Question

Cell membranes (the walled enclosure around a cell) are typically about 7.5 nm thick. They are partially permeable to allow charged material to pass in and out, as needed. Equal but opposite charge densities build up on the inside and outside faces of such a membrane, and these charges prevent additional charges from passing through the cell wall. We can model a cell membrane as a parallel-plate capacitor, with the membrane itself containing proteins embedded in an organic material to give the membrane a dielectric constant of about 10. (See \(\textbf{Fig. P24.48}\).) (a) What is the capacitance per square centimeter of such a cell wall? (b) In its normal resting state, a cell has a potential difference of 85 mV across its membrane. What is the electric field inside this membrane?

Short answer

(a) Capacitance is approximately 1.18 µF/cm². (b) Electric field is about 1.13 × 10⁷ V/m.

Step-by-step solution

Understanding the problem

We need to find the capacitance per square centimeter of a cell membrane modeled as a parallel-plate capacitor and then determine the electric field inside it at a given potential difference.

Identify the formula for capacitance

The formula for capacitance of a parallel-plate capacitor is \( C = \frac{\varepsilon A}{d} \), where \( \varepsilon \) is the permittivity of the dielectric material, \( A \) is the area, and \( d \) is the distance between the plates.

Calculate the permittivity

The relative permittivity (dielectric constant) \( \kappa \) is given as 10. The permittivity of the dielectric material is \( \varepsilon = \kappa \varepsilon_0 \),where \( \varepsilon_0 \approx 8.85 \times 10^{-12} \ \text{F/m} \) is the permittivity of free space.

Substitute values for capacitance

Given that \( \kappa = 10 \), \( \varepsilon_0 = 8.85 \times 10^{-12} \ \text{F/m} \), \( d = 7.5 \ \text{nm} = 7.5 \times 10^{-9} \ \text{m} \), substitute these values into the capacitance formula for unit area:\[ C = \frac{10 \times 8.85 \times 10^{-12} \ \text{F/m} \times 1 \ \text{cm}^2}{7.5 \times 10^{-9} \ \text{m}}. \]

Conversion of units

Convert \( 1 \ \text{cm}^2 \) to square meters: \( 1 \ \text{cm}^2 = 10^{-4} \ \text{m}^2 \). Substitute into the equation:\[ C = \frac{10 \times 8.85 \times 10^{-12} \times 10^{-4}}{7.5 \times 10^{-9}}. \]

Calculate the capacitance

Perform the calculations:\[ C \approx \frac{8.85 \times 10^{-15}}{7.5 \times 10^{-9}} = 1.18 \times 10^{-6} \ \text{F/cm}^2 = 1.18 \ \text{µF/cm}^2. \]

Identify the formula for electric field

The electric field \( E \) inside the membrane is given by \( E = \frac{V}{d} \) where \( V \) is the potential difference, and \( d \) is the thickness of the membrane.

Substitute values for electric field

Given \( V = 85 \ \text{mV} = 85 \times 10^{-3} \ \text{V} \) and \( d = 7.5 \times 10^{-9} \ \text{m} \), substitute into the formula:\[ E = \frac{85 \times 10^{-3}}{7.5 \times 10^{-9}}. \]

Calculate the electric field

Perform the calculation to find the electric field:\[ E \approx \frac{85 \times 10^{-3}}{7.5 \times 10^{-9}} \approx 1.13 \times 10^{7} \ \text{V/m}. \]

Key concepts

Parallel-Plate Capacitor

A cell membrane can be thought of as a parallel-plate capacitor. This model helps us understand how the membrane can store and separate charges. In a parallel-plate capacitor, two large plates face each other and are separated by a small distance.

• The plates hold equal and opposite charges.
• The distance between them is small compared to their size.

This arrangement allows for effective charging and energy storage. In the case of a cell membrane, the inner and outer surfaces of the membrane act as the plates. This configuration plays a crucial role in the membrane's ability to function in various biological processes, such as nerve impulses.

Dielectric Constant

The dielectric constant is an important factor when examining capacitors, including our cell membrane model. A dielectric is a non-conductive material placed between the plates of a capacitor. It increases the capacitance by reducing the electric field.

• It's represented by \( \kappa \) (kappa).
• A higher dielectric constant means a greater ability to store charge.

The cell membrane has a dielectric constant of about 10. This is due to its composition of proteins and organic material. The higher dielectric constant in the membrane enhances its capacity to store electrical energy, which is essential for its function as a biological capacitor.

Electric Field

Inside a cell membrane modeled as a capacitor, there exists an electric field. This field is crucial as it prevents further charges from crossing the membrane.

• The electric field (E) is calculated using the formula \( E = \frac{V}{d} \), where \( V \) is the potential difference, and \( d \) is the distance between the plates (thickness of the membrane).
• An increased electric field makes it more challenging for ions to pass through, controlling the movement of charged particles.

For the tested membrane, the electric field is approximately \( 1.13 \times 10^{7} \ \text{V/m} \). This strong electric field is a vital part of cellular functions like signalling and nutrient transport.

Potential Difference

In a cell membrane, the potential difference is the difference in electrical potential between its inner and outer surfaces. In our model, it directly affects the electric field strength across the membrane.

• This difference in potential is crucial for many biological processes, such as action potentials in neurons.
• For the cell membrane example, the potential difference is \( 85 \ \text{mV} \).

The potential difference indicates how much energy per charge is available to move across the membrane. It acts as a driving force that pushes ions from one side of the membrane to another, enabling essential cellular activities.

Permittivity of Free Space

Permittivity of free space is a fundamental constant in physics, denoted by \( \varepsilon_0 \). It's the measure of resistance encountered when forming an electric field in a vacuum.

• The value of \( \varepsilon_0 \) is approximately \( 8.85 \times 10^{-12} \ \text{F/m} \).
• In capacitors, it's used in calculating capacitance, such as \( C = \frac{\varepsilon A}{d} \), where \( \varepsilon \) is permittivity, \( A \) is area, and \( d \) is distance.

For our cell membrane, the permittivity of free space is multiplied by the dielectric constant to find the effective permittivity. This results in the ability of the membrane to store and separate charges effectively, an essential attribute for its biological functions.