ECE 2317
Applied Electricity and Magnetism
Spring 2014
Prof. David R. Jackson
ECE Dept.
Notes 13
1
Divergence – The Physical Concept
Start by considering a sphere of uniform volume charge density
The electric field is calculated using Gauss's law:
z
r < a:
v = v0
 D  nˆ dS
 Q encl
S
y
r
x
a
4 r D r   v 0
2
Dr 
4
3 
 r 
3

 v0 r
3
 C /m 2 


Sphere of uniform charge density
2
Divergence -- Physical Concept (cont.)
r > a:
4
2
3 
4 r D r   v 0   a 
3

z
Dr 
v0 a
3r
2
3
 C /m 2 


v = v0
y
a
r
x
3
Divergence -- Physical Concept (cont.)
Flux through a spherical surface:
 
 D  nˆ dS  4 r D r
2
S
Dr 
Recall that
Dr 
 v0 r
3
v0 a
3r
2
 C /m 
(r < a)
 C /m 2 


(r > a)
2
3
Hence
 
4
3
 
4
3
 r  v0
(r < a)
 a  v0
(r > a)
3
3
(increasing with distance inside sphere)
(constant outside sphere)
4
Divergence -- Physical Concept (cont.)
Observation:
More flux lines are added as the radius
increases (as long as we stay inside the charge).
 

D  nˆ dS  0
S
The net flux out of a
small volume V inside
the charge is not zero.
V
S
 

D  nˆ dS  0
S
Divergence is a mathematical way of describing this.
5
Gauss’s Law -- Differential Form
Definition of divergence:
div D  lim
V  0
1
V

D  nˆ dS
S
V
Note: The limit exists independent of the shape of the volume (proven later).
A region with a positive divergence acts as a “source.”
A region with a negative divergence acts as a “sink.”
6
Gauss’s Law -- Differential Form
Apply divergence definition to a small volume inside a region of charge:
div D  lim
V  0
V
S
v (r)
1
V

D  nˆ dS
S
Gauss's law:
r

D  nˆ dS  Q encl   v  r   V
S
Hence
div D  r   lim
V  0
1
V
   r  V 
v
 v r 
7
Gauss’s Law -- Differential Form (cont.)
The electric Gauss law in point (differential) form:
d iv D  r    v  r 
This is one of Maxwell’s equations.
8
Calculation of Divergence
z
div D  lim
  x / 2, 0, 0 
V  0
(0,0,0)
z
y

S
x
x
y
1
x y z
 D  nˆ dS
S
 x

D  nˆ d S  D x 
, 0, 0   y  z
 2

 x

 Dx  
, 0, 0   y  z
2


 y

 D y  0,
, 0  x z
2


Assume that the point of interest is at the
origin for simplicity (the center of the cube).
The integrals over the 6 faces are
approximated by “sampling” at the
centers of the faces.
y


 D y  0, 
, 0  x z
2


z

 D z  0, 0,
2


 x y

z 

 D z  0, 0, 
 x y
2 

9
Calculation of Divergence (cont.)
div D  lim
V  0

S
 x

D  nˆ d S  D x 
, 0, 0   y  z
 2

 x

 Dx  
, 0, 0   y  z
2


 y

 D y  0,
, 0  x z
2


y


 D y  0, 
, 0  x z
2


z

 D z  0, 0,
2


 x y

z 

 D z  0, 0, 
 x y
2 

1
x y z
 D  nˆ dS
S
1
x y z
 D  nˆ dS

S
 x

Dx 
, 0, 0   D x
 2

x
 x

, 0, 0 

2


 y 
D y  0,
,0   Dy
2



y
z 

D z  0, 0,
  Dz
2 


z
y 

0,

,0

2


z 

0,
0,



2 

10
Calculation of Divergence (cont.)
div D  lim
V  0
 x

Dx 
, 0, 0   D x
 2

x
 x


,
0,
0


2


 y 
D y  0,
,0   Dy
2



y
y 

0,

,0

2


z 

D z  0, 0,
  Dz
2 


z
z 

0,
0,



2 

Hence
div D 
D x
x

D y
y

D z
z
11
Calculation of Divergence (cont.)
Final result in rectangular coordinates:
div D 
D x
x

D y
y

D z
z
12
The “del” operator
  xˆ

x
 yˆ

y
 zˆ
Examples of derivative operators:
scalar
d
d
:
dx
dx
 sin x   cos x

z
This is a vector operator.
Note: The del operator is only
defined in rectangular coordinates.
  ˆ
  rˆ
vector
xˆ
d
dx
Note: Any vector can be
multiplied by a scalar,
or multiplied by another
vector in two different
ways (dot and cross).
:
xˆ
d
dx
 sin x  
 d
 xˆ
 dx



 ˆ
 zˆ
r

z



 ˆ
 ˆ
r


xˆ cos x
d

ˆ
ˆ
ˆ

x
sin
x

x

x

 sin x   cos x
 
dx

d
 d 
ˆ
ˆ
ˆ
ˆ
x

y
sin
x

x

y
 sin x   zˆ cos x


dx
dx




13
Example
V  x , y , z   xˆ  sin x   yˆ  3 y   zˆ  xy 
F in d   V
 x, y, z 
 

 
ˆ
ˆ
 V  x, y, z    x
 y
 zˆ
  xˆ  sin x   yˆ  3 y   zˆ  xy 
y
z 
 x


 



 V  x, y , z   
 sin x    3 y    xy  
y
z
 x

  V  x , y , z    cos x    3    0   3  cos x
14
Divergence Expressed with del Operator
Now consider:
 

 
  D   xˆ
 yˆ
 zˆ
  xˆ D x  yˆ D y  zˆ D z
y
z 
 x

 xˆ  xˆ
Hence
D x
x
D 
 yˆ  yˆ
D x
We therefore notice that
x

D y
y
D y
y

 zˆ  zˆ

D z
z
D z
z
  D  d iv D
15
Divergence with del Operator (cont.)
  D  d iv D
N o te th a t th e d o t a fte r th e d e l
o p e ra to r is im p o rta n t; a n y
sym b o l fo llo w in g it te lls
u s h o w it is to b e u se d
a n d h o w it is re a d :

 "g ra d ie n t"
   "d iv e rg e n ce "
   "cu rl"
16
Summary of Divergence Formulas
Rectangular:
D 
D x
x

D y
y

See Appendix A.2 in the
Hayt & Buck book for a
general derivation that holds
in any coordinate system.
D z
z
Cylindrical:
D 
1 
 
  D  
1  D
 

D z
z
Spherical:
D 
1 
r r
2
r
2
Dr  
1

r sin   
 D sin   
1
 D
r sin   
17
Maxwell’s Equations
(point or differential form)
E  
B
t
H  J 
D
  D  v
B  0
t
Faraday’s law
Ampere’s law
Electric Gauss law
Magnetic Gauss law
Divergence appears in two of Maxwell’s equations.
18
Example
Evaluate the divergence of the electric flux vector inside and outside a
sphere of uniform volume charge density, and verify that the answer is
what is expected from the electric Gauss law.
r<a:
  v0r 
D  rˆ 

3


z
v = v0
y
D 

  D  v0
2
2
 v0 1 
3 r

a
r r
 r Dr 
1   2   v0r  
 2
r 

r  r   3  
r
x
1 
1
r
2
2
 v0r
r 

r
3
2
Note: This agrees with the electric Gauss law.
19
Example (cont.)
r>a:
z
  v0a 3 
D  rˆ 

2
 3r 
D 
v = v0
r
2
r

r
2
Dr 
3
1   2   v0a
 2
r 
2
r r   3r
y
x
1 



3
1    v0a 
 2

0
r r  3 
a
D  0
Note: This agrees with the electric Gauss law.
20
Divergence Theorem
S
nˆ
V
   A dV
V


A  nˆ dS
S
In words:
The volume integral of "flux per volume" equals the total flux!
21
Divergence Theorem (cont.)
Proof
The volume is
divided up into
many small cubes.
V
rn
N
   A dV
V
 lim
V  0
   A
n 1
rn
V
Note: The point rn is the center of cube n.
22
Divergence Theorem (cont.)
From the definition of divergence:
  Ar
 lim
V  0
n

rn
1
V
 S n  surface of cube n
1
V


A  nˆ dS
Sn
A  nˆ dS
Sn
Hence:
N
   A dV
V
 lim
V  0
 V   A 
n 1
 1
 lim   V 
V  0
 V
n 1

N
rn

Sn
N

A  nˆ dS   lim   A  nˆ dS
  V  0 n 1
Sn

23
Divergence Theorem (cont.)
V
N
   A dV
rn
V
 lim
V  0
 
A  nˆ dS
n 1  S
n
Consider two adjacent cubes:
A  nˆ
nˆ 2
1
nˆ 1
is opposite on
the two faces
2
Hence, the surface integral cancels on all INTERIOR faces.
24
Divergence Theorem (cont.)
N
   A dV
V
 lim
V  0
V
 lim
rn
V  0
 
A  nˆ dS
 
A  nˆ dS
n 1  S
n
outside  S
n
face s
Hence
 A
dV  lim
V  0
V
As
V  0 :
 
outside  S
n
f aces
   A dV
V
A  nˆ dS 


A  nˆ dS
S

S
A  nˆ dS
(proof complete)
25
Example
Given: A  xˆ  3 x 
z
Verify the divergence theorem using this box.

A  nˆ dS  xˆ   xˆ 3  3    2     xˆ    xˆ 3  0    2 
S
1
 18
y
3
2
A
 Ax
x
x


x
   A dV   3 dV
V

Note: A is constant on the
front and back faces.
Ay
y

 Az
z
3x   3
 3 V  3 1  2  3   18
V
26
Note on Divergence Definition
div D  lim
V  0
1
V

Is this limit
independent of the
shape of the volume?
D  nˆ dS
S
nˆ
V
Use the divergence theorem for RHS:
r

S
D  nˆ dS 
S

  D dV
V
Small arbitrary-shaped volume
div D  lim
V  0
 lim
V  0
1
V
 D
dV
V
  D 

V
1
r
V
  D
r
Hence, the limit is the same regardless of the shape of the limiting volume.
27
Gauss’s Law (Differential to integral form)
We can convert the differential form into the integral form by
using the divergence theorem.
  D  v
Gauss’s law (differential form):
Integrate both sides over a volume:
D
dV 
V
Apply the divergence theorem to the LHS:
 D  nˆ
v
dV
V
 D  nˆ dS   
S
Hence

v
dV
V
dS  Q encl
S
28
Gauss’s Law (Summary of two forms)
 D  nˆ
dS  Q encl
Integral (volume) form of Gauss’s law
S
Definition of
divergence +
Gauss's law
Divergence theorem
  D  v
Differential (point) form of Gauss’s law
Note: All of Maxwell’s equations have both a point and an integral form.
29