= Solution
Use <Thomas-Fermi screening>, treating a homogeneous three-dimensional conductor in static <linear response>. For a positive test charge the screened <electric potential> and bare Coulomb potential are
$$
\boxed{V(r)=\frac{Q}{4\pi\varepsilon_b r}e^{-r/\lambda_{\rm TF}},\qquad
V_{\rm bare}(r)=\frac{Q}{4\pi\varepsilon_b r}.}
$$
Here $\varepsilon_b$ is the background <permittivity> before the mobile carriers screen; it is $\epsilon_0$ in the simplest free-electron model. Both curves have the same leading $1/r$ divergence near the origin, while the screened curve lies below the bare one and decays much faster.
A positive potential attracts <Electrons>. If $g(E_F)$ is their total <spin-summed density of states> per volume and energy, $\delta n=e g(E_F)V$ and $\rho_{\rm ind}=-e^2g(E_F)V$. Inserting this charge into <Poisson equation> gives the <Thomas-Fermi screening length>
$$
\boxed{\lambda_{\rm TF}=\sqrt{\frac{\varepsilon_b}{e^2g(E_F)}}.}
$$
A larger <density of states> lets the conductor rearrange more charge for the same potential change and shortens the screening length. For a parabolic <free electron model>, $g(E_F)=3n/(2E_F)$, hence $\lambda_{\rm TF}^2=2\varepsilon_bE_F/(3ne^2)$.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2019-q2-screening.png]
{title=Bare and Thomas-Fermi-screened potentials of a positive test charge}
{description=Original dimensionless model curves on the same axes, normalized by $Q/(4\pi\varepsilon_b\lambda_{\rm TF})$. The near-origin divergence is clipped for display. The local screening approximation omits microscopic nonlocal effects.}
The exponential form is the intended local screening model. A microscopic zero-temperature metal also has <Friedel oscillations>, so the exponential alone is not an exact description at every length scale.
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