2. Electrostatics
2A. Method of Images
Vanishing Potential from Two Point Charges
•
•
•
•
Consider two charges q and q' of opposite sign
Lets assume without loss of generality that |q|  |q'|
To simplify, put q at origin and q' on the z –axis at a distance d
Potential is


1 
q
q

given by:
  x 

2 
2
2
4 0  x 2  y 2  z 2
x  y z d 

• Where is the
2
2
2
2
2
2

q
x

y

z

d


q
x

y

z


potential zero?
q 2  x 2  y 2  z 2  2 zd  d 2   q2  x 2  y 2  z 2 
q
2
 q2  x 2  y 2  z 2   2q 2 dz  q 2 d 2  0
q'
d
q
Case 1: Planes
• Suppose the two charges are equal and
opposite, then the potential vanishes at
2q 2 dz  q 2 d 2  0
q
2
 q2  x 2  y 2  z 2   2q 2 dz  q 2 d 2  0
z  12 d
• This is a plane that is equidistant between the two points
How is this useful?
• Suppose we have a plane with Dirichlet
boundary conditions  = 0 on the plane
• If you pretend there is an “image charge” – q on the other side
of the Dirichlet boundary, the two charges will produce a potential
that cancels on the boundary
• It will also satisfy 2 = –/0 in the allowed region
–q
d
q
Sample Problem 2.1 (1)
An infinite line of charge with linear charge density  is at z = h, x = 0 above a plane
conductor at z = 0. Find the electric field everywhere for z > 0, and the surface charge.
• The conductor is at constant potential

h
• Since it extends forever, assume it has  = 0
• Add an opposite charge line charge on the other side of the

conductor
 ρ
– This guarantees that  = 0 on the conductor E   ρˆ

o
• Electric field from the original line charge
2 0  2 0  2
•  is a vector pointing from the line
ρ  xxˆ   z  h  zˆ
charge directly to the point of interest
• So the electric field from the original line is E   xxˆ   z  h  zˆ
o
• The image line charge causes an electric field
2 0 x 2   z  h 2
• So the total field is:
  xxˆ   z  h  zˆ xxˆ   z  h  zˆ 
 xxˆ   z  h  zˆ
E
 2
 2
Ei 
2
2 
2
2 0  x   z  h 
x   z  h  
2 0 x 2   z  h 
Sample Problem 2.1 (2)
An infinite line of charge with linear charge density  is at z = h, x = 0 above a plane
conductor at z = 0. Find the electric field everywhere for z > 0, and the surface charge.

E
2 0
 xxˆ   z  h  zˆ xxˆ   z  h  zˆ 
 2
 2
2
2 
x   z  h  
 x   z  h 

h

• To find the surface charge on the conductor, first find the electric field there
 hzˆ
  xxˆ  hzˆ xxˆ  hzˆ  
E z 0 
 2
2
2
2
2
2 


x

h


0
2  x  h
x h 
0
• Note that it is perpendicular to the surface there, as it must be


• Related to surface density by
 h
ˆ
E  n  zˆ


0
0
  x2  h2 
• Can show that, in this case, integrated surface charge cancels the original line
Case 2: Spheres
2
2
2
2
2
2
2 2

• Suppose |q| > |q'|
q

q
x

y

z

2
q
dz

q
d 0



then the potential
2
2 2
q
d
q
d
2
2
2
vanishes when
x  y  z 2 2
z 2
0
2
2
q  q
q  q
• Complete the square
2
 qd 
 qd 
qd
qd
x  y  z 2 2
z 2
 2
 2
2
2 
2
2 



q  q
q

q
q

q
q

q




2
2 2 2
2
4 2
2 2


q
q d
q
d
q
d
q
d
2
2

x  y z  2

 2
2
2 
2
2
2 2
2
2


q

q
q

q

  q  q 
 q  q 
2
2
2
2
2
2
2
• This is a sphere of radius:
• Distance from q to center:
• Distance from q' to center:
qqd
a 2
q  q2
q2d
r 2
q  q2
 q2

 1 d
r  r  d   2
2
 q  q

2
2
a
r'
q'
r
d
q
q2 d
r  2
q  q 2
Conducting Spheres
qqd
a 2
q  q2
q2d
r 2
q  q2
q2 d
r  2
q  q 2
• Let’s put these formulas
a  q r 


together in a useful way:
r
q
a
How is this useful?
• Suppose we had a grounded ( = 0) conducting sphere of radius a
• Plus a point charge q at distance r > a
• Pretend there is an image
2
aq
a
charge q' at distance r', where
q   , r  
r
r
• This will exactly cancel the potential from
the charge q on the surface of the sphere
• And it will satisfy 2 = 0 outside the sphere
• So we have solved the problem!
a
r'
q'
r
q
Sample Problem 2.2a
A conducting sphere of radius a with potential  = 0 at the origin has a charge q on
the z-axis at z = 2a. Find the potential (outside), and the total charge on the sphere.
2
aq
aq
a
• Add an image charge q  
  12 q
q   , r  
2a
r
r
2
of magnitude
a
1


r

• It will be at
2a
1
2a
 1

q
2
• The potential is then
 x 



1
4 0  x  2azˆ x  2 azˆ 
To get the charge:
• The hard way:
– Find the electric field from the potential
– Get the surface charge everywhere and integrate
• The easy way:
– Use Gauss’s Law on a surface surrounding the sphere
– Electric field looks like point charge –q/2
– So charge on sphere must be –q/2
Q   12 q
q
2a
– q/2
a/2
a
Sample Problem 2.2b
A neutral conducting sphere of radius a at the origin has
a charge q on the z-axis at z = 2a. Find the potential.
•
•
•
•
•
•
First guess: We can do this exactly the same way as Problem 2.2a
This is wrong because the sphere had charge – q/2
So far we have  = 0 on the sphere
To cancel this charge, add additional charge + q/2 on the sphere
It will distribute itself uniformly over the surface
This creates additional potential that looks like a point charge
of magnitude + q/2 at the center
• Total potential (outside) is sum of these three contributions
q  1
1
1 
  x 




1
ˆ
ˆ
4 0  x  2az 2 x  2 az 2 x 
q
2a
q/2 a
– q/2
a/2
Hollow Spheres
• We can use the same formulas for spherical cavity in a conductor
• Place an image charge at r' of magnitude q'
aq
a2
q   , r  
r
r
• Combination will cause vanishing potential
on the interior surface
• You don’t have to worry about whether
surrounding conductor is neutral, since excess
charge flows away from interior surface
• If you want to make   0 on interior surface,
just add a constant to it
– It is irrelevant anyway.
q
q'
Green’s Functions for Planes and Spheres
• Since we know how to get the potential for a point charge near a conducting
plane or sphere we can find the Green’s functions for them
• Green’s functions (with Dirichlet boundary conditions) satisfy
2G  x, x   4 3  x  x ' and G  x, x  xS  0
D
• For a plane at z = 0 for the region z > 0,
one source at x, one at xR = (x, y,–z)
G  x, x  
1
1

x  x x R  x
• For a sphere at origin with radius a, one point at x of size 1, one point at xR =
a x
1
a2x/x2 with magnitude –a/|x|
G  x, x  
 2 2
x  x  a x  x  x
• With some work
can rewrite this as
G  x, x  
1
x  x  2x  x
2
2

• This formula also works for interior of a sphere as well
a
a 4  x 2 x2  2a 2 x  x
Using Green’s Functions for Spheres (1)
1
a
• Probably easiest
G  x, x  

2
2
to use in
x  x  2x  x
a 4  x 2 x2  2a 2 x  x
spherical coordinates
1
a
G  x, x  

–  is angle
2
2
4
2 2
2



r

r

2
rr
cos

a

r
r

2
a
rr  cos 
between x and x'
• We can then use this formula for Dirichlet problems with spherical boundary
1
1

3




  x 
G
x
,
x

x
d
x


x
G  x, x  da








4 0 V
4 S
n
• The normal derivative is:
r   r cos 
ar 2 r   a 3r cos 



G  x, x   
G  x, x  
3/2
3/2
2
2
4
2 2
2
n
r 
 r  r  2rr cos    a  r r  2a rr cos  
• On the
a  r cos 
r 2 a 1  r cos 
a  r 2 a 1


3/2
3/2
3/2
surface  2
2
2
2
2
2
 r  a  2ra cos    a  r  2ra cos    r  a  2ra cos  
r' = a
this becomes
Using Green’s Functions for Spheres (1)
G  x, x  
  x 
1
4 0

V
1
r 2  r 2  2rr  cos 
G  x, x    x  d 3x 

1
4
a
a 4  r 2 r 2  2a 2 rr  cos 

S
  x 

G  x, x  da
n

a  r 2 a 1
G  x, x  xS 
3/2
2
2
n
 r  a  2ra cos  
• Finally, the surface integral will take the form:
• So we have

S
da  a 2  d 
  arˆ   d 
a 2
2
  x 
G  x, x    x  d x 
r a 

3/2

V
2
2
4 0
4
 r  a  2ar cos  
1
3
Sample Problem 2.3 (1)
The potential on the surface of a sphere of radius a centered at the
origin is given by  = Ez. Find the potential outside the sphere.
x
z
• We use our
a 2
Ez d 
2
 x 
r a 

formula for
3/2
2
2
4
 r  a  2ar cos  
the potential,
ignoring the  term
• For purposes of this integral, treat x as if it were the z-axis
– This makes  = '
• With this choice of z-axis, what we normally call z-direction becomes
–  is angle between x and ordinary z-axis
z  z cos  x sin 
• We are on the surface
z cos   x sin   a cos   cos   a sin   cos   sin 
of the sphere, so
• So we have:
• Set up the integrals
a2 E 2
cos  cos    sin  sin   cos  
2
  x 
r a 
d 

3/2
2
2
4
r

a
 2ar cos   

Sample Problem 2.3 (2)
The potential on the surface of a sphere of radius a centered at the
origin is given by  = Ez. Find the potential outside the sphere.
x
z
2
a E 2
cos  cos    sin  sin   cos  
2
  x 
r  a   d cos    d 

3/2
2
2
4
1
0
 r  a  2ar cos 
1
2
2
a
E 2
cos  cos  d cos  
2
• Do the ' integral

r a 

3/2
2
2
2
1  r  a  2ar cos   
2
2
• Let
w  r  a  2ar cos  
1
  x 
a E r  a
2
2
2  2ar 
2

r 2  a 2  2 ar
2

cos   r 2  a 2  w  w3/2 dw
r  a 2  2 ar
2
 r  a 2
Ea3
  x   2 cos 
r
 18 r 2 E  r 2  a 2  cos   2  r 2  a 2  w  2w  2

 r a 
2
2
2
2


E 2
r

a
r

a
2
 2  r  a  cos  

  r  a    r  a  
4r
ra
 r a

 12
1
2
Sample Problem 2.4
A neutral conducting sphere of radius a is placed in a background electric
field given by E  Ezˆ . Find the potential everywhere outside the sphere.
•
•
•
•
•
•
E  
Recall:
This suggests   x    Ez
But this can’t be right because it is not constant at r = a
Let’s try to find a solution of the form:   x    Ez    x 
Since there’s no charge outside, we must have
0   2   2
We need to cancel the potential from the first term, so
0    arˆ    Ez    arˆ 
  arˆ   Ez
• We already solved this problem:
a3
  E 2 cos 
r
• So the answer is:
a3
   Ez  E 2 cos 
r

a3 
  x   E cos   r  2 
r 

2B. Orthogonal Functions and Expansions
General Theory of Orthogonal Functions
• Let Un(x) be a set of functions in some region in d dimensions
– d may be less than 3, even if the problem is 3-dimensional!

• If we have enough functions Un(x), we can often approximate any function as linear combinations of these functions f  x    anU n  x 
n 1
• Often, finite sum is a good approximation for the infinite
• It is useful to normally arrange the Un’s to be orthogonal
*
d
– This can always be arranged
U
x
U
x
d
x  0 if m  n




m
n

V
– See quantum notes for details
*
d
• Assuming the integral is finite when m = n, you can
U
x
U
x
d
x   nm
m  n 

V
normalize them to make them orthonormal
• Given f(x), it is not hard to find the coefficients an:

V


U  x  f  x  d x   V amU  x  U m  x  d x   am nm
*
n
d
m 1
*
n
d
m 1
an   U n*  x  f  x  d d x
V
Completeness Relation, and Other Comments

f  x    anU n  x 
n 1
an   U n*  x  f  x  d d x
V
f x  

*
d


U
x
U
x
f
x
d
x






 n
n
• Substitute the expression
V
n 1
on the right into the left

*
• Since this is true for any function f(x),


U

n  x U n  x     x  x 
we must have the completeness relation
n 1
• I’m not sure what  means, because I’m not sure how many dimensions we’re in
Sometimes, we might not want all possible functions
• We might want functions that vanish
Un x S  0
at certain locations, so we’d choose
– Completeness still works in interior
• We might want functions that have
2

Un  x  0
vanishing Laplacian, so we’d choose
– This will ruin completeness
1D Discrete Fourier Transforms
• Consider functions on the region (0, a) with periodic boundary conditions, f(0) = f(a)
• It is well known that a complete set of functions
 2 inx 
that satisfy this equation are the Fourier modes: U n  x   exp  a 
a
– n =0, 1, 2, 3, 
*
U
0 m  x U n  x  dx  a nm
• These are not quite orthonormal

 2 inx 
• Any function can be written in terms of them:
f  x     n exp 

• Coefficients are
1 a
a
 2 inx 


n 
 n   f  x  exp  
dx

a 0
a


• Sometimes, it is easier to work with real functions cos  2 nx  , sin  2 nx 
 a 
 a 
– n = 0, 1, 2, 3, 
• Then we can write any


 2 nx 
 2 nx  
1
f  x   2 A0    An cos 
function in terms of them:
  Bn sin 

a
a




n 1 
• Coefficients are:
2 a
2 a
 2 nx 
 2 nx 
An   f  x  cos 
 dx , Bn  0 f  x  sin 
 dx
0
a
a
 a 
 a 
1D With Vanishing Boundaries
• If we instead demand that f(0) = 0 = f(a), we should use some different functions
• Use real functions, but want them to vanish at these points: U x  sin   nx 
n 


– n = 1, 2, 3, 
 a 
• Not quite orthonormal
a
  nx    mx 
0 sin  a  sin  a  dx  2  nm
a
• We can write an arbitrary function in the form:

  nx 
f  x    An sin 

a


n 1
• The coefficients are given by
2 a
  nx 
An   f  x  sin 
 dx
0
a
 a 
2D With Vanishing Boundaries
• Suppose we have a 2D problem with f(x,y) where 0 < x < a and 0 < y < b
• Let’s suppose we want f to vanish on all four boundaries
• Since f vanishes at x = 0 and x = a, we must be able to write

  nx 
f  x, y    An  y  sin 
b

 a 
n 1
• Since f vanishes at y = 0 and y = b, the same must be true of An
• Therefore, An must be writable in the form

  my 
An  y    Anm sin 

b


m 1
• We therefore have
• Coefficients are:
  nx    my 
f  x, y    Anm sin 
 sin 

a
b




n 1 m 1
b
4 a
  nx    my 
Anm 
dx
dy
f
x
,
y
sin
sin 







0
ab 0
a
b

 



a
Continuous Bases


*
U
 n  xU n  x     x  x
f  x    anU n  x 
n 1
n 1
an   U n*  x  f  x  d d x

V
U m*  x U n  x  d d x   nm
• It is sometimes the case that you need to use a basis where n takes on
continuous values
• We might write the basis functions as U(x)
• Then all sums become integrals
• The relations above get changed to:
*
U
   xU  x  d    x  x
f  x       U  x  d
     U  x  f  x  d x
*
d

V
U *  x U  x  d d x      
Continuous Fourier Transforms

• Theorem from quantum:


eikx dx  2  k 
• Consider the set of functions: U k  x  
•
•
•
•
•

1 ikx
e
2
1  i  k  k  x
They are orthonormal:
 U  x U k  x  dx  2  e dx    k  k 

1  i  x  x  k
*
Also complete
 U k  xU k  x  dk  2  e dk    x  x
1 
ikx
So we can write arbitrary
f  x 
A
k
e
dx




functions in terms of these
2
1 
Amplitude is given by
 ikx
Ak  
f
x
e
dx




2
Can find similar formulas in 3D:
1
 ik x 3
1
ik x 3
A
k

f
x
e
d k




f  x 
A
k
e
d
x
3/2 


3/2 
 2 
2

 
*
k
Decomposition in Different Coordinates
2   x      0
Consider the Laplacian in 3D in different coordinate systems:
2
2
2
• Cartesian:






2
  2  2  2
x
y
z
2
2


1


1




2
• Cylindrical:
 
 2

 2
2
      
z
• Spherical:
2
2
1

1


1




2 
r


r
sin


 2


2 
r r
r sin   
  r 2 sin 2   2
• Often good to solve problem explicitly by writing function as a complete
function in some of these variables, and unknown function in others
  nx    my 
f  x, y, z    Anm  z  sin 
 sin 

a
b

 

n 1 m 1


2C. Solving Problems in Cartesian Coordinates
The Method
• Suppose, for example, you have a box
with potential zero on some surfaces
  0, y, z     a, y, z     x,0, z     x, b, z   0
• And let’s say it’s
specified on others
  x, y , c    t  x, y 
t  x, y 
0
  x, y , 0    b  x, y 
• Write the potential in terms of complete functions on x and y
  nx    my 
  x, y, z    Anm  z  sin 
 sin 

a
b

 

n,m
• If there is no charge inside, then  2   0
0
0
0
b  x, y 
Modes That Satisfy Laplace’s Equation
•
•
•
•
•
  nx    ny 
2
  x, y, z    Anm  z  sin 
sin

0
 

 a   b 
n,m
Substitute this
 d 2  2 n 2  2 m2 
  nx    my 
into Laplace’s 0    2  2  2  Anm  z  sin 
 sin 

dz
a
a
a
b




n,m 

equation:
Since these wave functions are
 d 2  2 n 2  2 m2 
0   2  2  2  Anm  z 
independent, we must have
a
a 
 dz
Define
2 2
2 2

n

m
2
2
d
 nm  2  2
2
Then we have
A
z


A
z
a
b
nm  
nm nm  
2
dz
The general solution of this equation is Anm  z    nme nm z  nme nm z
  nx    my 
  x, y, z     nme nm z   nme nm z  sin 
 sin 

a
b

 

n,m
Matching the Boundary Equations
  x, y, z     nme
 nm z
  nme
n,m
 nm z
 nx    my 

 sin 
 sin 

 a   b 
2
 nm

 2 n2
a
2

 2 m2
b2
• We haven’t matched the boundary conditions
• On the top and bottom, we can also write the (known) potential as
  nx    my 
b  x, y    A sin 
 sin 

a
b

 

n,m
b
nm
• We can get the
coefficients
•
•
•
•
  nx    my 
t  x, y    A sin 
 sin 

a
b

 

n,m
t
nm
b
4 a
  nx    my 
A 
dx
dy

x
,
y
sin
sin 


b





0
ab 0
a
b

 

b
nm
To get it to match at z = 0, we must have
To get it to match at z = c, we must have
Two equations in two unknowns
Solve for nm and nm and we have 
b
Anm
  nm   nm
t
Anm
  nm e nmc   nm e  nm c
Sample Problem 2.5 (1)
A cube of size a has potential 0 on five faces and potential  = xy
on the last face. What is the potential everywhere, and at the center?
   xy
0
2
 nm
  2  n2  m2  a 2
• We have
0
  x, y, z     nme nm z  nme nm z  sin  nx a  sin  my a 
n,m
• We need it to vanish on z = 0, so nm = –nm
  x, y, z    2 nm sinh   nm z  sin  nx a  sin  my a 
0
0
0
n,m
• To match on z = a,
 xy   2 nm sinh   nm a  sin  nx a  sin  my a 
n,m
we have
2
4

a
a
a
nm
4

nx

my

 

• Find the
1

2 nm sinh   nm a   2  dx   xy sin 
sin 
2


0
a 0
coefficients
 a   a   nm
  x, y , z  
4 a 2
2
1
n  m sinh   nm z 
  nx    my 

1
sin
 


 sin 

nm
sinh

a
a
a
 nm  
 

n,m
Sample Problem 2.5 (2)
A cube of size a has potential 0 on five faces and potential  = xy
on the last face. What is the potential everywhere, and at the center?
2
 nm
  2  n2  m2  a 2
4 a
1
n  m sinh   nm z 
  nx    my 
  x, y , z   2 

1
sin
 

 sin 

 n ,m nm
sinh   nm a   a   a 
• Now just substitute in the center
  a, a, a  

1
2
4 a
2
2
1
2

n , m odd
4 a
2
2
1
 1

n , m nm

sinh 
sinh
nm
  1
nm sinh  n  m 
sinh

0
0
0
2
1
2
   xy
1
2
 n2  m2
2
2
1
2
0
0
 sin   n  sin   m 
n m 
 n2  m2
 n  m 2
2
2
1
2
1
2
2
 0.0416666666 a 2
with 62 terms
2D. Solving Probs. in Cylindrical Coordinates
Potential in a Sharp Corner
2
2


1


1




2
 
 2

 2
2
      
z
• Laplacian in cylindrical coordinates
• Let’s do a 2D problem
– Everything independent of z
• Consider a sharp conducting angle of size  with vanishing potential
• Presumably other sources/potentials somewhere else
• The potential must vanish at  = 0 and  = 
• We therefore can
  n 


,


R

sin
   n  

write the potential as


n


• Substitute into  2   0
1   
  2 n2    n 
0  
Rn   2 2 Rn  sin 










n 1 



 
• To make this vanish, we must have

  2 n2
1   
Rn   2 2 Rn  0

      
Solving the Radial Equation
  2 n2
• Second order differential equation should
1   
Rn   2 2 Rn  0

have two linearly independent solutions
      

• Guess solutions of the form Rn ~ 
2 2
• We therefore have 1       2 n 2 

n  2
2  2
   2  0
  2 2   0


   
  
• We therefore have
• The general solution is
   n 
Rn     An   n   Bn   n 
• If we want finite potential near  = 0, must choose Bn = 0
• Therefore:
   ,    A   n  sin  n  

n
• Largest contribution from n = 1:
Rn     An   n 
n
   ,    A1   sin   
Electric Field near a Sharp Corner
   ,    A1   sin   
• The electric field is the derivative of the potential

1 
  φˆ

E    ρˆ

 

1 
  
   


  A1  ρˆ sin    φˆ cos   

  
  


• This expression diverges at small  if  > 
• The sharper the corner, the faster it diverges
• In 3D, electric fields maximum at pointy places
• Lightning rods have large fields, cause charge to flow from atmosphere
• Drains away charge buildup in vicinity of the rod