Question 3.10: pertains to a very special charge configuration. I propose n...

V(r)=\frac{1}{4\pi\epsilon _{0}}\int{\frac{1}{\eta }\rho (\acute{r})d\acute{\tau } }.               (3.91)

Using the law of cosines,

where\alpha  is the angle between r and \acute{r}. Thus

\eta =r\sqrt{1+\epsilon } ,                           (3.92)

with

For points well outside the charge distribution, ε is much less than 1, and this invites a binomial expansion:

\frac{1}{\eta }=\frac{1}{r}\left(1+\epsilon \right)^{-1/2}=\frac{1}{r}\left(1-\frac{1}{2}\epsilon +\frac{3}{8}\epsilon ^{2}-\frac{3}{8}\epsilon ^{3}+… \right),                                (3.93)

or, in terms of r ,\acute{r}, and α:

In the last step, I have collected together like powers of (\acute{r} /r ); surprisingly, their coefficients (the terms in parentheses) are Legendre polynomials! The remarkable result^{16} is that

\frac{1}{\eta }=\frac{1}{r}\sum\limits_{n=0}^{\infty }{\left(\frac{\acute{r}}{r} \right)^{n} } P_{n}(\cos\alpha).                             (3.94)

Substituting this back into Eq. 3.91, and noting that r is a constant, as far as the
integration is concerned, I conclude that

V(r)=\frac{1}{4\pi\epsilon _{0}}\sum\limits_{n=0}^{\infty }{\frac{1}{r^{(n+1)}} }\int{(\acute{r})^{2}P_{n}(\cos\alpha)\rho (\acute{r})d\acute{\tau }},         (3.95)

or, more explicitly,

V(r)=\frac{1}{4\pi\epsilon _{0}}\left[\frac{1}{r}\int{\rho (\acute{r})d\acute{\tau +\frac{1}{r^{2}} }} \int{\acute{r}\cos\alpha\rho (\acute{r})d\acute{\tau }}+\frac{1}{r^{3}}\int{(\acute{r})^{2}\left(\frac{3}{2}\cos^{2}\alpha -\frac{1}{2} \right)\rho (\acute{r})d\acute{\tau }+…. } \right] ,              (3.96)

This is the desired result—the multipole expansion of V in powers of 1/r . The first term (n = 0) is the monopole contribution (it goes like 1/r ); the second (n = 1) is the dipole (it goes like 1/r^{2}); the third is quadrupole; the fourth octopole; and so on. Remember that α is the angle between r and \acute{r}, so the integrals depend on the direction to the field point. If you are interested in the potential along the \acute{z} axis (or—putting it the other way around—if you orient your \acute{r} coordinates so the z axis lies along r), then α is the usual polar angle \acute{\theta}. As it stands, Eq. 3.95 is exact, but it is useful primarily as an approximation scheme: the lowest nonzero term in the expansion provides the approximate potential at large r , and the successive terms tell us how to improve the approximation if greater precision is required.