= Solution
Expand the cloud's <potential energy> about a <stable equilibrium>: $U(z)=U(0)+U'(0)z+U''(0)z^2/2+\cdots$. Equilibrium gives $U'(0)=0$, stability gives $U''(0)=k>0$, and reflection symmetry also removes odd terms. For small displacements, $U-U(0)\simeq kz^2/2$ and $F=-dU/dz\simeq-kz$.
Including the electric and damping forces, <Newton's second law> is $m\ddot z=qE(t)-b\dot z-kz$. Dividing by $m$ gives the <Lorentz oscillator model>
$$
\boxed{\ddot z+2\gamma\dot z+\omega_0^2z=\frac qmE(t),\qquad \omega_0=\sqrt{\frac km},\quad \gamma=\frac b{2m}.}
$$
The specified convention uses $2\gamma$ as the damping coefficient per unit <mass>.
For the steady response, represent the field by $E_0e^{i\omega t}$ and the displacement by $\widetilde z e^{i\omega t}$, taking real parts for physical quantities. Substitution gives
$$
(\omega_0^2-\omega^2+2i\gamma\omega)\widetilde z=\frac{qE_0}{m},\qquad \widetilde z=\frac{qE_0/m}{\omega_0^2-\omega^2+2i\gamma\omega}.
$$
Define the nonnegative <amplitude> $z_0=|\widetilde z|$ and $\delta=\operatorname{atan2}(2\gamma\omega,\omega_0^2-\omega^2)$. Then
$$
\boxed{z_0=\frac{|q|E_0/m}{\sqrt{(\omega_0^2-\omega^2)^2+4\gamma^2\omega^2}},\qquad \phi=\arg q-\delta\pmod{2\pi}.}
$$
For positive $q$, $\phi=-\delta$. For electrons $q<0$, the <phase of a negatively charged driven oscillator> is $\phi=\pi-\delta$: the low-<frequency> displacement is opposite to the field, rather than in <phase> with it. A signed <amplitude> could absorb this extra $\pi$, but the <phase> convention must then be stated. The induced dipole $p=qz$ has <phase> $-\delta$ for either charge sign. Transient free oscillations have been omitted because the question asks for steady state.
For the electron cloud, $q/m=-e/m_e$ and
$$
\omega_0=\sqrt{\frac{k/n}{m_e}}=\sqrt{\frac{100}{9.109\times10^{-31}}}=1.05\times10^{16}\,\mathrm{rad\,s^{-1}}.
$$
The visible range corresponds to $\omega=2\pi c/\lambda$ between $2.35\times10^{15}$ and $4.71\times10^{15}\,\mathrm{rad\,s^{-1}}$, below this <resonance>. The resonant <wavelength> is about $180\,\mathrm{nm}$, in the ultraviolet.
The <acceleration> <amplitude> is $\omega^2z_0$. At fixed incident field <amplitude>, the scattered intensity is therefore proportional to
$$
\boxed{I_{\mathrm{sc}}(\omega)\propto\frac{\omega^4}{(\omega_0^2-\omega^2)^2+4\gamma^2\omega^2}.}
$$
Well below <resonance> and outside strong damping corrections, $z_0$ is nearly <frequency>-independent and this becomes the <Rayleigh scattering wavelength dependence> $I_{\mathrm{sc}}\propto\omega^4\propto\lambda^{-4}$. The shortest visible wavelengths are consequently scattered much more strongly into directions away from the direct solar beam than the longest ones, producing a blue sky. The simple fourth-power approximation gives a factor 16 between 400 and 800 nm at equal incident intensity. Retaining the undamped denominator with the given <resonance> raises that factor to about 22.7, since the upper end is not extremely far below <resonance>. The exact ratio needs the unspecified damping and incident solar spectrum; the stronger scattering of the blue end is the robust conclusion here.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017-bound-charge-scattering.png]
{title=Bound-charge response below an ultraviolet resonance favors short visible wavelengths}
The model describes <frequency>-dependent scattering, not the detailed perceived color after solar-spectrum weighting and human visual response. It also explains why treating the electrons as free at visible frequencies would miss the bound-charge fourth-power limit.
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