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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

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

Step by step solution

01

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.
02

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.
03

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.
04

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:\[ C = \frac{10 \times 8.85 \times 10^{-12} \ \text{F/m} \times 1 \ \text{cm}^2}{7.5 \times 10^{-9} \ \text{m}}. \]
05

Conversion of units

Convert \( 1 \ \text{cm}^2 \) to square meters: \( 1 \ \text{cm}^2 = 10^{-4} \ \text{m}^2 \). Substitute into the equation:\[ C = \frac{10 \times 8.85 \times 10^{-12} \times 10^{-4}}{7.5 \times 10^{-9}}. \]
06

Calculate the capacitance

Perform the calculations:\[ C \approx \frac{8.85 \times 10^{-15}}{7.5 \times 10^{-9}} = 1.18 \times 10^{-6} \ \text{F/cm}^2 = 1.18 \ \text{µF/cm}^2. \]
07

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.
08

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:\[ E = \frac{85 \times 10^{-3}}{7.5 \times 10^{-9}}. \]
09

Calculate the electric field

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

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Key Concepts

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

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.

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Most popular questions from this chapter

A parallel-plate capacitor has plates with area 0.0225 m\(^2\) separated by 1.00 mm of Teflon. (a) Calculate the charge on the plates when they are charged to a potential difference of 12.0 V. (b) Use Gauss's law (Eq. 24.23) to calculate the electric field inside the Teflon. (c) Use Gauss's law to calculate the electric field if the voltage source is disconnected and the Teflon is removed.

A budding electronics hobbyist wants to make a simple 1.0-nF capacitor for tuning her crystal radio, using two sheets of aluminum foil as plates, with a few sheets of paper between them as a dielectric. The paper has a dielectric constant of 3.0, and the thickness of one sheet of it is 0.20 mm. (a) If the sheets of paper measure 22 \(\times\)28 cm and she cuts the aluminum foil to the same dimensions, how many sheets of paper should she use between her plates to get the proper capacitance? (b) Suppose for convenience she wants to use a single sheet of posterboard, with the same dielectric constant but a thickness of 12.0 mm, instead of the paper. What area of aluminum foil will she need for her plates to get her 1.0 nF of capacitance? (c) Suppose she goes high-tech and finds a sheet of Teflon of the same thickness as the posterboard to use as a dielectric. Will she need a larger or smaller area of Teflon than of posterboard? Explain.

An air capacitor is made by using two flat plates, each with area A, separated by a distance \(d\). Then a metal slab having thickness \(a\) (less than \(d\)) and the same shape and size as the plates is inserted between them, parallel to the plates and not touching either plate (\(\textbf{Fig. P24.62}\)). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance C\(_0\) when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits \(a\) \(\rightarrow\) 0 and \(a\) \(\rightarrow\) d.

An air capacitor is made from two flat parallel plates 1.50 mm apart. The magnitude of charge on each plate is 0.0180 \(\mu\)C when the potential difference is 200 V. (a) What is the capacitance? (b) What is the area of each plate? (c) What maximum voltage can be applied without dielectric breakdown? (Dielectric breakdown for air occurs at an electric-field strength of 3.0 \(\times\) 10\(^6\) V/m.) (d) When the charge is 0.0180 \(\mu\)C, what total energy is stored?

A parallel-plate capacitor has capacitance \(C\) = 12.5 pF when the volume between the plates is filled with air. The plates are circular, with radius 3.00 cm. The capacitor is connected to a battery, and a charge of magnitude 25.0 pC goes onto each plate. With the capacitor still connected to the battery, a slab of dielectric is inserted between the plates, completely filling the space between the plates. After the dielectric has been inserted, the charge on each plate has magnitude 45.0 pC. (a) What is the dielectric constant \(K\) of the dielectric? (b) What is the potential difference between the plates before and after the dielectric has been inserted? (c) What is the electric field at a point midway between the plates before and after the dielectric has been inserted?

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