One-charged-wall electro-osmotic channel

With $\psi (0)=\zeta$, $\psi \prime (d)=0$ and no slip at either wall, linear screening gives $\psi /\zeta =\cosh [\kappa (d-x)]/\cosh (\kappa d)$. The velocity is $v/U=1-(1-\mathrm{sech}\kappa d)x/d-\cosh [\kappa (d-x)]/\cosh (\kappa d)$. This electrically neutral-wall boundary assumes no external normal field; it differs from a grounded wall.

Zero wall charge constrains the jump in normal electric displacement, while fixing a wall potential is a separate electrical condition. Taking the external normal field to vanish gives a zero normal derivative at an ideal uncharged wall. A grounded wall may carry induced charge even though its imposed potential is zero.

Differentiating the screened-channel velocity gives $x_{\max }=d-{\kappa }^{-1}\mathrm{arsinh}[(\cosh \kappa d-1)/(\kappa d)]$. For $\kappa d\gg 1$, $x_{\max }\simeq \log (\kappa d)/\kappa$. The outer flow declines almost linearly to the passive no-slip wall rather than forming a symmetric plug.