 Solution

Write $e=q>0$ for the elementary charge and $ϵ={ϵ}_0{ϵ}_r$ for the permittivity. Taking the bulk electric potential as zero, the Boltzmann distribution of point ions gives $c_i(x)=c_i^{\infty }\exp [-e{z}_i\varphi (x)/(k_{B}T)]$. Here $c_i$ is a number density, not a molar concentration. Combining this with Poisson's equation gives the planar Poisson-Boltzmann equationMultiplying by ${\varphi }^{\prime }$ and integrating from the neutral bulk, where $\varphi ={\varphi }^{\prime }=0$, yieldsFor the solution occupying $x>0$, with the field on the other side neglected, Gauss's law gives $\sigma =-ϵ{\varphi }^{\prime }(0)$. On the branch that decays to zero without crossing it,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 ${\varphi }^{\prime }(0)>0$ 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.