18.5

Coulomb’s Law

| The Force that Point Charges Exert on each Other

The electrostatic force that stationary charged objects exert on each other depends on the amount of charge on the objects and the distance between them. Experiments reveal that the greater the charge and the closer together they are, the greater is the force. To set the stage for explaining these features in more detail, Figure 18-10 shows two charged bodies. These objects are so small, compared to the distance r between them, that they can be regarded as mathematical points. The “point charges” have magnitudes* and . If the charges have unlike signs, as in part a of the picture, each object is attracted to the other by a force that is directed along the line between them; is the electric force exerted on object 1 by object 2 and is the electric force exerted on object 2 by object 1. If, as in part b, the charges have the same sign (both positive or both negative), each object is repelled from the other. The repulsive forces, like the attractive forces, act along the line between the charges. Whether attractive or repulsive, the two forces are equal in magnitude but opposite in direction. These forces always exist as a pair, each one acting on a different object, in accord with Newton’s action–reaction law.

Figure 18-10 Each point charge exerts a force on the other. Regardless of whether the forces are (a) attractive or (b) repulsive, they are directed along the line between the charges and have equal magnitudes.

The French physicist Charles Augustin de Coulomb (1736–1806) carried out a number of experiments to determine how the electric force that one point charge applies to another depends on the amount of each charge and the separation between them. His result, now known as Coulomb’s law, is stated below.

| Coulomb’s Law
| The magnitude F of the electrostatic force exerted by one point charge q1 on another point charge q2 is directly proportional to the magnitudes and of the charges and inversely proportional to the square of the distance r between them:

| (18.1)

where k is a proportionality constant: in SI units. The electrostatic force is directed along the line joining the charges, and it is attractive if the charges have unlike signs and repulsive if the charges have like signs.

It is common practice to express k in terms of another constant by writing is called the permittivity of free space and has a value of . Equation 18.1 gives the magnitude of the electrostatic force that each point charge exerts on the other. When using this equation, then, it is important to remember to substitute only the charge magnitudes (without algebraic signs) for and , as Example 2 illustrates.

| Example 2 | | | A Large Attractive Force

Two objects, whose charges are + 1.0 and −1.0 C, are separated by 1.0 km. Compared to 1.0 km, the sizes of the objects are small. Find the magnitude of the attractive force that either charge exerts on the other. Reasoning Considering that the sizes of the objects are small compared to the separation distance, we can treat the charges as point charges. Coulomb’s law may then be used to find the magnitude of the attractive force, provided that only the magnitudes of the charges are used for the symbols and that appear in the law.
Solution The magnitude of the force is

| (18.1)

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Interactive LearningWare18.1
An object of mass 215 kg is located just above the surface of the earth (). Suppose that the object and the earth have an identical charge q. Therefore, two forces–the gravitational force and the electrostatic force— act on the object. Assuming that the earth’s charge is located at the center of the earth, determine q (magnitude and sign) when the object is in equilibrium.
Related Homework: Problem 61

The force calculated in Example 2 corresponds to about 2000 pounds and is so large because charges of ± 1.0 C are enormous. Such large charges are encountered only in the most severe conditions, as in a lightning bolt, where as much as 25 C can be transferred between the cloud and the ground. The typical charges produced in the laboratory are much smaller and are measured conveniently in microcoulombs .
Coulomb’s law has a form that is remarkably similar to Newton’s law of gravitation . The force in both laws depends on the inverse square (1/r2) of the distance between the two objects and is directed along the line between them. In addition, the force is proportional to the product of an intrinsic property of each of the objects, the magnitudes of the charges and in Coulomb’s law and the masses m1 and m2 in the gravitation law. But there is a major difference between the two laws. The electrostatic force can be either repulsive or attractive, depending on whether or not the charges have the same sign; in contrast, the gravitational force is always an attractive force.
Section 5.5 discusses how the gravitational attraction between the earth and a satellite provides the centripetal force that keeps a satellite in orbit. Example 3 illustrates that the electrostatic force of attraction plays a similar role in a famous model of the atom created by the Danish physicist Niels Bohr (1885 – 1962).

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 3 | | | A Model of the Hydrogen Atom

In the Bohr model of the hydrogen atom, the electron (charge = −e) is in a circular orbit about the nuclear proton (charge = + e) at a radius of , as Figure 18-11 shows. The mass of the electron is . Determine the speed of the electron.

Figure 18-11 In the Bohr model of the hydrogen atom, the electron (− e) orbits the proton (+ e) at a distance of r = 5.29 × 10−11 m. The velocity of the electron is .

Reasoning Recall from Section 5.3 that a net force is required to keep an object such as an electron moving on a circular path. This net force is called the centripetal force and always points toward the center of the circle. It has a magnitude given by , where m and v are, respectively, the mass and speed of the electron and r is the radius of the orbit. This equation can be solved for the speed of the electron. Since the mass and orbital radius are known, we can calculate the electron’s speed provided that a value for the centripetal force can be found. For the electron in the hydrogen atom, the centripetal force is provided almost exclusively by the electrostatic force that the proton exerts on the electron. This attractive force points toward the center of the circle and its magnitude is given by Coulomb’s law. The electron is also pulled toward the proton by the gravitational force. However, the gravitational force is negligible in comparison to the electrostatic force.
Knowns and Unknowns The data for this problem are:

Description | Symbol | Value
Electron charge | −e |
Electron mass | m |
Proton charge | + e |
Radius of orbit | r |
Unknown Variable |  |
Orbital speed of electron | v | ?

Modeling the Problem

| Step 1 Centripetal Force
| An electron of mass m that moves with a constant speed v on a circular path of radius r experiences a net force, called the centripetal force. The magnitude Fc of this force is given by (Equation 5.3). By solving this equation for the speed, we obtain Equation 1 at the right. The mass and radius in this expression are known. However, the magnitude of the centripetal force is not known, so we will evaluate it in Step 2.

| (1)

| Step 2 Coulomb’s Law
| As the electron orbits the proton in the hydrogen atom, it is attracted to the proton by the electrostatic force. The magnitude F of the electrostatic force is given by Coulomb’s law as (Equation 18.1), where and are the magnitudes of the charges, r is the orbital radius, and . Since the centripetal force is provided almost entirely by the electrostatic force, it follows that . Furthermore, and . With these substitutions, Equation 18.1 becomes

All the variables on the right side of this expression are known, so we substitute it into Equation 1, as indicated in the right column.

Solution Algebraically combining the results of the modeling steps, we have

The speed of the orbiting electron is

Related Homework: Problems 16 and 20

Since the electrostatic force depends on the inverse square of the distance between the charges, it becomes larger for smaller distances, such as those involved when a strip of adhesive tape is stuck to a smooth surface. Electrons shift over the small distances between the tape and the surface. As a result, the materials become oppositely charged. Since the distance between the charges is relatively small, the electrostatic force of attraction is large enough to contribute to the adhesive bond. Figure 18-12 shows an image of the sticky surface of a piece of tape after it has been pulled off a metal surface. The image was obtained using an atomic-force microscope and reveals the tiny pits left behind when microscopic portions of the adhesive remain stuck to the metal because of the strong adhesive bonding forces.

Figure 18-12 After a strip of adhesive tape has been pulled off a metal surface, there are tiny pits (approximately one ten-millionth of a meter in diameter) in the stick surface of the tape, as this image shows. It was obtained using an atomic-force microscope. (Courtesy Louis Scudiero and J. Thomas Dickinson, Washington State University.)

The physics of adhesion.

| The Force on a Point Charge Due to Two or More Other Point Charges

Up to now, we have been discussing the electrostatic force on a point charge (magnitude ) due to another point charge (magnitude ). Suppose that a third point charge (magnitude ) is also present. What would be the net force on q1 due to both q2 and q3?. It is convenient to deal with such a problem in parts. First, find the magnitude and direction of the force exerted on q1 by q2 (ignoring q3). Then, determine the force exerted on q1 by q3 (ignoring q2). The net force on q1 is the vector sum of these forces. Examples 4 and 5 illustrate this approach when the charges lie along a straight line and on a plane, respectively.

| Example 4 | | | Three Charges on a Line

Figure 18-13a shows three point charges that lie along the x axis in a vacuum. Determine the magnitude and direction of the net electrostatic force on q1.

Figure 18-13 (a) Three charges lying along the x axis. (b) The force exerted on q1 by q2 is , while the force exerted on q1 by q3 is .

Reasoning Part b of the drawing shows a free-body diagram of the forces that act on q1. Since q1 and q2 have opposite signs, they attract one another. Thus, the force exerted on q1 by q2 is , and it points to the left. Similarly, the force exerted on q1 by q3 is and is also an attractive force. It points to the right in Figure 18-13b. The magnitudes of these forces can be obtained from Coulomb’s law. The net force is the vector sum of and .
Solution The magnitudes of the forces are

Since points in the negative x direction, and points in the positive x direction, the net force is

The plus sign in the answer indicates that the net force points to the right in the drawing.

Problem solving insight When using Coulomb’s law , remember that the symbols and stand for the charge magnitudes. Do not substitute negative numbers for these symbols.

| Example 5 | | | Three Charges in a Plane

Figure 18-14a shows three point charges that lie in the x, y plane in a vacuum. Find the magnitude and direction of the net electrostatic force on q1.

Figure 18-14 (a) Three charges lying in a plane. (b) The net force acting on q1 is = + . The angle that makes with the + x axis is .

Reasoning The force exerted on q1 by q2 is and is an attractive force because the two charges have opposite signs. It points along the line between the charges. The force exerted on q1 by q3 is and is also an attractive force. It points along the line between q1 and q3. Coulomb’s law specifies the magnitudes of these forces. Since the forces point in different directions (see Figure 18-14b), we will use vector components to find the net force.

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Interactive LearningWare18.2
A positive charge is placed on the x axis , and another positive charge is placed on the y axis . A third charge is placed at the coordinate origin. If the charge at the origin were free to move, what would be the magnitude of its acceleration? Related Homework: Problem 17

Solution The magnitudes of the forces are

The net force is the vector sum of and , as part b of the drawing shows. The components of that lie in the x and y directions are and , respectively. Our approach to finding is the same as that used in Chapters 1 and 4. The forces and are resolved into x and y components. Then, the x components are combined to give , and the y components are combined to give . Once and are known, the magnitude and direction of can be determined using trigonometry.

Force | x component | y component
| + 18 N | 0 N

The magnitude F and the angle of the net force are

Problem solving insight The electrostatic force is a vector and has a direction as well as a magnitude. When adding electrostatic forces, take into account the directions of all forces, using vector components as needed.

Check Your Understanding 2

The drawing shows three point charges arranged in different ways. The charges are + q, − q, and − q each has the same magnitude, one being positive and the other two negative. In each part of the drawing the distance d is the same. Rank the arrangements in descending order (largest first) according to the magnitude of the net electrostatic force that acts on the positive charge. (The answer is given at the end of the book.)

Background: The electrostatic force that one point charge exerts on another is described by Coulomb’s law. Like any force, this force is a vector quantity, and the rules of vector addition apply.
For similar questions (including calculational counterparts), consult Self-Assessment Test 18.1, which is described next.

Test your understanding of the material in Sections 18.1, 18.2, 18.3, 18.4 and 18.5 :

· | The Origin of Electricity
· | Charged Objects and the Electric Force
· | Conductors and Insulators
· | Charging by Contact and by Induction
· | Coulomb’s Law

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