23.2.3 The Electric Field of a Dipole

Two equal but opposite charges separated by a small distance form an electric dipole. FIGURE 23.4 shows two examples. In a permanent electric dipole, such as the water molecule, the oppositely charged particles maintain a small permanent separation. We can also create an electric dipole, as you learned in 《 SECTION 22.3, by polarizing a neutral atom with an external electric field. This is an induced electric dipole.

The figure contains two parts that illustrate the permanent and induced electric dipoles.

FIGURE 23.4 Full Alternative Text

FIGURE 23.5 shows that we can represent an electric dipole, whether permanent or induced, by two opposite charges $\pm q$ separated by the small distance s. The dipole has zero net charge, but it does have an electric field. Consider a point on the positive y-axis. This point is slightly closer to $+ q$ than to $- q$ , so the fields of the two charges do not cancel. You can see in the figure that $\vec{E}_{dipole}$ points in the positive y-direction. Similarly, vector addition shows that $\vec{E}_{dipole}$ points in the negative y-direction at points along the x-axis.

The figure illustrates the electric field caused by a dipole at two different points on a coordinate plane.

FIGURE 23.5 Full Alternative Text

Let’s calculate the electric field of a dipole at a point on the axis of the dipole. This is the y-axis in Figure 23.5. The point is distance $r_{+} = y - s / 2$ from the positive charge and $r_{-} = y + s / 2$ from the negative charge. The net electric field at this point has only a y-component, and the sum of the fields of the two point charges gives

\begin{array}{l} {(E_{dipole})}_{y} = {(E_{+})}_{y} + {(E_{-})}_{y} & = \frac{1}{4 π ϵ_{0}} \frac{q}{{(y - \frac{1}{2} s)}^{2}} + \frac{1}{4 π ϵ_{0}} \frac{(- q)}{{(y + \frac{1}{2} s)}^{2}} \\ = \frac{q}{4 π ϵ_{0}} [\frac{1}{{(y - \frac{1}{2} s)}^{2}} - \frac{1}{{(y + \frac{1}{2} s)}^{2}}] \end{array}

(23.6)

If we combine the two terms over a common denominator (and omit some of the algebraic steps), we find

(E_{dipole})_{y} = \frac{q}{4 π ϵ_{0}} [\frac{2 y s}{{(y - \frac{1}{2} s)}^{2} {(y + \frac{1}{2} s)}^{2}}]

(23.7)

In practice, we almost always observe the electric field of a dipole at distances $y ≫ s$ —that is, for distances much larger than the charge separation. In such cases, the denominator can be approximated ${(y - \frac{1}{2} s)}^{2} {(y + \frac{1}{2} s)}^{2} \approx y^{4}$ . With this approximation, Equation 23.7 becomes

(E_{dipole})_{y} \approx \frac{1}{4 π ϵ_{0}} \frac{2 q s}{y^{3}}

(23.8)

Because we assumed $y ≫ s$ , Equation 23.8 is called the far field of the dipole.

It is useful to define the dipole moment $\vec{p}$ , shown in FIGURE 23.6, as the vector

\vec{p} = (q s, from the negative to the positive charge)

(23.9)

The direction of $\vec{p}$ identifies the orientation of the dipole, and the dipole-moment magnitude $p = q s$ determines the electric field strength. The SI units of the dipole moment are $C m$ .

The figure illustrates the dipole moment between two opposite charges.

FIGURE 23.6 Full Alternative Text

We can use the dipole moment to write a succinct expression for the far field at a point on the axis of a dipole:

\vec{E}_{dipole} = - \frac{1}{4 π ϵ_{0}} \frac{2 \vec{p}}{r^{3}} (far field on the axis)

(23.10)

where r is the distance measured from the center of the dipole. We’ve switched from y to r because we’ve now specified that Equation 23.10 is valid only along the axis of the dipole. Notice that the electric field along the axis points in the direction of the dipole moment $\vec{p}$ .

A homework problem will let you calculate the electric field in the plane that bisects the dipole. This is the field shown on the x-axis in Figure 23.5, but it could equally well be the field on the z-axis as it comes out of the page. The field, for $r ≫ s$ , is

\vec{E}_{dipole} = - \frac{1}{4 π ϵ_{0}} \frac{\vec{p}}{r^{3}} (far field in the bisecting plane)

(23.11)

This field is opposite to $\vec{p}$ , and it is only half the strength of the on-axis field at the same distance.

NOTE Do these inverse-cube equations violate Coulomb’s law? Not at all. Coulomb’s law describes the force between two point charges, and from Coulomb’s law we found that the electric field of a point charge varies with the inverse square of the distance. But a dipole is not a point charge. The field of a dipole decreases more rapidly than that of a point charge, which is to be expected because the dipole is, after all, electrically neutral.

EXAMPLE 23.2 The electric field of a water molecule

The water molecule $H_{2} O$ has a permanent dipole moment of magnitude $6.2 \times 10^{- 30} C m$ . What is the electric field strength 1.0 nm from a water molecule at a point on the dipole’s axis?

MODEL The size of a molecule is $\approx 0.1 nm$ . Thus $r ≫ s$ , and we can use Equation 23.10 for the on-axis electric field of the molecule’s dipole moment.

SOLVE The on-axis electric field strength at $r = 1.0 nm$ is

\begin{matrix} E \approx \frac{1}{4 π ϵ_{0}} \frac{2 p}{r^{3}} & = & (9.0 \times 10^{9} N m^{2} / C^{2}) \frac{2 (6.2 \times 10^{- 30} C m)}{{(1.0 \times 10^{- 9} m)}^{3}} \\ = & 1.1 \times 10^{8} N / C \end{matrix}

REVIEW By referring to Table 23.1 you can see that the field strength is “strong” compared to our everyday experience with charged objects but “weak” compared to the electric field inside the atoms themselves. This seems reasonable.

23.2.3 The Electric Field of a Dipole

FIGURE 23.4 Permanent and induced electric dipoles.

FIGURE 23.5 The dipole electric field at two points.

FIGURE 23.6 The dipole moment.

EXAMPLE 23.2 The electric field of a water molecule