.
  • The formula for surface charge density (\(\sigma\)) in a parallel-plate capacitor is directly linked to the electric field \(E\):
\[ \sigma = E \cdot \varepsilon_0 \]Where:
  • \(\sigma\) = Surface charge density (C/m虏)
  • \(E\) = Electric field (N/C)
  • \(\varepsilon_0\) = Permittivity of free space
For nerve cells, where the membrane acts as a parallel plate, the charge density determines how ions are distributed across the membrane. Changes in charge density can impact nerve signal transmission, affecting how signals are communicated in our nervous system.
Permittivity of Free Space
The permittivity of free space, denoted as \(\varepsilon_0\), is a fundamental constant that describes how electric fields interact in a vacuum. This concept is crucial for understanding capacitors.
  • \(\varepsilon_0\) is approximately equal to \(8.85 \times 10^{-12} \, \text{C}^2/( ext{N} \cdot \text{m}^2)\).
  • It appears in equations like the one for electric field in capacitors, linking the electric field strength with surface charge density.
A higher permittivity means that the medium can "hold" more electric field lines, allowing more charge to be stored for a given potential difference.
In biological systems, such as nerve cell membranes, understanding permittivity helps explain how cells can integrate and transmit electrical signals efficiently, contributing to their complex functions.

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Most popular questions from this chapter

Four point charges are located at the corners of a square with sides of length \(a\). Two of the charges are \(+q\), and two are \(-q .\) Find the magnitude and direction of the net electric force exerted on a charge \(+Q\). located at the center of the square, for each of the following two arrangements of charge: (a) The charges alternate in \(\operatorname{sign}(+q,-q,+q,-q)\) as you go around the square; (b) the two positive charges are on the top corners, and the two negative charges are on the bottom corners.

The cell membrane in a nerve cell has a thickness of \(0.12 \mu \mathrm{m}\). (a) Approximating the cell membrane as a parallel-plate capacitor with a surface charge density of \(5.9 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2}\), find the electric field within the membrane. (b) If the thickness of the membrane were doubled, would your answer to part (a) increase, decrease, or stay the same? Explain.

An electron and a proton are released from rest in space, far from any other objects. The particles move toward each other, due to their mutual electrical attraction. (a) When they meet, is the kinetic energy of the electron greater than, less than, or equal to the kinetic energy of the proton? (b) Choose the best explanation from among the following: I. The proton has the greater mass. since kinetic energy is proportional to mass, it follows that the proton will have the greater kinetic energy. II. The two particles experience the same force, but the light electron moves farther than the massive proton. Therefore, the work done on the electron, and hence its kinetic energy, is greater. III. The same force acts on the two particles, Therefore, they will have the same kinetic energy and energy will be conserved.

Find the net charge of a system consisting of (a) \(6.15 \times 10^{6}\) electrons and \(7.44 \times 10^{6}\) protons or (b) 212 electrons and 165 protons.

When two identical ions are separated by a distance of \(6.2 \times 10^{-10} \mathrm{m},\) the electrostatic force each exerts on the other is \(5.4 \times 10^{-9} \mathrm{N} .\) How many electrons are missing from each ion?
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