 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 areHere ${\epsilon }_b$ is the background permittivity before the mobile carriers screen; it is ${ϵ}_0$ 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, $\delta n=eg(E_F)V$ and ${\rho }_{\mathrm{ind}}=-e^{2}g(E_F)V$. Inserting this charge into Poisson equation gives the Thomas-Fermi screening lengthA 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 ${\lambda }_{\mathrm{TF}}^2=2{\epsilon }_b{E}_F/(3n{e}^2)$.

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.