Transcript Document

Chapter 3 Special Techniques for Calculating Potential
3.1 Laplace’s Equation
3.2 The Method of Images
3.3 Separation of Variables
3.4 Multipole Expansion
3.1 Laplace’s Equation
3.1.1 Introduction
3.1.2 Laplace’s Equation in One Dimension
3.1.3 Laplace’s Equation in Two Dimensions
3.1.4 Laplace’s Equation in Three Dimensions
3.1.5 Boundary Conditions and Uniqueness
Theorems
3.1.6 Conducts and the Second Uniqueness
Theorem
3.1.1 Introduction
The primary task of electrostatics is to study the interaction
(force) of a given stationary charges.
since F  q
E
test
E 
1
4  0
ˆ
R
 ( 2 )  d
R
this integrals can be difficult (unless there is symmetry)
E  E x xˆ  E y yˆ  E z zˆ    
 we usually calculate
V  1  ( 1 )  d
4  0 R
E   V
This integral is often too tough to handle analytically.
3.1.1
In differential form
Eq.(2.21): Poisson’s eq.

 V 
0
2
 to solve a differential eq. we need boundary conditions.
 In case of ρ = 0, Poisson’s eq. reduces to
Laplace’s eq
or
2
 V 0
2
0 V 
 ( A 1 T )
1
i
hi  q i
 hk  q i
k
2
 V
x
2
2

 V
y
2
2

 V
z
2
0
The solutions of Laplace’s eq are called harmonic function.
3.1.2 Laplace’s Equation in One Dimension
2
d V
dx
2
 0  V  mx  b
m, b are to be determined by B.C.s
e.q., V ( x  1)  4
m  1
 
 V  x  5

V ( x  5)  0
b  5
V=4
V=0
1
5
x
1. V(x) is the average of V(x + R) and V(x - R), for any R:
V ( x) 
1
2
V ( x  R )  V ( x  R ) 
2. Laplace’s equation tolerates no local maxima or minima.
3.1.2 (2)
Method of relaxation:
 A numerical method to solve Laplace equation.
 Starting V at the boundary and guess V on a grid of interior points.
Reassign each point with the average of its nearest neighbors.
Repeat this process till they converge.
3.1.2 (3)
The example of relaxation
n V(x=0)V(x=1)V(x=2)V(x=3)V(x=4)
0
4
0
0
0
0
1
4
2
0
0
0
2
4
2
1
0
0
1
1
3
4
1
0
2
2
1
1
1
4
4
0
2
2
2
3
3
1
5
4
0
4
2
4
3
3
3
6
4
0
4
4
4
7
3
7
7
4
0
8
4
8
7
7
7
8
4
0
8
8
8
15
7
5
9
4
0
16
8
16
15
15
15
10
4
0
16
16
16
2
4
2
1
2
1
2
1
2
1
2
1
2
1
2
1
3
2
1
0
3.1.3 Laplace’s Equation in Two Dimensions
A partial differential eq. :
2
 V
x
2
2

 V
y
2
0
There is no general solution. The solution will be given in 3.3.
We discuss certain general properties for now.
1. The value of V at a point (x, y) is the average of those around
the point.
V ( x, y ) 
1
2 R
 circle V dl
2. V has no local maxima or minima; all
extreme occur at the boundaries.
3. The method of relaxation can be applied.
3.1.4 Laplace’s Equation in Three Dimensions
1. The value of V at point P is the average value of V
over a spherical surface of radius R centered at P:
V ( p) 
1
V 
2
2
2
r
 r  R  2 rR cos 
r
V ave
4 R
1 q,
4  0
1

2
 sphere V da
q
2
2
 [ r  R  2 rR cos 
1
] 2
2
R sin  d  d 
2
4  R 4  0
q
2
2

1

r  R  2 rR cos 
0
4  0 2 rR
q
q
1
1
  r  R    r  R   

 V at the center of the sphere
4  0 2 rR
4  0 r
The same for a collection of q by the superposition principle.
2. As a consequence, V can have no local maxima or minima;
the extreme values of V must occur at the boundaries.
3. The method of relaxation can be applied.
3.1.5 Boundary Conditions and Uniqueness Theorems
in 1D,
one end
V  mx  b
 Va
the other end
 Vb
V is uniquely determined by its value at the boundary.
First uniqueness theorem :
the solution to Laplace’s equation in some region is uniquely
determined, if the value of V is specified on all their surfaces;
the outer boundary could be at infinity, where V is ordinarily taken
to be zero.
3.1.5
Proof: Suppose V1 , V2 are solutions
2
2
 V2  0
 V1  0
V1  V 2 ?
V 3  V1  V 2
2
2
2
 V 3   V1   V 2  0
at boundary V1  V 2

V3 = 0
hence
 V 3  0 at boundary.
everywhere
V1 = V2
everywhere
3.1.5
The first uniqueness theorem applies to regions with charge.
Proof.

 V1  
0
2
2
 V2  

0
V 3  V1  V 2


 V 3   V1   V 2  

0
0 0
2
2
2
at boundary.
V 3  V1  V 2  0
 V 3  0,
i . e ., V1  V 2
Corollary : The potential in some region is uniquely determined if
(a) the charge density throughout the region, and
(b) the value of V on all boundaries, are specified.
3.1.6 Conductors and the Second Uniqueness Theorem
 Second uniqueness theorem:
In a region containing conductors and filled
with a specified charge density ρ,
the electric field is uniquely determined if the
total charge on each conductor is given.
(The region as a whole can be bounded by
another conductor, or else unbounded.)
Proof:
Suppose both E1 and E 2 are satisfied.
  E1 
and

1
0

E1  da1 
  E2 
1
0

1
0

Qi ,
E 2  da 2 
ith conducting
ith conducting
surface
surface

outer bourndary
E1  da  1 Q tot ,
0

outer bourndary
1
0
Qi
E 2  da  1 Q tot
0
3.1.6
define E 3  E1  E 2
  E 3    ( E1  E 2 )    E1    E 2 



0
0
0
 E 3  da   E1  da   E 2  da  0
for V3 is a constant over each conducting surface
2
  (V 3 E 3 )  V 3 (   E 3 )  E 3 (  V 3 )   E 3
volum e
2
E 3 d       (V 3 E 3 ) d    
 V 3
 E3  0
surface
 su rfa ce E 3  d a  0
i. e., E 1  E 2
V 3 E 3  da
3.2 The Method of Images
3.2.1 The Classical Image Problem
3.2.2 The Induced Surface Charge
3.2.3 Force and Energy
3.2.4 Other Image Problems
3.2.1 The Classical Image Problem
B.C.
z
 0  0
1.
V
2.
V  0 fo r x
2
 y
2
 z
2
What is V (z>0) ?
The first uniqueness theorem guarantees that
there is only one solution.
If we can get one any means, that is the only answer.
 d
2
3.2.1
Trick :
E   E z zˆ
Only care
z>0
 Ex  Ey  0


 at z  0

z<0
is not of
concern
or E  0
V(z=0) = const = 0
E  z  0  0
for z  0
V ( x, y, z ) 
in original
problem
1
4  0




q
x  y  z  d 
2
2
q

2
x  y  z  d
2
2

2




3.2.2 The Induced Surface Charge
Q
 E  da  
Eq.(2.49)
   0
V
z
Ez A 
z0
E 

1  
 

4  z 


0
x  y  z  d
2
0
 A
Ez  
0
zˆ    V  
q
2
at z  0

z
0
zˆ
q

2
V

x  y  z  d
2
2

2





z0
3.2.2

1
1
  2( z  d )
  2( z  d )

q 
2
2



3
3
4 
 x 2  y 2  ( z  d )2  2
 x 2  y 2  ( z  d )2  2
 



 qd
 
2  x  y  d


2
2
2
3/ 2
 qd

2  r  d


2
2
3/ 2
total induced surface charge
Q    da 
2
 qd

 0 0
2  r  d


2
1
  qd
r  d


2
2
1/ 2

0
 q
2
3/ 2
rdrd 
Q=-q






z0
3.2.3 Force and Energy
The charge q is attracted toward the plane.
The force of attraction is
F 
q q 
1
4  0  d    d  


2
zˆ  
1
q
2
4  0  2 d 
2
zˆ
With 2 point charges and no conducting plane, the energy is
W 
1
2
q iV i  p i 

2
Eq.(2.36)
i 1

1
1
  q  
2
4  0


1
q
2
4  0  2 d 
q 
1


 q  
 d d 
4  0






q
d  d 
2




 
3.2.3 (2)
For point charge q and the conducting plane at z = 0 the energy
is half of the energy given at above, because the field exist
only at z ≥ 0 ,and is zero at z < 0 ; that is
W 
W 
1
q
2
or
4  0  4 d 
d
 F  d l

1
4  0
 q2 




4   0  4 z 
1
d

d


q
2
2z 
1
2
dz
q
(
2
4  0  4 d

dl   dlzˆ )
3.2.4 Other Image Problems
Stationary distribution of charge
q1
 q1
 q2
q2
q3
 q3
3.2.4 (2)
Conducting sphere of radius R
R
Image charge q   
q
a
at b 
R
2
a
q 
q
V  r ,  
 

4  0 

1
r r

1 
q
q




4  0  rrˆ  asˆ
rrˆ  bsˆ 



1 
q
q




a
r
4  0 

ˆ
ˆ
ˆ
ˆ
r
r

s
b
r

s


r
b


1
r   r 2  a 2  2 ra cos   2
1
r    r 2  b 2  2 rb cos   2
3.2.4 (3)



1 
q
q
V  R ,  


0
a
R
4  0 

ˆ
ˆ
ˆ
ˆ
R
r

s
b
r

s


R
b


a

R
q
R
b
b

2
a
q
R
Force F 
R
q  
b
1
b
q
R
qq 
4  0  a  b 
2

R
q
a
2
1
4  0
q Ra
a
2
R
2

2
[Note : how about conducting circular cylinder?]
3.3 Separation of Variables
3.3.0 Fourier series and Fourier transform
3.3.1 Cartesian Coordinate
3.3.2 Spherical Coordinate
3.3.0 Fourier series and Fourier transform
Basic set of unit vectors in a certain coordinate can express
any vector uniquely in the space represented by the coordinate.
e.g.

V 
N
 V iˆ  V
i
x
xˆ  V y yˆ  V z zˆ
in 3D. Cartesian Coordinate.
i 1
V x , V y , V z are unique because xˆ , yˆ , zˆ are orthogonal.
i  j
iˆ  ˆj = 0
1 i  j
Completeness: a set of function f n  x  is complete.

if
x   
f
C n f n x 
for any function
n 1
Orthogonal: a set of functions is orthogonal
if

b
a
f n  x  f m  x dx  0
=const
for n  m
for n  m
f x 
3.3.0 (2)
A complete and orthogonal set of functions forms
a basic set of functions.
e.g.







sin kx sin nx dx 
cos kx cos nx dx 
  sin


0
if
if


0

kx cos nx dx  0
if
if
kn
k n
kn
k n
sin   kx    sin  kx 
odd
cos   kx   cos  kx 
even
k,n  N
sin(nx) is a basic set of functions for any odd function.
cos(nx) is a basic set of functions for any even function.
sin(nx) and cos(nx) are a basic set of functions for any functions.
3.3.0 (3)



f  x   f  x 
,
for any f  x  g  x  
odd ; h  x  
f  x   f  x 
2
f x   g x   hx 
2


odd
even
[sinkx] [coskx]
Fourier transform and Fourier series

(*)
f
x  
 An
sin nx  B n cos nx

n0
An 
Bn 
B0 
1

1


 

 
1
2

f
 x  s in  n x  d x
f
 x  cos  nx  dx


f
 x  dx

n  1, 2 ,    
A0  0



even
3.3.0 (4)
Proof

    *   sin  kx  dx


   f  x   sin kxdx



     An sin nx  B n cos nx  sin  kx  dx
n0



An
n0
 Ak 

Ak 

   sin  n x  sin  k x  d x
if k  0
1


0
f
 x  sin  kx  dx
3.3.0 (5)

    *   cos  kx  dx


   f  x  cos  kx  dx


     An sin nx  B n cos nx  cos  kx  dx

n0



n0


Bn 

1

Bn 
B0 

cos nx cos kx dx
   f  x  cos  nx  dx
1
2

   f  x d x

B0 2 for k = 0
Bk  for k = 1, 2, …
3.3.1 Cartesian Coordinate
Use the method of separation of variables to solve the Laplace’s eq.
Example 3
V ( y  0)  0
V (y   )  0
V ( x  0 )  V0 ( y )
V (x  )  0
Find the potential inside this “slot”?
2
 V
Laplace’s eq.
set
x
2
2

 V
y
0
2
V ( x , y )  X ( x )Y ( y )
2
Y
 X
x
2
2
 X
 Y
y
2
0
3.3.1 (2)
1  X
1  Y
2
2

X
 x

2
f ( x ) depedent
x only
0
Y  y

2
g ( y ) depedent
y only
f and g are constant
2
2
f ( x) 
1 d X
X dx
2
g( y) 
 C1
set C 1   C 2  k
Y dy
so
dx
2
2
C1  C 2  0
2
 C2
C 1 , C 2 are constant
f (x)  g( y)  0
2
2
d X
1 d Y
d Y
k X
2
X ( x )  Ae
dy
kx
 Be
V ( x , y )  ( Ae
kx
 kx
2
,
 Be
2
 k Y
Y ( y )  C sin ky  D cos ky
 kx
) ( C sin ky  D cos ky )
3.3.1 (3)
B.C. (iv) V ( x   )  0 
A  0,
k 0
V ( x , y )  e  kx ( C sin ky  D cos ky )
B.C. (i) V ( y  0)  0 
V ( x, y )  Ce
 kx
D 0
sin ky
B.C. (ii) V ( y   )  0 
sin k   0 k  1, 2, 3
The principle of superposition

V ( x, y )   C e
 kx
sin ky
k 1
B.C. (iii) V ( x  0 )  V 0 ( y )

V0 ( y )   C k e
 kx
sin ky
k 1
Ck
A fourier series for odd function
2 

0 V 0 ( y ) sin ky dy

3.3.1 (4)
For V 0 ( y )  V 0  const ant
Ck 
2V 0


0
sin ky dy
0
2V 0


(1  cos k  )   4V
0
k
 k 
V ( x, y ) 
4V 0


k  1,3,5,
1
k
e
k
if k  even
if k  odd
sin ky 
2V 0

tan
1
(
sin y
sinh x
)
3.4 Multipole Expansion
3.4.1 Approximate Potentials at Large distances
3.4.2 The Monopole and Dipole Terms
3.4.3 Origin of Coordinates
in Multipole Expansions
3.4.4 The Electric Field of a Dipole
3.4.1 Approximate Potentials at Large distances
Example 3.10 A dipole (Fig 3.27). Find the approximate potential
at points far from the dipole.
1
V ( p) 
r
2
4  0
2
2
d


1
r
d
(1
(
r
1
r+
cos  
r
 V ( p) 
d
2
1
r-
2
)
cos  )
cos  )

)
4r
d
2
r (1
r
rd cos 
r ( )
r
q

r
d 2
r  d
r
(
2
 r (1 
1
q
)

1
2
d
2
r

1
r
(1 
cos 
qd cos 
2
4  0
r
1
d
2r
cos  )
3.4.1
Example 3.10 For an arbitrary localized charge distribution.
Find a systematic expansion of the potential?
V ( p) 
r2
1
1
 r
4  0
 d
2
2
 r  r   2 rr  cos 
2
r 2
r
r
r
 r [1  ( )  2 ( ) cos  ]
rr
1 
r
r
r
r
  ( )(
 2 cos  )
3.4.1(2)
1

r

1

1

1
r
1
r
(1   )
[1 
r
r
1
2
1


2
r
r
r
r
( )(
1
r
(1 
1
2
 2 cos  ) 


3
8

2

3 r 2 r
8
r
r 2 3
r
r
( ) (
r
r
2

r n
 (r)
16

3

)
5
2
 2 cos  ) 
[1  ( ) cos   ( ) ( cos  
2
5
1
2
16
r 3 r
( ) (
r
r 3 5
r
3
)  ( ) ( cos  
r
3
 2 cos  ) 
3
2
2
2 3
1
2
2
]
cos  ) 
]
Pn (cos  )  d  
n0
V (r) 

1
4  0
1
4  0


n0
[
1
r
1
r
( n  1)
n
 ( r ) Pn (cos  )  d 
  d 
1
2
r
Monopole term
Multipole expansion
 r  cos   d  
Dipole term
1
3
r
 ( r ) ( cos  
Quadrupole term
)  d 
]
3.4.2 The Monopole and Dipole Terms
monopole
dipole
V m on ( p ) 
V dip ( p ) 

V dip ( p ) 
1
Q
dominates if r >> 1
4  0 r
1
1
4  0 r 2
 r  cos 
1
rˆ   r  d 
1
4  0 r 2
p  rˆ
4  0 r 2
1
P
 d
r  cos   rˆ  r 
dipole moment (vector)
n
p 
 r 
d 
 q i ri
i 1
A physical dipole is consist of a pair of equal and
opposite charge,  q
p  qr  qr  q ( r  r )  qd
3.4.3 Origin of Coordinates in Multipole Expansions
p 
 r 
d

 (r  d ) 

 r 
d
d  d   d
 p  dQ
if
Q 0
p  p
3.4.4 The Electric Field of a Dipole
A pure dipole
V dip ( r ,  ) 
Er 

E 

E 

V

r
1 V
r 
1
P  rˆ
4  0 r

2 P cos 
3
4  0 r

P sin 
3
4  0 r
V
0
r sin   
E dip ( r ,  ) 
2
P cos 
2
4  0 r
P
4  0 r
3
( 2 cos  rˆ  sin  ˆ )