# Why does the potential due to uniformly charged ring in entire space is given by solution of Laplace's Equation?

Suppose we have a ring of radius $$a$$, placed on $$xy$$-plane and electric charge $$q$$ is uniformly distributed over it. Now, what is the potential due to this charged ring at some point in space $$(r, \theta, \varphi)$$? I looked at different places and I found different answers. Like in 'Classical Electrodynamics' by Jackson, I found following expression:

$$V(r, \theta) = \frac{q}{4\pi\epsilon_0}\sum_{\ell = 0}^\infty \frac{1}{r_>}\left(\frac{r_<}{r_>}\right)^\ell P_\ell(\cos\theta)$$

where $$r_> := \max\{r, a\}$$ and $$r_< := \min\{r, a\}$$. I found this same solution at other places too. But if 'Mathematical Methods for Physicists' by Arfken, I found following solution:

$$V(r, \theta) = \frac{q}{4\pi\epsilon_0 r}\sum_{\ell = 0}^\infty (-1)^\ell \frac{(2\ell-1)!!}{(2\ell)!!}\left(\frac{a}{r}\right)^{2\ell} P_{2\ell}(\cos\theta) \hspace{1cm}(\text{for}\; r > a)$$

Now, first of all I don't understand how these two results are same. Secondly, in all places, they mentioned that the solution is found by solving the Laplace's Equation. But how the solution of Laplace's Equation will give potential due to charge distribution? The charge distribution $$\rho$$ is not zero in entire space. So how can we use Laplace equation? Why we are not using Poisson's Equation? And even after using Laplace equation, I don't understand how we are arriving at this first solution which is given in Jackson and at many other places?

• Your first field solution looks like that of a point charge, not a ring. The ring's potential is symmetric about the $z=0$ plane, and so only has even Legendre polys. Mar 28, 2023 at 12:44
• Concerning Laplace's equation vs. Poisson's equation: these solutions are defined separately for $r > a$ and $r < a$, in which regions $\rho = 0$. We then require that the two solutions approach the same limit (if it exists) as $r \to a$. Mar 28, 2023 at 16:50

Concerning your first question: Your first expression is not the potential for a charged ring, it is the potential for a point charge of magnitude $$q$$ at a point $$(x,y,z) = (0,0,a)$$. See Eq. (3.38) of Jackson. So there is no reason to expect the two results to be equivalent. The result for a charged ring is derived as Eq. (3.131) of Jackson; he does the derivation inside a grounded sphere of radius $$b$$, but if you take the limit as $$b \to \infty$$ it is easy to see that it reduces to the result you found in Arfken & Weber.
Concerning your second question: It is perfectly legitimate to solve Poisson's equation in the region $$r > a$$, since there is no charge in that region. It is also perfectly legitimate to solve it in the region $$r < a$$. The only places where $$\rho \neq 0$$ is at points with $$r = a$$, and so the general series form of the solution to Laplace's equation in spherical coordinates still applies (independently) for the two disconnected regions where $$r \neq a$$. We can then use other techniques to infer the values of the coefficients in this series expansion.