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For a vacuum electric potential, put $\Phi =R(r)Y(\theta )$ in Laplace's equation in spherical coordinates. Multiplication by $r^2/(RY)$ separates the radial and angular terms. With separation constant $\ell (\ell +1)$ and $x=\cos \theta$, the angular Legendre differential equation isRegularity at both polar axes selects nonnegative integers $\ell$ and the Legendre polynomials. The radial equation is $(r^2{R}^{\prime }{)}^{\prime }-\ell (\ell +1)R=0$, whose independent solutions are $r^{\ell }$ and $r^{-\ell -1}$. Hence the angularly regular axisymmetric vacuum solution isThe radial domain and its boundary conditions determine which terms are admissible.

For the given charge density, only $P_1(\cos \theta )=\cos \theta$ is forced. Write $\Phi =f(r)\cos \theta$. The interior radial Poisson equation is $f^{\prime \prime }+2f^{\prime }/r-2f/r^2=-4$. The identity for the Poisson equation with a spherical harmonic source gives the particular solution $f_p=-r^2$, since the radial operator takes $r^2$ to four. Finiteness at zero and decay at infinity therefore giveContinuity of the electric potential and its radial derivative at $a$ requires $Aa-a^2=B/a^2$ and $A-2a=-2B/a^3$. Solving gives $A=4a/3$, $B=a^4/3$, soAll other angular modes have zero coefficients after imposing regularity, matching and decay. At the origin the interior expression is $(4a/3)z-rz$, which extends with a continuous gradient and value zero. The angular charge density has no unique value exactly at that point; Poisson's equation holds classically for $r>0$ and as a weak solution through the origin. There is no additional point charge.