= Solution
Write $e=q>0$ for the <elementary charge> and $\epsilon=\epsilon_0\epsilon_r$ for the <permittivity>. Taking the bulk <electric potential> as zero, the <Boltzmann distribution> of point <ions> gives $c_i(x)=c_i^\infty\exp[-ez_i\phi(x)/(k_BT)]$. Here $c_i$ is a <number density>, not a molar <concentration>. Combining this with <Poisson's equation> gives the planar <Poisson-Boltzmann equation>
$$
\phi''(x)=-\frac e\epsilon\sum_i z_ic_i^\infty e^{-ez_i\phi(x)/(k_BT)}.
$$
Multiplying by $\phi'$ and integrating from the neutral bulk, where $\phi=\phi'=0$, yields
$$
\frac\epsilon2(\phi')^2=k_BT\sum_i c_i^\infty\left(e^{-ez_i\phi/(k_BT)}-1\right).
$$
For the solution occupying $x>0$, with the field on the other side neglected, <Gauss's law> gives $\sigma=-\epsilon\phi'(0)$. On the branch that decays to zero without crossing it,
$$
\boxed{\sigma=\operatorname{sgn}(\phi_0)\sqrt{2\epsilon k_BT\sum_i c_i^\infty\left(e^{-ez_i\phi_0/(k_BT)}-1\right)}.}
$$
\b[Source sign qualification:] the unsigned square root printed in the paper gives $|\sigma|$. A negative wall <electric potential>, such as the next part's $-73\,\mathrm{mV}$, has $\phi'(0)>0$ and hence $\sigma<0$, providing a concrete counterexample to interpreting that printed expression as signed <surface charge density>. This is the planar <Grahame equation>, with an explicit sign convention.
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