Charge distribution which yields harmonic oscillation
Perhaps this is more of a math question, or I may be using the wrong tools for a given problem. Any insight would be helpful.
Question
Assume <PERSON> "plum-pudding" model of the atom. Now take the hydrogen atom; we have a positive charge +e distributed over a sphere with radius $10^{-8}$cm. The electron is a point charge -e. Question: How should the positive charge be distributed over the sphere so as to lead to harmonic oscillations of the electron?
Note, we may neglect all mechanical interactions and the magnetic field of the moving electron.
Solution Attempt
Harmonic oscillations, $\vec{F} = -k_{s} \vec{x}$
Coulomb force, $\vec{F} = k \frac{q_{1} q_{2}}{r^{2}} \hat{r}_{1,2}$
Define a spherical coordinate system, $(r,\theta,\psi)$ with radius r, angle $\theta$ on the x-y axis, and angle $\psi$ from the z axis. Further: \begin{align} r&\in [-\infty, \infty] \ \theta &\in [0, \pi) \ \psi &\in[0,\pi) \end{align} [Note that I've limited the range of $\theta$ in favor of a +/- $r$]
Let the charge density of the sphere at a given coordinate be represented by $\rho(r',\theta',\psi')$ [C/m$^{3}$]. 649
The position of the electron is represented by $(r,\theta,\psi)$.
Assume harmonic oscillations will occur with the electron traveling along an axis through the center of the sphere. Rewrite harmonic oscillation equation $\vec{F} = -k_{s} \vec{r}$ (note the convenience of allowing negative $\vec{r}$). Set equal to force on electron due to $\rho$: \begin{align} F_{-e} &= -k_{s} r = \int -k e \frac{\rho(r',\theta',\psi')}{R^2} dV \ \text{where } R^2 &= r^2 + r'^2 - 2rr'(\cos\theta \cos\theta' \sin\psi \sin\psi' + \sin\theta \sin\theta' \sin\psi \sin\psi' + \cos\psi \cos\psi') \end{align}
Expanding: \begin{align} \frac{k_{s}}{k e} r = \int_0^\pi\int_0^\pi\int_0^{10^-10} \frac{\rho(r',\theta',\psi')}{R^2} r'^2 \sin\psi' dr' d\theta' d\psi' \end{align}
Simplification Attempt
From this point one needs to solve for $\rho(r',\theta',\psi')$, which I've made little headway in. By symmetry, we know $\rho(r',\theta',\psi')$ will really only depend on r', so $\rho(r',\theta',\psi') = \rho(r')$.
I've tried some simplifying assumptions to get started. In particular assuming the electron is only on the x-axis and thus simplifying the term $R^2$ by asserting $\psi,\theta = 0$. Now $R^2 = r^2 + r'^2 - 2rr'\cos\psi'$. 402