Equation 22.7 is precise, but it’s not terribly convenient. Furthermore, what happens if the source charge $q$ is negative? We need a more concise notation to write the electric field, a notation that will allow $q$ to be either positive or negative.

The basic need is to express “away from $q$” in mathematical notation. “Away from $q$” is a direction in space. To guide us, recall that we already have a notation for expressing certain directions—namely, the unit vectors $î\text{,}ĵ$, and $\widehat{k}$. For example, unit vector $î$ means “in the direction of the positive $x$-axis.” With a minus sign, $-î$ means “in the direction of the negative $x$-axis.” Unit vectors, with a magnitude of 1 and no units, provide purely directional information.

With this in mind, let’s define the unit vector $\widehat{r}$ to be a vector of length 1 that points from the origin to a point of interest. Unit vector $\widehat{r}$ provides no information at all about the distance to the point; it’s simply a pointer that specifies the direction.

FIGURE 22.26 shows unit vectors $\widehat{r}{}_1\text{,}\widehat{r}{}_2$, and $\widehat{r}{}_3$ pointing toward points 1, 2, and 3. Unlike $î$ and $ĵ$, unit vector $\widehat{r}$ does not have a fixed direction. Instead, unit vector $\widehat{r}$ specifies the direction “straight outward from this point.” But that’s just what we need to describe the electric field vector, which is shown at points 1, 2, and 3 due to a positive charge at the origin. No matter which point you choose, the electric field at that point is “straight outward” from the charge. In other words, the electric field $\vec{E}$ points in the direction of the unit vector $\widehat{r}$.

FIGURE 22.26 Using the unit vector $\widehat{r}$.

With this notation, the electric field at distance $r$ from a point charge $q$ is

where $\widehat{r}$ is the unit vector from the charge toward the point at which we want to know the field. Equation 22.8 is identical to Equation 22.7, but written in a notation in which the unit vector $\widehat{r}$ expresses the idea “away from $q$.”

Equation 22.8 works equally well if $q$ is negative. A negative sign in front of a vector simply reverses its direction, so the unit vector $-\widehat{r}$ points toward charge $q$. FIGURE 22.27 shows the electric field of a negative point charge. It looks like the electric field of a positive point charge except that the vectors point inward, toward the charge, instead of outward.

FIGURE 22.27 The electric field of a negative point charge.

We’ll end this chapter with three examples of the electric field of a point charge. Chapter 23 will expand these ideas to the electric fields of multiple charges and of extended objects.

EXAMPLE 22.7 Calculating the electric field

A $-1.0\text{nC}$ charged particle is located at the origin. Points 1, 2, and 3 have $(x\text{,}y)$ coordinates $(1\text{cm,}0\text{cm})\text{,}(0\text{cm}\text{,}1\text{cm})$, and $(1\text{cm,}1\text{cm})$, respectively. Determine the electric field $\vec{E}$ at these points, then show the vectors on an electric field diagram.

MODEL The electric field is that of a negative point charge.

VISUALIZE The electric field points straight toward the origin. It will be weaker at (1 cm, 1 cm), which is farther from the charge.

SOLVE The electric field is

$$\vec{E}={\displaystyle \frac{1}{4\pi {ϵ}_0}}{\displaystyle \frac{q}{r^2}}\widehat{r}$$

where $q=-1.0\text{nC}=-1.0\times {10}^{-9}\text{C}$. The distance $r_1=r_2$ is $1.0\text{cm}=0.010\text{m}$ for points 1 and 2; thus the magnitude of $\vec{E}$ at these two points is

$$\begin{array}{ccc}\hfill E_1& \hfill =\hfill & E_2={\displaystyle \frac{1}{4\pi {ϵ}_0}}{\displaystyle \frac{\left|q\right|}{r_1{}^2}}\hfill \\ \hfill & \hfill =\hfill & {\displaystyle \frac{(9.0\times {10}^9\text{N}{\text{m}}^2\text{/}{\text{C}}^2)(1.0\times {10}^{-9}\text{C})}{(0.010{\text{m)}}^2}}=90\text{,}000\text{N}\text{/}\text{C}\hfill \end{array}$$

The unit vector pointing toward point 1, on the $x$-axis, is $\widehat{r}{}_1=î$. Point 2 is on the $y$-axis, so the unit vector pointing toward point 2 is $\widehat{r}{}_2=ĵ$. Thus the electric fields at the first two points are

$$\begin{array}{lll}{\vec{E}}_1\hfill & =\hfill & -E_1{\widehat{r}}_1=-90\text{,}000î\text{ N/C}\hfill \\ {\vec{E}}_2\hfill & =\hfill & -E_2{\widehat{r}}_2=-90\text{,}000\widehat{j}\text{ N/C}\hfill \end{array}$$

where the minus sign comes from the sign of the charge.

Point 3 is at distance $r_3=\sqrt{2}\times 1\text{cm}=0.0141\text{m,}$ so the field has magnitude

$$\begin{array}{lll}{E}_3\hfill & =\hfill & \frac{1}{4\pi {ϵ}_0}\frac{\left|q\right|}{{{\displaystyle r}}_3^2}\hfill \\ \hfill & =\hfill & \frac{(9.0\times {10}^9{\text{ N m}}^2{\text{/C}}^2)(1.0\times {10}^{-9}\text{ C)}}{{(0.0141\text{ m)}}^2}=45\text{,}000\text{ N/C}\hfill \end{array}$$

The unit vector pointing toward point 3 is at a 45° angle, so

$$\widehat{r}{}_3=\text{cos}45\text{°}î+\text{sin}45\text{°}ĵ=0.707î+0.707ĵ$$

Thus

$$\vec{E}{}_3=-E_3\widehat{r}{}_3=(-31\text{,}800î-31\text{,}800ĵ)\text{N}\text{/}\text{C}$$

These vectors are shown on the electric field diagram of FIGURE 22.28.

FIGURE 22.28 The electric field diagram of a $-1.0\text{nC}$ charged particle.

FIGURE 22.28 Full Alternative Text

CHAPTER 22 CHALLENGE EXAMPLE A charge in static equilibrium

A horizontal electric field causes the charged ball in FIGURE 22.29 to hang at a $15\text{°}$ angle, as shown. The spring is plastic, so it doesn’t discharge the ball, and in its equilibrium position the spring extends only to the vertical dashed line. What is the electric field strength?

FIGURE 22.29 A charged ball hanging in static equilibrium.

FIGURE 22.29 Full Alternative Text

MODEL Model the ball as a point charge in static equilibrium. The electric force on the ball is $\vec{F}{}_{\text{E}}=q\vec{E}$. The charge is positive, so the force is in the same direction as the field.

VISUALIZE FIGURE 22.30 is a free-body diagram for the ball.

FIGURE 22.30 The free-body diagram.

FIGURE 22.30 Full Alternative Text

SOLVE The ball is in equilibrium, so the net force on the ball must be zero. With the field applied, the spring is stretched by $\text{Δ}x=L\text{sin}\theta =\left(0.60\text{m}\right)\left(\text{sin}15\text{°}\right)=0.16\text{m}$, where $L$ is the string length, and exerts a pulling force $F_{\text{Sp}}=k\text{Δ}x$ to the left.

Newton’s second law, with $\vec{a}=\vec{0}$ for equilibrium, is

$$\begin{array}{lll}\sum F_x\hfill & =\hfill & F_{\text{E}}-F_{\text{Sp}}-T\text{sin }\theta =0\hfill \\ \sum F_y\hfill & =\hfill & T\text{cos }\theta -F_{\text{G}}=T\text{cos }\theta -mg=0\hfill \end{array}$$

From the $y$-equation,

$$T={\displaystyle \frac{mg}{\text{cos }\theta }}$$

The $x$-equation is then

$$qE-k\text{Δ}x-mg\text{tan}\theta =0$$

We can solve this for the electric field strength:

$$\begin{array}{ccc}\hfill E& \hfill =\hfill & {\displaystyle \frac{mg\text{tan}\theta +k\text{Δ}x}{q}}\hfill \\ \hfill & \hfill =\hfill & {\displaystyle \frac{(0.0030\text{kg})\left(9.8\text{m}\text{/}{\text{s}}^2\right)\text{tan}15\text{°}+\left(0.050\text{N}\text{/}\text{m}\right)\left(0.16\text{m}\right)}{20\times {10}^{-9}\text{C}}}\hfill \\ \hfill & \hfill =\hfill & 7.9\times {10}^5\text{N}\text{/}\text{C}\hfill \end{array}$$

REVIEW We don’t yet have a way of judging whether this is a reasonable field strength, but we’ll see in the next chapter that this is typical of the electric field strength near an object that has been charged by rubbing.