# An electrostatic problem for two disks in $\mathbb{R}^2$ - how can the solution be represented?

The electrostatic Laplace problem for the exterior of a disk can be solved in a straightforward manner using separation of variables. Suppose we have a unit disk $\Omega$ with a charge density of $f$ on the boundary. Then the solution to the exterior problem, \begin{align} & \Delta v(x) = 0 \quad \quad x\in \mathbb{R}^2\setminus \overline{\Omega}, \\ & v |_{\partial D}(x) = f, \end{align} is given through separation of variables as $$v(x) = v(r,\theta) = \sum_{n=0}^\infty r^{-n}(a_n \cos(n\theta)+b_n \sin(n\theta)),$$ where $$a_n = \frac{1}{2\pi}\int_0^{2\pi} f(\theta)\cos(n\theta) d\theta, \\ b_n = \frac{1}{2\pi}\int_0^{2\pi} f(\theta)\sin(n\theta) d\theta.$$ But what about the case where we have two disks $\Omega_1$ and $\Omega_2$, and we want to determine the solution to the exterior Laplace problem for a charge density $f$, with $f$ defined on the union of the disks $D=\Omega_1\cup \Omega_2$? Say for example the disks both have radius $1$ and are symmetric on the $x_1$ axis with a distance of $6$ between them.

Can we just take define two new origins $O_1$ and $O_2$ at the center of the disks, with associated polar coordinates $(r_1,\theta_1)$ and $(r_2,\theta_2)$, and then say the following: $$v(x) = \sum_{n=0}^\infty r_1^{-n}(a_n \cos(n\theta_1)+b_n \sin(n\theta_1)) + \sum_{n=0}^\infty r_2^{-n}(a_n \cos(n\theta_2)+b_n \sin(n\theta_2)).$$ Is this a valid solution? If not, why not?

This is indeed a solution due to the linearity of the Lapacian. If I have 2 charge distributions $\rho_1$ and $\rho_2$ and