# Thomas-Fermi screening length and electronic compressibility

I am encountering a problem in understanding Thomas-Fermi screening theory. From Ashcroft & Mermin, it is defined as $k_0^{-1}$ such that the potential decays as $\phi(r) \sim e^{-k_0 r}/r$ in the metal. It can be related to the electronic compressibility through $$k_0^2 = 4 \pi e^2 \frac{\partial n}{\partial \mu}$$ where $n$ is the density and $\mu$ is the chemical potential. So the more compressible the electron gas is, the more efficient screening will be.

However this does also imply that I can apply an electric field inside a metal more efficiently as the compressibility is small (bad screening). Therefore a larger fraction of electrons in the bulk metal will be affected by the electric field. And so in principle the number of electrons (not the density) affected by a change of chemical potential in the metal will be greater, which contradicts this interpretation in terms of compressibility, don't you think ?

Any comment is welcome.

Assuming that we apply an electric field on a metal along the direction $x$, then one has $\phi(x) = \phi_0e^{-k_0 x}$ inside the metal. One can compute the total induced charge \begin{aligned} Q^{\rm ind} &= S\int_0^{\infty}\rho^{\rm ind}(x){\rm d}x\\ &= S\int_0^{\infty}-e^2\frac{\partial n}{\partial \mu}\phi_0e^{-k_0 x} {\rm d}x \\ &= S \phi_0 /4 \pi \times \int_0^{\infty} k_0^2e^{-k_0 x}{\rm d}x \\ &= k_0 \phi_0 S / 4 \pi \\ &= C_{\rm Q} \phi_0 \end{aligned} the capacitance $C_{\rm Q} = \epsilon_0 k_0 S$ (in S.I units now) is the quantum capacitance for the metallic interface. Therefore a better screening (higher $k_0$) also implies a higher total number of electrons $Q^{\rm ind}/(-e)$ affected by the potential, although the length on which these are spread is smaller.