# Learning Objectives

### Learning Objectives

By the end of this section, you will be able to do the following:

- Describe a force field and calculate the strength of an electric field due to a point charge
- Calculate the force exerted on a test charge by an electric field
- Explain the relationship between electrical force (
*F*) on a test charge and electrical field strength (*E*)

The information presented in this section supports the following AP® learning objectives and science practices:

**2.C.1.1**The student is able to predict the direction and the magnitude of the force exerted on an object with an electric charge*q*placed in an electric field*E*using the mathematical model of the relation between an electric force and an electric field: $\overrightarrow{F}=q\overrightarrow{E}$, a vector relation.**(S.P. 2.2)****2.C.1.2**The student is able to calculate any one of the variables—electric force, electric charge, and electric field—at a point given the values and sign or direction of the other two quantities.**(S.P. 2.2)****2.C.2.1**The student is able to qualitatively and semiquantitatively apply the vector relationship between the electric field and the net electric charge creating that field.**(S.P. 2.2, 6.4)****3.C.4.1**The student is able to make claims about various contact forces between objects based on the microscopic cause of those forces.**(S.P. 6.1)****3.C.4.2**The student is able to explain contact forces (tension, friction, normal, buoyant, spring) as arising from interatomic electric forces and that they therefore have certain directions.**(S.P. 6.2)**

Contact forces, such as between a baseball and a bat, are explained on the small scale by the interaction of the charges in atoms and molecules in close proximity. They interact through forces that include the Coulomb force. Action at a distance is a force between objects that are not close enough for their atoms to *touch.* That is, they are separated by more than a few atomic diameters.

For example, a charged rubber comb attracts neutral bits of paper from a distance via the Coulomb force. It is very useful to think of an object being surrounded in space by a force field. The force field carries the force to another object—called a test object—some distance away.

# Concept of a Field

### Concept of a Field

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.

In the same way, the Coulomb force field surrounding any charge extends throughout space. Using Coulomb's law, $F=k\mathrm{|}{q}_{1}{q}_{2}\mathrm{|}/{r}^{2}$, its magnitude is given by the equation
$F=k\mathrm{|}\mathrm{qQ}\mathrm{|}/{r}^{2}$, for a point charge, a particle having a charge *$Q$*, acting on a test charge $q$ at a distance $r$ (see Figure 1.20). Both the magnitude and direction of the Coulomb force field depend on *$Q$* and the test charge $q$.

To simplify things, we would prefer to have a field that depends only on *$Q$* and not on the test charge $q\text{.}$ The electric field is defined in such a manner that it represents only the charge creating it and is unique at every point in space. Specifically, the electric field $E$ is defined to be the ratio of the Coulomb force to the test charge

where $\mathbf{\text{F}}$ is the electrostatic force—or Coulomb force—exerted on a positive test charge $q$. It is understood that $\mathbf{\text{E}}$ is in the same direction as $\mathbf{\text{F}}\text{.}$ It is also assumed that $q$ is so small that it does not alter the charge distribution creating the electric field. The units of electric field are newtons per coulomb (N/C). If the electric field is known, then the electrostatic force on any charge $q$ is simply obtained by multiplying charge times electric field, or $\mathbf{\text{F}}=q\mathbf{\text{E}}\text{.}$ Consider the electric field due to a point charge $Q\text{.}$ According to Coulomb's law, the force it exerts on a test charge $q$ is $F=k\mathrm{|}\mathrm{qQ}\mathrm{|}/{r}^{2}\text{.}$ Thus the magnitude of the electric field, $E$, for a point charge is

Since the test charge cancels, we see that

The electric field is thus seen to depend only on the charge *$Q$* and the distance $r\text{;}$ it is completely independent of the test charge $q\text{.}$

### Example 1.2 Calculating the Electric Field of a Point Charge

Calculate the strength and direction of the electric field $E$ due to a point charge of 2.00 nC (nano-Coulombs) at a distance of 5.00 mm from the charge.

**Strategy**

We can find the electric field created by a point charge by using the equation $E=\text{kQ}/{r}^{2}\text{.}$

**Solution**

Here $Q=2\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\times \phantom{\rule{0.25em}{0ex}}{\text{10}}^{-9}$ C and $r=5\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\times \phantom{\rule{0.25em}{0ex}}{\text{10}}^{-3}$ m. Entering those values into the above equation gives

**Discussion**

This electric field strength is the same at any point 5.00 mm away from the charge *$Q$* that creates the field. It is positive, meaning that it has a direction pointing away from the charge *$Q\text{.}$*

### Example 1.3 Calculating the Force Exerted on a Point Charge by an Electric Field

What force does the electric field found in the previous example exert on a point charge of $\u20130.250\text{\hspace{0.22em}}\text{\mu C}$?

**Strategy**

Since we know the electric field strength and the charge in the field, the force on that charge can be calculated using the definition of electric field $\mathbf{\text{E}}=\mathbf{\text{F}}/q$ rearranged to $\mathbf{\text{F}}=q\mathbf{\text{E}}$.

**Solution**

The magnitude of the force on a charge $q=-0\text{.}\text{250}\phantom{\rule{0.25em}{0ex}}\text{\mu C}$ exerted by a field of strength $E=7\text{.}\text{20}\phantom{\rule{0.25em}{0ex}}\times \phantom{\rule{0.25em}{0ex}}{\text{10}}^{5}$ N/C is thus,

Because $q$ is negative, the force is directed opposite to the direction of the field.

**Discussion**

The force is attractive, as expected for unlike charges. The field was created by a positive charge and here acts on a negative charge. The charges in this example are typical of common static electricity, and the modest attractive force obtained is similar to forces experienced in static cling and similar situations.

### PhET Explorations: Electric Field of Dreams

Play ball! Add charges to the Field of Dreams and see how they react to the electric field. Turn on a background electric field and adjust the direction and magnitude.