Problem 77 - College Physics

Question

Potential in human cells. Some cell walls in the human body have a layer of negative charge on the inside surface and a layer of positive charge of equal magnitude on the outside surface. Suppose that the charge density on either surface is \(\pm 0.50 \times 10^{-3} \mathrm{C} / \mathrm{m}^{2},\) the cell wall is 5.0 \(\mathrm{nm}\) thick, and the cell-wall material is air. (a) Find the magnitude of \(\vec{E}\) in the wall between the two layers of charge. (b) Find the potential difference between the inside and the outside of the cell. Which is at the higher potential? (c) A typical cell in the human body has a volume of \(10^{-16} \mathrm{m}^{3} .\) Estimate the total electric-field energy stored in the wall of a cell of this size. (Hint: Assume that the cell is spherical, and calculate the volume of the cell wall.) (d) In reality, the cell wall is made up, not of air, but of tissue with a dielectric constant of \(5.4 .\) Repeat parts (a) and (b) in this case.

Step-by-Step Solution

Answer

(a) \( E = 5.65 \times 10^7 \mathrm{N/C} \), (b) \( \Delta V = 0.2825 \mathrm{V} \) with inside lower, (c) \( U = 7.3 \times 10^{-14} \mathrm{J} \), (d) \( E = 1.05 \times 10^7 \mathrm{N/C} \), \( \Delta V = 0.0525 \mathrm{V} \).

Step 1: Calculate Electric Field Without Dielectric

Using Gauss's Law, the electric field \( E \) inside a slab of air due to a single layer of charge in vacuum is given by \( E = \frac{\sigma}{\varepsilon_0} \), where \( \sigma = 0.50 \times 10^{-3} \mathrm{C/m}^2 \) and \( \varepsilon_0 = 8.85 \times 10^{-12} \mathrm{F/m} \) is the vacuum permittivity. Calculating this gives: \[ E = \frac{0.50 \times 10^{-3}}{8.85 \times 10^{-12}} = 5.65 \times 10^7 \mathrm{N/C}. \] This is the magnitude of the electric field between the charged layers.

Step 2: Calculate the Potential Difference Without Dielectric

The potential difference \( \Delta V \) between the cell walls can be calculated using \( \Delta V = E \cdot d \), where \( d = 5.0 \times 10^{-9} \mathrm{m} \) is the thickness of the cell wall. Thus, \[ \Delta V = 5.65 \times 10^7 \mathrm{N/C} \times 5.0 \times 10^{-9} \mathrm{m} = 0.2825 \mathrm{V}. \] The interior is at a lower potential because the electric field points from positive to negative.

Step 3: Calculate Volume of the Cell Wall

Assume the cell is spherical with volume \( V = \frac{4}{3} \pi r^3 = 10^{-16} \mathrm{m}^3 \). Solving for \( r \) gives \( r \approx 2.88 \times 10^{-6} \mathrm{m} \). The cell wall volume \( V_{\text{wall}} \approx 4\pi r^2 d \), so \( V_{\text{wall}} = 4\pi (2.88 \times 10^{-6})^2 \times 5 \times 10^{-9} = 5.2 \times 10^{-20} \mathrm{m}^3. \)

Step 4: Calculate Electric-Field Energy Without Dielectric

The energy stored in the cell wall is \( U = \frac{1}{2} \varepsilon_0 E^2 V_{\text{wall}} \). Substituting \( E \) and \( V_{\text{wall}} \), we get: \[ U = \frac{1}{2} \times 8.85 \times 10^{-12} \mathrm{F/m} \times (5.65 \times 10^7 \mathrm{N/C})^2 \times 5.2 \times 10^{-20} \mathrm{m}^3 \approx 7.3 \times 10^{-14} \mathrm{J}. \]

Step 5: Adjust Calculation for Dielectric Material

With a dielectric, \( E = \frac{\sigma}{\varepsilon_0 \cdot K} \), where \( K = 5.4 \) is the dielectric constant. Thus, \[ E = \frac{0.50 \times 10^{-3}}{8.85 \times 10^{-12} \times 5.4} = 1.05 \times 10^7 \mathrm{N/C}. \]

Step 6: Recalculate Potential Difference with Dielectric

Using the modified electric field and \( d = 5.0 \times 10^{-9} \mathrm{m} \): \[ \Delta V = 1.05 \times 10^7 \mathrm{N/C} \times 5.0 \times 10^{-9} \mathrm{m} = 0.0525 \mathrm{V}. \] The interior is still at a lower potential.

Key Concepts

Electric Field in Biology

In biology, electric fields play a crucial role in many cellular processes. Electric fields arise from the separation of charges across cell membranes. These charges create an `electric field`, which is a force field around a charged particle. In an organism, these fields are essential for functions such as nerve impulse transmission and muscle contraction; they are essentially the body's biological "circuitry".

In cell biology, the `electric field` can be calculated using \( E = \frac{\sigma}{\varepsilon_0} \) for a slab of air or vacuum where:

• \( E \) is the electric field.
• \( \sigma \) is the surface charge density.
• \( \varepsilon_0 \) is the vacuum permittivity.

This formula assumes a medium similar to air, not accounting for more complex biological materials like tissues. By understanding these basic principles, we can better understand and analyze how cells communicate and function at the atomic level.

Potential Difference in Cells

The potential difference across cell membranes is fundamental in maintaining cell function. The potential difference, also known as `membrane potential`, is the voltage difference between the inside and outside of a cell. It is determined by the concentration of ions, both inside and outside of the cell.

The formula to calculate the potential difference \( \Delta V \) is: \( \Delta V = E \cdot d \), where:

• \( \Delta V \) is the potential difference.
• \( E \) is the electric field.
• \( d \) is the thickness of the cell wall.

This equation shows how the potential difference is directly related to the electric field and the distance over which it acts.
This `potential difference` is crucial for a cell's signaling processes, such as neural transmission, where rapid changes in membrane potential result in signal propagation through the nervous system.

Dielectric Constant in Tissues

In biological tissues, the `dielectric constant` is a critical concept that modifies how an electric field behaves within materials. Unlike in vacuum or air, tissues have a property called the dielectric constant (denoted as \( K \)), which reflects how a material can store electrical energy in an electric field.

The presence of a dielectric reduces the electric field by the factor of the `dielectric constant`, calculated as: \( E = \frac{\sigma}{\varepsilon_0 \cdot K} \). Here, \( K \) represents the dielectric property of the tissue, affecting how easily electricity can pass through. For example, in the problem at hand, when \( K = 5.4 \), this results in a reduced electric field compared to air.

This modification in electric fields is vital for the stability and functionality of cells. It controls how signals are relayed and energy is stored in cellular membranes, ensuring appropriate cellular responses and activities.

Gauss's Law in Biology

Gauss's Law is a pivotal principle in electrostatics that is equally important in biological contexts. This law provides a method to calculate the electric field generated by a symmetric distribution of charges. Gauss's law states that the electric flux through a closed surface is proportional to the charge enclosed in that surface.

This law is succinctly expressed as \( \Phi = \frac{Q_{enc}}{\varepsilon_0} \), where \( \Phi \) is the electric flux and \( Q_{enc} \) is the charge enclosed.

In biology, this principle helps in understanding how membrane potentials are developed and maintained. It assists in computing electric fields that interact with cell organelles and drive essential biological processes such as passive ion movement across semi-permeable membranes. By applying Gauss's Law, one gains insights into electrostatic conditions within biological systems, which are fundamental to medical and biological research.