FIGURE 23.26a shows an electric dipole in a uniform external electric field $\vec{E}$ that has been created by source charges that are not shown. That is, $\vec{E}$ is not the field of the dipole but, instead, is a field to which the dipole is responding. In this case, because the field is uniform, the dipole is presumably inside an unseen parallel-plate capacitor.

FIGURE 23.26 A dipole in a uniform electric field.

Full Alternative Text

The net force on the dipole is the sum of the forces on the two charges forming the dipole. Because the charges $\pm q$ are equal in magnitude but opposite in sign, the two forces $\vec{F}{}_{+}=+q\vec{E}$ and $\vec{F}{}_{-}=-q\vec{E}$ are also equal but opposite. Thus the net force on the dipole is

There is no net force on a dipole in a uniform electric field.

There may be no net force, but the electric field does affect the dipole. Because the two forces in Figure 23.26a are in opposite directions but not aligned with each other, the electric field exerts a torque on the dipole and causes the dipole to rotate.

The torque rotates the dipole until it is aligned with the electric field, as shown in FIGURE 23.26b. In this position, the dipole experiences not only no net force but also no torque. Thus Figure 23.26b represents the equilibrium position for a dipole in a uniform electric field. Notice that the positive end of the dipole is in the direction in which $\vec{E}$ points.

FIGURE 23.27 shows a sample of permanent dipoles, such as water molecules, in an external electric field. All the dipoles rotate until they are aligned with the electric field. This is the mechanism by which the sample becomes polarized. Once the dipoles are aligned, there is an excess of positive charge at one end of the sample and an excess of negative charge at the other end. The excess charges at the ends of the sample are the basis of the polarization forces we discussed in Section 22.3.

FIGURE 23.27 A sample of permanent dipoles is polarized in an electric field.

It’s not hard to calculate the torque. Recall from Chapter 12 that the magnitude of a torque is the product of the force and the moment arm. FIGURE 23.28 shows that there are two forces of the same magnitude $(F_{+}=F_{-}=qE)$, each with the same moment arm $(d=\frac{1}{2}s\text{sin}\theta )$. Thus the torque on the dipole is

where $p=qs$ was our definition of the dipole moment. The torque is zero when the dipole is aligned with the field, making $\theta =0$.

FIGURE 23.28 The torque on a dipole.

Also recall from Chapter 12 that the torque can be written in a compact mathematical form as the cross product between two vectors. The terms p and E in Equation 23.35 are the magnitudes of vectors, and $\theta$ is the angle between them. Thus in vector notation, the torque exerted on a dipole moment $\vec{p}$ by an electric field $\vec{E}$ is

The torque is greatest when $\vec{p}$ is perpendicular to $\vec{E}$, zero when $\vec{p}$ is aligned with or opposite to $\vec{E}$.

EXAMPLE 23.9 The angular acceleration of a dipole dumbbell

Two 1.0 g balls are connected by a 2.0-cm-long insulating rod of negligible mass. One ball has a charge of $+10\text{nC,}$ the other a charge of $-10\text{nC}$. The rod is held in a $1.0\times {10}^4\text{N}\text{/}\text{C}$ uniform electric field at an angle of $30\text{°}$ with respect to the field, then released. What is its initial angular acceleration?

MODEL The two oppositely charged balls form an electric dipole. The electric field exerts a torque on the dipole, causing an angular acceleration.

VISUALIZE FIGURE 23.29 shows the dipole in the electric field.

FIGURE 23.29 The dipole of Example 23.9.

FIGURE 23.29 Full Alternative Text

SOLVE The dipole moment is $p=qs=(1.0\times {10}^{-8}\text{C})\times (0\text{.020 m})=2.0\times {10}^{-10}\text{C}\text{ m}$. The torque exerted on the dipole moment by the electric field is

$$\begin{array}{ccc}\hfill \tau & \hfill =\hfill & pE\text{sin}\theta =(2.0\times {10}^{-10}\text{C}\text{ m})(1.0\times {10}^4\text{N}\text{/}\text{C})\text{s}\text{i}\text{n}30\text{°}\hfill \\ \hfill & \hfill =\hfill & 1.0\times {10}^{-6}\text{N}\text{ m}\hfill \end{array}$$

You learned in Chapter 12 that a torque causes an angular acceleration $\alpha =\tau \text{/}I$, where I is the moment of inertia. The dipole rotates about its center of mass, which is at the center of the rod, so the moment of inertia is

$$I=m_1{r}_1{}^2+{\text{m}}_2{r}_2{}^2=2m{\left(\frac{1}{2}s\right)}^2=\frac{1}{2}m{s}^2=2.0\times {10}^{-7}{\text{kg m}}^2$$

Thus the rod’s angular acceleration is

$$\alpha ={\displaystyle \frac{\tau }{I}}={\displaystyle \frac{1.0\times {10}^{-6}\text{N}\text{ m}}{2.0\times {10}^{-7}\text{kg}{\text{m}}^2}}=5.0\text{rad}\text{/}{\text{s}}^2$$

REVIEW This value of $\alpha$ is the initial angular acceleration, when the rod is first released. The torque and the angular acceleration will decrease as the rod rotates toward alignment with $\vec{E}$.