Past exam of the soft matter and biological physics course of the University of Cambridge / 2015 / Paper 1 / 1 / Solution

Write $\epsilon ={\epsilon }_0{\epsilon }_r$ and take $n_0$ to be the reservoir density of each monovalent ion species. Equal electrochemical potentials in equilibrium give the Boltzmann ion densities in a symmetric electrolyte,Combining this with the electrostatic Poisson equation gives the planar Poisson-Boltzmann equation,The Debye length is ${\kappa }^{-1}$. If the symbol $n_0$ instead denotes the combined density of both species, the factor two is absent; the per-species convention above makes the normalization explicit. At $|e\psi |\ll k_{B}T$, the linearized planar screening solution obeys ${\psi }^{\prime \prime }={\kappa }^2\psi$. With $\psi (0)=\zeta <0$ and $\psi (\infty )=0$,

The positive ions are counterions and the negative ions are co-ions. At weak potential,They approach $n_0$ from above and below respectively, with equal first-order excess and deficit. For stronger negative potential, interpret the source's strong-potential comparison as a statement about $|e\psi |$, since $\psi$ itself is negative. Counterions become strongly concentrated near the wall, while co-ions are exponentially depleted and remain nonnegative. Linearization misses this asymmetry and would eventually predict unphysical negative co-ion density. Point-ion mean-field theory itself eventually needs finite-size and correlation corrections; it provides no steric cap at arbitrarily large counterion concentration.

For steady tangential electro-osmosis, with no imposed pressure gradient and negligible inertia, the Stokes equation is $\eta v^{\prime \prime }+{\rho }_{e}E=0$. Since ${\rho }_e=-\epsilon {\psi }^{\prime \prime }$,The conditions $v(0)=0$ and $v^{\prime }(\infty )=0$ giveThe printed constant speed is the bulk velocity outside the double layer, not the velocity at the no-slip wall. With negative $\zeta$ the positive counterion excess drives fluid along the applied field. The far-field value is the Helmholtz-Smoluchowski electro-osmotic velocity.^[ref]

For the finite channel, first model the added wall as an ideally uncharged insulating boundary with no normal field on its outer side: ${\psi }^{\prime }(d)=0$, while $\psi (0)=\zeta$. The one-charged-wall electro-osmotic channel then hasIntegrating the same force balance and enforcing $v(0)=v(d)=0$ givesFor $U>0$, the maximum occurs atIn that limit $v/U\simeq 1-x/d-e^{-\kappa x}$: the velocity rises rapidly from zero at the charged wall, reaches a maximum a few screening lengths away, then falls approximately linearly to zero at the neutral wall. One charged wall does not give a symmetric plug across a channel with no-slip at the other wall.

An alternative electrode condition $\psi (d)=0$ means a wall held at reservoir potential; it is not the same as zero surface charge. Under that convention,with $\cosh [\kappa (d-x_{\max })]=\sinh (\kappa d)/(\kappa d)$. Both boundary conventions have the same leading thin-double-layer profile above. An externally imposed pressure gradient would add a pressure-driven profile; none is specified here.

For dilute nonadsorbing uncharged polymers in equilibrium or sufficiently weak flow, the entropic polymer depletion near nonadsorbing walls suppresses chain configurations near each wall. A depletion layer of order the coil radius of gyration is expected at both walls, with a bulk-like central region when $d$ is much larger than the coil size. Strong confinement rounds this into a central enrichment. Neutrality removes a direct electrophoretic force, not excluded-volume entropy. Adsorption, concentration and strong-flow migration were not specified, so this is a conditional equilibrium sketch rather than a universal flowing-polymer profile.