18.4 Electric Field: Concept of a Field Revisited

Learning Objectives

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

• 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).

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 the 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 18.21). 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 $q$. 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:

$$\mathbf{\text{E}}=\frac{\mathbf{\text{F}}}{q},$$

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 $\mathbf{\text{F}}$. 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 $\mathbf{\text{F}}=q\mathbf{\text{E}}$. Consider the electric field due to a point charge $Q$. According to Coulomb’s law, the force it exerts on a test charge $q$ is $F=k\mathrm{|}\mathrm{qQ}\mathrm{|}/r^2$. Thus the magnitude of the electric field, $E$, for a point charge is

$$E=\left|\frac{F}{q}\right|=k|\frac{\text{qQ}}{{\mathrm{qr}}^2}|=k\frac{\left|Q\right|}{r^2}.$$

Since the test charge cancels, we see that

$$E=k\frac{\left|Q\right|}{r^2}.$$

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

Test Prep for AP Courses

Section Summary

• The electrostatic force field surrounding a charged object extends out into space in all directions.
• The electrostatic force exerted by a point charge on a test charge at a distance $r$ depends on the charge of both charges, as well as the distance between the two.
• The electric field $\mathbf{\text{E}}$ is defined to be

  $$\mathbf{\text{E}}=\frac{\mathbf{\text{F}}}{q,}$$

  where $\mathbf{\text{F}}$ is the Coulomb or electrostatic force exerted on a small positive test charge $q$. $\mathbf{\text{E}}$ has units of N/C.

• The magnitude of the electric field $\mathbf{\text{E}}$ created by a point charge $Q$ is

  $$\mathbf{\text{E}}=k\frac{\left|Q\right|}{r^2}.$$

  where $r$ is the distance from $Q$. The electric field $\mathbf{\text{E}}$ is a vector and fields due to multiple charges add like vectors.

Conceptual Questions

Problem Exercises