Equilibrium behavior of sessile drops under surface ... · The basic tenets of our analysis are: a total energy mini-mization, a phrasing of Maxwell’s electrostatic partial differ-ential

JOURNAL OF APPLIED PHYSICS VOLUME 93, NUMBER 9 1 MAY 2003

Equilibrium behavior of sessile drops under surface tension,applied external fields, and material variations

Benjamin Shapiroa)

Aerospace Engineering Department, University of Maryland at College Park, Maryland 20742

Hyejin MoonMechanical and Aerospace Engineering Department, University of California at Los Angeles (UCLA),Los Angeles, California 90095

Robin L. GarrellDepartment of Chemistry and Biochemistry, University of California at Los Angeles (UCLA), Los Angeles,California 90095

Chang-Jin ‘‘CJ’’ KimMechanical and Aerospace Engineering Department, University of California at Los Angeles (UCLA),Los Angeles, California 90095

~Received 18 November 2002; accepted 4 February 2003!

This article describes the equilibrium shape of a liquid drop under applied fields such as gravity andelectrical fields, taking into account material properties such as dielectric constants, resistivities, andsurface tension coefficients. The analysis is based on an energy minimization framework. A rigorousand exact link is provided between the energy function corresponding to any given physicalphenomena, and the resulting shape and size dependent force term in Young’s equation. Inparticular, the framework shows that a physical effect, such as capacitive energy storage in theliquid, will lead to 1/R ‘‘line-tension’’-type terms if and only if the energy of the effect isproportional to the radius of the liquid drop:E}R. The effect of applied electric fields on shapechange is analyzed. It is shown that a dielectric solid and a perfectly conducting liquid are all thatis needed to exactly recover the Young–Lippmann equation. A dielectric liquid on a conductingsolid gives rise to line tension terms. Finally, a slightly resistive liquid on top of a dielectric, highlyresistive solid gives rise to contact angle saturation and accurately matches the experimental datathat we observe in our electro-wetting-on-dielectric devices. ©2003 American Institute ofPhysics. @DOI: 10.1063/1.1563828#

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I. INTRODUCTION

The shape of a liquid drop on a surface is determinedthe composition of the liquid~solvent, and ionic and surfactant solutes! and by the composition and morphology of thunderlying solid. When an electric potential is applied acrthe liquid drop and the solid substrate, ions and dipolesdistribute in the liquid, in the solid, or in both dependingthe relative material properties. This redistribution can caa hydrophobic surface to behave in a hydrophyllic mannIn such a case, the liquid drop will change shape underapplied electric potential.

This electro-wetting phenomenon can be used to crefluid flow. 1–10 In practice, electro-wetting-based actuationaqueous solutions is limited by the onset of current flthrough the substrate and the solution, which leads to checal oxidation, the reduction of solutes, and to electroly~bubble formation!. It has recently been demonstrated thfluid actuation can be achieved without electrolysis by coing the conductor or semiconductor substrate withdielectric.1,3,6,10The dielectric serves both to block the ele

a!Author to whom correspondence should be addressed; [email protected]; http://www.glue.umd.edu/;benshap/

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tron transfer and to provide a hydrophobic surface thatables large changes in contact angle. This electro-wettingdielectric ~EWOD! driven actuation has been used to credroplets from reservoirs, as well as to cut, join, and mdrops on planar surfaces or in channels.3,10 Applications ofEWOD include microfluidics and biofluidic sensors and dvices.

In order to design and control such devices, we requiaccurate models of the underlying physics. First, we nsome way of deciding which physical mechanisms are donant in the devices: is the ionic double-layer more or leimportant then the dielectric energy stored in the liquiWhat percent of the energy is being stored/dissipated inliquid bulk, solid bulk, and at the interfaces? Second,need to understand the engineering limits: why does conangle saturate? What limits droplet switching speed? Tarticle addresses some of these needs.

A. Background

Prior modeling results are based on the classical workLippmann 11 and Young ~see, for example, Chap. 10 iProbstein12!. More recent articles include Refs. 1,7,10,1325. In particular, the total energy minimization framewoproposed by Digilov26 is similar to our starting point. How-il:

4 © 2003 American Institute of Physics

license or copyright, see http://jap.aip.org/jap/copyright.jsp

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5795J. Appl. Phys., Vol. 93, No. 9, 1 May 2003 Shapiro et al.

ever, our method of analysis and physical interpretationdifferent; plus, we go on to numerically solve the surfacenergy/electrostatic minimum energy conditions andstudy the properties of the solutions.

Due to its engineering importance, there have beelarge number of articles focused on electro-wetting limitiphenomena: why does the contact angle cease to changesome critical voltage is reached? To date, some of theposed physical mechanisms include: electrolysis,24 contactline electrostatic/capillary instabilities for pure water,6 ion-ization of air in the vicinity of the drop edge,6 chargetrapping,5 and a proposed zero surface/liquid energy lim4

Charge/ion adsorption from the liquid to the solid surfaand its effect on the solid/liquid surface energy, is anotpossible source of contact angle saturation.27,28 The matchbetween contact angle saturation theory and experimenoften inconclusive, and/or the model parameters have bchosen to fit one set of data but have not been validaagainst a different independent set of data.~A notable excep-tion is the work of Verheijen and Prins.5 These authors showgood agreement between experiment and theory andpresent a second independent test to show that chargeping is responsible for contact angle saturation in theirvices.! It is possible, in fact likely, that different limitingphenomena are important in different devices: Vallet, Vlade, and Berge6 see luminescence in their devices and argthat gas ionization is one of their dominant phenomena.do not see any luminescence in our devices but we haveable to accurately predict contact angle saturation for mtiple devices, without fitting, by including the small electricresistance found in the liquid.

There have also been a number of studies about the etrical and chemical details at the interfaces: Lyklema29 pre-sents a comprehensive discussion of ion double-layer thries; Chou30 presents an analytic solution for the liquid/gshape right at the triple point under an applied potentZimmerman, Dukhin, and Werner31 provide an experimentaand theoretical treatment ofz potentials and solid/liquid conductivities due to ion adsorption; and Koopal and Aven32

provide an excellent description of adsorption kinetics.do not consider such fine-scale spatial details here.

B. Current approach

Our analysis is aimed at quantifying how different phycal effects~gravity, electrical resistance, ionic double laye!influence the electro-wetting phenomena. In this articleare only interested in those physical phenomena that inence voltage induced shape change. Essentially, wesome way of deciding which physical effects are importand which are negligible. We do this by finding the enerassociated with each effect, by minimizing the energy to fiequilibrium conditions, and by rigorously converting that eergy minimum into a Young-type equation that describeschange in droplet shape as a function of applied voltageother physical parameters. This lets us compare the relasizes of different effects. In this sense, our analysis is simin spirit to Digilov.26 However, when necessary, we phraand solve Maxwell’s partial differential equations to find t


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electric fields, and thus the stored energies, inside the sand liquid materials. Moreover, we have been able to dea rigorous equation which takes any arbitrary energy teand analytically gives back the corresponding, sizdependent, force term in Young’s equation.

Section II A presents the mathematical framework thtakes the energy term for any physical effect and computhe resulting force term in Young’s equation. Sections IIIand III B verify this framework for two simple examplewhere the answer is known and is straightforward, resptively. New ground is covered in Secs. IV A 4 and IV Bculminating with the contact angle saturation exampleSec. IV C.

The basic tenets of our analysis are: a total energy mmization, a phrasing of Maxwell’s electrostatic partial diffeential equations~PDEs!, an analytical extraction of how thePDE solutions change with analytically accessible paraeters, and a numerical solution of the remaining, normalizshape-dependent PDEs to capture parametric dependethat are not available analytically.

a. A total energy minimization approach with a constaliquid volume constraint: We write down the energy due tliquid/gas, liquid/solid and solid/gas interfaces plus the eergy stored in the bulk due to applied external fields suchgravity and the imposed electrical potentials. The energminimized subject to the constraint that the liquid volummust remain the same. This gives rise to a Young-type eqtion that can account for any physical effects and whichcludes droplet size dependence.

Although the link between energetics and Young-tyformulations has been explored partially~see, for example,Chap. 10 in Probstein12 and Refs. 16 and 15! this argumenthas traditionally been applied for a pure translation ofliquid/gas front: no change in droplet size is considered. Tmeans that the radiusR does not appear in the formulationand so all the size information is lost. Using this approachis fundamentally impossible to recover size dependent telike the ‘‘line-tension’’ 1/R-type term debated in the literature. This term is usually included based on phenomenolocal considerations, not derived from first principles, henthe debate. Our analysis includes variations in bothR anduand analytically recovers the size dependent terms. Tgiven the energy due to any physical effect, we can analcally write down the corresponding force term in Youngequation. In particular, we can state when line tension teexist, and we can derive these terms from physical first pciples.

b. Solution scalings for the electrostatic partial differential equations (PDEs): In order to find the electrical energieswe first find the PDEs and the relevant boundary conditiothat describe the electric fields inside the liquid and sophases.~Typically, Maxwell’s equations are sufficient fophrasing the right set of PDEs. But there are cases whereconsider other coupled effects such as the thermal diffuseffects found in the ionic double layer.! Before solving theresulting PDEs, we perform an analytic scaling analysisextract as many parametric dependencies as possible. Bdoing, we find how the solutions, and also the electricalergies, scale with system parameters such as the applied


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5796 J. Appl. Phys., Vol. 93, No. 9, 1 May 2003 Shapiro et al.

ageV and liquid radiusR, and with intrinsic material coeffi-cients such as the resistivitiesr and dielectric constantse. Inmost instances, this type of analysis is sufficient to revealunderlying nondimensional numbers that determinestrength of the various physical phenomena. For examthe Bond numberB5rgR2/g lg which determines the size ogravity terms compared to surface tension effects can becovered from a scaling analysis.

c. Solving the shape-dependent normalized PDEs: Oncewe have extracted the dimensional parameters such asvoltages, radius, heights, dielectric constants and resisties, it remains to solve the PDEs for the shape, oru, depen-dence. This is done numerically.

d. Finding the energy minima: From the scaling analysiand the shape-dependent numerical results, we can findtotal electrical energies as a function of the applied fielmaterial coefficients and droplet shape. By minimizing tenergy, we can find the contact angle as a function of pareters. At the end, the result depends only on a few dimsionless numbers. In the case of gravity, the contact adepends on the nondimensional surface tension numbeGand the Bond numberB. In the case of a resistive liquid atoa dielectric solid, the contact angle depends on the surtension coefficientG, on the insulating solid electro-wettinnumberU, and on the nominal ratio of solid to liquid resistanceAo .

e. Predict key phenomena, including line-tension acontact angle saturation: This article essentially performscareful engineering analysis of the bulk electrical and surftension properties of a sessile drop. Using this approachhave been able to rigorously show that a dielectric liqleads to 1/R line tension terms, but a conducting liquid donot. We have been able to assess the electrical resiscapacitive (RC) charging time constants, and we have beable to quantitatively predict contact angle saturation indevices. It will be shown that saturation, at least in ourvices, is caused by the small amount of electrical resistafound in the liquid. This explains why we continue to sessentially the same contact angle saturation behaviordifferent dielectric coatings of different thicknesses: the saration is basically fixed by the net resistance of the liquwhich depends on its sizeR, shapeu, and its intrinsic resis-tivity r l .

II. ASSUMPTIONS AND THE MATHEMATICALFRAMEWORK

Our attention is restricted to a single, approximatespherical, sessile drop in equilibrium, under applied exterfields ~such as gravity and electric potentials!, with variablematerial properties~solvent, ion type and concentration, anthe dielectric constants of the liquid and solid!. For this case,the modeling framework and underlying assumptionslisted below.

a. An energy minimization approach: We phrase allphysical effects in terms of energies~not forces!. From atautological standpoint this is attractive because all knoforces are derivatives of a potential energy~see Vol. I, Chap.14, Sec. 4 in Feynman33!. Nonconservative forces, which ar


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not written as the derivatives of a potential energy, are uwhen it is not possible to track the details of all the undering conservative forces. An energetics framework is alsovantageous from a practical standpoint. It is not at all clhow ion diffusion gradients give rise to forces at the tripphase line, but it is~relatively! straightforward to find thepotential energy associated with an ion distribution field, ato then perform the energy differentiation described in SIII to find the associated term in Young’s equation.

b. Drop shape: The drop is assumed to be essentiallyperfect sphere truncated at the solid plane, as shown in1. High gravitational or electrical forces can squash a dbut we assume that the applied external fields are sufficiesmall that this distortion is negligible. We also neglect adroplet deformation right at the triple line because weonly interested in the bulk, not local, shape of the drop. Tmeans that the shape of our drop can be uniquely descrby two numbers: the radiusR and the contact angleu. Afterwe have solved for the electric, gravitational, and other fieas a function ofR andu, the liquid drop has only these twdegrees of freedom left. The constant liquid volume costraint tiesR andu together and thus reduces the problema single degree of freedom.

The methods in this article can be extended to nspherical drops and puddles. In such cases, the spirit ofdevelopment is exactly the same, but the associated mematics needed to find the larger number of parameterdescribe the minimal energy liquid shape is more sophicated. See Brakke34 for how to compute complex minimaenergy surfaces.

c. Equilibrium: Thus far we have only addressed thequilibrium shape of the liquid drop under applied fields amaterial variations. To include droplet dynamics, which aimportant for issues such as maximizing droplet switchspeed in the electro-wetting devices described in Cet al.,35 two extensions will be required.

First, we will have to consider the time varying naturethe electric fields. This is done partially in Secs. IV C 1 aIV C 2 where we find that our resistive-capacitive (RC)charging time constants are on the order of milliseconSecond, and more importantly, it will be necessary to incporate our results into fluid simulations that solve the loReynolds limit of the Navier-Stokes equations for two-phaflows. Two points are important. A common concern is tvalidity of the continuum assumption~see Beskok36 for agood overview! which is not an issue in our micrometesized devices. Also, there is an inconsistency betweenface tension contact angle and viscous no-slip fluid bound

FIG. 1. Spherical drop geometry is parametrized by radiusR and~apparent!contact angleu.


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conditions:15,17,19,37if both boundary conditions are enforceand if the fluid is a realistic fluid where discontinuous veloity fields are not possible, then the triple line cannot moResolution of this issue is an active area of research.

d. No roughness or hysteresis: No surface roughness efects are included in the current model. The contact anhysteresis that arises from surface heterogeneities or roness can be modeled by energy considerations,21 and thuscan be incorporated into the current framework.

e. No evaporation: The liquid volume of the dropleshown in Fig. 1 is assumed to remain constant. If we wanto include the liquid volume change associated with evaration, we would need to formulate the energies associawith phase change and let volume become a variable insof a fixed parameter.

Rigorous conversion from energy minimum to themodified Young’s equation

This section presents the mathematics for convertingsessile drop potential energy function~including energies foreffects such as ion concentrations, electric fields, and mrial variations! into a Young-type equation. This link is rigorous and exact: there are no approximations associatedthe conversion. All approximations reside within incompleknowledge of the energies, or within the assumption thatdrop is a perfect sphere completely described by radiusRandcontact angleu. In Sec. III we will find the total potentiaenergyE(R,u;p) of the drop for different physical scenarioAt the end of all computations, this energy will dependthe drop radiusR, the ~apparent! contact angleu, and rel-evant system parametersp such as applied voltageV, dielec-tric constantse, and es , and nominal liquid ion concentrations co .

At equilibrium, the drop will assume a shapeR,u thatminimizes this energyE. This means that the derivative othe energy with respect toR andu is zero

dE5F]E

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Equation~1! says that at an energy minimum, the infintesimal change in energy due to shape variations muszero, and that there are two possible shape variations: onR and the other inu. It is not possible to changeu withoutalso changingR; if u increases in Fig. 1,R must decrease tokeep the drop volume constant~neglecting evaporation!. Thedrop volume is given by

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where q(u)52@2 cos2(u/2)cot(u/2)#/@21cosu#. A similarequation is derived in Decamps and De Coninck.20 Using Eq.~4!, Eq. ~1! can now be rewritten to show how the enerchanges with contact angle

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In order to get the traditional Young termg lg cosu to appearin this equation, it is necessary to pre-multiply Eq.~5! by2(21cosu)/2pR2 sinu. This term is strictly negative for alpossible contact angles 0,u,p so there is no division byzero. Thus

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is exactly Young’s equation, although written in a new waIn the special case whenE only includes the energie

due to liquid/solid, liquid/gas, and solid/gas interfaces wconstant surface tension coefficients, as in Sec. III A, tequation becomes exactlyg lg cosu2(ggs2g ls)50. How-ever, this formulation can handle any potential energy fution E(R,u;p). If we include additional effects such as eletrical energy in the solid, electrical energy in the liquigravitational terms, or ion concentration effects, then Eq.~6!will rigorously produce additional terms in Young’s equatio

III. TWO EXAMPLES AND THE SIZE DEPENDENTTERMS

The first example is a drop that only has energies dueinterfaces. The purpose of this example is to verify tframework of Sec. II A and to show that we exactly recovthe traditional Young equation in this simple case. The sond example includes gravity. This example shows how beffects are included in the analysis, the electrical fieldsSec. IV are included in the same way, and it demonstrahow scaling arguments can be used to extract the relenondimensional parameters. We close this section with ssection III C which converts size dependent energy terinto the corresponding size dependent terms in Younequation. This subsection shows when 1/R line-tension termsare active.

A. Interfacial potential energy

We start with a trivial example. If we only consider thpotential energy due to the solid/liquid, solid/gas, and liqugas interfaces, and if we assume the surface tension cocients are constant, then the sessile drop interfacial poteenergy is given by Probstein12 in Chap. 10


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For the drop shown in Fig. 1, it follows from purelgeometrical reasoning that

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Alg~R,u!52pR2~12cosu!. ~9!

In consequence, the interfacial potential energy is

Eint~R,u!5R2@~g ls2ggs!p sin2u1g lg2p~12cosu!#.~10!

As expected, the interfacial potential energy term scales wdrop radius squared. If we plugEint into the conversion described by Eq.~15!, then, after some half angle trigonometridentities, we exactly recover the traditional Younequation12 g lg cosu2(ggs2g ls)50.

B. Gravitational potential energy

We now consider the potential energy due to gravThis case is presented because it demonstrates some okey concepts, such as solution scaling, for a simple andtuitive example. In reality, for most practical microfluiddevices, gravity is negligible.

It is possible to find the form of the gravitational potetial by a simple scaling analysis.

A liquid element of volumeDv, of density%, at height, above the solid reference plane, will have a potentialergy due to gravity ofDEgvty5mg,5%g,Dv, where m5%Dv is the mass of the element andg59.81m/s2 is theacceleration due to gravity. The total potential energy duegravity is the integral over all the liquid elements within thdrop shape. For a drop of radius one, the integral of%g,dvover the drop shape will give some function ofu only:Egvty(R51,u)5agvty(u). If we increase the size of the droby a factor ofR but keep the shape, meaningu, the same,then the integral will change by a factor ofR4 – the ‘‘num-ber’’ of elements remains the same, but there is one factoR for the change in, and three factors ofR for the cubicchange indv. Hence the potential energy of the drop duegravity must be

Egvty~R,u!5R4agvty~u!, ~11!

whereagvty(u) is the shape form factor. ThisR4 dependencewill create anR2 term in Young’s equation:g lg cosu5(ggs

2g ls)1R2b(u), as described by Eq.~15! in Sec. III C be-low.


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As necessary, this factor is zero whenu50 ~total spreadingcorresponds to an infinitely thin, infinitely large puddle ameans no potential energy due to gravity! is maximal whenu5p ~no wetting!, and is strictly positive for allu in be-tween.

Combining this result with the Eqs.~10! and ~11!, thepotential energy due to the interfacial and gravitational teris

E~R,u!5R2@~g ls2ggs!p sin2u1g lg2p~12cosu!#

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3@31cosu#sin6~u/2!. ~13!

The interfacial term is at a minimum whenu is equalto the no gravity equilibrium contact angle. The graviterm is at a minimum whenu50 and so it tends to flattenthe drop: its effect is more pronounced for largdrops where the Bond ratioB5R4%g/R2g5R2%g/gis substantial. A standard calculation shows that for0.1 mm sized drop of water, the Bond numberapproximately (1024 m)23(103 kg/m3)3(9.81 m/s2)/(g lg

57.331022 kg/s2)50.0013, which means that the gravipotential energy is only 0.1% of the interfacial energy.

Using Eq.~15! derived below, and dividing through bg lg , the dimensionless Young’s equation for a liquid drwith gravity is

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C. Rka„u… energy terms lead to RkÀ2b „u… Youngterms

As shown in the two examples above, many potenenergy terms scale asE(R,u)5Rkak(u) whereRk is the sizedependence andak is a shape factor. Interfacial energy termin Sec. III A scale asR2a2(u), gravity terms scale asR4a4(u) in Sec. III B, the conducting drop will haveR2a2(u) scaling ~Sec. IV A!, and the dielectric drop willdisplay aRa1(u) scaling~Sec. IV B!. Some physical effectslike the fixed electrode height resistivity effect of SeIV C 4, will lead to energies that do not scale simply as poers ofR. But even in this case we can expand such termsa power series inR, or we can just apply Eq.~6! directlywithout the additional analysis described below.

Using Eq. ~6!, we see that aE5Rkak(u) energy termgives a


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contribution in Young’s equation.R4 energy terms~e.g.,gravity! lead toR2 effects in Young’s equation,R2a2(u) en-ergy terms~e.g., interfacial areas or insulating dielectric soids! reduce to pureu terms, andRa1(u) terms lead to 1/Rline-tension variations. This means that the conducting liqdrop in Sec. IV A whose electrical energy scales asR2 willproduce a Young’s equation with noR dependence. Howevethe dielectric liquid drop whose energy scales asRa(u) willhave a line tension term, and the magnitude of this termbe determined by the energy derivation in Sec. IV B andEq. ~15!.

IV. THREE EXAMPLES WITH ELECTRICAL ENERGIES

Here we consider three examples that include electrfields. A conducting liquid atop a dielectric solid is discussin Sec. IV A: this recovers the traditional Lippmann–Younrelation. In this section we also address the role of the iodouble layer. A dielectric liquid atop a conducting solidanalyzed in Sec. IV B: this case leads to a 1/R line tensionterm. Section IV C considers a slightly resistive liquid atophighly resistive dielectric solid this case recovers the conangle saturation behavior we observe in our devices.each example, we find the total potential energy, extractnondimensional parameters, and find the dimensionlmodified Young’s equation.

A. Conducting liquid atop a dielectric solid

In bio-chip applications, the water will contain an apprciable number of ions and will be a good conductor of eltricity: see Probstein,12 Sec. 2.5, for a relation between ioconcentrations and the resistivity or conductivity of water.prevent current flow, the dielectric coatings in our EWOdevices35 are designed to act as insulators. Thus, to a fiapproximation, the experimental arrangement in EWODvices can be described as a conductive liquid above an ilating, dielectric solid. It will be shown that this conductinliquid/ insulating solid case exactly recovers the LippmanYoung relationg lg cosu5@ggs2g ls1esV

2/2h#, but it doesnot lead to contact angle saturation or any line tens1/R-type terms.

Figure 2 shows the relevant geometry. Because the liqis conductive, the potential at the solid/liquid interfaceequal to the applied voltage:fsl5V. There are three sourceof potential energy: the interfacial energy derived in SIII A, the dielectric energy stored in the solid, and the enestored in the externally applied charging source.


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1. Potential energy in the solid dielectric layer

For a dielectric solid element at location (x,y,z), of vol-ume Dv, with local electric fieldE(x,y,z); the electricalpotential energy isDEde5

12(D•E)Dv. Here D is the po-

larazibility vector field: it is the induced dipole momentthe solid per unit volume, see Feynman33 Volume II, Chap.10, Sec. 2. For an ideal dielectric, this moment is linearelated to the local electric field byD5esE wherees is thedielectric constant of the sold. HenceDEde5

12esuEu2Dv. Ne-

glecting edge effects, the electric field immediately undersolid/liquid contact area isE52(0,0,V/h); it points straightdown with a strength equal to the applied voltageV dividedby the dielectric thicknessh. The electric field everywhereelse is zero as illustrated in Fig. 2. Thus1

2esuEu2dv must beintegrated over the volumev5hAls and this gives, togethewith Eq. ~8!, the energy stored in the solid dielectric

Ede~R,u!51

2esS V

h D 2

hAls5esV

2

2hpR2 sin2u. ~16!

If there aren solid dielectric layers, as opposed to the singdielectric layer considered above, thenes /h is replaced bythe net in-series capacitance per unit area 1/(h1 /e11 . . .1hn /en).

2. Potential energy stored in the external chargingsource

The basic reason this term has to be included is tevery time the drop shape changes, the charged volumemediately under the solid/liquid contact area changes, anpacket of chargeDQ must be received from or pushed baagainst the fixed voltage source. This requires an amounwork, or minus potential energy,W5VDQ52E. It followsthat the energy stored in the charge source is twice againenergy stored in the dielectric but with opposite sign. A caful exposition of this result can be found in Vol. II, Chap.Sec. 2 of Feynman33 and also in Verheijen and Prins,5 and soit is not repeated. HenceEcs(R,u)52(esV

2/h)pR2 sin2u.

3. Total energy and the Young –Lippmann equation

Combining the interfacial energy of Sec. III A with thdielectric and external source energy derived above, the tenergy for the conducting drop system is

FIG. 2. Left: Conducting drop atop an insulating dielectric layer of thicnessh. The voltageV is applied between the bottom-most flat conductielectrode and the electrode inserted into the top of the drop. Right: Smatic showing resulting dipole moments6 in the dielectric immediatelyunder the liquid/solid contact area; here the electric fieldE52(0,0,V/h)points down as shown by the arrows. The electric field is zero everywhelse.


e

,

eum

olec

ththp-oltto

decteicubymb

dtrirele

0

th

ouaceic

sotis

-

t.-

einow

and

n-the

ndith

di-is

ctricverstsedthe

ili-

ayshealy-edaseto

nga

as a


E~R,u!5R2F S g ls2ggs2esV

2

2h Dp sin2u1g lg2p~12cosu!G .~17!

Note thates is the dielectric constant of the solid, not thliquid.

Equation ~17! is identical to Eq.~10! except thatg ls

2ggs has becomeg ls2ggs2esV2/2h. Using the results of

Eq. ~15!, and dividing through byg lg to nondimensionalizewe exactly recover the Lippmann–Young relation

cosu2S ggs2g ls

g lg1

esV2

2g lghD50. ~18!

This equation contains no line tension 1/R terms because thenergy stored in the dielectric scales as the charged volin the solid, and this volume scales asAlsh;R2h. Sinceh isconstant, this stored energy behaves just like a liquid/sinterfacial energy term. To get a line tension term, it is nessary to have a physical effect whose energy scales asR, notasR2 ~see Sec. IV B!.

4. Effect of ionic double layer

There are two basic physical effects associated withdouble layer. The first is the capacitive energy stored indouble layer: this effect is negligible in our devices. Lipmann theory treats the ionic double layer as a Helmhcapacitor. As pointed out in Ref. 24, this is equivalenttreating the ionic layer as yet another material layer~so inour case we would then have three layers: silicon dioxiTeflon, and the ionic layer!. Since the thickness of the ionilayer ~nm’s! is much smaller than the thickness of the marial coatings (mm8s), the dielectric energy stored in the iondouble layer is negligible. It is possible to make this argment precise even when nonlinear effects in the ionic doulayer are considered. For the standard fully dissociated, smetric salt situation discussed in Refs. 38 and 12, it canshown~see the Appendix! that the ratio of the energy storein the double layer to the energy stored in the solid dielecmust fall beloweslD /e lhs which is on the order of 0.001 foour devices. Heree denotes the dielectric constant in thliquid and solid,lD is the Deybe double layer length scawhich is typically on the order of nanometers, whilehs is theheight of the insulating solid layer and it ranges betweenand 10mm in our devices.

The second physical effect is the possible change inliquid/solid surface tension coefficientg ls due to voltage in-duced surface chemistry. This effect can be important. Indevices, protein adsorption/desorption to the Teflon surfis modified by the applied voltage, and the adsorbed protchange the surface tension properties of the Teflon apprebly.

Consider first a simpler case. For a standard fully disciated symmetric salt, the change in the positive and negaion concentrationDc6 at the solid/liquid interface dependexponentially on the applied voltage as

Dc65coe7~zF/RT!Vdl ~19!

whereco is the far field ion concentration,Vdl is the voltagedrop across the double layer, andzF/RT is the characteristic


e

id-

ee

z

,

-

-le

-e

c

.1

e

re

nsia-

-ve

potential.12,38 If, in turn, the solid/liquid surface tension coefficient g ls depends on the wall ion concentration,g ls

5g ls(c6), as stated in Butkus and Grasso,28 then gsl be-comes a function of the applied voltage. Ifg ls(c6) is knownexperimentally, say from Butkus and Grasso,28 then Eq.~19!together with a voltage balance givesg ls5g ls(V). This mustthen be substituted into Eq.~7! and the voltage dependeng ls(V) will then appear in Eq.~17! also. The methods of SecII A and Eq. ~6! will now return the modified Young’s equation for this case.

More complex situations, such as those involving protadsorption/desorption, raise two key issues. First, hstrongly does the liquid/solid surface tension coefficientgsl

depend on the species concentration at the wall? ButkusGrasso28 find a moderate change ingsl based on electrolyteconcentration. Van der Vegtet al.27 find a much strongervariation of both the solid/liquid and liquid/gas surface tesion coefficients. Second, what is the transport rate ofchemical species from the liquid bulk to the solid/liquid aliquid/gas interfaces? And how does this transport vary wapplied voltage? As noted in van der Vegtet al.,27 chemicalspecies transport is a complex and important issue.

B. Dielectric liquid atop a conducting solid

We now compute the electric potential energy for aelectric liquid drop with an applied voltage. This casetreated because we are interested in transporting dieleliquids such as silicone oil, and because this case recothe controversial 1/R line tension terms from physical firsprinciples. Such terms are included in Refs. 26 and 22 baon phenomenological grounds. Below it is assumed thatdrop is an insulator with dielectric constante l and that thesolid is a perfect conductor; for example, a droplet of scone oil atop a metal electrode. A voltageV is applied asshown in Fig. 3.

1. Electrical energy scaling

For this case, we assume the top electrode is alwpositioned so that it only penetrates the tip of the drop. Tanalysis for a fixed electrode case is analogous to the ansis carried out in Sec. IV C 3. The end result for the fixelectrode case is similar to the varying electrode height cdiscussed here. Like in the gravity example, it is possiblefind the form of the electrical potential energy by a scaliargument. As in Sec. IV A, the potential energy stored in

FIG. 3. A dielectric liquid drop with dielectric constante l atop a conductingsolid. The bottom plate has a zero ground potentialf50, but the liquidimmediately surrounding the tip of the electrode at the top of the drop hf5V potential.


d

getasteb

,

o-

o

,ce

--

a

th

t t

,tle

lt-the

cal

a-

linerst

-ows

in

all

he0 °.entand

rgy

der-

f

he


small volume Dv of ~ideal! dielectric material isDEde

5 12e l uEu2Dv, wheree l is the dielectric constant of the liqui

andE(x,y,z) is the electric field in the liquid.To see how the electrical energy scales with volta

radius, and the dielectric constant, we need to undershow the electric fieldE varies with these parameters. Firconsider a drop of unit radius with a unit applied voltagThe electric potential field within such a drop is describedPoisson’s equation¹2f(x,y,z)50, with boundary condi-tionsfbottom50 andf top5V51. ~Side boundary conditionswhich are independent ofR andV, do not affect the scalingargument.! The electric field is then the gradient of the ptential field:E52¹f52(]f/]x,]f/]y,]f/]z).

Consider the potential fieldf(x,y,z) inside a drop ofunit radius with applied unit voltage. If we double the sizethe liquid drop then the potential fieldf is stretched by afactor of 2: fR51(x,y,z) becomes fR52(x,y,z)5fR51(x/2,y/2,z/2). This means that the electric fieldwhich is the rate of change of the potential in spawill become half as strong. Thus ER(x,y,z)51/RER51(x/R,y/R,z/R). Conversely, if we double the applied voltageV then the electric field will be doubled. Therefore, if we know the electric field at position (x,y,z) for adrop of unit size with unit voltage, then the electric field(Rx,Ry,Rz) for a drop of radiusR with applied voltageV is

ER,V~Rx,Ry,Rz!5V

RER51,V51~x,y,z!. ~20!

To find the stored potential energy, we must integrateenergy per unit volumeDEde5

12e l uEu2Dv over the drop

shape. Namely

Ede51

2ER,V drope l uER,Vu2dv,

51

2ER,V drope l

V2

R2uER51,V51u2dv,

51

2e l

V2

R2R3E

R51,V51 dropuER51,V51u2dv,

where the last equation is a consequence of the fact thavolume v scales asR3. The integral in the last line onlydepends on the shapeu ~both R and V are fixed to unity!hence

Ede~R,u!51

2e lRV2E

R51,V51 dropuER51,V51u2dv

51

2e lRV2ade~u!. ~21!

In summary, the electric fieldE varies asV/R; it appearstwice in the potential energy giving aV2/R2 dependencewhile the volumev scales asR3. Together, they imply thathe stored electrical energy for a dielectric liquid drop scaas 1

2e lRV2ade(u).


,nd

.y

f

,

t

e

he

s

As in Sec. IV A, the potential energy stored in the voage source is twice again the capacitive energy stored indielectric, but with opposite sign. Hence the total electrienergy stored in the system is

Eelec~R,u!51

2e lRV2ade~u!2e lRV2ade~u!

521

2e lRV2ade~u!. ~22!

Equation~15! implies that theR dependence inside this termwill give rise to a line-tension-type effect in Young’s eqution:

g lg cosu5~ggs2g ls!11

Rb~u!.

Thus we have been able to derive the phenomenologicaltension term cited in Refs. 22 and 26 from physical fiprinciples by using Sec. II A and a scaling argument.

2. Shape factor a de„u…

To find the form factorade(u), we need to solve Poisson’s equation for all possible drop shapes. Figure 4 shthe electric potential fieldf(x,y,z)5c contours for contactanglesu5154 °, 114 °, 78 °, and 37 °.

Form factor results for 14 contact angles are shownFig. 5. Notice thatade(u) is nearly independent ofu forcontact angles between 50 ° and 140 °. This is becausethe high electric potential gradients¹f that make up themajority of the integral occur at the top of the drop, or at ttop and bottom when the contact angle is close to 18Hence only a very small angle can impact the high gradiregion at the top, and only a very large angle can createthen affect the high gradient region at the bottom.

Using the form factor of Fig. 5, together with Eq.~22!,the potential energy for the interfacial plus electrical eneis

E~R,u!5R2@~g ls2ggs!p sin2u1g lg2p~12cosu!#

2Re lV

2

2ade~u!. ~23!

DefineW5e lV2/Rg lg as the nondimensional dielectric liqui

electro-wetting number. It is exactly this number that detmines the size of the 1/R line-tension term. For aR50.1 mm drop of silicone oil with a dielectric constant oe l52.5e ~from CRC handbook39! wheree is the permittivity

FIG. 4. Four drops of equal radius but different contact anglesu5154 °,114 °, 78 °, and 37 °. The constant electric potential contoursf(x,y,z)5c are shown for a vertical slice through each of the four drops. Tcalculated form factor for each drop isade(u)5*R51,V51 dropu¹fu2dv50.0592, 0.0609, 0.0617, and 0.0640, respectively.


h

xasbe

ulfo

dg

han

r

id

ec

. It

ys a

an

ein

tro--

peitlly:er-

ngse

ro-su-thef

the

diu the

hattro-

hownhigh

e at


of vacuum, and an applied voltage of 100 V the ratio of tinterfacial to electrical energies is approximatelya/2W5a/2Re lV

2/R2g lg5a/2e lV2/Rg lg'(0.06/2)3(2.538.85

310212 C/Vm)3(100 V)2/(0.0001 m)3(0.02 J/m2)50.0033. Evidently, less than 1% of the energy of our eample drop is electrical energy. We would have to increthe voltage up to 1000 V before the electrical energycomes appreciable; in that casea/2W'0.4.

Creating such a high voltage for such a small drop colead to dielectric breakdown: the electric field generated1000 V across a 0.1 mm drop isuEu;V/R5107 V/m. Foroils, dielectric breakdown typically occurs right aroun107 V/m. In terms of the electric field, the electro-wettinnumber W scales ase lR

2(V/R)2/Rg lg5e lRuEu2/g lg , so itwould actually make more sense to pick an electric field tis high but is substantially below the dielectric breakdowand then to increase the drop radiusR until W approachesunity. Such an experiment should allow one to see appciable line-tension effects.

3. Young’s equation for a dielectric liquid: The‘‘line-tension’’ term

Applying Eq.~15! to Eq.~23! and dividing byg lg , givesthe nondimensional Young equation for a dielectric liqudrop in terms of the electro-wetting numberW5e lV

2/Rg lg

cosu2S ggs2g ls

g lgD2

1

2 S e lV2

Rg lgD F2

21cosu

2p sinu G3Fade~u!q~u!1

dade

du~u!G50. ~24!

Hereq(u) is defined immediately below Eq.~4! andade(u)is shown in Fig. 5. Notice the 1/R ‘‘line-tension’’ depen-dence. We write ‘‘line tension’’ in quotes because the eff

FIG. 5. Circles show the computed form factorade(u) for 14 differentcontact angles. The stored energy in the liquid dielectric, for a drop of ra

R with applied voltageV, is now given byEde512e lRV2ade(u). Since the

energy stored in a capacitor isEde512CV2 this also gives the liquid drop

capacitance asC(u)5e lRade(u). Using u in radians, the equation for thesolid line fit is ade(u)'0.059210.0012u10.0022 tan(1.712u) and it onlyholds for 0.4,u,3 in radians, or equivalently for 25 °,u,172 ° in de-grees.


e

-e-

dr

t,

e-

t

is not, in fact, due to a line tension in any physical sensearises because the drop volume scales asR3 and the electricfield scales asV/R. Upon integration of the dielectric energthis gives anR-type energy dependence, which become1/R force dependence via Sec. III C, Eq.~15!. The exactsame scaling argument gives, for a conducting drop oninsulating surface, andR2 energy dependence in Eq.~17! andno 1/R line tension in Eq.~18!.

Equation~24! cannot be solved analytically, but can bsolved numerically. Figure 6 shows the resulting variationcontact angle as a function of the nondimensional elecwetting parameterW5e lV

2/Rg lg . The contact angle decreases only gradually with increasingW. This means thatdielectric liquids on conducting solids will change shaonly slightly under applied electric fields. It is clear whywould be difficult to measure such an effect experimentathe effect is small and it is sensitive to the dielectric propties of the liquid.

C. Slightly resistive liquid atop a dielectric,highly-resistive solid implies contactangle saturation

In Sec. IV A we considered the case of a conductiliquid atop a perfectly insulating dielectric solid: this carecovered the Young–Lippmann Eq.~18!, and was a first-cutmodel of the physical situation encountered in our electwetting devices. However, the assumption of a perfect inlator is unrealistic, and so we introduce the resistance ofsolid ~which is large by design! and also a small amount oliquid resistance~which is unavoidable in practice!. Liquidresistivity depends on the number and type of ions in

sFIG. 6. For a dielectric liquid atop a conducting solid, this plot showscontact angle dependence on the electro-wetting numberW5e lV

2/Rg lg forsix nominal~zero voltage! contact angles. The analysis above predicts tthe drop shape will snap-to complete wetting past some critical elecwetting numberW* . The predicted snap-to limitW* is within the plot rangefor the three bottom curves. Three cautions are necessary: first, the ssnap-to situation for the bottom three curves corresponds to a veryelectric field~the drop is thin and the voltageV is high!; second, the fit forade(u) used to generate these results does not hold foru,26 °; third, wesuspect that other physical effects, like electrolysis, will become activhigh V/low u, and this snap-to total wetting will not occur.


solutionlid


FIG. 7. Left: A bulk circuit diagram for a liquid with a small amount of electrical resistanceRliq , atop a dielectric solid with capacitanceCsol and a largeamount of electrical resistanceRsol ~by design!. Middle: The corresponding~steady-state! PDE with boundary conditions. Here,f(x,y,z) is the electricpotential inside the three-dimensional drop;r is the resistivity~units V m) wherer5r l inside the liquid is small andr5rs inside the solid is large, and

¹(1/r¹f)50 includes the liquid/solid electric field jump conditions;n is the outward unit normal and so¹f•n50 is the no-flux external boundarycondition; finallyf5V andf50 are the top and bottom boundary conditions applied by the voltage source. Right: This figure shows an exampleof the PDE equations. The lines show 28 equally spaced contours of constantf(x,y,z)5c for a vertical slice through the three-dimensional liquid and sogeometry. Notice that almost all the voltage drop occurs across the solid but there is also a small amount of voltage drop in the liquid.

iss

enhathe

atd2

nt

d

eWu

tio

gobu

ng

y

--

of

in

s

-

en-

ance

low.

di-tial

sis-nt

liquid, see Probstein,12 Sec. 2.5. These features are all thatrequired to replicate the contact angle saturation that wein our devices.

We note that many different physical effects can pottially cause contact angle saturation. Any kind of loss mecnism will cause the reversible dielectric energy stored insolid to deviate away from the ideal Young–Lippmann valuVerheijen and Prins5 present a convincing argument thcharge trapping is the dominant loss mechanism in theirvices. Other mechanism are proposed in Refs. 4,6 andWe stress three points here. One, a reasonable amouliquid resistance will cause contact angle saturation~see thedevelopment below!. Two, the saturation predicted by liquiresistance accurately matches the experimental data wein our devices~see Fig. 13!. Three, liquid resistance is thleading cause of contact angle saturation in our devices.examined a large number of physical mechanisms and liqresistance was the only physically meaningful assumpthat was able to explain our experimental data.

1. Equivalent circuit diagram

To understand how liquid resistance affects contact ansaturation, first consider the bulk circuit diagram shownthe left side of Fig. 7. When the total resistance is largefinite, there is a small amount of current flowI through theliquid and solid. Following standard electrical engineeripractice, the relation between the voltageV and the currentIis most conveniently expressed in the frequency domain bcomplex impedancez(s)5V(s)/I (s). ~Heres is the Laplacevariable. For a sinusoidal signalV(t)5V cos(wt) of fre-quencyw, takes5 iw. Settingw50 gives back the steadystateV(t)5V case.! The total impedance for the circuit diagram shown in Fig. 7 is

V~s!

I ~s!5z~s!5

11Rliq

Rsol1sRliqCsol

sCsol11

Rsol

. ~25!

If the liquid resistance is set to zero (Rliq→0) and the solidresistance is set to infinity (Rsol→`) to model a perfect in-


ee

--

e.

e-4.of

see

eidn

lent

a

sulator, then the above impedancez(s) reduces toz(s)51/sCsol and we recover the pure solid capacitive caseSec. IV A.

As before, all the reversible electrical energy is storedthe solid capacitor and the voltage source.~The liquid andsolid resistance only cause a non-reversible energy loss.! Theenergy stored in the solid capacitor is stillEde5

12CsolVsol

2 ,where Vsol is the voltage drop across the solid~seeFeynman,33 Vol. II, Chap. 22, Sec. 5!. To find this voltagedrop, note that the impedance of the solid iszsol

51/(1/sCsol11/Rsol), that the current through the liquid ithe current through the solid is the total currentI liq5I sol

5I , and that Eq.~25! relatesV(s) and I (s), hence

Vsol~s!5zsol~s!I ~s!5zsol~s!

z~s!V~s!

5S 1

11Rliq

Rsol1sRliqCsol

D V~s!. ~26!

Thus in steady state, i.e., ass5 iw→0, the voltage and energy stored in the dielectric are

Vsol5S 1

11Rliq

Rsol

D V, Ede51

2CsolS 1

11Rliq

Rsol

D 2

V2,

~27!

whereV is the applied dc voltage. This is the same depdence as shown in Eq.~16! for the perfectly insulating solid~sinceCsol5esAls /h), except for the newRliq /Rsol term. Thekey observation is that the resistance of the liquid dropRliq isshape dependent, and it is this dependence of the resiston the contact angleRliq5Rliq(u) that is going to lead tocontact angle saturation. The mechanism is elucidated be

2. PDE’s and their solution

Our first task is to find the PDE’s and boundary contions that describe the steady-state electric potenf(x,y,z) inside the liquid and the solid.

We have assumed that the liquid is a resistor with retivity r l but that it has no capacitive effects. The curre


wire


FIG. 8. Solution scaling: Both pictures show the electric potential inside the liquid and solid with the same color scale: white denotes high electricpotential,black denotes zero potential, and the curves denote surfaces of constantf(x,y,z). The notch at the top represents the inserted wire electrode: theinsertion depthD is fixed and is taken into account in the scaling argument. The picture on the left shows a solution of Eq.~28! with Ro51, h50.2 and

Ao50.2rs /r l510. The picture on the right shows a solution forRo51.5, h50.15 and the same liquid/solid resistance ratioAo50.15rs/1.5r l510. Noticethat the solutions are essentially self-similar. There is a small discrepancy because the scaling argument ignores the horizontal stretching of theelectrical edgeseffects in the solid region immediately underneath the triple line.

iv

i

yi-

thorrti

p

ote

t

d

isre

op

es

r

ndhen

e toee

le

the

ds

thowge,aly-1,

eese-lid,

the

density in the liquid is given byj l5E/r l whereE is the localelectric field. By comparison, the solid has both a resistand capacitive component with resistivityrs and dielectricconstantes . The instantaneous current density in the solidgiven by j5E/rs1esdE/dt. At steady state, thedE/dt termgoes to zero and we are left withj5E/rs . Conservation ofcharge states that the divergence of the current densitzero:¹ j50. Moreover, the electric field is minus the gradent of the electric potentialE52¹f hence

¹S 1

r¹f D50, r5H r l resistivity in the liquid

rs resistivity in the solid~28!

is the PDE that describes the electric potential inside bothliquid and the solid at steady-state. This formulation crectly includes the conservation of current flow in the vecal direction across the solid/liquid interface, namely:j z

5(]f/]z)/r is a constant across the interface withr5r l inthe liquid andr5rs in the solid, hence the solid/liquid jumconditions arers]f l /]z5r l]fs /]z.

Boundary conditions for Eq.~28! are as follows. Thepotential at the bottom of the solid is fixed at a nominal~andarbitrary! f50 potential. An inserted electrode at the topthe liquid is held atf5V by the applied voltage source. Aall the liquid/gas and solid/gas boundaries we use a znormal electric field conditionE•n5¹f•n50, wheren isthe outward unit normal. This last condition is analogousthe liquid/gas jump condition, here (r l /s /rg)Eg•n5El /s•n,except that we further assume that the resistivity of airrg islarge compared to the resistivity of the liquidr l and solidrs , and soEl /s•n is essentially zero at the liquid/gas ansolid/gas boundary.

A summary of the PDE and its boundary conditionsshown in the middle of Fig. 7. The right side of the figushows a sample solution for au5114 ° contact angle withapplied voltageV51, liquid radiusR51, solid heighth50.2, and resistivity ratio randomly chosen atrs /r l5230.This solution should be understood as follows: if the drshape were to somehow be held atu5114 ° and a voltageV51 were suddenly applied, the electric potentialf(x,y,z)inside the liquid and solid would approach the field linshown on the right side of the figure at a rate of 1/t. Thistime constantt is the charging time for the solid capacito


e

s

is

e--

f

ro

o

based on the available current flow through the liquid asolid. From the preceding section, it can be shown and testimated that

t5Csol

1/Rsol11/Rliq'

esR2/h

R2/rsh1R/r l

. ~29!

For a R51 mm water drop, withr l'53104 V m, rs

'1012 V m, es516310212 C/Vm 39 and solid heighth51026 m, this time-constantt is on the order of 1023 s.Because this time constant is quite fast, it is reasonabltreat the potentialf(x,y,z) as a steady-state quantity. Oncthe potentialf is known, the dielectric energy stored in thsystem is given by the integral ofdEde5

12(D•E)dv

5 12(es¹f)•(¹f)dv over the solid geometry

Ede~R,u,h,V,rs /r l ,es!51

2Esolesu¹f~x,y,z!u2dv. ~30!

Equation ~30! mathematically captures the contact angshape dependence left unsaid in Eq.~27!. As previously, thetotal electrical energy is the sum of the energy stored indielectric and in the voltage source:Eelec5Ede1Evs5Ede

22Ede52Ede.

3. Electrical energy scaling

Equation~30! shows how the electrical energy depenon the geometry (R,u,h), the applied voltageV, and thematerial propertiesrs /r l andes . Our task now is to flush ouand simplify this dependence so that we can understandthe energy minimum varies with geometry, applied voltaand material properties. This can be done by a scaling ansis just like the one used in Secs. III B, IV A 1, and IV Bbut with one additional key assumption.

If we look at the electric potential solution shown on thright of Fig. 7, we see that the potential field surfacfs(x,y,z)5c inside the solid are horizontal except right blow the drop edges. This is because the height of the soh50.2, is small compared to the radius of the liquidR. Inour electro-wetting devicesh/R,1024, hence the energycontent of the edge effects is tiny, and we can assumeelectric field in the solid is essentially vertical:Es'2(0,0,]fs /]z).


is.

od

s

n

rsceio-

r

thle-

an

rs

nptal-

a-s.

ich-itialis

e

-

al

ram-


Then the basic scaling result is this. If we take an exing solution to Eq.~28! with the boundary conditions of Fig7, and we stretch the liquid in thex,y,z directions by a factorR, and stretch the solid by a factorR in the x,y directions,and byh in thez direction, the end result is still a solution slong as the resistance of the liquid or solid is also changeas to keep the resistance ratioA05hrs /Ror l at its previousvalue. Figure 8 shows this scaling idea graphically.

This means that if we know the solution to Eq.~28! withthe boundary conditions of Fig. 7 for a fixed liquid radiuR51, solid heighth50.2, applied voltageV51, and forany contact angleu, any normalized electrode penetratiodepthD/R and any resistivity ratiors / r l5A, then we alsoknow the solution for any combination of parameteR,h,D,V,u,rs andr l . We are going to assume that the eletrode is always at a fixed heightH above the solid becausthis is how the experiment is actually done. Thus the relatbetween the radiusR, contact angleu and the electrode penetration depthD is D512cosu2H/R whereH is fixed butR,u andD vary. ~The electrode insertion depthD is shown inFig. 8 but not in Fig. 7.! Using this relationD5D(R,u) wecan suppress further discussion of theD parameter. In math-ematical terms, if we letf u,A( x,y,z) be the known solutionfor R51,V51 andh50.2, ~and choosez such thatz50 atthe solid/liquid interface! then

f~x,y,z!55 f l~x,y,z!5Vf lu,AS x

R,

y

R,z

RD in the liquid

fs~x,y,z!5Vfsu,AS x

R,

y

R,hz

hD in the solid

~31!

is a solution for arbitraryR, h, V, u, rs , andr l whereA mustbe set tohA5rsh/r lR. For example, to find a solution foR51 mm, h50.2 mm, V550 V, u5120 °, and rs /r l

5327, we first find the nondimensional solutionf for R

51, h50.2, V51, u5120 °, andA50.327, then the di-mensional solution is given by Eq.~31!. ~For a proof of thisstatement, see the Appendix.!

Using the above scaling, and noting once again thattotal electric energy is minus the energy stored in the dietric ~see Secs. IV A 2 and IV A 3!, we find that the total electric energy is given by

Eelec~R,u,h,V,rs /r l ,es!

521

2ER,h solidesu¹f~x,y,z!u2dv

521

2 S esR2

h DV2h

3ER51,h5.2 solid

U¹fs~ x,y,z!Uu,A5rsh/h

r lRU2dv

521

2 S esR2

h DV2haS u,rsh/h

r lRD . ~32!


t-

so

-

n

ec-

By using scaling arguments, we have managed to takeenergy that depends on six variables (R,u,h,V,rs /r l ,es),and rewritten it in terms of two nondimensional numbe(u,A) times a simple dimensional quantity (esR

2V2/h). Itremains to find the shape factora(u,A). We do this numeri-cally in the next section.

4. Shape factor a „u,A „R…… and the constant volumeenergy minimum

At this stage, we are within the energy minimizatioframework outlined in Sec. II. For our slightly resistive droatop a highly resistive solid, we could note that the toenergy of the dropE(R,u) is given by a sum of the interfacial energyEint(R,u;p1) in Eq. ~10!, and the electrical en-ergy Eelec(R,u;p2) in Eq. ~32!. We could then computea(u,A) numerically and solve Eq.~5! with outside param-eters p15(g ls2ggs,g lg) and p25(h,V,rs /r l ,es). Thiswould yield the equilibrium contact angleu as a function ofR,p1 andp2.

However, this process is tedious for the following reson. The shape factora(u,A) here depends on two variableTo map it out accurately we would have to evaluatea for atleast 15 values ofu and 10 values ofA. This is 150 solutionsof the three-dimensional PDE Eq.~28!. To get a sufficientlyfine-scale solution takes about 15 min per simulation, whis a total of 37.5 h of run time.~Of course we could parallelize the computations, and take previous solutions as inconditions for subsequent solution, but still, doing it in thway is a significant computational burden.!

Instead, we are going to use a short-cut. The volumv5v(R,u) of the liquid drop is fixed. Inverting Eq.~2! yields

R~u!

Ro5r ~u!5 3A 4/3

2

32

3 cosu

41

cos 3u

12

, ~33!

whereRo5A3 @3v/4p is the nominal radius of a drop of volume v that is a perfect sphere~so for u5p). Under theconstant volume constraint, the shape factora only has audependence

a~u!5aS u,rsh/h

r lR~u!D 5aS u,

Ao

hr ~u!D ~34!

with Ao5rsh/r lRo . Using this relation for the radiusR interms ofu, the total energy can be written in nondimensionform as

E~u!

g lgRo2

5r 2~u!F S g ls2ggs

g lgDp sin2u12p~12cosu!G

21

2 S esV2

hg lgD r 2~u!F haS u,

rsh

r lRo

1/h

r ~u!D G . ~35!

Notice the dependence on the three nondimensional paeters

G5g ls2ggs

g lg5nondimensional surface tension coefficient,


t

iveuld


FIG. 9. The strength of the electric field inside the solid, and thus the amount of stored electrical energy, decreases as the liquid drop approaches toal wetting.Here we show a case where the resistivity of the liquid is 50 times smaller than the resistivity of the solid. All scaling is according to Eq.~31!, but with thefigures drawn to show a constant electrode height. The solid is colored by the strength of its electric fieldu¹fsu, with black denoting a low electric field, lightgray up to white representing a high field. Notice how the electric field strength in the solid decreases as the droplet spreads and there is a progressly longerliquid path from the bottom of the top electrode to the solid near the triple line.~If there was no liquid resistance, the size of electric field in the solid woremain the same for all contact angles.!

te

tro

hecetain

io

es

lecets

of

id

uidluw9.

ol-

tlysis-oridtion

U5esV

2

hg lg5electro-wetting number for dielectric solid,

Ao5rsh

r lRo5solid/liquid resistivity ratio.

These three nondimensional parameters will uniquely demine the contact angle.

Figure 9 shows sample potential field solutionsf(x,y,z)for Ao510 for three values ofu. Here the solid is colored bythe magnitude of the local electric fieldu¹fs(x,y,z)u. Theconstant liquid volume shape factorha@u,Ao /hr (u)# is nowcomputed by numerically integratingu¹fs(x,y,z)u over thesolid ~edge effects are truncated!. Results are plotted againsu for four values ofAo in Fig. 10. As is necessary, the zeliquid resistance~infinite Ao case! reduces to the Sec. IV Ascenario withha(u,1000/h) indistinguishable fromp sin2u@compare with Eq.~16!#. As the resistance is increased, tform factor a begins to fall away from the zero resistancase, reflecting the fact that there is now a substantial voldrop across the liquid and less capacitive energy is bestored in the solid.

5. Detailed explanation of contact angle saturation

We can now precisely explain contact angle saturatthrough Figs. 11 and 12.

Figure 11 shows the net electrical energy~when 12U

51) as a function of the contact angleu for a liquid drop of

FIG. 10. The constant liquid volume shape factor of Eqs.~32! and ~34!. Ifthere is no liquid resistance, the form factor is proportional to the liqusolid area:a(u,`)5p sin2u. As the liquid resistance increases (Ao de-creases! the energy stored in the solid falls away from the ideal zero liqresistance case. Points on the graph above are found by a numerical soof Eq. ~28!. Whenu reaches 66 ° the top of the liquid drop has fallen belothe bottom tip of the inserted electrode: this effect can be seen in Fig.


r-

geg

n

constant volume. Different curves are shown for four valuof the solid to liquid resistance ratioAo5rsh/r lRo . Thesolid/liquid interface energy curveEsl(u) is shown for com-parison. For zero liquid resistance, theAo5` curve is themirror image of theEsl curve: Eelec(u)52Esl(u). In thiscase, Eq.~35! becomes

E5GEsl1Elg2 12UEsl5~G2 1

2U!Esl1Elg . ~36!

It is as if the applied voltage inU5esV2/hg lg were directly

changing the surface tension coefficients inG5(g ls

2ggs)/g lg . So this says that if we increaseU high enough~up to G2 1

2U521) then we would drive the contact angto u50. The left side of Fig. 12 shows this scenario notihow asU increases, the energy curve unbends, and aU53 ~whenG2 1

2U5 122 1

233521) the contact angle arrivesmoothly at total spreading.

But there is always some liquid resistance:Ao5” `. Asthis liquid resistance increases (Ao decreases! the electricalenergyEelec(u,Ao) deviates away from the ideal2Esl(u)value as shown in Fig. 11. This is just a consequence

/

tion

FIG. 11. This figure shows electrical energy curves for a fixed liquid vume. When there is no liquid resistance (Ao5`), the electrical energyexactly balances the solid/liquid interfacial energy:Eelec(u,Ao5`)52(2G/U)Els(u). This implies that the imposed electric energy can perfeccancel the energy due to the liquid/solid interface. When the liquid retance is nonzero (Ao,`), the electric energy deviates away from the mirrimage ofEsl and it is not possible to cancel the effect of the solid/liquenergy by driving up the voltage. This leads to the contact angle saturashown in Figs. 12 and 13.


ndwgreatermum


FIG. 12. Left: Total energy curves for a constant liquid volume as a function of contact angleu when the liquid resistance is zero. Different curves correspoto different electro-wetting numbersU5esV

2/hg lg5@0,0.5, . . . ,2.5,3#. The contact angle slides smoothly to zero asU increases. Right: The same plot, noincluding a small amount of liquid resistance:A5rsh/r lRo5100. At lower contact angles, there is a greater net liquid resistance, hence there is aenergy loss, and hence the applied electric field cannot drive the contact angle to zero. Consequently, the contact angle is caught in an energy miniaroundu'75 °.

fq

y

gehuta

ta

ctoe

Butto

ityi-

theon-of

thebe

solving Maxwell’s Eq.~28! with the boundary conditions oFig. 7 and using the constant liquid volume constraint of E~33!. This numerical result can be explained intuitively. Asudecreases, the radius of the drop increases~to keep the vol-ume constant! and in addition the liquid edges pull awafrom the fixed electrode~as shown in Fig. 9! this means thatthe ions in the liquid have to travel a longer distance tofrom the electrode at the top to the solid at the bottom. Tthe effective resistance of the liquid increases as conangle decreases. For greater liquid resistivitiesr l , the resis-tance first starts to increase appreciably at larger conanglesu.

We note that even a small amount of liquid resistanimplies that it is not possible to drive the contact anglezero with an applied voltage of any size. For a fixed volumthe interfacial area, and hence the energy, of Eq.~35! goes toinfinity as contact angle goes to zero at a rate ofr (u)


.

tsct

ct

e

,

;u22/3 due to an area versus volume scaling argument.the electrical energy goes to infinity at a slower rate duethe 1/r (u) term inside the shape factora in Eq. ~35!. Hencefor sufficiently smallu, the interfacial energy will alwaysbeat the electrical energy, the total energy will go to infinasu goes to zero, and sou50 can never be an energy minmum, no matter the applied voltage.

V. CONTACT ANGLE SATURATION MODEL VERSUSEXPERIMENTS

The experimental setup is as shown in Fig. 2. Forexperiments cited here, the insulating dielectric layer csists of either a single layer of Teflon or a double layerTeflon and silicon dioxide~see Fig. 13!. Silicon is used forthe bottom electrode, and a metal wire is employed fortop inserted electrode. More experimental details can

ainstry for

FIG. 13. Left: Measured contact angle vs applied voltage for four different Teflon/silicon oxide coatings. Right: The same data is re-plotted agthenondimensional electro-wetting numberU5esV

2/hg lg . The thin solid line shows the Young–Lippmann prediction. The two dashed lines show our theoa low and high liquid resistance. The thick solid line shows our prediction when we take the resistance ratioAo5100. Since for our experimentsh/R;1024, this corresponds to a liquid resistivity 1026 times smaller than the solid resistivity. We have not yet been able to measureAo experimentally~we haveto measure the resistance across the solid and in the liquid! but Ao5100 is of the right order of magnitude for our high resistance dielectric coatings.


hows thee, the rasize


TABLE I. Summarizes the examples of Sec. III. For each physical effect: column two lists the energy associated with that effect; column three sresulting term that appears on the right-hand side of Young’s equation; and column four gives the relevant nondimensional number. For exampltiobetween surface tension and gravity forces is given by the bond numberB. If there are many competing effects, then each effect will enter with acorresponding to its nondimensional number.

Physical Resulting energy term: Term on right in Nondimensionaleffect E(R,u,p1 ,p2 , . . . )5 Young: cosu5 . . . number Comments

Interfacialenergy

(g ls2ggs)Als1g lgAlg

see Eqs.~8!, ~9!, ~10!cosu52G

G5g ls2ggs

g lg

Exactly recoversYoung’s equation

Gravity R4%g2p

3@31cosu#sin6Su2D 2BFcosu

32

cos 2u

122

1

4G B5R2%g

g lg

Usually small

Dielectricsolid

2esV

2Als

2h52

esV2

2hR2p sin2u

see Eqs.~16! and ~17!1

12U U5

esV2

hg lg

Recovers theLipp–Young Eq.

Ion layercapacitance

2elAlsVdl

2

2lD>2

e lAls

2lDF2lDes

he lVG2

Vdl voltage across ion layer

<12U34D

D5lDes

he lvery small

Is negligible,see Sec. IV A 4

Dielectricliquid 2

12e lRV2ade(u)

See Eq.~22!, Fig. 51

12Wb(u)

for b, see Eq.~24!W5

e lV2

Rg lgNote the

1R

line

tension inW

Liquidresistance 2

esAls

2h S V

11Rliq /RsolD 2

52esR

2V2

2ha(u,Ao /R)

Not found explicitly,see Sec. IV C 4. Ao5

rsh

r lRois large

Liquid resistance leads tocontact angle saturation

egs

en

iapeacn

s

ta

m,avthinioe

f

rler-rmsa

rce

ents

Fig.thethe

his

orere

found in Moonet al.24 Experimental results are shown on thleft of Fig. 13. Results are plotted for four different coatinas contact angle versus voltage.

The first step is to re-plot this data against the nondimsional electro-wetting numberU5esV

2/hg lg of Eq. ~35!,then, as seen on the right of the figure, all the data essentfall on a single master curve. The theory we have develoin the preceding sections predicts this master curve. Weable to match all the data if we take a solid/ liquid resistanratio of A5100. Since the liquid radius in our devices is othe order of 10 000 times greater than the solid thicknesh,this corresponds to a liquid/solid resistivity ratior l /rs

;1026: this is all that is necessary to cause a 75 ° conangle saturation.

VI. RESULTS SUMMARY

a. Minimum total energy and Young’s equation: All theresults in this article are based on a minimum energy frawork. This in itself is not new, see Chap. 10 in Probstein12

and Refs. 15, 16, and 26 for example. However, we hmade a careful effort to extract as much information fromenergetics framework as is possible. We have explicitlycluded size dependence in the energy minimum formulatand have found an analytic relation between the changcontact angledu and the change in radiusdR necessary tokeep the liquid volume constant@see Eqs.~2! and ~4!#. Thisleads to Eq.~6! which is in fact exactly Young’s equation iwe consider the energy due to interfacial effects only~see


-

llydree

ct

e-

ee-n,in

Sec. III A!; but it further allows the inclusion of any otheenergy terms~due to gravity, capacitive effects, the doublayer, etc.!. Specifically, we have found a simple and inteesting link between energy scalings and the associated tein Young’s equation. Any physical effect that gives rise toE}Rk energy size dependence, will give aRk22 term inYoung’s equation~see Sec. III C!.

b. Summary of physical examples: Table I summarizesthe examples of Sec. III.

c. A triple-line force balance is insufficient: Much of theearly literature analyzed surface tension by phrasing a fobalance at the triple line only~see Fig. 14! The limitation ofthis viewpoint has been recognized in some recarticles.5,6,26 Essentially, if we have internal bulk forces aoccur in the case of gravity~the simplest example! or be-cause of internal electric fields such as the one shown in4, then we must balance the bulk volume forces againstinterfacial effects. To do so, one must either considerforces everywhere~not just at the triple line! or one must

FIG. 14. The left diagram shows a force balance at the triple line only. Tmodel cannot capture the effect of internal forces~shown schematically onthe right! such as gravity or the forces due to internal electric fields. Fexample, the electric field of Sec. IV B, Fig. 4 will create forces everywhinside the liquid.


.

uofigosi-ae

gylsn-ash

orthe

gte

y/est

t

arsth.tu

se

ecfiren

cteals

di

oft tlinecaf

ew

this

n:al-b-

de-rgytionm-

at

is-derr.

alall

ec.le

e

lem

l,n-the

n

e

d

di-


minimize the total system energy as we have done hered. Only consider gross liquid shape: We have ignored

local details of the liquid shape, meaning we do not accofor liquid pinching at the top electrode or for the detailsthe shape at the three phase line. Instead, we have asstwo numbersR andu which parametrize the gross shapethe liquid drop as shown in Fig. 1. For all the different phycal scenarios discussed in Sec. III, at the end we have alwexpressed the total energy in terms of these two numbhave then relatedR to u through Eq.~4! @or more directlythrough Eq.~33!# and have then found the minimum enercontact angleu. Our basic point is that including the detaiof the shape in the vicinity of the triple line is computatioally expensive, difficult to check experimentally, and, at lein our devices, unnecessary to explain phenomena sucline tension and contact angle saturation.

It is possible to extend our framework to account fdroplet deformation at the top electrode, and also forshape of droplets between two planar electrodes, or betwmultiple electrodes of any shape. Instead of considerintruncated sphere whose shape is uniquely described bytwo parametersR and u, we consider a drop whose shapis described by a longer list of parametersr5(r 1 ,r 2 , . . . ,r n). For example, if the drop is rotationallsymmetric,r j could be a list of points that define the liquidgas curve in the vertical plane. If the droplet is not symmric, then ther j ’s will define a discretized surface. To recathe analysis of Sec. II A, we find the energyE in terms of thisshape vectorr and physical parametersp. This involvessolving Maxwell’s equations as a function of the shaper . Wethen minimizeE(r ,p) with respect tor , subject to a constanvolume constraint]v(r )/]r 50, to find the minimum energyshaper* . Thus our semianalytic formulation is replaced bypurely numerical optimization. This formulation recovedroplet pinching at inserted electrodes, and it predictsshapes of drops squashed between two planar electrodestailed shape results for such cases will be presented in fupublications.

e. Numerical solution of the electrostatic PDEs pluscaling arguments:For cases that involve electric fields, whave solved the Maxwell’s PDEs that give rise to the eltrostatic energy terms. Moreover, in each case we haveused a scaling argument to elucidate how the energy depon parameters such as drop radiusR, insulating solid heighth, applied voltageV and material parameters like the condutivity and dielectric constants. Only after we have extracall possible parametric dependencies, do we numericsolve Poisson’s Eq.~28! for theu shape dependence. It turnout that the scaling arguments~the liquid electric field goesas the voltage over radius, the liquid volume scales as racubed, the solid volume scales as radius squared timesheight of the solid! can provide a tremendous amountinformation. In fact, scaling arguments alone are sufficienshow when line tension terms do and do not exist. Scaarguments reveal the underlying nondimensional numbthat capture the relative strength of the different physieffects, and scaling arguments can also be used to takeadvantage of a limited set of numerical solutions. Howevto predict the details of theu shape changes we need to kno


nt

nedf

ysrs,

tas

eenahe

t-

eDe-re

-stds

-dly

usthe

ogrslullr,

how the electrical energy changes with contact angle andrequires numerical solutions of Eq.~28! for varying dropletshapes.

f. Liquid resistance leads to contact angle saturatioFor our devices, we have found that including a small reistic amount of liquid resistance is sufficient to explain oserved contact angle saturation data. Basically, the shapependent resistance of the liquid drop leads to lower enestorage in the solid dielectric at small contact angles. SecIV C 5 provides a detailed analysis. Section V shows a coparison with experimental data.

ACKNOWLEDGMENTS

Dr. Shapiro would like to thank Dr. Elisabeth Smelathe University of Maryland for helpful discussions. Dr. Kimand Dr. Garrell would like to thank Dr. Junghoon Lee, whonow at Northwestern University, for his initial electrowetting modeling work. This work has been supported unDARPA Grant No. FCPO.0205GDB191, contact monitor DAnantha Krishnan.

APPENDIX A: MATHEMATICAL DETAILS

Equations~8!, ~9!, and~12! are all derived by partition-ing the spherical drop into infinitesimally thin horizontdisks of varying radii and performing an integration overthe disks.

The double layer capacitive energy ratio result of SIV A 4 is proved as follows. The Gouy–Chapman doublayer theory outlined in Hiemenz and Rajagopalan38 ~Sec.11.6! can be solved analytically for the potential in thdouble layer. Specifically, using normalized (ˆ) variables

f~ y!52 lnF 1212exp~Vl /2!

11exp~Vl /2!e2 y

1112exp~Vl /2!

11exp~Vl /2!e2 yG , ~A1!

wheref5zF/RTf is the normalized potential in the doublayer, y5y/lD is the normalized vertical distance away frothe y50 wall, lD5Ae lRT/2F2z2co is the Deybe lengthscale,f( y50)5Vl is the normalized potential at the walandz,F,R,T,co ande l are the charge number, Faraday costant, the gas constant, the far field ion concentration, anddielectric constant of the liquid. Differentiating Eq.~A1!

with respect toy gives the nondimensional electric field ithe liquid El52df/dy. Specifically, at they50 wall

El~ y50!52df

dyu y50522 sinh~Vl /2!. ~A2!

In Sec. IV A 4 we have a solid dielectric layer under thliquid ion layer. This layer has a dielectric constantes and avoltage dropVs . The total voltage drop across the liquid anthe solid must equal the applied voltageV5Vs1Vl . More-over, the electric field must satisfy the standard jump contion esEs5e lEl , Feynman,33 Vol. II, Chap. 10, whereEs is


ye

e

he

n

a

.ina

ore

n

eis

a12loto-u

as

in

ua-

e.

ster,

d

ins

e:

w

di-

-

tors


the electric field in the solid aty50. If the dielectric hasheighths , then by virtue of the fact thatEs5Vs /hs we re-cover ~after normalization!

22 sinh~Vl /2!5es /e l

hs /lDVs . ~A3!

This can be inverted and then bounded from above

Vl52 sinh21S 2es /e l

2hs /lDVsD<2 sinh21S 2

es /e l

2hs /lDVD ,

~A4!

whereV is the voltage applied across the ionic double laand the solid. The inequality follows fromVs5V2Vl<V.The energy of a single chargeq located at heighty above thewall is

u~y!5ezRT

zFf~y/ld!, ~A5!

wheree51.631029 C is the elementary unit of charge. Thcharge per unit volume in the double layer is

n~y!5NA@c1~y!2c2~y!#5NAco@exp~2f !2exp~f !#

522NAco sinhf~ y!, ~A6!

whereNA is Avogadro’s number. Multiplying Eqs.~A5! and~A6!, and simplifying the dimensional coefficients, gives tnet capacitive energy stored in the ionic double layer as

EDL cap51

2

e l

lDS RT

zFD 2E0

`

22f~ y!sinhf~ y!dy. ~A7!

The key point is that usingf( y) from Eq. ~A1! and theupper bound of Eq.~A4! it can be shown that the integral iEq. ~A7! is bounded by (eslDV/e lhs)

2. Thus the capacitiveenergy stored in the ionic double layer is much smaller ththe capacitive energy stored in the solid dielectric:EDL cap

<(eslD /e lhs)EDEs . This is the result stated in Sec. IV A 4Numerical solutions of Maxwell’s equations used

Figs. 4, 5, 7, 8, 9, 10, 11, 12, and 13 are carried outfollows. Poisson’s equation are phrased in cylindrical codinates with an assumed rotational symmetry about thzaxis: ¹2f(r ,a,z)5]2f/]r 211/r ]f/]r 1]2f/]z2. In allcases, we take¹fsides•n50 where n is the outward unitnormal at the liquid/gas or liquid/solid boundary. This codition assumes that the dielectric constant of aireg is muchsmaller than that of the liquide l or that of the solides .Boundary conditions for the remaining surface are outlinin the main text. The partial differential equations are dcretized and solved usingFEMLAB software ~www.femlab-.com!. Adaptive meshing is used because very high accuris required of the numerical solutions. Specifically, in Fig.we need to accurately find the energy minima inside shalwells. Even a 1% error in the numerical solution will leada significant lateralu error in the energy minimum placement. The numerical solutions shown in Fig. 12, and thalso Fig. 13, are accurate to within 0.01%.

The solution scaling of Eq.~31! for the slightly resistiveliquid atop a highly resistive dielectric solid is provedfollows. The proof proceeds by assuming thatf is a valid


r

n

s-

-

d-

cy

w

s

solution of the PDE and boundary conditions presentedSec. IV C 2, and then showing thatf is also a valid solution.The original liquid solution f l is multiplied by V andstretched by a factor ofR in all three directions. A stretchedand multipled field still satisfies the necessary Laplace eqtion ¹2f l50 ~within the liquid regionr5r l is constant andmay be moved outside the gradient operator!, the voltage atthe top of the drop goes fromf l(top)5V51 to f l(top)5V, the edges of the liquid solution are moved fromR51to R and ¹f l•n remains zero at the liquid/gas interfacLikewise, the solid potential fieldfs only has az component~approximately!, so if it is stretched byR in the x,y direc-tions and byh/h in the z direction then it still satisfies¹2fs5]2fs /]z250; thex,y scaling ensures that points juabove and below the liquid/solid interface move togethand the multiplication of bothf l and fs by V means thatf(x,y,z) remains continuous across thez50 liquid/solid in-terface; finally f(bottom)50 remains true. So the scalefield f l is a permissable solution in the liquid region, andfs

is a permissable solution in the solid region; it only remato show that the liquid/solid matching conditionrs]f l /]z5r l]fs /]z still holds. A stretching and magnifying of thpotential fields creates the following scaled electric fields

E~x,y,z!

55 ¹f l~x,y,z!5V

R¹f l

u,AS x

R,

y

R,z

RD in the liquid

¹fs~x,y,z!5hV

h¹fs

u,AS x

R,

y

R,hz

hD in the solid.

~A8!

Hence the liquid/solid electric field jump condition is nowritten

rs

]f l

]z5

rsV

R

]f l

] z5

r l hV

h

]fs

] z5r l

]fs

]z. ~A9!

But rs]f l /] z5 r l]fs /] z with rs / r l5A, thus ]fs /] z

5A]f l /] z, substituting this into equation~A9! gives, afterrearrangement and cancelation of the]f l /] z term, A

5 rs / r l5rsh/h/r lR. So the last necessary boundary contion is still satisfied whenA is chosen in this way.

1B. Berge, Acad. Sci. Paris C.R.317, 157 ~1993!.2M. Vallet, B. Berge, and L. Vovelle, Polymer37, 2465~1996!.3C. Kim and J. Lee,Proceedings Conference of IEEE Micro Electro Mechanical Systems Workshop, Heidelberg, Germany, 1998, pp. 538–543.

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Equilibrium behavior of sessile drops under surface ... · The basic tenets of our analysis are: a total energy mini-mization, a phrasing of Maxwell’s electrostatic partial differ-ential

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