This document discusses techniques for calculating electric potential, including: 1. Laplace's equation and its solutions in 1D, 2D, and 3D, including boundary conditions. 2. The method of images, which uses fictitious "image" charges to solve problems involving conductors. The classical image problem and induced surface charge on a conductor are examined. 3. Other techniques like multipole expansion, separation of variables, and numerical methods like relaxation are mentioned but not explained in detail.

1. 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

2. 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. 3.1.1 Introduction
The primary task of electrostatics is to study the interaction
(force) of a given stationary charges.
since
this integrals can be difficult (unless there is symmetry)
we usually calculate
This integral is often too tough to handle analytically.
2
0
ˆ1 ( )
4
F q Etest
RE d
R
ρ τ
πε
=
= ∫
v v
v
ˆˆ ˆx y zE E x E y E z= + + +×××
v
0
1 1( )
4
V d
R
E V
ρ τ
πε
= ∫
= −∇
v
∴
Q

4. In differential form
Eq.(2.21): Poisson’s eq.
Laplace’s eq
or
The solutions of Laplace’s eq are called harmonic function.
3.1.1
2
0
V
ρ
ε
∇ = −
2
0V∇ =
2
2 2 2
2 2 2
1 10 ( )
0
i
k i i i
k
TV A
h q h q
V V V
x y z
∂ ∂= ∇ =
∂ ∂∏
∂ ∂ ∂
+ + =
∂ ∂ ∂
 to solve a differential eq. we need boundary conditions.
 In case of ρ = 0, Poisson’s eq. reduces to

5. 3.1.2 Laplace’s Equation in One Dimension
2
2
0
d V
V mx b
dx
= ⇒ = +
m, b are to be determined by B.C.s
e.q.,
1. V(x) is the average of V(x + R) and V(x - R), for any R:
2. Laplace’s equation tolerates no local maxima or minima.
( 1) 4 1
5
( 5) 0 5
V x m
V x
V x b
= = = − 
⇒ ⇒ = − + 
= = = 
V=4 V=0
1 5 x
[ ]1
2
( ) ( ) ( )V x V x R V x R= + + −

6. Method of relaxation:
3.1.2 (2)
 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.

7. 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
3 4 1 0
4 4 0
5 4 0
6 4 0
7 4 0
8 4 0
9 4 0
10 4 0
4 3 2 1 0

8. 3.1.3 Laplace’s Equation in Two Dimensions
A partial differential eq. :
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.
2. V has no local maxima or minima; all
extreme occur at the boundaries.
3. The method of relaxation can be applied.
2 2
2 2
0
V V
x y
∂ ∂
+ =
∂ ∂
1( , )
2 circle
V x y Vdl
Rπ
= ∫Ñ

9. 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:
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.
2
1( )
4 sphere
V p Vda
Rπ
= ∫Ñ
2 2 2
0
1 , 2 cos
4
q
V r R rR θ
πε
= = + −
r r
1
2 2 22
2
0
1 [ 2 cos ] sin
44
ave
q
V r R rR R d d
R
θ θ θ ϕ
πεπ
−
= + −∫
2 2
0
1 2 cos 04 2
q
r R rR
rR
πθ
πε
= + −
( ) ( ) at the center of the sphere
0 0
1 1
4 2 4
q q
r R r R V
rR rπε πε
 = + − − = = 

10. 3.1.5 Boundary Conditions and Uniqueness Theorems
in 1D,
•
one end
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.
V mx b= +

11. 3.1.5
Proof: Suppose V1 , V2 are solutions
at boundary at boundary.
V3 = 0 everywhere
hence V1 = V2 everywhere
2
1 0V∇ = 2
2 0V∇ = 1 2 ?V V=
3 1 2V V V= −
2 2 2
3 1 2 0V V V∇ = ∇ −∇ =
1 2 3 0V V V= ⇒ =
∴

12. Proof.
at boundary.
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.
3.1.5
The first uniqueness theorem applies to regions with charge.
2
1
0
V
ρ
ε
∇ = − 2
2
0
V
ρ
ε
∇ = −
3 1 2V V V= −
2 2 2
3 1 2
0 0
0V V V
ρ ρ
ε ε
∇ = ∇ −∇ = − + =
3 1 2
3
0
0, . ., 1 2
V V V
V i e V V
= − =
∴ = =

13. 3.1.6 Conductors and the Second Uniqueness Theorem
 Second uniqueness
theorem:
Proof:
Suppose both and are satisfied.
and
0
1
1E ε
ρ∇× =
v
0
1
2E ε
ρ∇× =
v
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.)
2E
v
1E
v
0 0
1 1
1 1 2 2,
ith conducting ith conducting
surface surface
i iE da Q E da Qε ε
× = × =∫ ∫
v vv v
Ñ Ñ
0 0
1 1
1 2,
outer bourndary outer bourndary
tot totE da Q E da Qε ε
× = × =∫ ∫
v vv v
Ñ Ñ

14. for V3 is a constant over each conducting surface
define
3.1.6
3 1 2E E E= −
v v v
0 0
3 1 2 1 2( ) 0E E E E E
ρ ρ
ε ε
∇× = ∇× − = ∇× −∇× = − =
rv v v v
3 1 2 0E da E da E da× = × − × =∫ ∫ ∫
v v vv v v
Ñ Ñ Ñ
2
3 3 3 3 3 3 3( ) ( ) ( )V E V E E V E∇× = ∇× + ∇ = −
v v v
2
3 3 3 3 3( )
volume surface
E d V E d V E daτ τ= − ∇× = − ×∫ ∫ ∫
v v v
Ñ
3 3 0
surface
V E da= − × =∫
v v
Ñ
3 1 20E E E∴ = =
v v v
i.e.,

15. 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

16. 3.2.1 The Classical Image Problem
What is V (z>0) ?
( )
2 2 2 2
1. 0 0
2. 0
V z
V for x y z d
= =
→ + + >>
B.C.
The first uniqueness theorem guarantees that
there is only one solution.
If we can get one any means, that is the only answer.

17. Trick :
Only care
z > 0
z < 0
is not of
concern
3.2.1
in original
problem
for
V(z=0) = const = 0
( )0 0E z < =
v
0z ≥
( ) ( )2 22 2 2 20
1
( , , )
4
q q
V x y z
x y z d x y z dπε
 
− = +
 
+ + − + + +  
ˆ
0
0
0
z
x y
E E z
E E
at z
or E
= −
= = 
 ÷
= 
=P
v

18. 3.2.2 The Induced Surface Charge
Eq.(2.49)
0
ˆˆ
V
E z V z
z
σ
ε
∂
= = −∇ = −
∂
v
0
0z
V
z
σ ε
=
∂
= −
∂
0 0
z z
A
E A E
σ σ
ε ε
×
− = = −
0
0
Q
E da at z
ε
× = ≅∫
v v
QÑ
( ) ( )2 22 2 2 2
0
1
4
z
q q
z x y z d x y z d
σ
π
=
 
− ∂  
= − 
∂  + + − + + + 

19. total induced surface charge
3.2.2
Q=-q
3 3
2 2 2 2 2 22 2
0
1 1
2( ) 2( )
2 2
4
( ) ( )
z
z d z dq
x y z d x y z d
π
=
 
− × − − × + −  
= − 
    + + − + + +     
3/2 3/22 2 2 2 2
2 2
qd qd
x y d r d
σ
π π
− −
= =
   + + +
   
01/22 2
1
qd q
r d
∞−
= − = −
 +
 
2
3/20 0 2 2
2
qd
Q da rdrd
r d
π
σ θ
π
∞ −
= =
 +
 
∫ ∫ ∫

20. 3.2.3 Force and Energy
The charge q is attracted toward the plane.
The force of attraction is
With 2 point charges and no conducting plane, the energy is
Eq.(2.36)
( )
( ) ( )
2
2 2
0 0
1 1
ˆˆ
4 4 2
q q q
F z z
dd dπε πε
−
= = −
 − − 
v
( )
( ) ( )
( )
( )
2
1
20 0
2
0
1
2
1 1 1
2 4 4
1
4 2
i i i
i
W q V p
q q
q q
d d d d
q
d
πε πε
πε
=
=
  
    = × − + − ×    +  − −   
= −
∑

21. 3.2.3 (2)
or
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
( )
2
0
1
4 4
q
W
dπε
= −
( )
( )
2
2
0
2 2
0 0
1
4 2
1 1
4 4 4 4
d d
d
q
W F dl dz
z
q q
z d
πε
πε πε
∞ ∞
∞
= × =
 
= − = − ÷ ÷
 
∫ ∫
vv
ˆ( )dl dlz= −
v
Q

22. 3.2.4 Other Image Problems
Stationary distribution of charge
1
q
2
q
3
q
1
q−
2
q−
3
q−

23. 3.2.4 (2)
Conducting sphere of radius R
Image charge
at
R
q q
a
′ = −
2
R
b
a
=
( )
0
0
0
1
,
4
1
ˆˆˆˆ4
1
4 ˆˆˆˆ
q q
V r
q q
rr as rr bs
q q
a r
r r s b r s
r b
θ
πε
πε
πε
′ 
= + ÷
 
 ′
= + ÷
− − 
 
 ÷′
= + ÷
 ÷− − ÷
 
′r r
1
2 2 22 cosr a ra θ = + −
 
r
1
2 2 22 cosr b rb θ = + −
 ′r

24. Force
[Note : how about conducting circular cylinder?]
3.2.4 (3)
( )
0
1
, 0
4
ˆˆˆˆ
q q
V R
a R
R r s b r s
R b
θ
πε
 
 ÷′
= + = ÷
 ÷− − ÷
 
q q
R b
′
= −
b R
q q q
R a
′ = − = −
a R
R b
=
2
R
b
a
=
( ) ( )
2
2 22 20 0
1 1
4 4
qq q Ra
F
a b a R
πε πε
′
= = −
− −

25. 3.3 Separation of Variables
3.3.0 Fourier series and Fourier transform
3.3.1 Cartesian Coordinate
3.3.2 Spherical Coordinate

26. ˆˆ ˆ, ,x y zare unique because
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.
Completeness: a set of function ( )xfn
( ) ( )xfCxf n
n
n∑
∞
=
=
1
( )xf
is complete.
if for any function
Orthogonal: a set of functions is orthogonal
if ( ) ( ) 0=∫ dxxfxf m
b
a
n
mn ≠
=const
for
for mn =
in 3D. Cartesian Coordinate.zVyVxViVV zyx
N
i
i ˆˆˆˆ
1
++== ∑=

zyx VVV ,, are orthogonal.
ji ˆˆ ⋅ = i j=
i j≠0
1

27. 3.3.0 (2)
A complete and orthogonal set of functions forms
a basic set of functions.
e.g.
Nnk ∈,
( ) ( )sin sinkx kx− = −
( ) ( )cos coskx kx− =
odd
{ nkif
nkifdxnxkx ≠
=
−∫ = 0
coscos π
π
π
0cossin∫−
=
π
π
dxnxkx
{ nkif
nkifdxnxkx ≠
=
−∫ = 0
sinsin π
π
π
even
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.

28. 




for any ( )xf ( ) ( ) ( )
2
xfxf
xg
−−
= ( ) ( ) ( )
2
xfxf
xh
−+
=odd
( ) ( ) ( )xhxgxf +=
even
3.3.0 (3)
,
↑
odd
↑
even
[sinkx] [coskx]
Fourier transform and Fourier series
( ) ( )
0
(*) sin cosn n
n
f x A nx B nx
∞
=
= +∑
( ) ( )
1
sinnA f x nx dx
π
ππ −
= ∫
( ) ( )
1
cosnB f x nx dx
π
ππ −
= ∫
( ) 0
2
1
00 == ∫−
AdxxfB
π
ππ
} ∞⋅⋅⋅⋅= ,2,1n
;

29. 3.3.0 (4)
Proof
( ) ( )
( )
* sin
sin
kx dx
f x kxdx
π
π
π
π
−
−
×
⇒ ×
∫
∫
( ) ( )
0
sin cos sinn n
n
A nx B nx kx dx
π
π
∞
−
=
= +∑ ∫
( ) ( )
0
sin sinn
n
A nx kx dx
π
π
∞
−
=
= ∑ ∫
( ) ( )0
1
sinkA f x kx dx
π
π
∴ = ∫
if 0kA kπ= ≠

30. 0
cos cosn
n
B nx kx dx
π
π
∞
−
=
= ∑ ∫
( ) ( )
( ) ( )
* cos
cos
kx dx
f x kx dx
π
π
π
π
−
−
×
⇒
∫
∫
( ) ( )
0
sin cos cosn n
n
A nx B nx kx dx
π
π
∞
−
=
= +∑ ∫
( )0
1
2
B f x dx
π
ππ −
= ∫
{=
( ) ( )
1
cosnB f x nx dx
π
ππ −
∴ = ∫
3.3.0 (5)
B0 2π for k = 0
Bk π for k = 1, 2, …

31. 3.3.1 Cartesian Coordinate
Use the method of separation of variables to solve the Laplace’s eq.
Example 3
0)(
)()0(
0)(
0)0(
0
→∞→
==
==
==
xV
yVxV
yV
yV
π
Find the potential inside this “slot”?
Laplace’s eq.
2 2
2 2
0
V V
x y
∂ ∂
+ =
∂ ∂
set )()(),( yYxXyxV =
2 2
2 2
0
X Y
Y X
x y
∂ ∂
+ =
∂ ∂

32. 0
11
)(
2
2
)(
2
2
=
∂
∂
+
∂
∂

onlyydepedentygonlyxdepedentxf
y
Y
Yx
X
X
f and g are constant
1 2 0C C+ =
12
2
1
)( C
dx
Xd
X
xf ==
2
22
1
( )
d Y
g y C
Y dy
= =
2
21 kCC =−=set
so Xk
dx
Xd 2
2
2
=
2
2
2
d Y
k Y
dy
= −
kyDkyCyYBeAexX kxkx
cossin)(,)( +=+= −
=),( yxV )( kxkx
BeAe −
+ )cossin( kyDkyC +
3.3.1 (2)
1 2, are constant
( ) ( ) 0
C C
f x g y+ =

33. 3.3.1 (3)
B.C. (iv) ( ) 0 0, 0V x A k→ ∞ → ⇒ = >
B.C. (i) ( 0) 0 0V y D= = ⇒ =
=),( yxV kx
e−
)cossin( kyDkyC +
( , ) sinkx
V x y Ce ky−
=
B.C. (ii) ( ) 0 sin 0 1,2,3V y k kπ π= = ⇒ = = L
B.C. (iii) )()0( 0 yVxV ==
=)(0 yV
1
sinkx
k
k
C e ky
∞
−
=
∑ A fourier series for odd function
00
2
( )sinkC V y ky dyπ
π
= ∫
The principle of superposition =),( yxV
1
sinkx
k
Ce ky
∞
−
=
∑

34. 3.3.1 (4)
For antconstVyV == 00 )(
0
0
0
4 0
2
sin
0
2
(1 cos )
k
V
k
V
C ky dy
if k even
V
k
k if k odd
π
π
π
π
π
=
=

= − = 
=
∫
sin10 0
sinh
1,3,5,
4 21
( , ) sin tan ( )
yk
x
k
V V
V x y e ky
kπ π
− −
=
= =∑
L

35. 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

36. 1 1
2
1 cos
24 0
( ) cos
( )
d
r
qd
r
V p
θ
πε
θ− ≅
≅
r r+ -
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
4 0
( ) ( )
q q
V p
πε + −
= −
r r
2 2 2
2
( ) cos
d
r rd θ± = + mr
2
2
24
(1 cos )
d d
r r
r θ= ± +
2
(1 cos )
r d
d
r
r θ
>>
≅ m
1
1 1 12
2
(1 cos ) (1 cos )
d d
r r r r
θ θ
−
±
≅ ≅ ±m
r

37. 3.4.1
Example 3.10 For an arbitrary localized charge distribution.
Find a systematic expansion of the potential?
0
1 1
( )
4
V p dρ τ
πε
= ∫ r
2 2 2
2 2
2 cos
[1 ( ) 2( )cos ]
r r
r r
r r rr
r
θ
θ
′ ′
′ ′= + −
= + −
r
1 ( )( 2cos )
r r
r r
r θ
′ ′
= +∈ ∈= −r

38. 3.4.1(2)
1 1 3 52 2 3 3
2 8 16
[1 ( )( 2cos ) ( ) ( 2cos ) ( ) ( 2cos ) ]
r r r r r r
r r r r r r r
θ θ θ
′ ′ ′ ′ ′ ′
= − − + − − − +L
1 3 1 5 32 2 3 3
2 2 2 2
[1 ( )cos ( ) ( cos ) ( ) ( cos cos ) ]
r r r
r r r r
θ θ θ θ
′ ′ ′
= + + − + − +L
1
0
( ) (cos )
r n
nr r
n
P dθ ρ τ
∞
′
=
′= ∑
Monopole term Dipole term Quadrupole term
Multipole expansion
1
1 1 1 1 3 52 32
2 8 16
(1 ) (1 )
r r
−
∈ ∈ ∈= +∈ = − + − +L
r
1
( 1)4 0 0
1 1 1 1 3 12
2 34 2 20
1
( ) ( ) (cos )
[ cos ( ) ( cos ) ]
n
nn
n
r r r
V r r P d
r
d r d r d
πε
πε
θ ρ τ
ρ τ θ ρ τ θ ρ τ
∞
+
=
′=
′ ′= + + − +
∑ ∫
∫ ∫ ∫ L

39. 3.4.2 The Monopole and Dipole Terms
monopole
dipole
dominates if r >> 1
dipole moment (vector)
A physical dipole is consist of a pair of equal and
opposite charge,
1
4 0
( )
Q
mon r
V p
πε
=
1 1
24 0
1 1
24 0
( ) cos
ˆ
dip
r
r
P
V p r d
r r d
πε
πε
θ ρ τ
ρ τ
′=
′= ×
∫
∫
v
v
14243
ˆcosr r rθ′ ′= ×
v
ˆ1
24 0
1
( )
p r
dip
r
n
i i
i
V p
p r d q r
πε
ρ τ
×
=
=
′ ′= = ∑∫
v
v v v
q±
( )p qr qr q r r qd+ − + −′ ′ ′ ′= − = − =
vv v v v v

40. 3.4.3 Origin of Coordinates in Multipole Expansions
if
( )
p r d
r d d
r d d d
p dQ
ρ τ
ρ τ
ρ τ ρ τ
′=
′= −
′= −
= −
∫
∫
∫ ∫
vv
vv
vv
0Q p p= =
v

41. 3.4.4 The Electric Field of a Dipole
A pure dipole
ˆ cos
2 24 40 0
2 cos
34 0
1 sin
34 0
1
sin
34 0
( , )
0
ˆˆ( , ) (2cos sin )
P r P
dip
r r
V P
r r r
V P
r r
V
r
P
dip
r
V r
E
E
E
E r r
θ
πε πε
θ
πε
θ
θ θ πε
ϕ θ ϕ
πε
θ
θ θ θ θ
×
∂
−
∂
∂
−
∂
∂
−
∂
= =
= =
= =
= =
= +
v
v