Question

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${}^{-3}$ C/m${}^2$, the cell wall is 5.0 nm thick, and the cell- wall material is air. (a) Find the magnitude of $\overrightarrow{E}$ S 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${}^{-16}$ 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.

Short answer

(a) 5.65 × 10⁷ N/C; (b) 0.283 V, inside higher; (c) Use energy formula; (d) With dielectric: E'=1.046 × 10⁷ N/C, ΔV'=0.0523 V.

Step-by-step solution

Understanding Charge Density and E-Field Relationship

We need to calculate the electric field ($E$) in the wall between the two charged layers. First, we use Gauss's law, which relates electric charge and electric field. For an infinite plane with surface charge density $\sigma$, the magnitude of the electric field just outside the surface is $E={\displaystyle \frac{\sigma }{{\epsilon }_0}}$, where ${\epsilon }_0=8.85\times {10}^{-12}$ C${}^2$/N m${}^2$ is the permittivity of free space.

Calculating the Electric Field in the Wall

Given charge density $\sigma =0.50\times {10}^{-3}$ C/m${}^2$, the magnitude of the electric field inside the wall is$$E={\displaystyle \frac{0.50\times {10}^{-3}}{8.85\times {10}^{-12}}}=5.65\times {10}^7\text{ N/C}.$$

Finding the Potential Difference

The potential difference $\mathrm{\Delta }V$ is given by $\mathrm{\Delta }V=Ed$, where $d$ is the thickness of the cell wall (5.0 nm or $5.0\times {10}^{-9}$ m). Therefore,$$\mathrm{\Delta }V=5.65\times {10}^7\times 5.0\times {10}^{-9}=0.283\text{ V}.$$ The inside surface has a higher potential because positive charges reside on the outside surface.

Estimating Total Electric-field Energy

The energy stored in an electric field per unit volume is given by $u={\displaystyle \frac{1}{2}}{\epsilon }_0{E}^2$. For a spherical shell, the volume $V$ of the wall is calculated using the outer and inner radius. Assume a single spherical cell, with radius $r$ from volume $\frac{4}{3}\pi r^3={10}^{-16}$. The thickness is 5.0 nm, so the energy is $U=u\times V$.

Adjusting for a Dielectric Layer

When the dielectric constant $\kappa =5.4$ is considered, the electric field inside the wall changes to $E^{\prime }=E/\kappa =5.65\times {10}^7/5.4=1.046\times {10}^7\text{ N/C}$. The potential difference is now$$\mathrm{\Delta }{V}^{\prime }=E^{\prime }d=1.046\times {10}^7\times 5.0\times {10}^{-9}=0.0523\text{ V}.$$ The inside surface remains at a higher potential.

Key concepts

Gauss's Law

Gauss's law is a fundamental principle in electrostatics that connects electric fields and electric charges. The law states that the total flux of an electric field through a closed surface is proportional to the enclosed electric charge. In mathematical terms, Gauss's law is expressed as: $$\mathrm{\Phi }=\oint \overrightarrow{E}\cdot d\overrightarrow{A}={\displaystyle \frac{Q_{enc}}{{\epsilon }_0}}$$ Here, $\mathrm{\Phi }$ is the electric flux, $\overrightarrow{E}$ is the electric field, $d\overrightarrow{A}$ is the differential area vector on the surface, $Q_{enc}$ is the enclosed charge, and ${\epsilon }_0$ is the permittivity of free space. For practical calculations, Gauss's law is particularly useful in determining electric fields for symmetrical charge distributions, such as infinite planes, spheres, or cylinders. In this exercise, we apply Gauss's law to calculate the electric field between two parallel charge layers with known surface charge density.

Electric Field

The electric field (E) represents the force that a charge experiences in a region due to other electric charges. It is a vector quantity, which means it has both magnitude and direction. An important equation for electric fields with uniform surface charge density $\sigma$ is given by: $$E={\displaystyle \frac{\sigma }{{\epsilon }_0}}$$ This equation indicates that the electric field is directly proportional to the charge density on the surface and inversely proportional to the permittivity of free space. In our exercise, this formula is used to find the magnitude of the electric field within the cell wall, exhibiting a parallel layer of negative and positive charges.

• The electric field points from the positive to the negative charge.
• Higher charge density results in a stronger electric field.

Understanding how electric fields operate between different charges helps explain how ions move across cell membranes in biological systems.

Potential Difference

The potential difference, often referred to as voltage ($\mathrm{\Delta }V$), describes the work needed to move a unit charge from one point to another in an electric field. It can be calculated using the electric field and the distance between charges: $$\mathrm{\Delta }V=E\times d$$ Where $E$ is the electric field strength and $d$ is the distance between the points, known as the thickness of the cell wall in this exercise. A higher potential indicates more energy per charge at a specific point, and in the context of the exercise, the inside of the cell is at a higher potential. This is essential in understanding how cells maintain ion gradients. Points to consider:

• Potential difference determines the direction of charge flow.
• A higher potential on the inner wall means charges will tend to move outwards.

Dielectric Constant

The dielectric constant, also known as relative permittivity ($\kappa$), measures a material's ability to reduce the electric field within it compared to the vacuum. It is defined as: $$\kappa ={\displaystyle \frac{\epsilon }{{\epsilon }_0}}$$ Where $\epsilon$ is the permittivity of the material and ${\epsilon }_0$ is the permittivity of free space. Materials with a higher dielectric constant effectively reduce the electric field's strength, impacting how charge behaves within the material. In our exercise, when the cell wall is modeled with a dielectric constant of 5.4, the electric field inside is reduced, thereby decreasing the potential difference. This reflects the effect of the tissue on ion movement in biological cells:

• High dielectric constant materials better insulate against electric fields.
• Reduced electric field strength leads to lesser potential energy storage.