In the derivation of the Poisson-Boltzmann equation, my textbook arrives at the expression
$$ \rho_i = c_+ z_+ F + c_-z_-F = c_+^oz_+^oFe^{-z_+e\phi_i/kT} + c_-^oz_-Fe^{-z_-e\phi_i/kT} $$
where $c_\pm$ is the concentration (of positive or negative ions), $z_\pm$ is the valence of the ions, $F$ is the Faraday constant ($=eN_A$) and $c_\pm^o$ is the concentration in the bulk. The rest is standard notation.
The book now writes that
Because the average electrostatic interaction energy is small compared with $kT$, we may write
$$ \rho_i = (c_+^o z_+ + c_-^oz_-)F - (c_+^oz_+^2 + c_-^oz_-^2)\frac{F^2\phi_i}{kN_AT} + \cdots $$
How do they know that $ze\phi \ll kT$?
To make an estimate, I find the electric potential of a proton (or an $\text{Na}^+$ ion) in vacuum at 1 nm away.
$$ V^{vac} = \frac{1}{4\pi\epsilon_0}\frac{+1e}{1nm} \approx 1.43 \text{V}. $$
In water ($\epsilon = \epsilon_0\epsilon_r = \epsilon_078$),
$$ V^{sol} = \frac{1}{4\pi78}\frac{+1e}{1nm} \approx 0.02 \text{V}. $$
So in solution a $\text{Cl}^-$ ion would have the energy
$$ w=(-1e)V^{sol} \approx -3.2 \cdot 10^{-21}J $$
while at room temperature $kT \approx 3.9 \cdot 10^{-21}J$.
Is this roughly the idea behind the above assumption?