Axisymmetric Charge Distributions

For the case of an axisymmetric charge distribution (i.e., a charge distribution that is independent of the azimuthal angle $\varphi$ ), we can neglect the spherical harmonics of non-zero order (i.e., the non-axisymmetric harmonics) in Equation (335), which reduces to the following expression for the general axisymmetric Green's function:

$$G({\bf r},{\bf r}') = -\frac{1}{4\pi} \sum_{l=0,\infty}\frac{r_<^{\,l}}{r_>^{\,l+1}}\,P_l(\cos\theta')\,P_l(\cos\theta).$$ (364)

Here, use have been made of the fact that [see Equation (309)]

$$Y_{l,0}(\theta,\varphi) = \left(\frac{2\,l+1}{4\pi}\right)^{1/2} P_l(\cos\theta).$$ (365)

In this case, the general solution to Poisson's equation, (337), reduces to

$$\phi({\bf r}) = \frac{1}{4\pi\,\epsilon_0}\sum_{l=0,\infty}\left[r^{\,l}\,p_{l}(r)+ \frac{q_{l}(r)}{r^{\,l+1}}\right]P_l(\cos\theta),$$ (366)

where

$$p_{l}(r) = \int_r^\infty \int_0^\pi \frac{1}{r'^{\,l+1}} \,\rho(r',\theta)\,\,P_l(\cos\theta)\,2\pi\,r'^{\,2}\,\sin\theta\,d\theta\,dr',$$ (367)

$$q_{l}(r) = \int_0^r \int_0^\pi r'^{\,l}\, \rho(r',\theta)\,P_l(\cos\theta)\,2\pi\,r'^{\,2}\,\sin\theta\,d\theta\,dr'.$$ (368)

Consider the potential generated by a charge $q$ distributed uniformly in a thin ring of radius $a$ that lies in the $x$ -$y$ plane, and is centered at the origin. It follows that

$$\rho(r,\theta)\,2\pi\,r^{\,2}\,\sin\theta\,d\theta\,dr \rightarrow q\,\delta(r-a)\,\delta(\theta-\pi/2)\,d\theta\,dr.$$ (369)

Hence, for $r<a$ we obtain $q_l=0$ and $p_l=q\,P_l(0)/a^{\,l+1}$ . On the other hand, for $r>a$ we get $p_l=0$ and $q_l=q\,a^{\,l}\,P_l(0)$ . Thus,

$$\phi(r,\theta) = \frac{q}{4\pi\,\epsilon_0}\sum_{l=0,\infty} \frac{r_<^{\,l}}{r_>^{\,l+1}}\,P_l(0)\,P_l(\cos\theta),$$ (370)

where $r_<$ represents the lesser of $r$ and $a$ , whereas $r_>$ represents the greater.