Dipolar Forces

April 2025

Forces due to dipoles (such as magnetic forces) drop off quicker than forces due to monopoles. Indeed, it’s easy to show that they drop off as \(r^{-3}\).

Consider two equal and opposite charges which individually cause an inverse-square force. Let the charges be located a distance \(w\) from each other, and wlog let’s say the midpoint of the two is the origin and the y-axis is aligned along the line connecting the two charges.

The derivation is easy enough if we consider the forces directly but it’s slightly algebraically simpler to consider the potential, then convert this back to a force. Like the forces, the potentials of the two charges just add.

Let’s consider a point \(\mathbf r\). In polar coordinates \(\mathbf r =(r,\theta)\). The distances from \(\mathbf r\) to the two charges will differ slightly depending on the angle \(\theta\). If \(r >\!\!> w\) then the lines from the two charges to \(\mathbf r\) are approximately parallel, and both make an angle \(\theta\) with the y-axis. This means the difference in path-length to \(\mathbf r\) will be \(\Delta r = w\sin\theta\). This is shown in the diagram below:

Dipole Geometry
Geometry of the dipole.

By superposition, the total potential is given by

$$\begin{align} V(\mathbf r) &= V_{+}\big(r - \tfrac{\Delta r}{2}\big) + V_{-}\big(r + \tfrac{\Delta r}{2}\big) \\ &= \frac{+q}{r - \tfrac{\Delta r}{2}} + \frac{-q}{r + \tfrac{\Delta r}{2}} \end{align}$$

where \(q\) has swallowed any constant of proportionality of the inverse square law (e.g. \(1/4\pi\varepsilon_0\) for Coulomb’s law  external link .) Now, combine the two fractions by multiplying each by the denominator of the other, then simplify

$$\begin{align} \frac{V(\mathbf r)}{q} &= \frac{(r + \Delta r/2) - (r - \Delta r/2)}{(r - \Delta r/2)(r + \Delta r/2)} \\ &= \frac{\Delta r}{r^2 - \Delta r^2/4} \\ &= \frac{w\sin \theta}{r^2 - \frac{w^2}{4} \sin^2\theta} \end{align}$$

Since \(r \gg w\) the denominator is approximately \(r^2\) and we get

$$V(\mathbf r) = -\frac{qw\sin\theta}{r^2}$$

We can see the potential drops off faster than for an inverse-square force (for which the potential goes as \(r^{-1}\).) The force is given by minus the gradient of the potential. In polar coordinates this is

$$\begin{align} \mathbf F(\mathbf r) &= -\nabla V = -(\partial_r V, \frac{1}{r}\partial_\theta V) \\ &= -qw \Big(\frac{2\sin\theta}{r^3}, -\frac{\cos\theta}{r^3}\Big) \end{align}$$

which indeed drops off in strength as \(r^{-3}\).

We considered the case where \( 0 < w \ll r.\) If we let \(w\) approach zero but increase \(q\) so that the product \(qw\) remains finite, we get a “point dipole” and our result is exact.

Plotting the potential (scalar field) & force (vector field) gives the following:

Dipole Force Field
Potential & force field due a point dipole.

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