Question 4.31: A point charge Q is “nailed down” on a table. Around it, at ...

In cylindrical coordinates (in the z = 0 plane), p =p \hat{\phi}, p \cdot \nabla=p \frac{1}{s} \frac{\partial}{\partial \phi}, \text { and } E =\frac{1}{4 \pi \epsilon_{0}} \frac{Q}{s^{2}} \hat{ s } , so 

F =( p \cdot \nabla) E =\left(\frac{p}{s} \frac{\partial}{\partial \phi}\right) \frac{1}{4 \pi \epsilon_{0}} \frac{Q}{s^{2}} \hat{ s }=\frac{p Q}{4 \pi \epsilon_{0} s^{3}} \frac{\partial \hat{ s }}{\partial \phi}=\frac{p Q}{4 \pi \epsilon_{0} s^{3}} \hat{\phi}=\frac{Q}{4 \pi \epsilon_{0} R^{3}} p .

Qualitatively, the forces on the negative and positive ends, though equal in magnitude, point in slightly di↵erent directions, and they combine to make a net force in the “forward” direction:

To keep the dipole going in a circle, there must be a centripetal force exerted by the track (we may as well take it to act at the center of the dipole, and it is irrelevant to the problem), and to keep it aiming in the tangential direction there must be a torque (which we could model by radial forces of equal magnitude acting at the two ends). Indeed, if the dipole has the orientation indicated in the figure, and is moving in the \hat{\phi} direction, the torque exerted by Q is clockwise, whereas the rotation is counterclockwise, so these constraint forces mustactually be larger than the forces exerted by Q, and the net force will be in the “backward” direction—tending to slow the dipole down. [If the motion is in the -\hat{\phi} direction, then the electrical forces will dominate, and the net force will be in the direction of p, but this again will tend to slow it down.]