代做Electrostatics、代写Java/c++设计、代做liquid droplet

日期：2018-12-03 10:50

Project 2. Electrostatics of a Water Droplet in Air

Liquid water droplets suspend in the air have been widely studied in a lot of

science and industrial applications, attracting up-to-date interest. The method of

image charges can be a very useful tool to understand the physics of liquid droplets

theoretically or computationally.

In this project, we study a spherical drop of water in the air. The drop is of

radius a (dimensionless unit). The water and air are treated as uniform dielectric

media of dielectric constant εw = 80 and εa = 1, respectively. When both r and

r

′ are not within the spherical cavity (Standing on the surface is OK), the Green’s

function of this problem is, G(r, r

′

) = 1/(εa|r r

′

|) + F(r, r

′

), which can be solved

by either the spherical harmonic series or the method of images (see the notes of

Lectures 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 =

1

2

q

2

AF(rA, rA)

as a function of radial distance rA by truncating the harmonic series and the method

of images. Test the accuracy, and do comparison of the results between the two

methods.

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 by

Upair = 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 quadratures

and Legendre polynomials.)

Reading materials:

A. Diehl et al., Surface tension of an electrolyte-air interface: a Monte Carlo

study, J. Phys.: Condens. Matter 24(2012) 284115

D. J. Tobias et al., Simulation and theory of ions at atmospherically relevant

aqueous liquid-air interfaces, Annu. Rev. Phys. Chem., 64(2013), 339-359

Note: Please print your project report with A4 paper. The due date to hand

in your project is Thursday, November 29, 2018.

1

1 Boundary-Value Problems and Interface Problems,

I

Boundary-value problems: Point charge in the presence of a grounded

conducting sphere. As illustrated in the figure below, we consider a point charge

q located at y outside a grounded conducting sphere of radius a and centering at

origin. We seek the potential Φ(x) such that Φ(|x| = a) = 0. By asymmetry, suppose

that only one image q

′

at y

′

is needed and that is lies on the ray from the origin to

the 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/4π0

Naturally, the following choices satisfy Φ(x = a) = 0,

2

This gives the expression for the magnitude and position of the image charge,

q

′ =

a

y

q, y′ =

a

2

y

.

Polarization and electric displacement. The collective response of the constituent

molecules of a material to an external electric field can be described the

phenomenological quantity, the susceptibility χ, which measures the displacement

of permanent dipole moment in polar molecules or the creation of induced dipole

moments in non-polar molecules. This quantity connects the polarization density P

per 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-field

approximation of dipole, the polarization potential is,

ΦP (r) = 1

We have,

ΦP (r) = 1

,

where the integration by part has been used, and S = ?V and n is the outward unit

normal vector to the surface.

The linear superposition with the potential due to free charge ρ(r),

ΦF (r) = 1

leads us to

Φ(r) = 1

We see the volume integral is just the expression for the potential caused by a charge

density (ρ · P). Since E =Φ, the equation for the electric field reads,

· E =

1

0

(ρ  · P).

3

Now 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 constant

or relative electric permittivity.

Interface conditions. In the presence of different media, boundary conditions

at the interfaces between media should be considered, which are the normal components

of D and the tangential components of E satisfies the continuities:

(D2 D1) · n21 = σ

(E2 E1) × n21 = 0

where n21 is a unit normal to the surface, directed from region 1 to region 2, and σ

is the macroscopic surface-charge density on the boundary surface. For the electric

potential, the continuous condition is, (Φ2 ? Φ1) = 0.

Laplace equation in spherical coordinates.

In spherical coordinates (r, θ, φ), the Laplace equation is,

4

We are looking for the solution of the form,

Φ = U(r)

r

P(θ)Q(φ).

Substituting it into the Laplace equation multiplied by r

2

sin2

θ/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, we

have 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 P

becomes,

which is the generalized Legendre equation, and its solutions are the associated

Legendre Polynomials Plm(x). In the case of m = 0, it is the ordinary Legendre

polynomials, which can be represented by the Rodrigues’ formula,

Pl(x) = 1

More 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.

5

Interface problems with azimuthal symmetry. For a problem with azimuthal

symmetry m = 0, the general solution of the Laplace equation is,

Φ(r, θ) = X∞

l=0

Alr

l + 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 rs

outside a dielectric sphere. The dielectric constants inside and outside of this sphere

are 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 the

Laplace equation, Φ = 0, for r < a. By the azimuthal symmetry, the Laplace

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 described

in terms of spherical harmonics,

Φ(r) = X∞

n=0

Anr

nPn(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,

6

where Φcoul = q/εo|r?rs| is the direct Coulomb potential, and F is the polarization

potential, also known as the image correction, due to the dielectric jump on the

spherical surface. Applying the multipole expansion of the reciprocal distance, the

general form of the potential Φ(r) reads

Φ(r) = q

n+1Pn(cos η),

where we have used the harmonic property of Φim, and r<(r>) is the smaller (larger)

one of r and rs.

Expansion of the free-space Green’s function. (Homework!) Prove the following

expansion holds,

Pn(cos η).

Now we choose the constant coefficients An and Bn so that the potentials inside

and outside the sphere satisfy the boundary conditions on the interface. The boundary

conditions are the continuities of the potential and the dielectric displacement,

Φin = Φout, and εi

Φin

r = εo

Φout

r ,

which lead to a set of two linear equations for each n,

n+1

an+2Bn,

and its solution is given by,

The polarization potential is then obtained:

F(r) = q

εo

X∞

n=0

a

2n+1

(rrs)

n+1

n(εo εi)

nεi + (n + 1)εo

Pn(cos η).

7

2 Boundary-Value Problems and Interface Problems,

II

Line image. Last lecture, we considered the following interface problem with a

unit source charge at rs outside a dielectric sphere. The dielectric constants inside

and outside of this sphere are εi and εo. We solve the potential Φ, which satisfies the

equation, Φ(r) = 4πδ(r rs), with interface conditions and zero boundary

condition at infinity.

We found the solution can be represented by Φ(r) = Φcoul(r) + F, where the

polarization potential is expressed as a harmonic series:

F(r) = 1

εo

X∞

n=0

a

2n+1

(rrs)

n+1

n(εoεi)

nεi + (n + 1)εo

Pn(cos η).

When εi → ∞, the polarization potential is a point image. When εi → εo, the

potential vanishes.

Now we define rK = rsa

2/r2

s

, x = xrs/rs and γ = (εiεo)/(εi + εo). Then, in F,

a

2n+1

(rrs)

n+1 =1

arn+1K

rn+1 ,

and

n(εo  εi)

nεi + (n + 1)εo

= γ +

γ(1 γ)

1γ + 2n

.

We have rK < a < r, and then

F(r) = 1

n+1Pn(cos η) + 1

r

n+1

γ(1 γ)

1γ + 2n

Pn(cos η) = S1 + S2.

8

We have

S1 =

rKγ

aεo|r rK|

.

On the other hand, we have the identity,

Z rK

Then by letting σ =

where the strengths of the Kelvin image and the line image are given by,

qK =

γa

rs

, and qline(x) = γσ

a

When a → ∞, the line image vanishes, and we have the results of a planar

interface.

The I-point Gauss-Legendre quadrature is used to approximate the line integral,

leading us to,

Φim(r) = qK

εo|r  rK|

+

X

I

m=1

qm

εo|r  xm|

,

with the charge strengths qm =

ωm2γa

rs

, and locations xm = rK

1sm

2

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