23.7.2 Dipoles in a Nonuniform Field

Suppose that a dipole is placed in a nonuniform electric field, one in which the field strength changes with position. For example, FIGURE 23.30 shows a dipole in the nonuniform field of a point charge. The first response of the dipole is to rotate until it is aligned with the field, with the dipole’s positive end pointing in the same direction as the field. Now, however, there is a slight difference between the forces acting on the two ends of the dipole. This difference occurs because the electric field, which depends on the distance from the point charge, is stronger at the end of the dipole nearest the charge. This causes a net force to be exerted on the dipole.

The figure illustrates a dipole in the electric field of a positive point charge.

FIGURE 23.30 Full Alternative Text

Which way does the force point? Once the dipole is aligned, the leftward attractive force on its negative end is slightly stronger than the rightward repulsive force on its positive end. This causes a net force toward the point charge.

In fact, for any nonuniform electric field, the net force on a dipole is toward the direction of the strongest field. Because any finite-size charged object, such as a charged rod or a charged disk, has a field strength that increases as you get closer to the object, we can conclude that a dipole will experience a net force toward any charged object.

EXAMPLE 23.10 The force on a water molecule

The water molecule $H_{2} O$ has a permanent dipole moment of magnitude $6.2 \times 10^{- 30} C m$ . A water molecule is located 10 nm from a ${Na}^{+}$ ion in a saltwater solution. What force does the ion exert on the water molecule?

VISUALIZE FIGURE 23.31 shows the ion and the dipole. The forces are an action/reaction pair.

The figure illustrates interaction between a positive ion and a dipole.

FIGURE 23.31 Full Alternative Text

SOLVE A ${Na}^{+}$ ion has charge $q = + e$ . The electric field of the ion aligns the water’s dipole moment and exerts a net force on it. We could calculate the net force on the dipole as the small difference between the attractive force on its negative end and the repulsive force on its positive end. Alternatively, we know from Newton’s third law that the force $\vec{F}_{dipole on ion}$ has the same magnitude as the force $\vec{F}_{ion on dipole}$ that we are seeking. We calculated the on-axis field of a dipole in Section 23.2. An ion of charge $q = e$ will experience a force of magnitude $F = q E_{dipole} = e E_{dipole}$ when placed in that field. The dipole’s electric field, which we found in Equation 23.10, is

E_{dipole} = \frac{1}{4 π ϵ_{0}} \frac{2 p}{r^{3}}

The force on the ion at distance $r = 1.0 \times 10^{- 8} m$ is

F_{dipole on ion} = e E_{dipole} = \frac{1}{4 π ϵ_{0}} \frac{2 e p}{r^{3}} = 1.8 \times 10^{- 14} N

Thus the force on the water molecule is $F_{ion on dipole} = 1.8 \times 10^{- 14} N$ .

REVIEW While $1.8 \times 10^{- 14} N$ may seem like a very small force, it is $\approx 10^{11}$ times larger than the size of the earth’s gravitational force on these atomic particles. Forces such as these cause water molecules to cluster around any ions that are in solution. This clustering plays an important role in the microscopic physics of solutions studied in chemistry and biochemistry.

CHAPTER 23 CHALLENGE EXAMPLE An orbiting proton

In a vacuum chamber, a proton orbits a 1.0-cm-diameter metal ball 1.0 mm above the surface with a period of $1.0 μ s$ . What is the charge on the ball?

MODEL Model the ball as a charged sphere. The electric field of a charged sphere is the same as that of a point charge at the center, so the radius of the ball is irrelevant. Assume that the gravitational force on the proton is extremely small compared to the electric force and can be neglected.

VISUALIZE FIGURE 23.32 shows the orbit and the force on the proton.

The figure illustrates a proton of charge e orbiting a metal ball.

FIGURE 23.32 Full Alternative Text

SOLVE The ball must be negative, with an inward electric field exerting an inward electric force on the positive proton. This is exactly the necessary condition for uniform circular motion. Recall from Chapter 8 that Newton’s second law for uniform circular motion is $(F_{net})_{r} = m v^{2} / r$ . Here the only radial force has magnitude $F_{elec} = e E$ , so the proton will move in a circular orbit if

e E = \frac{m v^{2}}{r}

The electric field strength of a sphere of charge Q at distance r is $E = Q / 4 π ϵ_{0} r^{2}$ . From Chapter 4, orbital speed and period are related by $v = circumference / period = 2 π r / T$ . With these substitutions, Newton’s second law becomes

\frac{e Q}{4 π ϵ_{0} r^{2}} = \frac{4 π^{2} m}{T^{2}} r

Solving for Q, we find

Q = \frac{16 π^{3} ϵ_{0} m r^{3}}{e T^{2}} = 9.9 \times 10^{- 12} C

where we used $r = 6.0 mm$ as the radius of the proton’s orbit. Q is the magnitude of the charge on the ball. Including the sign, we have

Q_{ball} = - 9.9 \times 10^{- 12} C

REVIEW This is not a lot of charge, but it shouldn’t take much charge to affect the motion of something as light as a proton.

23.7.2 Dipoles in a Nonuniform Field

FIGURE 23.30 An aligned dipole is drawn toward a point charge.

EXAMPLE 23.10 The force on a water molecule

FIGURE 23.31 The interaction between an ion and a permanent dipole.

CHAPTER 23 CHALLENGE EXAMPLE An orbiting proton

FIGURE 23.32 An orbiting proton.