= Solution
Resolve the angular <surface charge density> into <Legendre polynomials>:
$$
\sin^2\theta=\frac23[1-P_2(\cos\theta)],\qquad \sigma_0=\frac{2A}{3},\quad\sigma_2=-\frac{2A}{3}.
$$
For a single degree $\ell$, write the common surface potential as $V_\ell P_\ell$. Regularity and decay require radial factors $(r/R)^\ell$ inside and $(R/r)^{\ell+1}$ outside. Their exterior-minus-interior radial derivative at $R$ is $-(2\ell+1)V_\ell P_\ell/R$. Thus <spherical harmonic matching with a derivative jump> gives $V_\ell=R\sigma_\ell/(2\ell+1)$.
Only degrees zero and two occur, yielding
$$
\boxed{\begin{aligned}
\Psi_{\rm in}(r,\theta)&=\frac{2AR}{3}-\frac{2A}{15R}r^2P_2(\cos\theta),&&r<R,\\
\Psi_{\rm out}(r,\theta)&=\frac{2AR^2}{3r}-\frac{2AR^4}{15r^3}P_2(\cos\theta),&&r>R.
\end{aligned}}
$$
Both expressions have surface value $2AR/3-2AR P_2/15$, and their derivative difference is $-2A/3+(2A/3)P_2=-A\sin^2\theta$, checking both interface conditions. The interior quadratic is regular at the origin, and the exterior potential tends to zero.
The leading far-field term is the monopole term. Indeed the total charge is $Q=R^2\int\sigma\,d\Omega=8\pi AR^2/3$, and the normalization of the specified jump gives $Q/(4\pi r)=2AR^2/(3r)$. No extra permittivity factor is inserted into the supplied normalization. Regularity here means finite piecewise smooth potential; its normal derivative necessarily jumps at the charged surface.
Back to article page
