Assume that we have some non-constant electric field $E(x,t)$ and a point-dipole at a position $q$ with a constant dipole moment $\vec{p}$. How would you describe the time evolution, i.e. the motion of such a dipole?

  • $\begingroup$ When you say constant dipole moment $vec{p}$, do you mean the orientation is also fixed, or just the magnitude? $\endgroup$ – lionelbrits Dec 7 '13 at 0:37

When an electric dipole is placed in a uniform electric field making an angle with the direction of the field as shown in the figure.

enter image description here

Force on charge $-q=-q\overrightarrow{E}$ (opposite to $\overrightarrow{E}$)

Force on charge $+q=q\overrightarrow{E}$ (along $\overrightarrow{E}$)

Thus, electric dipole is under the action of two equal and unlike parallel forces, which give rise to a torque on the dipole.

So, in a uniform electric field, an electric dipole experiences only a torque. But, when the electric field is non-uniform as you say, it experiences torque as well as net force.


The energy of a dipole in an electric field is just the sum of the energies of it's constituents, and is easily seen to be $U = -\mathbf{p} \cdot \mathbf{E}$. From this you can derive the dynamics of the dipole through the generalized forces $-\boldsymbol\nabla_\mathbf{p} U$ and $-\boldsymbol\nabla_\mathbf{x} U$.

This assumes that the dipole's constituents are much closer together than the length scale over which the field varies.

From the above you can see that the dipole will tend to orient itself along the electric field. Furthermore, if the field is not uniform in space, the dipole will move in the direction of increasing/decreasing field strength (depending on the instantaneous orientation).

  • 2
    $\begingroup$ It sounds like in the statement of the problem the electric field may vary in time which would generate some magnetic field. Wouldn't this add some forces that will affect the dipole dynamics? $\endgroup$ – Maxim Umansky Dec 6 '13 at 4:59
  • $\begingroup$ There should be no Lorentz force acting on the dipole COM because it is neutral, and similarly it has no magnetic dipole moment. In either case, we make the assumption that the fields vary over distances much larger than the dipole itself. Otherwise there will be a Lorentz force propotional to the gradient of $\mathbf{B}$. But in that case the electric field term would also be inaccurate. $\endgroup$ – lionelbrits Dec 6 '13 at 16:35
  • $\begingroup$ @MaximUmansky, on second reading, in Thomas' dipole, the charges are infinite and infinitesimally separated, so I shouldn't be so quick to say that the magnetic dipole moment vanishes. $\endgroup$ – lionelbrits Dec 7 '13 at 0:45
  • $\begingroup$ Yes, I think the same way as there is a finite electric force on a point (infinitesimal) dipole ~pgrad(E), there should be a finite net Lorentz force on an electric dipole moving with speed V, something like ~Vp*grad(B). $\endgroup$ – Maxim Umansky Dec 7 '13 at 0:58
  • $\begingroup$ Apparently a moving electric dipole aquires a magnetic dipole moment $\mathbf{m} = - \mathbf{v} \times \mathbf{p}$ (to lowest order in v/c) (D. J. Griffiths, Introduction to Electrodynamics, 3rd ed. (Prentice Hall, Upper Saddle River, 1999), Problem 12.62.) $\endgroup$ – lionelbrits Dec 7 '13 at 1:37

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