18.6

The Electric Field

| Definition

As we know, a charge can experience an electrostatic force due to the presence of other charges. For instance, the positive charge q0 in Figure 18-15 experiences a force , which is the vector sum of the forces exerted by the charges on the rod and the two spheres. It is useful to think of q0 as a test charge for determining the extent to which the surrounding charges generate a force. However, in using a test charge, we must be careful to select one with a very small magnitude, so that it does not alter the locations of the other charges. The next example illustrates how the concept of a test charge is applied.

Figure 18-15 A positive charge q0 experiences an electrostatic force due to the surrounding charges on the ebonite rod and the two spheres.

| Example 6 | | | A Test Charge

The positive test charge shown in Figure 18-15 is and experiences a force in the direction shown in the drawing. (a) Find the force per coulomb that the test charge experiences. (b) Using the result of part (a), predict the force that a charge of would experience if it replaced q0.
Reasoning The charges in the environment apply a force to the test charge q0. The force per coulomb experienced by the test charge is . If q0 is replaced by a new charge q, then the force on this new charge is the force per coulomb times q.
Solution

a. | The force per coulomb of charge is

b. | The result from part (a) indicates that the surrounding charges can exert 2.0 newtons of force per coulomb of charge. Thus, a charge of would experience a force whose magnitude is

The direction of this force would be the same as that experienced by the test charge, since both have the same positive sign.

The electric force per coulomb, , calculated in Example 6(a) is one illustration of an idea that is very important in the study of electricity. The idea is called the electric field. Equation 18.2 presents the definition of the electric field.

| Definition of the Electric Field
| The electric field that exists at a point is the electrostatic force experienced by a small test charge* q0 placed at that point divided by the charge itself:

| (18.2)

The electric field is a vector, and its direction is the same as the direction of the force on a positive test charge.
SI Unit of Electric Field: newton per coulomb (N/C)

Equation 18.2 indicates that the unit for the electric field is that of force divided by charge, which is a newton/coulomb (N/C) in SI units.
It is the surrounding charges that create an electric field at a given point. Any positive or negative charge placed at the point interacts with the field and, as a result, experiences a force, as the next example indicates.

| Example 7 | | | An Electric Field Leads to a Force

In Figure 18-16a the charges on the two metal spheres and the ebonite rod create an electric field at the spot indicated. This field has a magnitude of 2.0 N/C and is directed as in the drawing. Determine the force on a charge placed at that spot, if the charge has a value of (a) and (b) .

Figure 18-16 The electric field that exists at a given spot can exert a variety of forces. The force exerted depends on the magnitude and sign of the charge placed at that spot. (a) The force on a positive charge points in the same direction as while (b) the force on a negative charge points opposite to .

Reasoning The electric field at a given spot can exert a variety of forces, depending on the magnitude and sign of the charge placed there. The charge is assumed to be small enough that it does not alter the locations of the surrounding charges that create the field.
Solution

a. | The magnitude of the force is the product of the magnitudes of q0 and :

| (18.2)

Since q0 is positive, the force points in the same direction as the electric field, as part a of the drawing indicates.
b. | In this case, the magnitude of the force is

| (18.2)

The force on the negative charge points in the direction opposite to the force on the positive charge—that is, opposite to the electric field (see part b of the drawing).

At a particular point in space, each of the surrounding charges contributes to the net electric field that exists there. To determine the net field, it is necessary to obtain the various contributions separately and then find the vector sum of them all. Such an approach is an illustration of the principle of linear superposition, as applied to electric fields. (This principle is introduced in Section 17.1, in connection with waves.) Example 8 emphasizes the vector nature of the electric field, and Example 9 illustrates that a charged particle accelerates in an electric field.

| Example 8 | | | Electric Fields Add as Vectors Do

Figure 18-17 shows two charged objects, A and B. Each contributes as follows to the net electric field at point P: directed to the right, and directed downward. Thus, and are perpendicular. What is the net electric field at P?

Figure 18-17 The electric field contributions and , which come from the two charge distributions, are added vectorially to obtain the net field at point P.

Reasoning The net electric field is the vector sum of and : . As illustrated in Figure 18-17, and are perpendicular, so is the diagonal of the rectangle shown in the drawing. Thus, we can use the Pythagorean theorem to find the magnitude of and trigonometry to find the directional angle .
Solution The magnitude of the net electric field is

The direction of is given by the angle in the drawing:

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 9 | | | A Proton Accelerating in an Electric Field

In a vacuum, a proton is moving parallel to a uniform electric field that is directed along the + x axis (see Figure 18-18). The proton starts with a velocity of and accelerates in the same direction as the electric field, which has a value of . Find the velocity of the proton when its displacement is + 2.0 mm from the starting point.

Figure 18-18 A proton, moving to the right, accelerates in the presence of the electric field .

Reasoning Since we know the initial velocity and displacement of the proton, we can determine its final velocity from an equation of kinematics, provided the proton’s acceleration can be found. The acceleration is given by Newton’s second law as the net force acting on the proton divided by its mass. The net force is the electrostatic force, since the proton is moving in an electric field. The electrostatic force depends on the proton’s charge and the electric field, both of which are known.
Knowns and Unknowns The data for this problem are listed as follows:

Description | Symbol | Value | Comment
Proton charge | + e |  |
Proton mass | m |  |
Initial velocity of proton | v0x |  |
Electric field | Ex |  |
Displacement of proton | x | + 2.0 mm |
Unknown Variable |  |  |
Final velocity of proton | vx | ? |

Modeling the Problem

| Step 1 Kinematics
| To obtain the final velocity vx of the proton we employ Equation 3.6a from the equations of kinematics: . We have chosen this equation because two of the variables, the initial velocity v0x and the displacement x, are known. Taking the square root of each side of this relation and choosing the + sign, since the proton is moving in the + x direction (see Figure 18-18), we arrive at Equation 1 in the right column. Although the acceleration ax is not known, we will obtain an expression for it in Step 2.

| (1)

| Step 2 Newton’s Second Law
| Newton’s second law, as given in Equation 4.2a, states that the acceleration ax of the proton is equal to the net force acting on it divided by the proton’s mass . Only the electrostatic force Fx acts on the proton, so it is the net force. Setting in Newton’s second law gives

This expression can be substituted into Equation 1, as indicated at the right. The electrostatic force is not known, so we proceed to Step 3 to evaluate it using the concept of the electric field.

| Step 3 The Electric Field
| Since the proton is moving in a uniform electric field Ex, it experiences an electrostatic force Fx given by (Equation 18.2), where q0 is the charge. Setting q0 = e for the proton, we have

All the variables on the right side of this equation are known, so we substitute it into Equation 2, as shown in the right column.

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

The final velocity of the proton is

where the + sign denotes that the final velocity points along the + x axis.
Related Homework: Problems 38, 40 and 66

| Point Charges

A more complete understanding of the electric field concept can be gained by considering the field created by a point charge, as in the following example.

| Example 10 | | | The Electric Field of a Point Charge

There is an isolated point charge of in a vacuum at the left in Figure 18-19a. Using a test charge of , determine the electric field at point P, which is 0.20 m away.

Figure 18-19 (a) At location P, a positive test charge q0 experiences a repulsive force due to the positive point charge q. (b) At P the electric field is directed to the right. (c) If the charge q were negative rather than positive, the electric field would have the same magnitude as in (b) but would point to the left.

Reasoning Following the definition of the electric field, we place the test charge q0 at point P, determine the force acting on the test charge, and then divide the force by the test charge.
Solution Coulomb’s law (Equation 18.1), gives the magnitude of the force:

Equation 18.2 gives the magnitude of the electric field:

The electric field points in the same direction as the force on the positive test charge. Since the test charge experiences a force of repulsion directed to the right, the electric field vector also points to the right, as Figure 18-19b shows.

The electric field produced by a point charge q can be obtained in general terms from Coulomb’s law. First, note that the magnitude of the force exerted by the charge q on a test charge q0 is . Then, divide this value by to obtain the magnitude of the field. Since is eliminated algebraically from the result, the electric field does not depend on the test charge:

| (18.3)

As in Coulomb’s law, the symbol |q| denotes the magnitude of q in Equation 18.3, without regard to whether q is positive or negative. If q is positive, then is directed away from q, as in Figure 18-19b. On the other hand, if q is negative, then is directed toward q, since a negative charge attracts a positive test charge. For instance, Figure 18-19c shows the electric field that would exist at P if there were a charge of − q instead of + q at the left of the drawing. Example 11 reemphasizes the fact that all the surrounding charges make a contribution to the electric field that exists at a given place.

| Example 11 | | | The Electric Fields from Separate Charges May Cancel

Two positive point charges, and , are separated in a vacuum by a distance of 3.0 m, as Figure 18-20 illustrates. Find the spot on the line between the charges where the net electric field is zero.

Figure 18-20 The two point charges q1 and q2 create electric fields and that cancel at a location P on the line between the charges.

Reasoning Between the charges the two field contributions have opposite directions, and the net electric field is zero at the place where the magnitude of equals that of . However, since q2 is smaller than q1 this location must be closer to q2, in order that the field of the smaller charge can balance the field of the larger charge. In the drawing, the cancellation spot is labeled P, and its distance from q1 is d.
Solution At P, , and using the expression , we have

Rearranging this expression shows that . Taking the square root of each side of this equation reveals that

The plus and minus signs on the right occur because either the positive or negative root can be taken. Therefore, there are two possible values for d: + 2.0 m and + 6.0 m. The value + 6.0 m corresponds to a location off to the right of both charges, where the magnitudes of and are equal, but where the directions are the same. Thus, and do not cancel at this spot. The other value for d corresponds to the location shown in the drawing and is the zero-field location: .

Problem solving insight Equation 18.3 gives only the magnitude of the electric field produced by a point charge. Therefore, do not use negative numbers for the symbol |q| in this equation.

When point charges are arranged in a symmetrical fashion, it is often possible to deduce useful information about the magnitude and direction of the electric field by taking advantage of the symmetry. Conceptual Example 12 illustrates the use of this technique.

| Conceptual Example 12 | | | Symmetry and the Electric Field

Figure 18-21 shows point charges fixed to the corners of a rectangle in two different ways. The charges all have the same magnitudes, but they have different signs. Consider the net electric field at the center C of the rectangle in each case. Which field is larger?

Figure 18-21 Charges of identical magnitude, but different signs, are placed at the corners of a rectangle. The charges give rise to different electric fields at the center C of the rectangle, depending on the signs of the charges.

Reasoning and Solution In Figure 18-21a, the charges at corners 1 and 3 are both + q. The positive charge at corner 1 produces an electric field at C that points toward corner 3. In contrast, the positive charge at corner 3 produces an electric field at C that points toward corner 1. Thus, the two fields have opposite directions. The magnitudes of the fields are identical because the charges have the same magnitude and are equally far from the center. Therefore, the fields from the two positive charges cancel.
Now, let’s look at the electric field produced by the charges on corners 2 and 4 in Figure 18-21a. The electric field due to the negative charge at corner 2 points toward corner 2, and the field due to the positive charge at corner 4 points the same way. Furthermore, the magnitudes of these fields are equal because each charge has the same magnitude and is located at the same distance from the center of the rectangle. As a result, the two fields combine to give the net electric field shown in the drawing.
In Figure 18-21b, the charges on corners 2 and 4 are identical to those in part a of the drawing, so this pair gives rise to the electric field labeled . The charges on corners 1 and 3 are identical to those on corners 2 and 4, so they give rise to an electric field labeled , which has the same magnitude as . The net electric field is the vector sum of and and is also shown in the drawing. Clearly, this sum is greater than alone. Therefore, the net field in part b is larger than that in part a.
Related Homework: Conceptual Question 12, Problem 34

Need more practice?

Interactive LearningWare18.3
A constant electric field exists in a region of space. The field has a magnitude of 1600 N/C and points due north. A point charge of is then placed in this electric field. (a) What is the magnitude and direction of the net electric field at a spot 13 cm due east of the charge? Specify the direction as an angle relative to due east. (b) If a charge were placed at this spot, what would be the electrostatic force (magnitude and direction) exerted on it?
Related Homework: Problem 59

| The Parallel Plate Capacitor

Equation 18.3, which gives the electric field of a point charge, is a very useful result. With the aid of integral calculus, this equation can be applied in a variety of situations where point charges are distributed over one or more surfaces. One such example that has considerable practical importance is the parallel plate capacitor. As Figure 18-22 shows, this device consists of two parallel metal plates, each with area A. A charge + q is spread uniformly over one plate, while a charge − q is spread uniformly over the other plate. In the region between the plates and away from the edges, the electric field points from the positive plate toward the negative plate and is perpendicular to both. It can be shown (see Example 16 in Section 18.9) that the electric field has a magnitude of

| (18.4)

where is the permittivity of free space. In this expression the Greek symbol sigma () denotes the charge per unit area () and is sometimes called the charge density. Except in the region near the edges, the field has the same value at all places between the plates. The field does not depend on the distance from the charges, in distinct contrast to the field created by an isolated point charge.

Figure 18-22 A parallel plate capacitor.

Check Your Understanding 3

A positive point charge + q is fixed in position at the center of a square, as the drawing shows. A second point charge is fixed to either corner B, corner C, or corner D. The net electric field at corner A is zero. (a) At which corner is the second charge located? (b) Is the second charge positive or negative? (c) Does the second charge have a greater, a smaller, or the same magnitude as the charge at the center? (The answers are given at the end of the book.)

Background: The electric field is a vector quantity. Its magnitude and direction depend on the magnitudes and algebraic signs of the charges that create it.
For similar questions (including calculational counterparts), consult Self-Assessment Test 18.2, which is described at the end of Section 18.9.

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