Why should a dipole have zero net charge?

Question

Why can a dipole not have two unequal charges separated by a distance? Is there any significance for the dipole being defined as electrically neutral?

Answer 1

The concept of a dipole moment, and other moments such as a monopole, quadrupole, etc, comes from the process of writing a field as a sum of components called multipoles. This is known as a multipole expansion of the field. The reason we do this is that it can make calculations quicker and easier because it allows us to approximate a complicated field by a simpler sum of multipoles.

A single isolated point charge produces a field that is a pure monopole field, and two equal and opposite charges produce a field that is approximately a dipole field (it is exactly a dipole only in the limit of the distance between the charges becoming zero). So if you add a single charge to a pair of equal and opposite charges you get a total field that is a sum of monopole and dipole fields.

And this is what happens in the example you give. Suppose we have two charges $+2Q$ and $-Q$, then this is equivalent to a single charge $+Q$ and a pair of charges $+\tfrac32 Q$ and $-\tfrac32Q$. The field from the charges would be the vector sum of a monopole field from the $+Q$ charge and a dipole field from the $\pm\tfrac32Q$ pair.

So the reason a dipole cannot have two unequal charges is simply because such an arrangement would be a sum of a monopole and dipole, and not just a dipole.

Answer 2

If we have several charges distributed in space, then the total potential is

$$ \phi = k\sum_{n = 1}^{N} \frac{q_{n}}{\left|\mathrm{r} - \mathrm{r}_{n}\right|} $$

If charges are concentrated in a narrow region of space (narrow compared to the distance to the observer and, if there are EM waves, to the wavelength) then we can expand the fraction linearly assigning $r_{n} = r_{0} + \Delta_{n}$:

$$ \frac{1}{\left|\mathrm{r} - \mathrm{r}_{n}\right|} = \frac{1}{\sqrt{(\mathrm{r} - \mathrm{r}_{0} - \mathrm{\Delta}_{n})^2}} \approx \frac{1}{\sqrt{(\mathrm{r} - \mathrm{r}_0)^2 - 2 (\mathrm{r} - \mathrm{r}_0) \cdot \mathrm{\Delta}_{n}}} \approx \\ \frac{1}{\left|\mathrm{r} - \mathrm{r}_{0}\right|} \frac{1}{1 - \frac{(\mathrm{r} - \mathrm{r}_0) \cdot \mathrm{\Delta}_{n}}{(\mathrm{r} - \mathrm{r}_0)^2}} \approx \frac{1}{\left|\mathrm{r} - \mathrm{r}_{0}\right|} + \frac{1}{\left|\mathrm{r} - \mathrm{r}_{0}\right|^2}\frac{\left(\mathrm{r} - \mathrm{r}_{0}\right)}{\left|\mathrm{r} - \mathrm{r}_{0}\right|} \cdot \mathrm{\Delta}_{n} $$

For the first approximation I neglected quadratic term of $\Delta$, for the second i used Taylor series for the square root, for the third I expanded the fraction in Taylor series.

Substituting it into the equation for the potential, we get

$$ \phi \approx \frac{k}{\left|\mathrm{r} - \mathrm{r}_{0}\right|}\sum_{n = 1}^{N} q_{n} + \frac{k}{\left|\mathrm{r} - \mathrm{r}_{0}\right|^2}\frac{\left(\mathrm{r} - \mathrm{r}_{0}\right)}{\left|\mathrm{r} - \mathrm{r}_{0}\right|} \cdot \sum_{n = 1}^{N} q_{n} \mathrm{\Delta}_{n} = \frac{k Q }{\left|\mathrm{r} - \mathrm{r}_{0}\right|} + \frac{k}{\left|\mathrm{r} - \mathrm{r}_{0}\right|^2}\frac{\left(\mathrm{r} - \mathrm{r}_{0}\right)}{\left|\mathrm{r} - \mathrm{r}_{0}\right|} \cdot \mathrm{d} $$

Here $Q$ is the total charge, and $d$ is the dipole moment. As you see, the first term decays as $1/r$ and the second one as $1/r^2$. If the first one is present (i. e. $Q\ne0$) then the second term can be, usually, neglected when we are far away form the system of charges.