Transcript notes13 2317 - University of Houston

ECE 2317
Applied Electricity and Magnetism
Prof. D. Wilton
ECE Dept.
Notes 13
Notes prepared by the EM group,
University of Houston.
Divergence -- Physical Concept
Start by considering a sphere of uniform volume charge density
The electric field is calculated using Gauss's law:
r < a:
z
 D  n dS  Q
v = v0
encl
S
y
r
x
a
4 3
4 r Dr   v 0   r 
3

 r
C/m 2 
Dr  v 0
3
2
Divergence -- Physical Concept (cont.)
r > a:
4 3
4 r Dr   v 0   a 
3

v 0 a3
2


Dr 
C/m
2

3r
2
z
v = v0
y
a
r
x
Divergence -- Physical Concept (cont.)
Flux through a spherical surface:
   D  nˆ dS  4 r 2 Dr
S
Dr 
Dr 
v 0 r
3
v 0 a 3
3r
2
C/m 2 
(r < a)
C/m2 
(r > a)
4 3
   r v 0
3
(r < a)
4 3
   a v 0
3
(r > a)
Divergence -- Physical Concept (cont.)
Observation:
More flux lines are added as the radius increases
(as long as we stay inside the charge region).
 
 D  n dS  0
S
The net flux out of a
small volume V inside
the charge region is not
zero.
V
S
 
 D  n dS  0
S
Divergence is a mathematical way of describing this.
Gauss’s Law -- Differential Form
Definition of divergence:

1
div D  lim
 lim
V 0 V
V 0 V
 D  nˆ dS
S
V
Note: the limit exists independent of the shape of the volume (proven later).
Gauss’s Law -- Differential Form
Apply divergence definition to small volume inside a region of charge
1
div D  lim
V 0 V
V
 D  n dS  Q
 D  n dS
S
encl
 v  r  V
S
v (r)
Qencl
div D  r   lim
V 0
V
 v  r 
Gauss’s Law -- Differential Form (cont.)
Alternatively,

Qencl
div D  r   lim
 lim
V  0 V
V  0
V
1
 lim
 v dV

V  0 V V
 lim  vavg
V  0
 v  r 
div D  r   v  r 
The electric Gauss law: This is one of Maxwell’s equations.
Example
z
v = v0
V
y
Choose V to be small sphere of radius r:
r
a
x
4 3
V   r
3
Verify that the differential form of Gauss’s law gives
the correct result at the origin for the example of a
sphere of uniform volume charge density.
1
div D  lim
V 0 V
 D  n dS
S
 v 0 r 
3


4

r
2
2
S D  n dS  Dr  4 r    3   4 r   v0  3 
 v 0 4 r 3 
div D  lim
 lim v 0  v 0


V 0 4
3
 V 0
 r3 
3
1
Calculation of Divergence
z
1
V 0 x y z
div D  lim
(0,0,0)
z
y
x
y
x
Assume point of interest is at the
origin for simplicity.
The integrals over the 6 faces are
approximated by “sampling” the
integrand at the centers of the
faces.
 D  n dS
S
 x

D

n
dS

D
,
0,
0
x
 y z
S
2


 x

 Dx   , 0, 0  y z
 2

 y 
 Dy  0, , 0  x z
 2 
y 

 Dy  0, 
, 0  x z
2 

z 

 Dz  0, 0,  x y
2 

z 

 Dz  0, 0,   x y
2 

Calculation of Divergence (cont.)
1
V 0 x y z
div D  lim
 x

D

n
dS

D
,
0,
0
x
 y z
S
 2

 x

 Dx   , 0, 0  y z
 2

 y 
 Dy  0, , 0  x z
 2 
y 

 Dy  0, 
, 0  x z
2


z 

 Dz  0, 0,  x y
2 

z 

 Dz  0, 0,   x y
2 

 D  n dS
S
1
x y z
 D  n dS 
S
 x

 x

Dx  ,0,0   Dx   ,0,0 
 2

 2

x
y 
 y 

D y  0, ,0   D y  0, 
,0 
2
2





y
z 
z 


Dz  0,0,   Dz  0,0, 

2 
2 



z
Calculation of Divergence (cont.)
div D  lim
V 0
 x

 x

Dx  , 0, 0   Dx   , 0, 0 
 2

 2

x
y 
 y 

Dy  0, , 0   Dy  0, 
,0
2
2





y
z 
z 


Dz  0, 0,   Dz  0, 0,  
2 
2 

 
z
For arbitrary origin, just add x,y,z to
coordinate quantities in parentheses!
Hence
Dx Dy Dz
div D 


x
y
z
Calculation of Divergence (cont.)
Dx Dy Dz
div D 


x
y
z
1
div D  lim
V 0 V
 D  n dS
S
The divergence of a vector is its “flux per unit volume”
“del operator”



x y z
x
y
z
The “del” operator is a vector
differential operator
Examples of derivative operators:
d
:
dx
d
vector x
:
dx
scalar
vector->scalar
vector->vector
d
 sin x   cos x
scalar -> scalar
dx
d
x  sin x   x cos x scalar -> vector
(1)
dx
d
 d 
 x   x sin x  x  x  sin x   cos x (2)
dx
 dx 
d
 d 
 x   y sin x  zˆ  sin x   zˆ cos x (3)
dx
 dx 




Example
V  x, y, z   x  sin x   y  3 y   z  xy 
Find V

 


V  x, y, z    x  y  z   x  sin x   y  3 y   z  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

“del operator” (cont.)
Now consider:

 

 
  D   x  y  z   xDx  yDy  zDz
y
z 
 x
Dy
Dx
Dz
 xx
 y y
 zz
x
y
z
Hence
so
Dx Dy Dz
 D 


x
y
z
 D  div D

 Note: No unit vectors appear!

Gauss's law :
  D  v
Summary of Divergence Formulas
Rectangular:
Dx Dy Dz
 D 


x
y
z
Note the dot after the del
Cylindrical:
1 
1 D Dz
 D 
 D  


 
 
z
operator is important; any
symbol following it tells
us how to use and read it:
  "gradient"
   "divergence"
   "curl"
Spherical:
1  2
1 
1 D
  D  2  r Dr  
 D sin  
r r
r sin  
r sin  
The divergence of a vector is its “flux per unit volume”
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:
z
v =
v0
y
r
x
a
 v 0 r 
D  r

3


1
D  2
r
1
 2
r

  D  v 0
 2
r Dr 

r
  2  v 0 r  
r 

r   3  
1
2

r
v0
r2
Note: This agrees with the electric Gauss law.
Example (cont.)
r>a:
z
v =
v0
y
x
a
 v 0 a 3 
D  r
2 
 3r 
1   2  v 0 a3  
 D  2 r  2 
r r   3r  
1   v 0 a 3 
 2 
0
r r  3 
 D  0
Note: This agrees with the electric Gauss law.
Maxwell’s Equations
B
 E  
t
D
 H  J 
t
  D  v
B  0
Faraday’s law
Ampere’s law
electric Gauss law
magnetic Gauss law
Divergence Theorem
S
V
   A dV   A  n dS
V
S
In words, for a vector
A  A( x, y, z ) an arbitrary
vector function of position
A:
The volume integral of “flux per unit volume” equals the total flux!
Divergence Theorem (cont.)
Proof:
V
N
  A dV  lim   A
V
V 0
n 1
rn is the center of cube n
rn
V
Divergence Theorem (cont.)
From the definition of divergence:
1
V  0 V
   Ar  lim
n
1

V


A  n dS
Sn
A  n dS
S n
Hence:
N
  AdV  lim  V  A
V
V 0
n 1
N
rn
 lim 
V 0

n 1 Sn
A  n dS
Divergence Theorem (cont.)
N
  AdV  lim  
V
V 0
V
A  n dS
n 1 Sn
Consider two adjacent cubes:
A n
1
n2
n1
2
Hence: the surface integral cancels on all INTERIOR faces.
is opposite on
the two faces
Divergence Theorem (cont.)
n
N
   A dr  lim  
V 0
V
V
n
 lim
V 0
Hence:
  A dV  lim  
V 0
V
Therefore:
A  n dS 
outside Sn
faces
   A dV   A  n dS
V
A  n dS
n 1 Sn
 
A  n dS
outside Sn
faces
 A  n dS
S
(proof complete)
S
The vol. integral of the “flux per unit volume” is the “flux”
Example
Given:
z
A  x 3x
Verify the divergence theorem using the box.
 A  n dS  x   x 33  2   x   x 3  0  2
S
 18
1 [m]
y
3 [m]
2 [m]
x
Ax Ay Az
 A 


x
y
z

  3x   3
x
   A dV   3 dV  3V  3 1 2  3  18
V
V
Validity of Divergence Definition
1
div D  lim
V 0 V
Is this limit
independent of the
shape of the volume?
S
S
nˆ
V
From the divergence
theorem:

D  n dS
r
1
div D  lim
V 0 V
1
 lim
V 0 V
   D dV
V
   D 
r

V    D r
Hence, the limit is the same regardless of the shape of the limiting volume.
Gauss’s Law (Conversion between forms)
 D  n dS  Q
encl
V
S
Divergence theorem:
Qencl
    D  dV   D  n dS
V
This is valid for any volume,
so let V  V (a small
volume inside the original
volume)
v
V 0
Hence:
S
S
 Qencl   v dV
V
    D  dV   
V
V
v
dV
   D  V  v V
 D  v
v
Gauss’s Law (Summary of two forms)
 D  n dS  Q
encl
Integral (volume) form of Gauss’s law
S
Divergence
definition
Divergence theorem
 D  v
Differential (point) form of Gauss’s law