Poisson’sEquationinElectrostatics

1 poisson s Equation in ElectrostaticsJinn-Liang LiuInstitute of Computational and Modeling Science, NationalTsing Hua University,Hsinchu 300, Taiwan. E-mail: 2010, 2011, 2012, 2017 AbstractPoisson s equation is derived from Coulomb s law and Gauss s theorem. It is a par-tial differential equation with broad utility in electrostatics, mechanical engineer-ing, and theoretical physics. It is named after the French mathematician, geometerand physicist Sim eon-Denis poisson (1781-1840). CharlesAugustin Coulomb (1736-1806) was a French physicist who discovered an inverse relationship on the forcebetween charges and the square of its distance. Karl Friedrich Gauss (1777-1855)was a German mathematician who also proved the fundamental theorems of algebraand Coulomb s Law, Electric Field, and Electric PotentialElectrostaticsis the branch of physics that deals with the forces exertedby a static ( unchanging) electric field upon charged objects [1].

2 The ba-sic electrical quantity is charge (e= 10 19[C] electron charge incoulombC). In a medium, an isolated chargeQ >0 located atr0= (x0,y0, z0)produces an electric fieldEthat exerts a force on all other charges. Thus, achargeq >0 located at a different pointr= (x,y, z) experiences a force fromQgiven byCoulomb s law[2] asF=qE=qQ4 r2r r0|r r0|[N] ,r=|r r0|.( )Adimensiondefines some physical characteristics. For example, length[L],mass [M], time [T], velocity [L/T], and force [N=ML/T2] (in newton). Aunitis a standard or reference by which a dimension can be expressed numeri-cally. In SI (the International System of Units or in the French name SystemeInternationale d Unites), themeter[m],kilogram[kg],second[s],amper e[A],19 April 2017kelvin[K], andcandela[cd] are the base units for the six fundamental di-mensions of length, mass, time, electric current, temperature, and luminousintensity [3].

3 The electric permittivity is conventionally expressed as = r 0, r= the dielectric constant or relative electric permittivity, 0= permittivity of vacuum = 10 12[C/(V L)] ,V= air at atmospheric pressure r= Air= Other dielectric constants Water= 80, Si= , InAs= , and GaAs= in room field E(r) (aforce per unit charge) produced byQatr0andfelt by the unit charge atris thus defined byE(r) =Q4 r2r r0|r r0|[N/C] .( )By moving a chargeqagainst the field between the two pointsaandbwith adistance x, work is done. That is, for 1D case,F x=qE x : =E x=forcecharge distance =workcharge[V=JC,J=NLjoule].( )This work per charge defines theelectric potential difference betweenthe pointsaandb. Here the Greek letter delta means a difference. Thissymbol will also be used in another meaning for the Laplace operator energy exists whenever an object has chargeqand is placed atrin an electric fieldEwhich is produced by the other chargeQlocated , the electric field can be defined by the potential asE(x) = x= lim x 0 (x+ x) (x) x= d (x)dx[in 1D]E(r) = grad (r) = (r) [in 2 Dor 3D]= ( x, y, z) (r).

4 ( )2 The negative sign above reminds us that moving against the electric fieldresults in positive work. Theelectric potential (r) is therefore a scalarfunction ofrand is the (positive) work per unit charge that we must pay formoving the unit charge from infinity to the positionrwithin the electric fieldgenerated Gauss s TheoremGauss s Theorem is a 3D generalization of the Fundamental Theorem of Cal-culus in 1D and Green s Theorem in 2D. The following theoremscan be foundin standard Calculus 1 (Fundamental Theorem of Calculus)Iffis a differentiablefunction on[a, b], then baf (x)dx=f(b) f(a).( )Line Integral (1D)=Point Evaluation (0D)Theorem 2 (Divergence Theorem of Gauss (1832))LetBbe an openbounded domain inR3with a piecewise smooth boundary B. Letu(x,y,z)bea differentiable vector function inB.

5 Then Bdivudr= B udr= Bu ndS( )Domain Integral (3D)=Boundary Integral (2D)Total Mass Change inB=Mass Flows across B,wherenis an outward unit normal vector on B. The integral Bu ndSisalso called theflux uacross the surface 3 (Mean Value Theorem (MVT))Iffis continuous on[a,b],then there exists a numberc [a, b]such that baf(x)dx=f(c) badx=f(c) (b a).( )33 poisson s EquationAssume that the electric fieldE(r) is differentiable in its domain R3. Tosimplify our presentation of using Gauss s theorem, we consider any subsetB as a ball with radiusrcentered atr0, ,B={r :|r r0|< r}whose boundary Bis a sphere. The following argument can be generalizedto any closed bounded domain as required by Causs s theorem with, of course,more technicalities [2]. We may think for the moment that thevariabler0is fixed.

6 Applying ( ) to the electric field in ( ), there existsrc Bsuchthatlimr 0 B E(r)dr= limr 0 B (r)dr( )= (rc)[limr 0 Bdr]by MVT( )= (rc)[limr 04 r33]( )= limr 0 B E(r) ndS.( )Eq. ( ) is valid if (r) is continuous onB. SinceBis infinitesimallysmall, the existing point guaranteed by MVT is nothing butr0asr 0, 0( )The surface integral = B E ndS[C]( )is called theelectric fluxofEthrough the sphere B. We can imagine theflux as the sum (total charge) of infinitely many small charges that havebeen flowing through infinitesimal surfacesdS(note the unit in Coulomb).Note also that ( ) gives us a hint that = 4 r2 E=Q( )is the total chargeQpassing through the sphere B. There are several situa-tions we need to be more specific about the limiting flux in ( ).Case charges withinBand at is passing through the sphere B.

7 Before taking the limit in both4( ) and ( ), we have zero for ( ) which implies that (r0) = 0 for allr0 ( )provided that any ball containingr0does not have any charge in it. Let us as-sume for the moment that the domain contains nothing and thepermittivityis constant. We then have theLaplace equation (r) = 0 for allr ,( )where = = 2 x2+ 2 y2+ 2 z2is the Laplace operator. Note that wereplacer0byrsince they both represent the same position notation for thedomain . However, when specifying charges at various locations, we must bevery careful about which charge is producing the fieldEatr0and what is thepotential felt by another charge at what locationr. One may wonder thatthe solution of ( ) is identically zero. Yes or no, it depends on the boundarycondition of ( ). Note carefully that we apply Gauss s theorem for anopendomain and say nothing about the charges on the boundary.

8 This isone of examples that we must be very rigorous on the precise statement ofmathematical definitions and theorems. In general, theLaplace problemcanbe stated as to find a solution (potential) such that (r) = 0, r ,( ) (r) = D(r), r D,( ) (r) n= N(r), r N,( )where = D N, Dand Nare given functions. ( ) is usuallycalled aDirichlet(or essential) boundary condition whereas ( ) calledaNeumann(or natural) boundary condition. Here (r) n= (r) nis adirectional derivative with the direction of the outward normal unitnto a single chargeQ0atr0and contains finitely manychargesQiatri,i= 1, ( ) and ( ), we have (r0)[limr 04 r33]= limr 0Q0=Q0which yields a very strange equation (r0) =Q0limr 04 r33( )5 The termQ0limr 04 r33has no physical meaning and can only beabstractedmath-ematically by means of the Dirac delta function [1] (a distribution as calledby mathematicians) (r r0) = ,r=r00,r =r0; R3 (r r0)dr= 1.

9 ( )For any continuous functionQ(r), the delta function has the fundamentalproperty that R3Q(r) (r r0)dr=Q(r0)( )Since the volume limr 04 r33is infinitely small, we can write limr 04 r33=drand define (r) as a continuous function such that (r) =Q0dr[CL 3]( )This function is thus called acharge density function. This is not a mathe-matically perfect way to define the density function for a point charge becausethe volumedris not definite. A more general definition for all kinds of chargesincluding the point charge is the following. We define the density function fora point chargeQ0atr0by means of delta function as (r) =Q0 (r r0)( )By ( ), we thus have R3 (r)dr= R3Q0 (r r0)dr=Q0( )Consequently, we have the followingPoisson equationfor a point charge (r) =Q0 (r r0)( )It should be noticed that the delta function in this equationimplicitly definesthe density which is important to correctly interpret the equation in actualphysical the domain contains isolated chargesQiatri,i= 1,2, , n, thePoissonequationbecomes (r) =n i=1Qi (r ri)( )6 Case contain infinitely many charges expressed by a densityfunction (r).

10 It is clear now that the total flux charges of ( ) flowing through the sphere Bis equal to the total charges residing in the open ballB, , by ( ), ( ),and ( ) = B E ndS= B (r)dr= B (r)dr( )These equalities hold for any arbitrary domainB. The last equality thereforeimplies thePoisson problem (r) = (r), r ,( ) (r) = D(r), r D,( ) (r) n= N(r), r N.( )4 poisson s Equation for Hard SpheresIn biophysics, charge particles are ions that are generallymodeled as hardspheres with finite diameters [?], namely, the conditionrc=r0asr 0( )is no longer feasible in such [1] Wikipedia: The Free Encyclopedia, [2] J. D. Jackson, Classical Electrodynamics, 3rded., Wiley, 1999.[3] J. D. Kraus and D. A. Fleisch, Electromagnetics with Applications, 5thed.,McGraw-Hill, 1999.[4] B. Eisenberg, Crowded charges in ion channels.