BoundaryValue Problems in Electrostatics IReading: Jackson 1.10, 2.1 through 2.10
We seek methods for solving Poisson's eqn with boundary conditions. The mathematical techniques that we will develop have much broader utility in physics.
Consider a grounded conducting plane at z = 0 and a point charge q at
q
z = 0
Dirichlet boundary conditions: (z = 0) = 0 and 0 as r ∞ in upper halfspace
What is for z > 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,
For z = 0, ( x, y, z) = 0. And, 0 as r ∞.
The 1st term satisfies Poisson's eqn in upper halfspace: it's the potential ofa point charge at
Explicitly:
The 2nd term satisfies Laplace's eqn in upper halfspace:
= 0 in upper halfspace (since does not lie in upper halfspace).
Thus, in upper halfspace and satisfies the boundaryconditions for the original problem, with the conducting plane at z = 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 placedat the location of the mirror image of q (for this simple geometry).
The attractive force on q,
The work done in bringing charge q from ∞ to z = z' is half the work done in bringing q and q to distance 2 z' :
Explicit calculation:
Suppose, instead of a single point charge, there were a charge distributionin the upper halfspace. 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 plane z = 0 is more complicated. For example, = V inside a circle of radius a centered 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 have already found the Green function
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Thus, the physical meaning of the Dirichlet Green function is:due to (1) a point charge at
plus (2) whatever external charges are needed to make = 0 onthe boundary (e.g.—charges induced on the surfaces, if they are conductors).
points AWAY from the region of interest)
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On the boundary z' = 0:
If = V inside a circle of radius a (centered on the origin) and = 0 otherwise, then we should adopt cylindrical coords (r, , z).
when there is no charge in the upper halfplane. Otherwise, add
to this.
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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 S is infinite)
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 plane z' = 0, use the “antiimage”:q
q (same sign)
Ez = 0
(the last equality holds when z' = 0)
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for z' = 0
For the infinite plane, S = ∞, so we want to enforce GN/n = 0 everywhere.
Let's check that for this example.
First, replace the upper half plane with a hemispherical volume, with radius R,and place on the zaxis. We already know that for z' = 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 all R > z and the infinite plane corresponds to R ∞.
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To explicitly verify that again consider a hemisphere with radius R.
On the plane z' = 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 a centered on the originand is zero outside. If there is no charge in the upper halfplane, 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 G
N
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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 q outside a grounded, conductingsphere of radius a
Suppose q is located at Can we find an image charge inside the spheresuch that = 0 on the sphere's surface By symmetry, it must lieon the xaxis. So, take an image charge q located 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 q for = 0 on surface)
Dirichlet Green fcn:
with ; ;
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Caution: I'm followingJackson's notation, wherex' is mag of vector, not x'component!
(Note: and b can easily be recalled with dimensional reasoning)
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 the zaxis:
Suppose = V (const) on the sphere.
, which checks.
Symmetry => zaxis is equivalent to any other axis =>
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Suppose = +V on “Northern” hemisphere
and = V on “Southern” hemisphere
Splitting integral into 2 pieces:
(check: = V when z = a)
Note: Given the solns for = const and = V in the Northern/ Southern hemispheres, superposition yields soln for any 2 different potentials in the 2 hemispheres.
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SP 2.1—2.3
Orthogonal Functions and Expansions; Fourier Series
This mathematical interlude is preparation for the next method of solvingelectrostatic boundary value problems—separation of variables.
Consider a set of functions Un() (n = 1, 2, 3, ...)
They are orthogonal on interval (a, b) if
* denotes complex conjugation:
Normalize so that the integral with n = m is unity. Then the functions are orthonormal:
The orthonormal functions Un() are “complete” if any wellbehaved fcn
f () can be expressed with arbitrarily small error as a series of them.
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In this case,
The coefficients an can be found using this trick:
So,
Thus,
(“completeness relation”)
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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 fcnis expressed in terms of them, it's called a “Fourier series”.
Useful trig identities:
Consider the interval
If n ≠ m,
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Similarly for integrals of
Normalization:
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Thus, the Fourier series expansion for a fcn f (x) is:
with
The ½ is included in front of A0 so that the above formula for A
m will
also work with m = 0. The interval (a/2, a/2) may be shifted by anyconstant a
0: (a
0 a/2, a
0 + a/2). e.g.: (0, a).
Note: The constant term is needed for completeness. The proof ofcompleteness can be extremely difficult, depending on how wellbehaved we require the fcn f (x) to be; we will omit it. (This is true of all thecomplete sets of orthogonal fcns that we will consider.)
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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, sinceare even functions and f (x) = x is odd.
So,
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SP 2.4, 2.5
Separation of Variables
Some solns of the Laplace eqn are of the form
For example:
[But not all solns are of this form; e.g., = A (x + y + z)]
Each of the 3 terms is a fcn of just 1 variable—x, y, or z.
=> Each term must be a constant.
Otherwise, if the sum = 0 for one value of x, then it won't for a differentvalue of x, since only the first term would have changed. But Laplace eqnmust hold for all x. Likewise for y and z.
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So:
The problem got vastly easier, since a PDE was replaced with 3 ODEs!The solutions are sines, cosines, and exponentials. When the boundary surfaces 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 single coordinate 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, c inthe x, y, z directions and one corner at the origin.
x
y
z z=c
y=b
x=a
= 0 on all the faces except z = c, where = V (x,y).Find everywhere inside the box.
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Since = 0 at both ends in x and y, X and Y need to be sines; nodifference of exponentials can vanish for 2 different values of x.
=> C1 and C
2 are negative
Similarly for Y:
Solns are
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Thus, separated solns that satisfy 5 of the 6 boundary conditions have the
form
Since the Laplace eqn is linear, linear combinations of solns are alsosolns. Can we find a linear combination of solns of the above separatedform 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.
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