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Axisymmetric Charge Distributions
For the case of an axisymmetric charge distribution (i.e., a charge distribution that is independent of the azimuthal angle
), 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:
![$\displaystyle G({\bf r},{\bf r}') = -\frac{1}{4\pi} \sum_{l=0,\infty}\frac{r_<^{\,l}}{r_>^{\,l+1}}\,P_l(\cos\theta')\,P_l(\cos\theta).$](img815.png) |
(364) |
Here, use have been made of the fact that [see Equation (309)]
![$\displaystyle Y_{l,0}(\theta,\varphi) = \left(\frac{2\,l+1}{4\pi}\right)^{1/2} P_l(\cos\theta).$](img816.png) |
(365) |
In this case, the general solution to Poisson's equation, (337), reduces to
![$\displaystyle \phi({\bf r}) = \frac{1}{4\pi\,\epsilon_0}\sum_{l=0,\infty}\left[r^{\,l}\,p_{l}(r)+ \frac{q_{l}(r)}{r^{\,l+1}}\right]P_l(\cos\theta),$](img817.png) |
(366) |
where
Consider the potential generated by a charge
distributed uniformly in a thin ring of radius
that lies in the
-
plane, and
is centered at the origin. It follows that
![$\displaystyle \rho(r,\theta)\,2\pi\,r^{\,2}\,\sin\theta\,d\theta\,dr \rightarrow q\,\delta(r-a)\,\delta(\theta-\pi/2)\,d\theta\,dr.$](img822.png) |
(369) |
Hence, for
we obtain
and
. On the other hand, for
we get
and
. Thus,
![$\displaystyle \phi(r,\theta) = \frac{q}{4\pi\,\epsilon_0}\sum_{l=0,\infty} \frac{r_<^{\,l}}{r_>^{\,l+1}}\,P_l(0)\,P_l(\cos\theta),$](img827.png) |
(370) |
where
represents the lesser of
and
, whereas
represents the greater.
Next: Dirichlet Problem in Spherical
Up: Potential Theory
Previous: Multipole Expansion
Richard Fitzpatrick
2014-06-27