Electric Field due to a Dipole

A quick-and-dirty way of seeing why the fall-off rate should be $r^{-3}$ is as follows: Suppose you are at a point $P$ on the $x$-axis, a distance $r$ from the origin. There is a positive charge $+q$ at the origin, and a negative charge $-q$ at $x = d$. What is the total field?

The field from the positive charge is $E_x = q/r^2$, and the field from the negative charge is $E_x = - q/(r+d)^2$. Define $f(r) = q/r^2$. Then the total electric field is $$ E_x = f(r) - f(r+d) = - d left(frac{f(r+d) - f(r)}{d} right) approx - d f'(r) $$ in the limit where $d$ is sufficiently small. But $f'(r) = - 2 q /r^3$, so $E_x approx - 2 q d / r^3$.

The main idea here is that the fields of the positive and negative charges almost cancel out (but not quite); and the lack of cancellation between them is due to the slight difference in their distances. But the difference in any function (such as the electric field) due to a small difference in its argument (such as the distance) can be approximated by the derivative of that function.

The field of an arbitrary localized charge distribution can be expanded as a Laurent series in inverse powers of the distance: $1/r^2$, $1/r^3$, $1/r^4$, etc. This is called a multipole expansion.

The total charge or “monopole moment” gives the strength of the $1/r^2$ term. The “dipole moment” is related to the strength of the $1/r^3$ term. The “quadrupole moment” is related to the strength of the $1/r^4$ term. The “octupole moment” is related to the strength of the $1/r^5$ term. Etc.

What introductory treatments call “a dipole” — two equal and opposite charges, in the limit that their charge goes to infinity while their separation goes to zero — is simply an idealized charge distribution where only the $1/r^3$ term happens to be nonzero.

The multipole expansion is usually expressed on terms of the potential rather than the field. That expansion starts with a $1/r$ monopole term and a $1/r^2$ dipole term.

The key mathematics behind it is the identity

$$frac{1}{|mathbf{r}-mathbf{r}’|}=frac1rsum_{ell=0}^inftyleft(frac{r’}{r}right)^ell P_ell(cosgamma)$$

where $P_ell(x)$ is a Legendre polynomial and $gamma$ is the angle between the vectors $mathbf{r}$ and $mathbf{r}’$. (Legendre just expanded the left side for $rgg r’$ and found that the terms involved polynomials in $hat{mathbf{r}}cdothat{mathbf{r}}’$, which have come to be named after him.)

Using this expansion in the integral for the potential of an arbitrary charge distribution,

$$varphi(mathbf{r})=frac{1}{4piepsilon_0}intfrac{rho(mathbf{r}’)d^3mathbf{r}’}{|mathbf{r}-mathbf{r}’|},$$

gives one kind of multipole expansion for the potential.

The physical significance of the dipole moment is that if the net charge (i.e., the monopole moment) is zero, then the dipole term in the multipole expansion is the dominant term at a large distance away: the additional terms terms are negligible in comparison as $rtoinfty$. In other words, you can approximate any charge distribution that is overall neutral as a dipole at large distances.