Section Learning Objectives
By the end of this section, you will be able to do the following:
 Calculate the strength of an electric field
 Create and interpret drawings of electric fields
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Teacher Support
The learning objectives in this section will help your students master the following standards:
 (5) The student knows the nature of forces in the physical world. The student is expected to:
 (C) describe and calculate how the magnitude of the electrical force between two objects depends on their charges and the distance between them.
Section Key Terms
electric field  test charge 
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Teacher Support
Ask students whether they have seen movies that use the concept of fields as in force fields. Have them describe how such fields work. Describe how gravity can be thought of as a field that surrounds a mass and with which other masses interact. Explain that electric fields are very similar to gravitational fields.
You may have heard of a force field in science fiction movies, where such fields apply forces at particular positions in space to keep a villain trapped or to protect a spaceship from enemy fire. The concept of a field is very useful in physics, although it differs somewhat from what you see in movies.
A field is a way of conceptualizing and mapping the force that surrounds any object and acts on another object at a distance without apparent physical connection. For example, the gravitational field surrounding Earth and all other masses represents the gravitational force that would be experienced if another mass were placed at a given point within the field. Michael Faraday, an English physicist of the nineteenth century, proposed the concept of an electric field. If you know the electric field, then you can easily calculate the force (magnitude and direction) applied to any electric charge that you place in the field.
An electric field is generated by electric charge and tells us the force per unit charge at all locations in space around a charge distribution. The charge distribution could be a single point charge; a distribution of charge over, say, a flat plate; or a more complex distribution of charge. The electric field extends into space around the charge distribution. Now consider placing a test charge in the field. A test charge is a positive electric charge whose charge is so small that it does not significantly disturb the charges that create the electric field. The electric field exerts a force on the test charge in a given direction. The force exerted is proportional to the charge of the test charge. For example, if we double the charge of the test charge, the force exerted on it doubles. Mathematically, saying that electric field is the force per unit charge is written as
where we are considering only electric forces. Note that the electric field is a vector field that points in the same direction as the force on the positive test charge. The units of electric field are N/C.
If the electric field is created by a point charge or a sphere of uniform charge, then the magnitude of the force between this point charge Q and the test charge is given by Coulomb’s law
$$F=\frac{k\leftQ{q}_{\text{test}}\right}{{r}^{2}}$$
where the absolute value is used, because we only consider the magnitude of the force. The magnitude of the electric field is then
This equation gives the magnitude of the electric field created by a point charge Q. The distance r in the denominator is the distance from the point charge, Q, or from the center of a spherical charge, to the point of interest.
If the test charge is removed from the electric field, the electric field still exists. To create a threedimensional map of the electric field, imagine placing the test charge in various locations in the field. At each location, measure the force on the charge, and use the vector equation $\overrightarrow{E}=\overrightarrow{F}/{q}_{\text{test}}$ to calculate the electric field. Draw an arrow at each point where you place the test charge to represent the strength and the direction of the electric field. The length of the arrows should be proportional to the strength of the electric field. If you join together these arrows, you obtain lines. Figure 18.17 shows an image of the threedimensional electric field created by a positive charge.
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Teacher Support
[BL][OL]Point out that all electric field lines originate from the charge.
[AL]Point out that the number of lines crossing an imaginary sphere surrounding the charge is the same no matter what size sphere you choose. Ask whether students can use this to show that the number of field lines crossing a surface per unit area shows that the electric field strength decreases as the inverse square of the distance.
Just drawing the electric field lines in a plane that slices through the charge gives the twodimensional electricfield maps shown in Figure 18.18. On the left is the electric field created by a positive charge, and on the right is the electric field created by a negative charge.
Notice that the electric field lines point away from the positive charge and toward the negative charge. Thus, a positive test charge placed in the electric field of the positive charge will be repelled. This is consistent with Coulomb’s law, which says that like charges repel each other. If we place the positive charge in the electric field of the negative charge, the positive charge is attracted to the negative charge. The opposite is true for negative test charges. Thus, the direction of the electric field lines is consistent with what we find by using Coulomb’s law.
The equation $E=k\leftQ\right/{r}^{2}$ says that the electric field gets stronger as we approach the charge that generates it. For example, at 2 cm from the charge Q (r = 2 cm), the electric field is four times stronger than at 4 cm from the charge (r = 4 cm). Looking at Figure 18.17 and Figure 18.18 again, we see that the electric field lines become denser as we approach the charge that generates it. In fact, the density of the electric field lines is proportional to the strength of the electric field!
Electricfield maps can be made for several charges or for more complicated charge distributions. The electric field due to multiple charges may be found by adding together the electric field from each individual charge. Because this sum can only be a single number, we know that only a single electricfield line can go through any given point. In other words, electricfield lines cannot cross each other.
Figure 18.19(a) shows a twodimensional map of the electric field generated by a charge of +q and a nearby charge of −q. The threedimensional version of this map is obtained by rotating this map about the axis that goes through both charges. A positive test charge placed in this field would experience a force in the direction of the field lines at its location. It would thus be repelled from the positive charge and attracted to the negative charge. Figure 18.19(b) shows the electric field generated by two charges of −q. Note how the field lines tend to repel each other and do not overlap. A positive test charge placed in this field would be attracted to both charges. If you are far from these two charges, where far means much farther than the distance between the charges, the electric field looks like the electric field from a single charge of −2q.
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Teacher Support
Ask students to interpret the electric field maps. Where is the field strongest? Where is the field weakest? In which direction is the field increasing or decreasing? Where is the field the most uniform? Can they verify that the magnitude of the charges is the same in a given panel? How does the field for the two negative charges differ from that of the positive and negative charges?
Virtual Physics
Probing an Electric Field
This simulation shows you the electric field due to charges that you place on the screen. Start by clicking the top checkbox in the options panel on the righthand side to show the electric field. Drag charges from the buckets onto the screen, move them around, and observe the electric field that they form. To see more precisely the magnitude and direction of the electric field, drag an electricfield sensor, or Efield sensor from the bottom bucket, and move it around the screen.

It is constant everywhere.

It is zero near each charge.

It is zero halfway between the charges.

It is strongest halfway between the charges.
Watch Physics
Electrostatics (part 2): Interpreting electric field
This video explains how to calculate the electric field of a point charge and how to interpret electricfield maps in general. Note that the lecturer uses d for the distance between particles instead of r. Note that the point charges are infinitesimally small, so all their charges are focused at a point. When larger charged objects are considered, the distance between the objects must be measured between the center of the objects.
Grasp Check
True or false—If a point charge has electric field lines that point into it, the charge must be positive.
 true
 false
Worked Example
What is the charge?
Look at the drawing of the electric field in Figure 18.20. What is the relative strength and sign of the three charges?
Strategy
We know the electric field extends out from positive charge and terminates on negative charge. We also know that the number of electric field lines that touch a charge is proportional to the charge. Charge 1 has 12 fields coming out of it. Charge 2 has six field lines going into it. Charge 3 has 12 field lines going into it.
The electricfield lines come out of charge 1, so it is a positive charge. The electricfield lines go into charges 2 and 3, so they are negative charges. The ratio of the charges is ${q}_{1}:{q}_{2}:{q}_{3}=+12:6:12$ . Thus, magnitude of charges 1 and 3 is twice that of charge 2.
Although we cannot determine the precise charge on each particle, we can get a lot of information from the electric field regarding the magnitude and sign of the charges and where the force on a test charge would be greatest (or least).
Worked Example
Electric field from doorknob
A doorknob, which can be taken to be a spherical metal conductor, acquires a static electricity charge of $q=\mathrm{1.5}\phantom{\rule{0.3em}{0ex}}\text{nC.}$ What is the electric field 1.0 cm in front of the doorknob? The diameter of the doorknob is 5.0 cm.
Strategy
Because the doorknob is a conductor, the entire charge is distributed on the outside surface of the metal. In addition, because the doorknob is assumed to be perfectly spherical, the charge on the surface is uniformly distributed, so we can treat the doorknob as if all the charge were located at the center of the doorknob. The validity of this simplification will be proved in a later physics course. Now sketch the doorknob, and define your coordinate system. Use $+x$ to indicate the outward direction perpendicular to the door, with $x=0$ at the center of the doorknob (as shown in the figure below).
If the diameter of the doorknob is 5.0 cm, its radius is 2.5 cm. We want to know the electric field 1.0 cm from the surface of the doorknob, which is a distance $r=2.5\phantom{\rule{0.3em}{0ex}}\text{cm}+1.0\phantom{\rule{0.3em}{0ex}}\text{cm}=3.5\phantom{\rule{0.3em}{0ex}}\text{cm}$ from the center of the doorknob. We can use the equation $E=\frac{k\leftQ\right}{{r}^{2}}$ to find the magnitude of the electric field. The direction of the electric field is determined by the sign of the charge, which is negative in this case.
Inserting the charge $Q=\mathrm{1.5}\phantom{\rule{0.3em}{0ex}}\text{nC}=\mathrm{1.5}\phantom{\rule{0.25em}{0ex}}\times \phantom{\rule{0.25em}{0ex}}{10}^{\mathrm{9}}\text{C}$ and the distance $r=3.5\phantom{\rule{0.3em}{0ex}}\text{cm}=0.035\phantom{\rule{0.3em}{0ex}}\text{m}$ into the equation $E=\frac{k\leftQ\right}{{r}^{2}}$ gives
Because the charge is negative, the electricfield lines point toward the center of the doorknob. Thus, the electric field at $x=3.5\phantom{\rule{0.3em}{0ex}}\text{cm}$ is $(1.1\phantom{\rule{0.25em}{0ex}}\times \phantom{\rule{0.25em}{0ex}}{10}^{4}\text{N/C})\widehat{x}$ .
This seems like an enormous electric field. Luckily, it takes an electric field roughly 100 times stronger ( $3\phantom{\rule{0.25em}{0ex}}\times \phantom{\rule{0.25em}{0ex}}{10}^{6}\text{N/C}$ ) to cause air to break down and conduct electricity. Also, the weight of an adult is about $70\phantom{\rule{0.3em}{0ex}}\text{kg}\phantom{\rule{0.25em}{0ex}}\times \phantom{\rule{0.25em}{0ex}}9.8{\text{m/s}}^{2}\approx 700\phantom{\rule{0.3em}{0ex}}\text{N}$, so why don’t you feel a force on the protons in your hand as you reach for the doorknob? The reason is that your hand contains an equal amount of negative charge, which repels the negative charge in the doorknob. A very small force might develop from polarization in your hand, but you would never notice it.
Practice Problems
What is the magnitude of the electric field from 20 cm from a point charge of q = 33 nC?
 7.4 × 10^{3} N/C
 1.48 × 10^{3} N / C
 7.4 × 10^{12} N / C
 0
A −10 nC charge is at the origin. In which direction does the electric field from the charge point at x + 10 cm ?
 The electric field points away from negative charges.
 The electric field points toward negative charges.
 The electric field points toward positive charges.
 The electric field points away from positive charges.
Check Your Understanding
When electric field lines get closer together, what does that tell you about the electric field?
 The electric field is inversely proportional to the density of electric field lines.
 The electric field is directly proportional to the density of electric field lines.
 The electric field is not related to the density of electric field lines.
 The electric field is inversely proportional to the square root of density of electric field lines.
If five electricfield lines come out of a +5 nC charge, how many electricfield lines should come out of a +20 nC charge?
 five field lines
 10 field lines
 15 field lines
 20 field lines