Project 2. Electrostatics of a Water Droplet in AirLiquid water droplets suspend in the air have been widely studied in a lot ofscience and industrial applications, attracting up-to-date interest. The method ofimage charges can be a very useful tool to understand the physics of liquid dropletstheoretically or computationally.In this project, we study a spherical drop of water in the air. The drop is ofradius a (dimensionless unit). The water and air are treated as uniform dielectricmedia of dielectric constant εw = 80 and εa = 1, respectively. When both r andr′ are not within the spherical cavity (Standing on the surface is OK), the Green’sfunction of this problem is, G(r, r′) = 1/(εa|r r′|) + F(r, r′), which can be solvedby either the spherical harmonic series or the method of images (see the notes ofLectures 2 and 3).1. Suppose a unit point charge, qA = 1, lies at rA outside the droplet surface.Calculate the self energy of this charge,Uself =12q2AF(rA, rA)as a function of radial distance rA by truncating the harmonic series and the methodof images. Test the accuracy, and do comparison of the results between the twomethods.2. Suppose there is another unit charge, qB = 1, on the surface of the droplet.Calculate the interaction energy between charges A and B, expressed byUpair = qAqBG(rA, rB)for different locations of rA. Use the two different methods for comparison.3. Analyze and discuss your results.(Hint: You can use the subroutines from Numerical Recipes for Gauss quadraturesand Legendre polynomials.)Reading materials:A. Diehl et al., Surface tension of an electrolyte-air interface: a Monte Carlostudy, J. Phys.: Condens. Matter 24(2012) 284115D. J. Tobias et al., Simulation and theory of ions at atmospherically relevantaqueous liquid-air interfaces, Annu. Rev. Phys. Chem., 64(2013), 339-359Note: Please print your project report with A4 paper. The due date to handin your project is Thursday, November 29, 2018.11 Boundary-Value Problems and Interface Problems,IBoundary-value problems: Point charge in the presence of a groundedconducting sphere. As illustrated in the figure below, we consider a point chargeq located at y outside a grounded conducting sphere of radius a and centering atorigin. We seek the potential Φ(x) such that Φ(|x| = a) = 0. By asymmetry, supposethat only one image q′at y′is needed and that is lies on the ray from the origin tothe charge q. Then,Φ(x) = q/4π0|x y|+q′/4π0|x y′|.Let x = xn and y = yn′ where n and n′ are unit vectors. Then,Φ(x) = q/4π0|xn yn′|+q′/4π0|xn y′n′|.Now the potential at x = a becomes,Φ(x = a) = q/4π0Naturally, the following choices satisfy Φ(x = a) = 0,2This gives the expression for the magnitude and position of the image charge,q′ = ayq, y′ =a2y.Polarization and electric displacement. The collective response of the constituentmolecules of a material to an external electric field can be described thephenomenological quantity, the susceptibility χ, which measures the displacementof permanent dipole moment in polar molecules or the creation of induced dipolemoments in non-polar molecules. This quantity connects the polarization density Pper unit volume and the electric field E by,P(r) = 0χE(r).For a given polarization density P within a volume V , and for r 6∈ V , by the far-fieldapproximation of dipole, the polarization potential is,ΦP (r) = 1We have,ΦP (r) = 1,where the integration by part has been used, and S = ?V and n is the outward unitnormal vector to the surface.The linear superposition with the potential due to free charge ρ(r),ΦF (r) = 1leads us toΦ(r) = 1We see the volume integral is just the expression for the potential caused by a chargedensity (ρ · P). Since E =Φ, the equation for the electric field reads, · E =10(ρ · P).3Now define the electric displacement,D = 0E + P,then the equation becomes, · D = ρ.As P(r) = 0χE(r), the displacement is proportional to E,D = E,where= 0(1 + χ)is the electric permittivity. The ratio 0 = 1+χ is called the dielectric constantor relative electric permittivity.Interface conditions. In the presence of different media, boundary conditionsat the interfaces between media should be considered, which are the normal componentsof D and the tangential components of E satisfies the continuities:(D2 D1) · n21 = σ(E2 E1) × n21 = 0where n21 is a unit normal to the surface, directed from region 1 to region 2, and σis the macroscopic surface-charge density on the boundary surface. For the electricpotential, the continuous condition is, (Φ2 ? Φ1) = 0.Laplace equation in spherical coordinates.In spherical coordinates (r, θ, φ), the Laplace equation is,4We are looking for the solution of the form,Φ = U(r)rP(θ)Q(φ).Substituting it into t代做Electrostatics作业、代写Java/c++课程设计作业、代做liquid droplet作业 代写留学生he Laplace equation multiplied by r2sin2θ/UP Q gives,The last term must be a constant, called (?m2), which has solutions,Q = e±imφ,where m must be an integer if the full azimuthal range is allowed. Similarly, wehave equation for P(θ) and U(r),where l(l + 1) is another real constant. The latter equation has solution U =Arl+1 + Brl.Legendre polynomials. By a transformation of x = cos θ, the equation for Pbecomes,which is the generalized Legendre equation, and its solutions are the associatedLegendre Polynomials Plm(x). In the case of m = 0, it is the ordinary Legendrepolynomials, which can be represented by the Rodrigues’ formula,Pl(x) = 1More conveniently, the polynomial is calculated by the recursive formula,(l + 1)Pl+1 (2l + 1)xPl + lPl?1 = 0,where P0 = 1 and P1 = x.5Interface problems with azimuthal symmetry. For a problem with azimuthalsymmetry m = 0, the general solution of the Laplace equation is,Φ(r, θ) = X∞l=0Alrl + Blr(l+1)Pl(cos θ).The coefficients are determined by boundary conditions.Now we consider the following interface problem with a source charge q at rsoutside a dielectric sphere. The dielectric constants inside and outside of this sphereare εi and εo. We are solving the potential Φ, satisfies the equation,· Φ(r) = 4πδ(r rs),with interface conditions and zero boundary condition at infinity.The absence of the source charge inside the sphere reduces the equation to theLaplace equation, Φ = 0, for r equation can be written as,where η is the angle between r and rs, and satisfies cos η = cos θ cos θs+sin θ sin θs cos(φ?φs). The general solution for the potential inside the sphere can be then describedin terms of spherical harmonics,Φ(r) = X∞n=0AnrnPn(cos η),where Pn(·) is Legendre polynomial of order n.Outside the dielectric sphere, the solution can be described by,Φ(r) = Φcoul(r) + F,6where Φcoul = q/εo|r?rs| is the direct Coulomb potential, and F is the polarizationpotential, also known as the image correction, due to the dielectric jump on thespherical surface. Applying the multipole expansion of the reciprocal distance, thegeneral form of the potential Φ(r) readsΦ(r) = qn+1Pn(cos η),where we have used the harmonic property of Φim, and r) is the smaller (larger)one of r and rs.Expansion of the free-space Green’s function. (Homework!) Prove the followingexpansion holds,Pn(cos η).Now we choose the constant coefficients An and Bn so that the potentials insideand outside the sphere satisfy the boundary conditions on the interface. The boundaryconditions are the continuities of the potential and the dielectric displacement,Φin = Φout, and εiΦinr = εoΦoutr ,which lead to a set of two linear equations for each n,n+1an+2Bn,and its solution is given by,The polarization potential is then obtained:F(r) = qεoX∞n=0a2n+1(rrs)n+1n(εo εi)nεi + (n + 1)εoPn(cos η).72 Boundary-Value Problems and Interface Problems,IILine image. Last lecture, we considered the following interface problem with aunit source charge at rs outside a dielectric sphere. The dielectric constants insideand outside of this sphere are εi and εo. We solve the potential Φ, which satisfies theequation, Φ(r) = 4πδ(r rs), with interface conditions and zero boundarycondition at infinity.We found the solution can be represented by Φ(r) = Φcoul(r) + F, where thepolarization potential is expressed as a harmonic series:F(r) = 1εoX∞n=0a2n+1(rrs)n+1n(εoεi)nεi + (n + 1)εoPn(cos η).When εi → ∞, the polarization potential is a point image. When εi → εo, thepotential vanishes.Now we define rK = rsa2/r2s, x = xrs/rs and γ = (εiεo)/(εi + εo). Then, in F,a2n+1(rrs)n+1 =1arn+1Krn+1 ,andn(εo εi)nεi + (n + 1)εo= γ +γ(1 γ)1γ + 2n.We have rK F(r) = 1n+1Pn(cos η) + 1rn+1γ(1 γ)1γ + 2nPn(cos η) = S1 + S2.8We haveS1 =rKγaεo|r rK|.On the other hand, we have the identity,Z rKThen by letting σ =where the strengths of the Kelvin image and the line image are given by,qK = γars, and qline(x) = γσaWhen a → ∞, the line image vanishes, and we have the results of a planarinterface.The I-point Gauss-Legendre quadrature is used to approximate the line integral,leading us to,Φim(r) = qKεo|r rK|+XIm=1qmεo|r xm|,with the charge strengths qm =ωm2γars, and locations xm = rK1sm2�1/σ, where{ωm, sm, m = 1, · · · , I} are the Gauss weights and locations on the interval [1, 1].Homework exercise! Derive image charges for a charge within a dielectric sphere.Additional reading: Cai, Deng, and Jacobs, Journal of Computational Physics,223, (2007), Pages 846–864转自：http://ass.3daixie.com/2018120413534730.html
