Skip to main content
Sandbox Physics
Electrochemistry and ion transport

L3 · Chapter 5

How surface ions drag the surrounding liquid

Sandbox Physics ·

An electric field pulls excess ions near a surface, and those ions drag liquid with them. A very thin charge layer can move liquid far beyond it. If the field also induces the charge, reversing the field reverses both factors, leaving a flow proportional to field strength squared.

Begin with a stationary flat wall

Take wall-normal coordinate yy into the liquid, and tangential coordinate ss along the positive tangential field EsE_s. The wall is stationary with no slip. Let ψ(y)\psi(y) be potential relative to the outer liquid. Its surface value ζ\zeta is the zeta potential, measured in volts.

Electrical force per volume is ρeEs\rho_eE_s. With negligible inertia and no imposed pressure gradient, this balances the gradient of viscous stress. Replace charge density by the second derivative of potential through Poisson’s equation. Treat the tangential field as constant across the thin layer and integrate twice.

η∂y2us+ρeEs=0,ρe=−ε∂y2ψ,us(∞)=−εζηEs\eta\partial_y^2u_s+\rho_eE_s=0,\quad\rho_e=-\varepsilon\partial_y^2\psi,\quad u_s(\infty)=-\frac{\varepsilon\zeta}{\eta}E_s
Use zero wall velocity and vanishing shear outside the layer. Viscosity η\eta is in Pa s\mathrm{Pa\,s}. The outer effective slip is the Helmholtz–Smoluchowski relation.

Reading: MIT 10.626 · Squires & Bazant

Check direction and magnitude

A negative zeta potential has excess cations beside the surface. A rightward field pulls them and the liquid rightward. With permittivity 6.95×10−10  F m−16.95\times10^{-10}\;\mathrm{F\,m^{-1}}, viscosity 10−3  Pa s10^{-3}\;\mathrm{Pa\,s}, ζ=−10.0  mV\zeta=-10.0\;\mathrm{mV} and Es=103  V m−1E_s=10^3\;\mathrm{V\,m^{-1}}, the outer speed is 6.95  μm s−16.95\;\mathrm{\mu m\,s^{-1}} to the right.

This velocity is defined outside the charge layer. The solid surface still obeys no slip. Effective slip compresses the action of an unresolved thin layer into a boundary condition. It requires separation of layer and outer length scales, with corrections such as surface conduction small.

Let a cylinder induce its own charge layer

Place a fixed, initially neutral metal cylinder of radius aa in an infinite planar electrolyte. Set metal potential to zero and distant field strength to E0E_0. Polar angle θ\theta increases anticlockwise from the applied-field direction; rr measures distance from the axis. No Faradaic reaction occurs. After charging, normal outer current vanishes at the cylinder.

The neutral outer solution obeys Laplace’s equation. A uniform distant field, cylinder symmetry and zero surface normal field determine its dipole correction. Subtract outer surface potential from metal potential to obtain zeta potential. Differentiate that same potential around the circumference to get tangential field.

Φ=−E0(r+a2r)cos⁡θ,ζ=2E0acos⁡θ\Phi=-E_0\left(r+\frac{a^2}{r}\right)\cos\theta,\qquad\zeta=2E_0a\cos\theta
For a weak-potential thin layer, metal charge is qm=εζ/λDq_m=\varepsilon\zeta/\lambda_D and diffuse charge is qd=−qmq_d=-q_m. They have opposite signs.
Eθ(a)=−2E0sin⁡θ,uθ(a)=2U0sin⁡2θ,U0=εaE02ηE_\theta(a)=-2E_0\sin\theta,\qquad u_\theta(a)=2U_0\sin2\theta,\quad U_0=\frac{\varepsilon aE_0^2}{\eta}
Use zeta potential and tangential field from the same solution in the slip law. One power of field establishes charge and the other drives it.

Reading: Squires & Bazant · Bazant & Squires

Follow liquid from poles to equator

Liquid approaches the cylinder near its poles, moves along the surface toward the two equators, then moves outward. The four quadrants form a quadrupolar flow. Reversing the applied field reverses both induced charge and tangential field, so their product drives the same flow.

The symmetric fixed neutral cylinder selects no net pumping direction. Electrophoresis would require a moving object and overall force and torque conditions. Pumping requires broken symmetry in boundaries, charge or shape. Here the question is electro-osmosis around a stationary surface.

ur=2U0a(a2−r2)r3cos⁡2θ,uθ=2U0a3r3sin⁡2θu_r=2U_0\frac{a(a^2-r^2)}{r^3}\cos2\theta,\qquad u_\theta=2U_0\frac{a^3}{r^3}\sin2\theta
This analytic outer Stokes field has zero radial surface velocity, the required tangential slip and vanishing distant velocity. Together the two components satisfy incompressibility.

Reading: Squires & Bazant

Calculate the default experimental scales

Take a=10.0  μma=10.0\;\mathrm{\mu m}, E0=0.450  V mm−1E_0=0.450\;\mathrm{V\,mm^{-1}} and λD=0.100  μm\lambda_D=0.100\;\mathrm{\mu m} with the water properties above. Maximum induced zeta potential is 9.00  mV9.00\;\mathrm{mV}, or 0.3500.350 thermal voltages. The layer-to-radius ratio is 0.01000.0100.

The velocity scale is U0=1.41  μm s−1U_0=1.41\;\mathrm{\mu m\,s^{-1}}. At forty-five degrees on the surface, slip reaches 2.81  μm s−12.81\;\mathrm{\mu m\,s^{-1}}. With D=10−9  m2 s−1D=10^{-9}\;\mathrm{m^2\,s^{-1}}, charging scale λDa/D\lambda_Da/D is about 1.00  ms1.00\;\mathrm{ms}. The flow is the charged steady solution; tracer playback may be accelerated, so physical speed and animation time must be read separately.

Doubling field quadruples this ideal velocity and doubles the induced potential. The second change progressively challenges weak-potential screening. Check the charge model’s assumptions along with the field-squared law.

Try the prediction

Reverse the charge and test the flow

  1. Predict what a half-turn of the field will do to surface charge and flow.
  2. Rotate the field and compare pole colours with tracer streamlines. Then change only field magnitude.
  3. Record maximum slip and test field-squared scaling. Use induced potential to judge whether extrapolation remains appropriate.

Exercises and explanations

  1. Induced charge is largest at the cylinder poles. Why is surface slip zero there?

    Read the explanation

    Tangential field vanishes at the poles. Electro-osmotic slip needs that field to act on charge. Maximum slip lies near forty-five degrees, where both induced charge and tangential field are nonzero.

  2. Double radius at fixed field. How do speed, maximum zeta potential and charging time change? Which assumption improves and which worsens?

    Read the explanation

    All three double. At fixed Debye length, the relative layer becomes thinner, improving thin-layer separation. Induced potential grows, making weak-potential screening less accurate. The useful speed increase ends as induced potential leaves the weak-field limit.

Sources and calculation scope

  1. MIT 10.626, Electrokinetics, 2011 Lecture 29

    Thin-double-layer electro-osmosis and viscous force balance. The chapter derives slip with the wall normal, zeta-potential sign and tangential field defined before use.

  2. Squires & Bazant, Induced-charge electro-osmosis, arXiv:physics/0304090v2 (9 February 2004)

    Frozen preprint of Journal of Fluid Mechanics 509, 217–252: section 3.1, PDF equations (3.4)–(3.9), Figure 3(a) and Table 1. The limited reproduction calculates the steady field, charge, slip and outer Stokes flow for a fixed neutral cylinder. The HTML conversion renumbers equations consecutively.

  3. Bazant & Squires, Induced-charge Electrokinetic Phenomena: Theory and Microfluidic Applications, arXiv:physics/0306100v2 (2004)

    Physical Review Letters 92, 066101, especially equation (1) and Figure 1(c). This short paper supplies the physical picture; it is distinct from the longer JFM paper. Its original figures are linked, not redistributed.

FeedbackSomething broken, confusing, or missing on this page?Tell us
What kind of feedback?

Sent with this page’s address, its Lab ID, the site language and your device type. Please leave out personal details. Feedback is kept for up to 12 months.Privacy policy