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Boundary-Value Problems in Electrostatics I

Reading: Jackson 1.10, 2.1 through 2.10

We seek methods for solving Poisson's eqn with boundary conditions. Themathematical techniques that we will develop have much broader utility in

physics.

Consider a grounded conducting plane atz= 0 and a point charge qat

q

z= 0

Dirichlet boundary conditions: (z= 0) = 0

and 0 as rin upper half-space

What is forz> 0?

It's not just since charge is induced on the conductor.

Consider, for a moment, a completely different problem: no conducting

plane and 2 point charges:

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For this,

Forz= 0, (x,y,z) = 0. And, 0 as r .

The 1stterm satisfies Poisson's eqn in upper half-space: it's the potential of

a point charge at

Explicitly:

The 2ndterm satisfies Laplace's eqn in upper half-space:

= 0 in upper half-space (since does not lie in upper

half-space).

Thus, in upper half-space and satisfies the boundary

conditions for the original problem, with the conducting plane atz= 0.

=> This is the unique soln to the original problem.

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This is called the method of images,since the image charge is placed

at the location of the mirror image of q(for this simple geometry).

The attractive force on q,

The work done in bringing charge qfrom toz=z'is half the work

done in bringing qand -qto distance 2z' :

Explicit calculation:

Suppose, instead of a single point charge, there were a charge distributionin the upper half-space. Then there is an image charge for each part of

the distribution (works for both discrete and continuous dists).

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Suppose the potential on the planez= 0 is more complicated. For example,

= V inside a circle of radius acentered on the origin and = 0 outside

the circle. In this case, we can use a powerful method developed by Green.

Introduce Green functions which satisfy

Recall Green's Thm:

=>

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For the Dirichlet problem,choose such that

for ANY Dirichlet boundary conditions [i.e., any on the surface]!

Returning to the image problem with the conducting plane: We havealready found the Green function

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Along the circle axis, r= 0:

How to treat Neumann boundary conditions? Recall:

It is tempting to choose

Also:

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(unless surface area Sis infinite)

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Instead, take on surface, where S= total

surf. area of boundary

= the average of over the surface.

If the region of interest is infinite and 0 as r, then

To find for the infinite planez'= 0, use the anti-image:

q

q(same sign)

Ez= 0

(the last equality holds whenz'= 0)

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forz'= 0

For the infinite plane, S= , so we want to enforce GN/n= 0 everywhere.

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Let's check that for this example.

First, replace the upper half plane with a hemispherical volume, with radiusR,

and place on thez-axis. We already know that forz'= 0,so the flat part of the hemisphere does not contribute to the surface integral.

To evaluate the derivative at r'=R, express GN

in spherical coords.

This holds for allR>z and the infinite plane corresponds to R.

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To explicitly verify that again consider a hemisphere with radiusR.

On the planez'= 0,

as R.

For the curved part of the hemisphere, as R.

Total surface area = 3R2

as R.

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within a circle of radius acentered on the origin

and is zero outside. If there is no charge in the upper half-plane, then

On the circle axis (r= 0):

Check:

Symmetry of Green functions:

Green's Thm:

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=>

For Dirichlet boundary conditions,

Green's Reciprocity Theorem

plus external (.e.g.,induced) charge needed to satisfy boundary conditions. Reciprocity Thm

are interchangeable!

Obviously true for an isolated charge with no boundaries except at .

Remarkably, it remains true in the presence of conductors with fixed .

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For Neumann boundary conditions,

Redefine

So, if the original GN

satisfied and

then the new GN

will, too. Thus, we can choose GN

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Note: The freedom to add a constant to GN

arises because only

is specified on the boundary.

Another classic image problem: a charge qoutside a grounded, conductingsphere of radius a

Suppose qis located at Can we find an image charge inside the sphere

such that = 0 on the sphere's surface By symmetry, it must lie

on thex-axis. So, take an image charge qlocated at

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Must be true for all =>

Image charge must be inside sphere => b< d => 2< 1 => must choose

neg sign

(image charge must have opp sign as qfor = 0 on surface)

Dirichlet Green fcn:

with ; ;

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Caution: I'm following

Jackson's notation, wherex'is mag of vector, not

x'-component!

(Note: and bcan easily be recalled

with dimensional reasoning)

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In spherical coords (with the polar angle):

Law of Cosines:

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(normal points away from region of interest,

i.e., from exterior to interior of sphere)

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Soln of Laplace eqn is

Because of the complicated dependence of cos on , ', , ', the

general case does not yield analytic results.

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On thez-axis:

Suppose = V (const) on the sphere.

, which checks.

Symmetry => z-axis is equivalent to any other axis =>

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Suppose = +Von Northern hemisphere

and = -V on Southern hemisphere

Splitting integral into 2 pieces:

(check: = Vwhenz= a)

Note: Given the solns for = const and = Vin the Northern/

Southern hemispheres, superposition yields soln for any 2 differentpotentials in the 2 hemispheres.

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SP 2.12.3

O th l F ti d E i F i S i 22

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Orthogonal Functions and Expansions; Fourier Series

This mathematical interlude is preparation for the next method of solving

electrostatic boundary value problemsseparation of variables.

Consider a set of functions Un() (n= 1, 2, 3, ...)

They are orthogonalon interval (a, b) if

* denotes complex conjugation:

Normalize so that the integral with n = mis unity. Then the functions are

orthonormal:

The orthonormal functions Un() are completeif any well-behaved fcn

f () can be expressed with arbitrarily small error as a series of them.

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In this case,

The coefficients ancan be found using this trick:

So,

Thus,

(completeness relation)

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St i htf d li ti t lti l di i F l f 24

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Straightforward generalization to multiple dimensions. For example, for

variables

with

Sines and cosines form a complete set of orthogonal functions. When a fcn

is expressed in terms of them, it's called a Fourier series.

Useful trig identities:

Consider the interval

If nm,

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Similarly for integrals of

Normalization:

Th h F i i i f f f ( ) i26

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Thus, the Fourier series expansion for a fcn f (x) is:

with

The is included in front ofA0so that the above formula forA

mwill

also work with m= 0. The interval (-a/2, a/2) may be shifted by any

constant a0: (a

0- a/2, a

0+ a/2). e.g.: (0, a).

Note: The constant term is needed for completeness. The proof of

completeness can be extremely difficult, depending on how well-behaved

we require the fcn f (x) to be; we will omit it. (This is true of all the

complete sets of orthogonal fcns that we will consider.)

The following alternative Fourier series are also complete: 27

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The following alternative Fourier series are also complete:

1) with

on interval (0, a). NB: Not orthogonal on (-a/2, a/2)!

Fourier sine series

2) with

on interval (0, a). NB: Not orthogonal on (-a/2, a/2)!

3) with

Example:

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=> none of the cosines contribute. As expected, since

are even functions and f(x) =x is odd.

So,

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SP 2.4, 2.5

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So: 33

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The problem got vastly easier, since a PDE was replaced with 3 ODEs!

The solutions are sines, cosines, and exponentials. When the boundarysurfaces are planes, a sum of separated solns may satisfy the boundary

conditions.

There are 11 known coord systems for which the Laplace eqn is separable

(i.e., for which can be expressed as a product of 3 functions of a singlecoordinate with the Laplace eqn reducing to an ODE for each function).

We'll discuss spherical and cylindrical coords later.

Example in Cartesian coords: Rectangular box with lengths a, b, cin

thex,y,zdirections and one corner at the origin.

x

y

z z=c

y=b

x=a

= 0 on all the faces exceptz = c,

where = V(x,y).

Find everywhere inside the box.

Since = 0 at both ends inxandy, Xand Yneed to be sines; no 34

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y, ;

difference of exponentials can vanish for 2 different values ofx.

=> C1and C

2are negative

Similarly for Y:

Solns are

Thus, separated solns that satisfy 5 of the 6 boundary conditions have the 35

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form

Since the Laplace eqn is linear, linear combinations of solns are also

solns. Can we find a linear combination of solns of the above separated

form that will satisfy the 6th boundary condition? We require

This is a 2D Fourier sine series. Any V(x,y) can be expanded this way.