# Electric Dipole Moment

Imagine a blob of charged jelly.

The jelly can be pictured as a continuous distribution of point charges; to work out the electrostatic potential field the jelly generates, we add up the contribution of each point charge.

You may know that the potential $d\varphi$ at position $\bold{r}$ generated by a point charge $dq$ at position $\bold{r}$‘ is, according to Coulomb’s law,

$\displaystyle d\varphi(\bold{r})=\frac{dq(\bold{r}')}{4\pi\epsilon_0|\bold{r}-\bold{r}'|}$

For a continuous charge distribution, each point charge $dq$ is equivalent to a small element of volume $d$3$\bold{r}$‘ multiplied by the charge density $\rho(\bold{r}')$ at that point:

$dq\equiv\rho(\bold{r}')\,d^3\bold{r}'$

To find the field established by the entirety of the jelly, we add up the contribution from each part of it, which for a continuous distribution amounts to doing an integral:

$\displaystyle \varphi(\bold{r})=\frac{1}{4\pi\epsilon_0}\int\frac{\rho(\bold{r}')}{|\bold{r}-\bold{r}'|}\,d^3\bold{r}'$

Evaluating this integral is about as tough as it looks. The fact that we have to integrate over three spatial variables is troublesome, but the real culprit is that denominator. Let’s take a closer look at it. By definition,

$|\bold{r}-\bold{r}'|=\sqrt{(\bold{r}-\bold{r}')\cdot(\bold{r}-\bold{r}')}$

We expand the brackets:

$|\bold{r}-\bold{r}'|=\sqrt{r^2+r'^2-2rr'\cos\theta}$

Here, we’ve made use of the fact that $\bold{r}\cdot\bold{r}$‘ = $rr$$\cos$ $\theta$, where $\theta$ is the angle made between $\bold{r}$ and $\bold{r}$‘. We take out a factor of $r$2 from the root:

$\displaystyle |\bold{r}-\bold{r}'|=r\sqrt{1+\left(\frac{r'}{r}\right)^2-2\left(\frac{r'}{r}\right)\cos\theta}$

$\displaystyle \frac{1}{|\bold{r}-\bold{r}'|}=\frac{1}{r}\left(1+\left(\frac{r'}{r}\right)^2-2\left(\frac{r'}{r}\right)\cos\theta\right)^{-\frac{1}{2}}$

Now suppose $r$$r$‘ ∀ $r$‘. That is, the position at which we’re evaluating the potential field is displaced by a huge distance from the jelly. In this regime, we can Taylor expand the square root, giving

$\displaystyle \frac{1}{|\bold{r}-\bold{r}'|}\approx\frac{1}{r}\left(1-\frac{1}{2}\left[\left(\frac{r'}{r}\right)^2-2\left(\frac{r'}{r}\right)\cos\theta\right]\right)$

In fact, we’re going to go one step further and throw away the quadratic term:

$\displaystyle \frac{1}{|\bold{r}-\bold{r}'|}\approx\frac{1}{r}\left(1+\frac{r'}{r}\cos\theta\right)$

Let’s put this back into the expression for the electrostatic potential:

$\displaystyle \varphi(\bold{r})\approx\frac{1}{4\pi\epsilon_0 r}\int\rho(\bold{r}')\,d^3\bold{r}'+\frac{1}{4\pi\epsilon_0 r^2}\int\rho(\bold{r}')\,r'\cos\theta\,d^3\bold{r}'$

The second integral looks a bit unwieldy with the cosine in the integrand, so let’s reinstate the dot product:

$\hat{\bold{r}}\cdot\bold{r}'=r'\cos\theta\quad\text{where}\quad\bold{r}=r\,\hat{\bold{r}}$

Then

$\displaystyle \varphi(\bold{r})\approx\frac{1}{4\pi\epsilon_0 r}\int\rho(\bold{r}')\,d^3\bold{r}'+\frac{1}{4\pi\epsilon_0 r^2}\int\rho(\bold{r}')\,\hat{\bold{r}}\cdot\bold{r}'\,d^3\bold{r}'$

That $\hat{\bold{r}}$ we can actually take outside the integral, because it does not depend on any of the dummy spatial variables:

$\displaystyle \varphi(\bold{r})\approx\frac{1}{4\pi\epsilon_0 r}\int\rho(\bold{r}')\,d^3\bold{r}'+\frac{1}{4\pi\epsilon_0 r^2}\,\hat{\bold{r}}\cdot\int\rho(\bold{r}')\,\bold{r}'\,d^3\bold{r}'$

We give the two integrals names:

$\displaystyle \int\rho(\bold{r})\,d^3\bold{r}=q$

$\displaystyle \int\rho(\bold{r})\,\bold{r}\,d^3\bold{r}=\bold{p}$

In summary then,

$\displaystyle \varphi(\bold{r})=\left(\frac{q}{4\pi\epsilon_0 r}\right)+\left(\frac{\bold{p}}{4\pi\epsilon_0 r^2}\right)\cdot\hat{\bold{r}}+\mathcal{O}\left(\frac{1}{r^3}\right)$

What are these two terms?

The first term looks like the potential field from a point charge. Why should this be? Imagine taking the limit that $r$$\infty$. In this limit, the first term dominates, so

$\displaystyle \varphi(\bold{r})\to\frac{q}{4\pi\epsilon_0 r}$

That this is the case should not be surprising. If we are extremely far from the jelly, its complex internal structure is impossible to discern; all we see is a point-like charge with a spherically symmetric potential field. And $q$? The quantity $q$ is indeed the jelly’s net electric charge, since it is just the integral over space of the jelly’s charge density. From a sufficiently great distance, any charge distribution looks like a point charge.

If we go closer to the jelly, however, the second term comes into play. What can we say about it? We first note that there is an inherent ‘direction’ built into it by the vector $\bold{p}$. What is $\bold{p}$? We find $\bold{p}$ by taking a kind of weighted integral of the charge density:

$\displaystyle \bold{p}=\int\rho(\bold{r})\,\bold{r}\,d^3\bold{r}$

The vector $\bold{p}$ is called the electric dipole moment of the charge distribution. A good question is: why should we care about it? The second term’s magnitude decreases as one over distance squared, so its effect is practically negligible compared to that of the first term. Isn’t the effect negligible in the far field? Certainly, unless the jelly is electrically neutral.

If $q$ = 0, the first term in the expansion vanishes, and we are left with, to leading order,

$\displaystyle \varphi(\bold{r})=\left(\frac{\bold{p}}{4\pi\epsilon_0 r^2}\right)\cdot\hat{\bold{r}}$

Note that $q$ = 0 does not imply $\bold{p}$ = 0, and vice versa. For electrically neutral objects, the so-called dipole field generally gives the leading order contribution to the potential field in the far-field limit.

The name dipole most likely conjures up an image of a bar magnet, and indeed the shape of the magnetic dipole field it generates is almost identical to that of the electric dipole field the jelly generates. Something cute: in the same way that the generator of a magnetic dipole field is colloquially called a magnet, that which generates an electric dipole field is called an electret. Good name for a Pokémon.

As a concrete example, consider a water molecule. Each hydrogen atom is bound to the central oxygen atom by a covalent bond. Covalent bonds are characterised by the approximately equal sharing of valence electrons between two ions. But in a covalent bond between different elements, there is always one ion that is a little hungrier for electrons. This hunger is called electronegativity. Out of oxygen and hydrogen, oxygen is the hungrier; the two hydrogen ions are largely robbed of their valence electrons, and are left feeling a little naked. The accumulation of valence electron density around the oxygen ion means the H2O molecule has an asymmetric charge density. Consequently, $\bold{p}$ ≠ 0. Water molecules are described as polar, because they have a permanent dipole moment.

Note that an electric dipole does not necessarily comprise a pair of equal and opposite charges, as is often pictured. Most any asymmetrical charge distribution has an associated dipole moment. For example, an electrically charged blob in the shape of a tear-drop has a non-zero dipole moment. Not every spherically-asymmetric distribution has a dipole moment, however; if

$\rho(\bold{r})=\rho(-\bold{r})$,

the dipole moment can be shown to vanish.

Here’s a diagram showing the dipole field. To clarify, this is the associated electric field $\bold{E}(\bold{r})$, obtained from the potential field by $\bold{E}=-\nabla\varphi$.

The large central arrow indicates the direction of the electric dipole moment $\bold{p}$ (and not the electret’s physical extent!). The high density of field lines around the two poles suggests the electric field is strongest there. Remember that this field is accurate only in the far-field limit; the electret would be too small to see at this scale.

It is possible that both $q$ and $\bold{p}$ vanish, in which case we would have to go to the next term in the expansion of $|\bold{r}-\bold{r}'|^{-1}$. The third term in the expansion of $\varphi(\bold{r})$ would go like $r$-3, and it is called the quadrupole field. The potential can be expanded to arbitrary precision, in increasingly large powers of $r$-1. But that’s a discussion for another time!

I aim to write more posts about electromagnetism soon, hopefully of significantly greater quality than this one …