Problem 40 - University Physics with Modern Physics

Question

B10 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 \(\vec{\boldsymbol{E}}\) 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) Atypical 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) 5.65e7 N/C, (b) 0.2825 V (outside higher), (c) 7.51e-14 J, (d) 1.05e7 N/C, 0.0525 V (outside higher).

Step 1: Calculate the Electric Field (E) in the Cell Wall (air)

Using Gauss's Law, we know that the electric field between the two layers of charge is determined by the formula:\[E = \frac{\sigma}{\varepsilon_0}\]where \(\sigma = 0.50 \times 10^{-3} \mathrm{C/m^2}\) is the surface charge density and \(\varepsilon_0 = 8.85 \times 10^{-12} \mathrm{C^2/(N \, m^2)}\) is the vacuum permittivity. Substituting these values gives us:\[E = \frac{0.50 \times 10^{-3}}{8.85 \times 10^{-12}} \approx 5.65 \times 10^7 \, \mathrm{N/C}\]

Step 2: Determine the Potential Difference (Air)

The potential difference \(\Delta V\) between two points can be determined using:\[\Delta V = E \cdot d\]where \(E = 5.65 \times 10^7 \mathrm{N/C}\) (from Step 1) and \(d = 5.0 \times 10^{-9} \, \mathrm{m}\) is the thickness of the cell wall. Hence, the potential difference is:\[\Delta V = 5.65 \times 10^7 \times 5.0 \times 10^{-9} = 0.2825 \mathrm{V}\]The inside of the cell is at a lower potential (more negative), while the outside is at a higher potential (more positive).

Step 3: Estimate the Electric Field Energy in the Cell Wall

Assuming the cell is spherical, we approximate the cell membrane volume. The radius \(r\) of a cell with volume \(10^{-16} \, \mathrm{m^3}\) is calculated as:\[\frac{4}{3}\pi r^3 = 10^{-16} \, \Rightarrow r \approx 2.88 \times 10^{-6} \mathrm{m}\]The volume of the cell wall (thickness \(d = 5.0 \times 10^{-9}\)) is:\[4\pi r^2 \, d = 4\pi (2.88 \times 10^{-6})^2 \times 5.0 \times 10^{-9}onumber\]This volume is approximately \(5.2 \times 10^{-20} \, \mathrm{m^3}\). The energy \(U\) in the electric field is:\[U = \frac{1}{2} \varepsilon_0 E^2 \cdot \text{volume}\]Substituting for \(E\) and volume, we find:\[U = \frac{1}{2} \times 8.85 \times 10^{-12} \times (5.65 \times 10^7)^2 \times 5.2 \times 10^{-20} \approx 7.51 \times 10^{-14} \, \mathrm{J}\]

Step 4: Repeat Part (a) with Dielectric Material (tissue)

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

Step 5: Repeat Part (b) with Dielectric Material

Using the new electric field \(E = 1.05 \times 10^7 \, \mathrm{N/C}\) and the same distance \(d\), the potential difference is:\[\Delta V = E \cdot d = 1.05 \times 10^7 \times 5.0 \times 10^{-9} = 0.0525 \, \mathrm{V}\]Again, the outside of the cell is at a higher potential than the inside.

Key Concepts

Electric Field in Cell Membrane

The electric field in the cell membrane is a critical concept in understanding how cells operate at the molecular level. The cell membrane, or cell wall, is a thin layer that separates the interior of the cell from its external environment. Electric fields in this membrane due to charge differences between the inner and outer surfaces are key to many biological processes.
In cellular biology, membranes can have various charged molecules embedded within them. These charges can create an electric field that affects the movement of ions and molecules, which is crucial for cell signaling and maintaining homeostasis.

In our exercise, the electric field between the layers of charge on a cell membrane is calculated using Gauss's Law. According to this law, the electric field (\( E \)) 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 \( \varepsilon_0 \) is the permittivity of free space. By substituting values like a charge density of \( 0.50 \times 10^{-3} \, \mathrm{C/m^2} \), we find an electric field of approximately \( 5.65 \times 10^7 \, \mathrm{N/C} \) for a cell membrane represented by air. This calculation highlights how the electric field environment inside and around the cell is intentionally structured to facilitate biological functions.

Dielectric Constant in Tissues

The dielectric constant is an important property when it comes to understanding how cells manage electric fields in tissue environments. Biological tissues have dielectric properties that differ significantly from those in air, impacting the behavior of electric fields in the body.
The dielectric constant (\( \kappa \)) of a material describes how much it can reduce the electric field compared to a vacuum. In the exercise, the cell wall is more accurately represented with a dielectric constant of 5.4, typical of various biological tissues. A dielectric material, like a membrane, allows cells to withstand higher electric fields without breakdown.

In terms of calculations, the effective permittivity (\( \varepsilon \)) is altered when the dielectric constant is included, changing the electric field calculation to \[ E = \frac{\sigma}{\kappa \varepsilon_0} \]. With a dielectric constant of 5.4, the effective electric field is reduced to approximately \( 1.05 \times 10^7 \, \mathrm{N/C} \). This showcases how tissue composition, reflected in its dielectric constant, influences the movement of ions and the electrical stability within cells, underlining the importance of accurate biological modeling.

Gauss's Law in Biophysics

Gauss's Law is a fundamental principle underpinning much of electromagnetic theory, extending into biophysics where it is used to calculate electric fields in biological systems. The law relates the distribution of electric charge to the resulting electric field, describing how charged particles within a volume create a field, which can be measured across a surface surrounding that volume.

• Gauss's Law can be actualized in biological contexts to determine how cells generate electric fields based on their charge distributions.
• For the exercise, it helped in determining the electric field across the cell membrane by linking it to the surface charge density and the permittivity of the medium surrounding the cells.

This principle is vital for understanding how cells communicate and process various signals using electrochemical gradients. By calculating the electric field via Gauss's Law, we can predict how ions will move across cell membranes, which are crucial for functions such as nerve impulse transmission and muscle contractions.
Gauss's Law can be efficiently applied to predict physiological changes and reactions within a cell, making it a significant tool for insights into cellular operations and biophysical modeling.