Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2016 / S16 / 2 / 1 / Solution

For weak potentials, Boltzmann electrons give $\delta n_e=n_{e}e\varphi /(k_{B}T)$ against fixed ions. Poisson's equation becomes ${\nabla }^2\varphi =\varphi /{\lambda }_D^2$ outside a test charge, giving a Yukawa potential with $\boxed{{\lambda }_D=\sqrt{{ϵ}_0{k}_{B}T/(n_e{e}^2)}}$. For a small electron displacement $x$ relative to fixed ions, charge separation produces restoring field $E=n_{e}ex/{ϵ}_0$, so $m\ddot{x}=-n_e{e}^{2}x/{ϵ}_0$. Hence $\boxed{{\omega }_p=\sqrt{n_e{e}^2/({ϵ}_0{m}_e)}}$. The two formulas respectively encode thermal screening and electron inertia.