Electric Charge and Electric Field

4 Coulomb’s Law

Learning Objectives

  • State Coulomb’s law in terms of how the electrostatic force changes with the distance between two objects.
  • Calculate the electrostatic force between two charged point forces, such as electrons or protons.
  • Compare the electrostatic force to the gravitational attraction for a proton and an electron; for a human and the Earth.

 

Two spiral galaxies show the strong gravitational attraction between them as their arms appear to reach out toward one another.
Figure 4.1: This NASA image of Arp 87 shows the result of a strong gravitational attraction between two galaxies. In contrast, at the subatomic level, the electrostatic attraction between two objects, such as an electron and a proton, is far greater than their mutual attraction due to gravity. (credit: NASA/HST)

By the late eighteenth century, scientists had identified several key properties of electric charge: there are two types of charge, like charges repel, unlike charges attract, and the force between charges decreases as the distance between them increases. Careful experiments eventually led to a precise mathematical relationship describing this interaction. This relationship is known as Coulomb’s law, named after the French physicist Charles Coulomb (1736–1806), who performed detailed measurements of electric forces.

The electrostatic force plays a central role in biology and medicine. At the molecular scale, the structure of proteins, the folding of DNA, and the behavior of ions in cells are all strongly influenced by electrostatic interactions. These forces determine how charged molecules attract or repel one another in aqueous environments such as blood plasma and intracellular fluid.

Coulomb’s Law

[latex]F=k\frac{|{q}_{1}{q}_{2}|}{{r}^{2}}[/latex]

Coulomb’s law gives the magnitude of the force F between two point charges [latex]{q}_{1}[/latex] and [latex]{q}_{2}[/latex] separated by a distance [latex]r[/latex]. The constant [latex]k[/latex] in SI units is

[latex]k=8.988×{10}^{9}\frac{\text{N}\cdot{\text{m}}^{2}}{{\text{C}}^{2}}\approx8.99×{10}^{9}\frac{\text{N}\cdot{\text{m}}^{2}}{{\text{C}}^{2}}[/latex]

The electrostatic force is a vector quantity and acts along the line connecting the two charges. As shown in Figure 4.2, the forces on the two charges are equal in magnitude and opposite in direction, consistent with Newton’s third law.

Although Coulomb’s law appears simple, establishing it experimentally was challenging. Coulomb used a torsion balance to measure extremely small forces. Modern experiments have confirmed that the force varies with the inverse square of distance,

[latex]F\propto\frac{1}{r^{2}}[/latex]

with extremely high precision. This inverse-square dependence is also found in gravitation and plays a fundamental role in many areas of physics and engineering.

Figure 4.2: The magnitude of the electrostatic force between point charges is given by Coulomb’s law. Like charges repel, and unlike charges attract. The forces on the two charges are equal and opposite.

How Strong is the Coulomb Force Relative to the Gravitational Force?

Compare the electrostatic force between an electron and proton separated by

[latex]0.530×10^{-10}\text{ m}[/latex]

with the gravitational force between them. This distance corresponds to their average separation in a hydrogen atom.

Strategy

We calculate the electrostatic force using Coulomb’s law and the gravitational force using Newton’s law of gravitation. Then we compare their magnitudes by taking a ratio.

Solution

[latex]F=k\frac{|q_1q_2|}{r^2}[/latex]
[latex]F=(8.99×10^9)\frac{(1.60×10^{-19})(1.60×10^{-19})}{(0.530×10^{-10})^2}[/latex]
[latex]F=8.19×10^{-8}\text{ N}[/latex]

The gravitational force is:

[latex]F_G=G\frac{mM}{r^2}[/latex]
[latex]F_G=3.61×10^{-47}\text{ N}[/latex]
[latex]\frac{F}{F_G}=2.27×10^{39}[/latex]

Discussion

This enormous ratio shows that electrostatic forces dominate interactions at atomic and molecular scales. In biological systems, this dominance explains why ionic bonds, hydrogen bonding, and membrane potentials are governed primarily by electrostatic interactions rather than gravity.

At macroscopic scales, gravitational forces dominate because most objects are nearly electrically neutral. Positive and negative charges within matter largely cancel, leaving gravity as the net force between large bodies such as the Earth and a person. In contrast, at the scale of cells and molecules, electrostatic forces determine how ions move across membranes, how proteins bind, and how electrical signals propagate in nerves.

Section Summary

  • Frenchman Charles Coulomb was the first to publish the mathematical equation that describes the electrostatic force between two objects.
  • Coulomb’s law gives the magnitude of the force between point charges:
    [latex]F=k\frac{|q_1q_2|}{r^2}[/latex]
  • This Coulomb force is responsible for electrostatic effects and underlies many molecular interactions in biological systems.
  • The Coulomb force is extraordinarily strong compared with gravitational force at atomic scales.
  • The electrostatic force between two subatomic particles is far greater than the gravitational force between them.

Conceptual Questions

  1. Figure 4.3 shows the charge distribution in a water molecule, which is called a polar molecule because it has an inherent separation of charge. Given water’s polar character, explain what effect humidity has on removing excess charge from objects.
    A schematic representation of the outer electron cloud of a neutral water molecule is shown. Three atoms are placed on the vertices of a triangle. The hydrogen atom has positive q charge and the oxygen atom has minus two q charge, and the angle between the line joining each hydrogen atom with the oxygen atom is one hundred and four degrees. The cloud density is shown more at the oxygen atom.
    Figure 4.3 Schematic representation of the outer electron cloud of a neutral water molecule. The electrons spend more time near the oxygen than the hydrogens, giving a permanent charge separation as shown. Water is thus a polar molecule. It is more easily affected by electrostatic forces than molecules with uniform charge distributions.

  2. Using Figure 4.3, explain, in terms of Coulomb’s law, why a polar molecule (such as in Figure 4.3) is attracted by both positive and negative charges.
  3. Given the polar character of water molecules, explain how ions in the air form nucleation centers for rain droplets.

Problems & Exercises

  1. What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC?
  2. (a) How strong is the attractive force between a glass rod with a [latex]0.700\phantom{\rule{0.25em}{0ex}}\mu \text{C}[/latex] charge and a silk cloth with a [latex]–0.600\phantom{\rule{0.25em}{0ex}}\mu \text{C}[/latex] charge, which are 12.0 cm apart, using the approximation that they act like point charges? (b) Discuss how the answer to this problem might be affected if the charges are distributed over some area and do not act like point charges.
  3. Two point charges exert a 5.00 N force on each other. What will the force become if the distance between them is increased by a factor of three?
  4. Two point charges are brought closer together, increasing the force between them by a factor of 25. By what factor was their separation decreased?
  5. How far apart must two point charges of 75.0 nC (typical of static electricity) be to have a force of 1.00 N between them?
  6. If two equal charges each of 1 C each are separated in air by a distance of 1 km, what is the magnitude of the force acting between them? You will see that even at a distance as large as 1 km, the repulsive force is substantial because 1 C is a very significant amount of charge.
  7. A test charge of [latex]+2\phantom{\rule{0.25em}{0ex}}\mu \text{C}[/latex] is placed halfway between a charge of [latex]+6\phantom{\rule{0.25em}{0ex}}\mu \text{C}[/latex] and another of [latex]+4\phantom{\rule{0.25em}{0ex}}\mu \text{C}[/latex] separated by 10 cm. (a) What is the magnitude of the force on the test charge? (b) What is the direction of this force (away from or toward the [latex]+6\phantom{\rule{0.25em}{0ex}}\mu \text{C}[/latex] charge)?
  8. Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated protons separated by 2.00 nm (a typical distance between gas atoms). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics.
  9. (a) By what factor must you change the distance between two point charges to change the force between them by a factor of 10? (b) Explain how the distance can either increase or decrease by this factor and still cause a factor of 10 change in the force.
  10. Suppose you have a total charge [latex]{q}_{\text{tot}}[/latex] that you can split in any manner. Once split, the separation distance is fixed. How do you split the charge to achieve the greatest force?
  11. (a) Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force can be great enough to support the top piece’s weight. Assuming equal point charges (only an approximation), calculate the magnitude of the charge if electrostatic force is great enough to support the weight of a 10.0 mg piece of tape held 1.00 cm above another. (b) Discuss whether the magnitude of this charge is consistent with what is typical of static electricity.\
  12. (a) Find the ratio of the electrostatic to gravitational force between two electrons. (b) What is this ratio for two protons? (c) Why is the ratio different for electrons and protons?
  13. At what distance is the electrostatic force between two protons equal to the weight of one proton?
  14. A certain five cent coin contains 5.00 g of nickel. What fraction of the nickel atoms’ electrons, removed and placed 1.00 m above it, would support the weight of this coin? The atomic mass of nickel is 58.7, and each nickel atom contains 28 electrons and 28 protons.
  15. (a) Two point charges totaling [latex]8.00\phantom{\rule{0.25em}{0ex}}µ\text{C}[/latex] exert a repulsive force of 0.150 N on one another when separated by 0.500 m. What is the charge on each? (b) What is the charge on each if the force is attractive?
  16. Point charges of [latex]5.00\phantom{\rule{0.25em}{0ex}}µ\text{C}[/latex] and [latex]–3.00\phantom{\rule{0.25em}{0ex}}µ\text{C}[/latex] are placed 0.250 m apart. (a) Where can a third charge be placed so that the net force on it is zero? (b) What if both charges are positive?
  17. Two point charges [latex]{q}_{\text{1}}[/latex] and [latex]{q}_{\text{2}}[/latex] are [latex]3.00 m[/latex] apart, and their total charge is [latex]20\phantom{\rule{0.25em}{0ex}}µ\text{C}[/latex]. (a) If the force of repulsion between them is 0.075N, what are magnitudes of the two charges? (b) If one charge attracts the other with a force of 0.525N, what are the magnitudes of the two charges? Note that you may need to solve a quadratic equation to reach your answer.

Glossary

Coulomb’s law
the mathematical equation calculating the electrostatic force vector between two charged particles
Coulomb force
another term for the electrostatic force
electrostatic force
the amount and direction of attraction or repulsion between two charged bodies

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Introductory Physics for the Health and Life Sciences II Copyright © 2012 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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