1. Electrostatics
1A. Coulombs Law, El. Field, and Gauss’s Law
Coulomb’s Law
• Charges are measured in units called Coulombs
k1qqrˆ
• The force on a charge q at x from another charge q' at x':
F
2
x  x
• The unit vector points from x' to x rˆ  x  x
• We rewrite the unit vector as
x  x
1
• For reasons that will make some sense later, we rewrite constant k1 as k1 
4 0
• So we have Coulomb’s Law:
qq x  x
F
• For complicated reasons having
4 0 x  x 3
to do with unit definitions, the
constant 0 is known exactly:
 0  8.854 1012 C2 /N/m 2
• This constant is called the
permittivity of free space
Multiple Charges, and the Electric Field
• If there are several charges q'i,
you can add the forces:
• If you have a continuous distribution
of charges (x), you can integrate:
qqi x  xi
F
3
4

i
0 x  xi
F
q
4 0
qq x  x
F
4 0 x  x 3
3


x
d
x



x  x
3
x  x
• In the modern view, such “action at a distance” seems unnatural
• Instead, we claim that there is an
1
x  x
3
E x 
  x  d x
3

electric field caused by the other charges
4 0

xx
– Electric field has units N/C or V/m
• It is the electric field that
then causes the forces
F  qE
Gauss’s Law: Differential Version
x  x
E
  x  d x 
3

4 0
x  x
1
3
• Let’s find the divergence
of the electric field:
x  x
  E x 
  x  d x 
3

4 0

xx
• From four slides ago:
x  x
3


4

 x  x 
3
x  x
1
3
1
3
3
• We therefore have:   E  x   1



  x  d x 4  x  x     x 

4 0
0
• Gauss’s Law (differential version):
• Notice that this equation is local
 0 E  x     x 
Gauss’s Law: Integral Version
 0 E  x     x 
• Integrate this formula over   E  x  d 3x    x  d 3x
0 V

V
an arbitrary volume
• Use the divergence theorem:
 0  nˆ  E  x  da  q V 
S
– q(V) is the charge inside the volume V
• Integral of electric field over area is called electric flux
Why is it true?
• Consider a charge in a region
• Electric field from a charge inside a region
produces electric field lines
q
• All the field lines “escape” the region
somewhere
• Hence the total electric flux escaping must be
proportional to amount of charge in the region
Sample Problem 1.1 (1)
A charge q is at the center of a cylinder of radius r and height 2h. Find the electric
flux out of all sides of the cylinder, and check that it satisfies Gauss’s Law
• Let’s work in cylindrical coordinates
• Electric field is:
q  ρˆ  zzˆ
q x
q  ρˆ  zzˆ

 0E 
3 
3
4   2  z 2 3/2
4 x
4  ρˆ  zzˆ
E
q
• Do integral over top surface:  0  E  nˆ da 
top
4

q
4 0
2
r
 d 
0
0

h d 
2
h

2 3/2

• By symmetry, the
integral over the bottom
surface is the same
2 qh
4  2  h 2
r
0
ρ̂


z da
2
z
h
h
q
h
z
r

q
h
 1 

2
2
2
r h 
 0  E  nˆ da   0  E  nˆ da 
top

2 3/2
bot
E
ẑ

q
qh

2 2 r 2  h2
r
Sample Problem 1.1 (2)
A charge q is at the center of a cylinder of radius r and height 2h. Find the electric
flux out of all sides of the cylinder, and check that it satisfies Gauss’s Law
q  ρˆ  zzˆ
 0E 
4   2  z 2 3/2
q
qh
ˆ
 0  E  nda  
top
2 2 r 2  h2
• Do integral over lateral surface:
q
 da
q
 0  E  nˆ da 

lat
4    2  z 2 3/2
4
2 q
z

4 r 2  z 2
2
h
 rd 
0
h
r
h
h
 0  E  nˆ da 
lat
• Add in the top and bottom surfaces:
 0  E  nˆ da   0  E  nˆ da  2 0  E  nˆ da 
tot
lat
top
r
2
z
E
ρ̂

2 3/2
E
ẑ

h
h
q
h
z
r
dz
qh
r 2  h2
qh
r h
2
2
r
q
qh
r h
2
2
q
Using Gauss’s Law in Problems
 0  nˆ  E  x  da  q V 
S
Gauss’s Law can be used to solve three types of problems:
• Total electric flux out of an enclosed region
– Simply calculate the total charge inside
• Electric flux out of one side of a symmetrical region
– Must first argue that the flux out of each side is the same
• Electric field in a highly symmetrical problem
– Must deduce direction and symmetry of electric field from other arguments
– Must define a Gaussian Surface to perform the calculation
– Generally use boxes, cylinders or spheres
Sample Problem 1.2
A line with uniform charge per unit length  passes through the long diagonal of
a cube of side a. What is the electric flux out of one face of the cube?
• The long diagonal of d  a2  a2  a2  a 3
the cube has a length
• The charge inside the cube is therefore
Q  d  a 3
• The total electric flux
 0  nˆ  E da  Q  a 3
out of the cube is
S
• If we rotate the cube 120 around the axis, the three faces at
one end will interchange
– So they must all have the same flux around them
• If we rotate the line of charge, the three faces at one
end will interchange with the three faces in back
– So front and back must be the same
a 3
1
• Therefore, all six
nˆ  E da   nˆ  E da 

one
face
faces have the same flux
6 0
6 S
Sample Problem 1.3
A sphere of radius R with total charge Q has its charge spread
uniformly over its volume. What is the electric field everywhere?
• By symmetry, electric field points directly away from the center
• By symmetry, electric field depends only on distance from origin
Outside the sphere:
E  x   E  r  rˆ
• Draw a larger sphere of radius r
• Charge inside this sphere is q(r) = Q
• By Gauss’s Law,
Q   0  E  x   nˆ da   0  E  r  da  4 r 2 0 E  r 
S
S
Inside the sphere:
Q
V
r


E r  
• Draw a smaller sphere of radius r q  r   Q
2
 Qr 3 R 3
4

r
0
V  R
• Charge inside this sphere is only
• By Gauss’s Law,
Qr 3 R3   0  E  x   nˆ da   0  E  r  da  4 r 2 0 E  r 
S
S
• Final answer:
Qrˆ
E
4 0
 r 2 if r  R,
 3
if r  R .
rR
E r  
Qr
4 0 R 3
Sample Problem 1.4
An infinite line of charge has charge per unit
length . What is the electric field everywhere?
• Which direction does the electric field point?
– Surely, it is radially away from the line
• What can it depend on?
E  x   E    ρˆ
– Only distance  from the line
• What Gaussian shape should we use?
– Cylinder, length L, radius R, centered on the line
L q
• Gauss’s law says:
  E  nˆ da  E    ρˆ  nˆ da  E  R  da

S
S
lat
0 0
• Enclosed charge is L
• End caps have zero dot product
 2 RLE  R 
• Lateral surface is a (rolled up) rectangle length L width 2R
• Solve for E:


E  R 
E x 
ρˆ
2 0 R
2 0 
1B. Electric Potential
Curl of the Electric Field

x
0
• From homework problem 0.1:
3
x
• Generalize to origin at x':
x  x

0
3
x  x
• Consider the curl of the electric field:
1
x  x
1
x  x
3
3


E 
    x  d x
  x  d x  
3 
3

4 0
4 0
x  x

xx
• Using Stokes’ theorem, we can get
an integral version of this equation:
 E  dl  0
3

E
d
x0


S
 E  0
Electric Field: Discontinuity at a Boundary
Consider a surface (locally flat) with a surface charge 
• How does electric field change across the boundary?
A
L
• Consider a small thin box of area A
–L
crossing the boundary
• Since it is small, assume E is constant over top surface and bottom surface
A q
• Use Gauss’s Law

  E  x   nˆ da  Et  nˆ t A  Eb  nˆ b A  A  Et  Eb   nˆ t
on this small box
0
0
• Charge inside the box is A
• Since box is thin, ignore lateral surface

ˆ
 E   n 
• Consider a small loop of length L penetrating the surface
0
• Use the identity
0   E  dl  Et  L  Eb  L
• Ends are short, so

only include the lateral part  E  L  0
E  nˆ
0
• So the change in E across the boundary is
The Electric Potential
• In general, any function that has curl zero can be written as a gradient
 E  0
– Proven using Stokes’ Theorem
E  
• We therefore write:
–  is the potential (or electrostatic potential)
– Unit is volts (V)
It isn’t hard to find an expression for :
1
x  x
1
1 rˆ

x
• First note that


   2  3
3

x

x

• Generalize by shifting: x
r
r
x

x
x
• If we write:
  x 
1
4 0

  x  d 3x
x  x
• Then it follows that:
1
1
1
x  x
3
3


  x  
 x  d x 

  x  d x
3  E  x



4 0
x  x 4 0
x  x
Working with the Potential
1
  x  d 3x
  x 
E  
Why is potential useful?
4 0  x  x
• It is a scalar quantity – easier to work with
• It is useful when thinking about energy
– To be dealt with later
How can we compute it?
• Direct integration of charge density when possible
• We can integrate the electric field
2

    0









E


• It satisfies the Poisson equation:
0
0
Solving this equation is one of the main goals of the next couple chapters
• There is an ambiguity about , because it is an integral of the electric field
– Constant of integration is ambiguous
• Normally, resolved by demanding () = 0
Sample Problem 1.5
Find the potential and electric field at all points from a line charge with
charge per unit length  stretching from z = a to z = b along the z-axis.
• Easiest to work in cylindrical coordinates: x  ρˆ  zzˆ
x  z zˆ
• Find the potential:
b
  x  d 3x
1
 dz
1
 b
dz

  x 



4 0 a  z  z  zˆ  ρˆ 4 0 a  z  z 2   2
4 0
x  x
• Maple: > Phi:= integrate(1/sqrt((zp-z)^2+rho^2),zp=a..b);
 
2
2
2
2 
 x 
ln
b

z

b

z



ln
a

z

a

z







4 0 
• To get electric field, use E    zˆ   ρˆ 
z

• Maple: > -expand(simplify(diff(Phi,z)));-diff(Phi,rho);


 
1
1

Ez 

2
2 
4 0   b  z 2   2
a

z







 

Sample Problem 1.6
A sphere of radius R with total charge Q has its charge spread
uniformly over its volume. What is the potential everywhere?
Qrˆ
E
4 0
 r 2 if r  R,
 3
if r  R .
rR
• We already found the electric field
• It makes sense that potential
depends only on r
   r 

• Relation between potential and electric field E    rˆ
r
2
• So
if r  R,
d  Q  r

1
we
 3


r
 C1
if r  R,

Q
dr 4 0 rR
if
r

R
.

have
 1 2 3
4

 C2 if r  R .
0 2 r R
• Integrate this in the two regions:
• We choose () = 0, so C1  0
1
3
1
2 3
1
C


R

R

C

R
R

C
2
2
1
2
• Want potential continuous at r = R, so
2
• Put it together
r 1
if r  R,
Q 

 3 1 1 2 3
4 0  2 R  2 r R
if r  R .
Conductors
• A conductor is any material that has charges in it that can move freely
• If an electric field is present inside a conductor, then:
– Charges will shift in response
– These shifting charges will create electric fields
– They will stop only when all electric fields are cancelled
• Therefore, (perfect) conductors have E = 0 inside them
• Recall that E  
  constant
• Hence potential must be constant in a conductor
• Consider Gauss’s law for any shape contained within the interior of a conductor
• Since there is no electric field, there is no charge in the interior
– There can be charge density  on the surface of the conductor
• Recall the discontinuity in the electric field at the surface

E  nˆ
• But it vanishes inside
0
• Therefore, the electric field at the surface of a conductor is

E  nˆ
0
Sample Problem 1.7a
A neutral solid conducting sphere of radius R has a cubical cavity
of side a inside it, with a charge q at its center. What is the electric
field outside the sphere?
• Consider any Gaussian surface surrounding the cavity
• No electric field, so no charge inside
• So the charge q must be balanced by charge –q from the conductor on
the inside walls of the cavity
• But there must be no net charge on the conductor, so charge + q on the
outer surface of the conductor
• This surface charge feels no force from the other charges, so it will
distribute itself uniformly over the surface
• This creates the same field as a point source at x = 0
qrˆ
– Nothing to do with where the charge is
E
2
4

r
0
– Nothing to do with shape/size, etc. of cavity
q
–q
+q
Sample Problem 1.7b
A neutral solid conducting sphere of radius R has a cubical cavity
of side a inside it, with a charge q outside. What is the electric
field inside the cavity?
q
• The conductor is at constant potential,  = 0.
• The potential on the interior surface of the cavity is  S  0
• There is no charge inside the cavity, so in the cavity,  2      0  0
• As we will demonstrate shortly, the value on the boundary plus the
value of the Laplacian is sufficient to determine the potential
• I’m so good, I can get this solution by guess and check:    0
– Check it yourself
E0
• Therefore, the electric field inside is
E  
• This has nothing to do with the shape of the cavity, the position of q, etc.
Potential and Potential Energy
F  qE  q    q 
• Consider the formula:
• Consider the relationship between energy and potential energy F  W
• By comparison, we see that
W  q
•
•
•
•
•
q1q2
For example, for two charges q1 and q2 at x1 and x2: W 
4 0 x1  x2
If you have a lot of charges,
qi q j
qi q j
1
– The i < j to avoid double counting W  
 
2 i  j 4 0 xi  x j
i  j 4 0 xi  x j
This is equivalent to
If you have a continuum of charges, this can be written as
1   x    x  3 3
3
W  
d xd x
  x  d x
2 4 0 x  x
Recall   x   
4 0 x  x
• We therefore have:
W
1
2
3

x

x
d
     x
Sample Problem 1.7
A sphere of radius R with total charge Q is being approached by a
charge q and mass m with speed v from infinity. How fast must q
be moving to make it to the center of the sphere?
Q
R
• Use conservation of energy
1

r
if r  R,
• We already know
Q

 3 1 1 2 3
the potential
4 0  2 R  2 r R
if r  R .
• At infinity, the energy is
E  W  U  12 mv2  q     12 mv 2
• At the origin, the charge stops, so
3Qq
E  W  U  m0  q  0 
8 0 R
• Equating these, we have
3Qq
mv 2
3Qq
v

4 0 Rm
2
8 R
1
2
0
2
v
q
1C. Boundary Value Problems
Definition of the Problem
• Sometimes, we don’t know, don’t care, or don’t want to
do the work of figuring out the potential everywhere
• Sometimes, just want potential in a volume V
– Which may be infinite
2

    0
• Poisson equation alone is not sufficient alone to determine potential
• You need additional information about what happens to the potential at boundary
• This information can take one of two types:
  x  S    SD 
D
– On Dirichlet boundary D, the potential is specified
• Assume this happens somewhere perhaps x = 

– On Neumann boundary N, the normal derivative
nˆ  S 
 SN 
N
n
S

SD
N
/n
• Goal: find solution to these equations
Uniqueness of Solution?
2     0
  x  S    SD 
D

n

SN

 SN 
n
SD 
•
•
•
•
Could there be two solutions that satisfy all three equations?
If 1 and 2 are both solutions of these, then define: U  1   2
SN
It follows that
/n
 2U  0
nˆ U S  0
U  x S  0
N
D
Consider the integral:
2 3
2
3
3


ˆ


U


U

U

U
d
x
U

U

n
da

   U  d x
0 


  U U  d x V    
S
•
•
•
•
•
•
V
V
Also, do the integral by divergence theorem
On each surface, one of the factors vanishes 0  U or nˆ U  0
The integrand is never negative
Therefore, the only way to make it vanish is for the integrand to vanish U  0
Therefore U is constant
If anywhere we have Dirichlet boundary conditions, then U  0  1   2
Finding a “Simple” Solution
2     0
  x  S    SD 
D

n

SN

 SN 
n
SD 
• Consider the problem with

x    x  x 

a simple point source at x:
SN
• Let’s also let all the

  SD  
 SN   0
/n
boundary conditions vanish:
n
• We know the solution is unique:  x  x 
1


x

• If we had no boundaries, (except at ), solution would be
x 
4 0 x  x
• Let’s call the general solution (with zero boundaries):
1
– These are called Green functions

x  x  
G  x, x 
• This solution will have the properties:
4 0
2G  x, x  4 3  x  x G  x, x    0
x S
D

G  x, x 
0
n
xS N
How to Find or Check a Green’s Function
2G  x, x  4 3  x  x G  x, x  xS  0  G  x, x 
0
D
n
xS N
• Green functions
1
generally look like G  x, x   x  x  F  x, x 
1
2

 4 3  x  x 
• The first term satisfies:
x  x
• It follows that
SD 
SN
/n
2 F  x, x  0
• The F term is just there to make the boundary conditions work out right
• It is nonetheless usually hard to find
If we think we have G, how do we check it?
• Check that the Laplacian satisfies its equation
• Check boundary conditions on Dirichlet boundaries (including G(x,) = 0)
• Check boundary conditions on any Neumann boundaries
Sample Problem 1.8
Consider the half-space z > 0 with Dirichlet boundary conditions.
Show that the function below is the correct Green’s function
SD 
1
1
G  x, x  

, where x R   x, y,  z 
x  x x R  x
• Since x' = 0 is included in the space, we should have: G  x,    0
• We also have the boundary z' = 0, at which we should have:
1
1

0  G  x, x  z0 
2
2
2
2
2
2
 x  x    y  y     z  0 
 x  x    y  y      z  0 
• Finally, we have to check:
1
1
2
2
2
 G  x, x   
 
 4 3  x  x  4 3  xR  x
x  x
x R  x
• We are only interested in points within the allowed region
– That means both z > 0 and z' > 0
2
3
• The second delta function can never vanish, so  G  x, x  4  x  x
Using the Green’s Function
2     0

n
  x  S    SD 
D
2G  x, x  4 3  x  x
G  x, x  xS  0
D

SN

 SN 
n

G  x, x 
0
n
xS N
SD 
SN
/n
• G is the problem for a single point source
• For a more general source, you would think you could then add the charges to get
the general solution, so long as boundary conditions are still zero
• Surprisingly, you can use G to get the general solution even when the boundary
conditions aren’t zero
• We will now demonstrate this
Potential from Green’s Functions (1)
• Consider any two functions f and g of x
• Consider the
following identity:   gf   g2 f  g   f 
• Swap f and g:
  f g   f 2 g   f   g 
  gf  f g   g2 f  f 2 g
• Subtract them:
• Integrate using
2
2
3
ˆ
n

g

f

f

g
da

g

f

f

g
d
x






S
V
divergence theorem:
• Turn it around:
 
 
2
2
3
– Normal derivatives: V  g  f  f  g  d x  S  g n f  f n g  da
2
2
3








G
x
,
x


x


x

G
x
,
x
d








• Now, change variable x to x', V 
 x
let f be (x'),



 





G
x
,
x

x


x
G
x
,
x
and let g be G(x,x'):
S    n     n   da
Potential from Green’s Functions (2)
2G  x, x  4 3  x  x
2     0
G  x, x  xS  0
D
G  x, x     x     x   G  x, x   d x
V

2
2
3







  G  x, x 
 x    x 
G  x, x   da
S
n
n



G  x, x 
0
n
xS N
SD 
SN
/n
• Use the Laplacians
3
3






4


x

x

x

G
x
,
x

x

d
x 








0



above on left side: V
• Do the first integral
1



 
3







4  x    G  x, x    x  d x   G  x, x 
 x    x 
G  x, x  da
V
S
0
n
n


• Break surface integral into Dirichlet part and Neumann part:


4  x    G  x, x    x  d x   G  x, x 
  x  da     x 
G  x, x  da
V
S
S
N
D
0
n
n
1
3
Sample Problem 1.9 (1)
Consider the half-space z > 0 conditions. Find the potential everywhere if
there is no charge, and its boundary value is as given by   x, y,0   V   x 
• Since we are given potential, we need Dirichlet Green function:
1
1
G  x, x  

, where x R   x, y,  z 
x  x x R  x
• The general solution (dropping the Neumann term):
1
1
  x 
G  x, x    x  d x 

V
4 0
4
3
SD


SD   x n G  x, x da
• The normal derivative is out of the region, so  n   z
3 2
2
2
• We find: 
2
G  x, x  z0  2 z  x  x    y  y   z 


z 
• No charge so 4  x     x  2 z  x  x 2   y  y 2  z 2  3 2 da
S




 2zV  dx   x   x  x    y  y   z 




2
2
2
3 2
dy 
Sample Problem 1.9 (2)
Consider the half-space z > 0 conditions. Find the potential everywhere if
there is no charge, and its boundary value is as given by   x, y,0   V   x 


3 2
4  x   2zV  dx   x   x  x    y  y   z  dy
SD 




3 2

2
2
2
 2zV   x   y  y   z  dy

 
2
2
• Can complete the integral by hand using a trig substitution: y  y  x  z tan 
• Substituting in, we have
1
2
2
2
2
2
1
2
zV
sec
 d
x

z
sec

d

2
 2
4  x   2 zV  1
2 
3
3/2
 2
2
2
2
2
2
x

z
sec

1
 2
 x  z   x  z  tan  


1
zV
4zV
2 zV 2

x
,
y
,
z




 2
cos  d
2
2
2 
x z
  x2  z 2 
x  z  1
2
2
2
2
1D. Capacitance and Energy
Capacitance
• Suppose you have conductors (and nothing else) in free space
• Potential is constant on each   Si   Vi
 2  i  x   0,
• For each i, set Vi = 1 and Vj = 0 for j  i, then
there is some unique solution to this equation (x):  i  S j    ij .
• For arbitrary Vi the solution will be:
  x   Vi i  x 
• The charge on any conductor is then
i
Qi     x  da   0  nˆ   x  da    0V j  nˆ  j  x  da
Si
Si
j
Si
• Define the capacitance: Cij   0 nˆ  j  x  da

Si
– Units are Farads (F)
• Then we see that
Qi   CijV j
j
• Can argue: Cij  C ji
S1
S2
V1
V2
Sample Problem 1.10
What’s the capacitance of the Earth?
N
Qi   CijV j
• We have only one object, so we write Q  CV
j 1
• Treat the Earth as a spherical conductor of radius R = 6370 km
• If we place a charge Q on it, the charge distributes itself uniformly over
the surface
• Electric field will be radially outward and depend only on r
ˆ
• By Gauss’s law, electric field will look like from a point charge: E  Qr 2
4 0 r
• The potential outside will also
Q
 x 
look like a point charge:
4 0 r
• By continuity, the potential
Q
V S 
on the surface is then just
4 0 R
• Solve for the
C  4 0 R
capacitance: C  Q V
• Substitute numbers in:
C  4  8.854 1012 C 2 /N/m 2  6.370 106 m   7.087 104 C2 /J  709  F
Energy of a System of Conductors
N
Qi   CijV j
• Consider the formula for energy:
W
1
2
j 1
3

x

x
d
x





V2
• Only charge is on the surfaces, so
W
N
1
2
1
1

x

x
da





 S
2  Vi S   x  da  2  Vi Qi
i 1
V1
i
i
i
• We therefore have:
W
1
2
 C VV
ij i
i
j
j
i
Another Formula for Energy
•
•
•
•
•
•
•
•
•
Previous formula for potential energy: W  12    x    x  d 3x
Recall:  0  E  
W  12  0    E  x     x  d 3x
So we have:
Now use the identity:   E  x    x      E  x     x   E  x    x 

 

We therefore have:
W  12  0    E  x    x    E  x    x  d 3x
On first term, use divergence theorem
On second term, use E(x) = –(x) W  12  0  nˆ  E  x    x  da  12  0  E2  x  d 3x
S
At infinity, E and  both vanish
We therefore have
W  12  0  E2  x  d 3x


• Also equivalent to W   0    x  d 3x
1
2
2
• Think of this expression as energy density: w  1  0 E2
2
• Modern viewpoint: The energy density is in the electric field
1E. Variational Method
Minimization of Energy
• Consider a problem with Dirichlet boundaries
     0
2
  x  S    SD 
SD 
  x
D
• We have an intuitive sense that
systems try to minimize their energy
W   0    x  d 3x
1
2
2
• Can we find  by minimizing this expression?
• Not exactly, because we have to make sure  is large
when  is large
2
• For any function (x), S    x    V 12   x      x    x   0 d 3x
consider the functional
• We will pick  so it has
  x  S    SD 
D
correct boundary conditions


• We will demonstrate that if we pick  = , we minimize this expression
 Minimizes the Functional
S    x    
V
  x  S    x  S    SD 
D

1
2

  x      x    x   0 d 3x
2
2     0
D
SD 
  x
• Suppose that  and  differ by f(x):     f
– Note that f vanishes on the boundary
• Then we have
2
1
S    S   f    2   f      f    0 d 3x
V


2
2
3
1
2  3
1




S



f



f
d
x





 S    f   2 f   f   d x
2



V
V

 S   
S
f nˆ  da  12   f  d 3x  S    12   f  d 3x
2
V
2
V
• The last term is always positive
• So this functional is minimized for the choice  = 
• If we are close (f is small), we will get close (order f2)
S     S 
The Variational Method
S    x    
V

1
2

  x      x    x   0 d 3x
2
  x
SD 
• Rather than picking a single function, pick a lot with the right boundary conditions
α  x 
α  x  S    x  S
• For each of them, find the functional:
S α   
V

1
2
D

 α  x      x   α  x   0 d 3x
2
• Minimize it with respect to all the parameters

S α  α  0
min
 i
• Then the approximation for the potential is
  x   αmin  x 
D
The Variational Method and Capacitance
S    x    
V
•
•
•
•
•

1
2

  x      x    x   0 d 3x
2
Suppose we have a capacitor with voltage V
Set V = 1 and try to find the potential 
α  x 
Use the variational approach
Minimize
2 3
The energy for a capacitor is
V  α  x  d x
W  12 CV 2  12 C
• The energy is also given by
W   0    x  d x   0   α min  x   d 3x
1
2
• We therefore have
2
3
2
1
2
C   0  αmin  x  d 3x
V
• This is always an overestimate
2
V
α  x S  1
Sample Problem 1.11
Estimate the capacitance for a cube of side 2a
3


C



x
d
x
0 V 
αmin   
Place the cube at the origin
Guess the functional form for potential
max  x , y , z   a
– Must be 1 on the surface
– Want it to fall off as 1/r at infinity
b

x

b 
By symmetry, assume |x| is largest and x > 0
max  x , y , z   a  b
– Multiply by 6 to account for all the regions
We therefore have
2

x
x
2 2

2 3
b
24

x
b


0
 0    x   d x  6 0  dx  dy  dz  
dx
  a
4
V
 xba 
 x  b  a
a
x
x
2
•
•
•
•
 8 0  a 2 b  a  b 
ba
• Minimize with respect to the parameter b 0   a 2 b2  1
• Substitute back in
C  8 0  a 2 a  a  a 
C  24 0 a
1F. Relaxation Method
Potential Can Be Found from Nearby Points
2
• Consider the potential near a point x with

0
 x S    S 
no charge density and Dirichlet boundaries
• Let’s work in 2D (not sure why)
• For a point offset in x-direction or y-direction will be approximately

1 2 2
1 3 3
4
  x  hxˆ     x   h   x   h

x

h

x

O
h






2
3
x
2 x
6 x
2

1 2 
1 3 3
4
  x  hyˆ     x   h   x   h

x

h

x

O
h






2
3
y
2 y
6 y
• Add these four points
  x  hxˆ     x  hxˆ     x  hyˆ     x  hyˆ   4  x   h 2 2  x   O  h 4 
• We
therefore
have
  x   14   x  hxˆ     x  hxˆ     x  hyˆ     x  hyˆ    O  h 4 
Using a Grid
  x   14   x  hxˆ     x  hxˆ     x  hyˆ     x  hyˆ    O  h 4 
• Set up a rectangular grid:
• Label points by i and j
• Potential on boundaries is fixed
• In the interior, calculate using
 ij 
1
4

i 1, j
  i 1, j   i , j 1   i , j 1 
• Repeat until it converges
• Comparably, you can use
ij 
1
4

i 1, j 1
  i 1, j 1   i 1, j 1   i 1, j 1 
• Can show a more accurate method would be
ij  54 ij  15 ij
Sample Problem 12
A square of side a has potential 0 on three sides and potential 1 on y = a. What is the potential at (x,y) = (a/4 , a/4)
 1
0
0
• Let’s set up a grid of size a/4
0
– Need 55 size grid
• Set it up in an appropriate program
– I used Excel
• Type in all the boundary values

1

• Put in one of the formulas
ij  4   i 1, j   i 1, j   i , j 1   i , j 1 
• Recalculate repeatedly (F9)

1


ij
4   i 1, j 1   i 1, j 1   i 1, j 1   i 1, j 1 
until it converges
• Increase grid size if you want
ij  54 ij  15 ij