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The membrane surrounding a living cell consists of an inner and an outer wall that are separated by a small space. Assume that the membrane acts like a parallel plate capacitor in which the effective charge density on the inner and outer walls has a magnitude of \(7.1 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2} \cdot\) (a) What is the magnitude of the electric field within the cell membrane? (b) Find the magnitude of the electric force that would be exerted on a potassium ion (K \(+\); charge \(=+e\) ) placed inside the membrane.

Short Answer

Expert verified
(a) The electric field magnitude is approximately \( 8.02 \times 10^{5} \ \mathrm{N/C} \). (b) The electric force on a potassium ion is approximately \( 1.28 \times 10^{-13} \ \mathrm{N} \).

Step by step solution

01

Understand the problem

We're given a biological cell membrane acting as a parallel plate capacitor with a charge density and need to find (a) the electric field and (b) the force on a potassium ion.
02

Recall formulas

For a parallel plate capacitor, the magnitude of the electric field \( E \) is related to the charge density \( \sigma \) by the formula: \( E = \frac{\sigma}{\varepsilon_0} \), where \( \varepsilon_0 \) is the vacuum permittivity \( 8.85 \times 10^{-12} \ \mathrm{C}^2/\mathrm{N}\cdot\mathrm{m}^2 \).
03

Calculate the electric field

Given \( \sigma = 7.1 \times 10^{-6} \ \mathrm{C/m}^2 \), use the formula \( E = \frac{\sigma}{\varepsilon_0} \) to find the electric field. Plug in the values: \( E = \frac{7.1 \times 10^{-6}}{8.85 \times 10^{-12}} \ \mathrm{N/C} \).
04

Solve for the electric field

Calculate \( E = \frac{7.1 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 8.02 \times 10^{5} \ \mathrm{N/C} \). Thus, the electric field inside the membrane is approximately \( 8.02 \times 10^{5} \ \mathrm{N/C} \).
05

Determine the electric force on a potassium ion

Use the formula for electric force \( F = qE \), where \( q \) represents charge and \( e = 1.6 \times 10^{-19} \ \mathrm{C} \) for a potassium ion. Substituting values gives \( F = (1.6 \times 10^{-19} \mathrm{C})(8.02 \times 10^{5} \ \mathrm{N/C}) \).
06

Solve for the electric force

Calculate \( F = (1.6 \times 10^{-19})(8.02 \times 10^{5}) \approx 1.28 \times 10^{-13} \ \mathrm{N} \). Hence, the force on the potassium ion is approximately \( 1.28 \times 10^{-13} \ \mathrm{N} \).

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Key Concepts

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

Parallel Plate Capacitor
In a biological context, cell membranes can be likened to a parallel plate capacitor. This is due to their structure, consisting of two charged surfaces separated by a small gap. A parallel plate capacitor is a device used in electronics to store electrical energy, with two conductive plates placed parallel to each other. The cell membrane's charged inner and outer surfaces mimic this setup. This unique structure is crucial for various cellular functions and allows the storage and management of ions across the membrane.
The electric field between the plates, or in this case the membrane, is uniform and depends on the charge density and other factors, allowing ions to be regulated effectively in living cells.
Electric Force on Ions
The electric force on ions within a cell membrane can be understood by considering how charged particles interact with electric fields. When a charged ion, such as a potassium ion (K+), is placed in the electric field generated by a cell membrane, it experiences a force. This force is calculated using the formula:
  • \( F = qE \)
Here, \( F \) is the force, \( q \) is the charge of the ion, and \( E \) is the electric field strength.
For a potassium ion, with a charge of \( +e \), this force dictates how the ion moves across the membrane, influencing cellular activities and signaling.
Charge Density
Charge density refers to the quantity of electric charge per unit area on the surface of the cell membrane. It is symbolized by \( \sigma \), and in this context, it has a magnitude of \( 7.1 \times 10^{-6} \, \mathrm{C/m}^2 \).
This density is pivotal in determining the electric field in a parallel plate capacitor, where a higher density results in a stronger electric field. Charge density influences the behavior of ions and how they are stored or released across the membrane, playing a significant role in cellular processes such as nerve impulses.
Vacuum Permittivity
Vacuum permittivity, represented by \( \varepsilon_0 \), is a fundamental physical constant that measures the ability of a vacuum to permit electric field lines. Its value is approximately \( 8.85 \times 10^{-12} \, \mathrm{C}^2/\mathrm{N}\cdot\mathrm{m}^2 \).
This constant is crucial when calculating the electric field in a parallel plate capacitor, as shown in the equation:
  • \( E = \frac{\sigma}{\varepsilon_0} \)
This formula illustrates how vacuum permittivity mediates the relationship between charge density and the resulting electric field. Understanding this concept helps in analyzing how electric fields operate in various environments, including biological systems.

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Most popular questions from this chapter

A metal sphere has a charge of \(+8.0 \mu \mathrm{C}\). What is the net charge after \(6.0 \times 10^{13}\) electrons have been placed on it?

Interactive Solution \(\underline{18.37}\) at provides a model for problems of this kind. A small object has a mass of \(3.0 \times 10^{-3} \mathrm{~kg}\) and a charge of \(-34 \mu \mathrm{C}\). It is placed at a certain spot where there is an electric field. When released, the object experiences an acceleration of \(2.5 \times 10^{3} \mathrm{~m} / \mathrm{s}^{2}\) in the direction of the \(+x\) axis. Determine the magnitude and direction of the electric field.

Water has a mass per mole of \(18.0 \mathrm{~g} / \mathrm{mol}\), and each water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) has 10 electrons. (a) How many electrons are there in one liter \(\left(1.00 \times 10^{-3} \mathrm{~m}^{3}\right)\) of water? (b) What is the net charge of all these electrons?

Two charges are located along the \(x\) axis: \(q_{1}=+6.0 \mu \mathrm{C}\) at \(x_{1}=+4.0 \mathrm{~cm}\), and \(q_{2}=+6.0 \mu \mathrm{C}\) at \(x_{2}=-4.0 \mathrm{~cm}\). Two other charges are located on the \(y\) axis: \(q_{3}=+3.0 \mu \mathrm{C}\) at \(y_{3}=+5.0 \mathrm{~cm}\), and \(q_{4}=-8.0 \mu \mathrm{C}\) at \(y_{4}=+7.0 \mathrm{~cm} .\) Find the net electric field (magnitude and direction) at the origin.

Suppose you want to determine the electric field in a certain region of space. You have a small object of known charge and an instrument that measures the magnitude and direction of the force exerted on the object by the electric field. How would you determine the magnitude and direction of the electric field if the object were (a) positively charged and (b) negatively charged?

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