Transcript DISP-2003: Introduction to Digital Signal Processing

Dr. Hugh Blanton
ENTC 3331
Gauss’s Law
• Recall
• Divergence literally means to get farther
apart from a line of path, or
• To turn or branch away from.
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
3
• Consider the velocity vector of a
cyclist not diverted by any thoughts or
obstacles:
Goes straight ahead at
constant velocity.
 (degree of) divergence  0
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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Now suppose they turn with a constant velocity
 diverges from original direction
(degree of) divergence  0
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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Now suppose they turn and speed up.
 diverges from original direction
(degree of) divergence >> 0
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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Current of water
 No divergence from original direction
(degree of) divergence = 0
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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Current of water
 Divergence from original direction
(degree of) divergence ≠ 0
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Source
• Place where something originates.
• Divergence > 0.
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Sink
• Place where something disappears.
• Divergence < 0.
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Derivation of Divergence Theorem
• Suppose we have a cube that is infinitesimally small.
y
Vector field, V(x,y,z)
n̂ i
x
z
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
one of six faces
11
• Need the concept of flux:
• water through an area
• current through an area
A
Â
ĵ
• water flux per cross-sectional area (flux
density implies
• (total) flux =
ˆj  A
ˆ = scaler.
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Let’s assume the vector, V(x,y,z),
represents something that flows, then
• flux through one face of the cube is:

V  nˆ i
• For example n̂ i might be:
• and
nˆ yz  dydz xˆ
Vx xˆ  dydz xˆ  Vx dydz
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• The following six contributions for each
side of the cube are obtained:
Vx dydz
 Vx dydz
Vy dxdz
 V y dxdz
Vz dxdy
 Vz dxdy
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Now consider the opposite faces of the
infinitesimally small cube.
vector magnitude on the input
side.
Vx1
Vx 2  Vx1 
dx
x
y
n̂ i
Vx1
z
differential change of Vx over dx
Vx 2
x
dx
• This holds equivalently for the two other pairs of faces.
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Flux in the x-direction.
Vx
Vx 2  Vx1 
dx
x
y
and
n̂ i
Vx1
z
Vx xˆ  dydz xˆ  Vx dydz
Vx 2
x
dx
Vx1 
Vx 


Vx1  x dx  dydz   Vx 2  x dx  dydz




Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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Divergence Theorem
• Divergence Theorem
• Gauss’s Theorem
• Valid for any vector field
• Valid for any volume,
• Whatever the shape.

 


divV    V  Vx  V y  Vz
x
y
z
Note that the above only applies to the
Cartesian coordinate system.
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Since Gauss’s law can be applied to
any vector field, it certainly holds for
the electric field, Ex, y, z and the
electric flux density, D x, y, z .


   DdV   D  d sˆ
V
S


• The use of D in this context instead of E
is historical.
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• If Gauss’s law is true in general, it should
be applicable to a point charge.
• Constuct a virtual sphere around a positive
charge with radius, R.
+
q
dŝ D

• D must be radially outward along the unit
vector, R̂ .
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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
ˆ R
ˆ d s  Dd s
ˆ
D

d
s

D
R



S
S
S
S
S
2
Dd
s

D
d
s

D
R


 sin dd
DR
2


0
2DR
2
S
2
sin d  d
0


0
sin d
2DR  cos 0  2DR 2  1 1  4DR 2
2

Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• What about the volume integral?

1 
1

1 D
2




D
dV

R
D

D
sin


R

V
R 2 R
R sin  
R sin  



D
1  2
1

D sin    1
D  2
R DR 
R R
R sin  
R sin  


DR  0, D  0, D  0

• D only has a component along the
radius vector
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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
1 
D  2
R 2 DR
R R



R 1

2

2
2
V   DdV  0 R2 R R DR R dR0 sin  d 0 d


4
 4 
R
0
R 
1 
2
2
2
R
D
R
dR

4

R
R
2
0 R DR dR
R R




What is this?
DR   o ER   o
1
q
q
2

D
R

R
4o R 2
4
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Throw in some physics!
DR  ER  
4 
R
0
1
q
4 R 2
R 
R
  2 q 
R
dR

q
dR

q
 1 q

2 


0
0
R  4R 
R
integration and
differentiation cancel out
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• So what?


4DR   D  dsˆ     D dV  q
 2S
V
4 oER  q

1 q
E
2
4 o R
2
• Coulomb’s law and Gauss’s law are
equivalent for a point charge!
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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

4DR   D  dsˆ     D dV  q
2
S
V
divergence
theorem
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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

4DR   D  dsˆ     D dV  q
2
S
V
Gauss’s Law
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Because of its greater mathematical
versatility, Gauss’s law rather than
Coulomb’s law is a fundamental postulate
of electrostatics.
• A postulate is believed to be true, although no
proof may be possible.


 D  dsˆ     D dV  Q
S
V
• Any surface of an arbitrary volume.
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Note


 V dV  Q     D dV   D  dsˆ
V
V
definition of
charge distribution
S
Gauss’s Law
• which infers

  D  V
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
Differential form of
Gauss’s Law
28
• Maxwell Equation

  D  V
• One of two Maxwell equations for
electrostatics.
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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Electric flux density or
Displacement Field [C/m2]
Charge Density [C]

D  

Magnetic Induction [Weber/m
B  0
or Tesla]]


B
 E  
Time [s]
t
Electric Field [V/m]

  D
 H  J 
t
Magnetic Field [A/m]
2
Current Density [A/m2]
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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Page 139

  ( 0 r E )  

  (0 r H )  0


( 0  r H )
 E  
t

   ( 0 r E )
 H  J 
t
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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Page 139
• Use Gauss’s law to obtain an
expression for the E-field from an
infinitely long line of charge.
l  constant
0

E(r )
X
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Symmetry Conditions
• Infinite line of charge
• D  E  0
• Dz  Ez  0
• Dr  Dr r̂
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Gauss’s law considers a hypothetical
closed surface enclosing the charge
distribution.
• This Gaussian surface can have any shape,
but the shape that minimizes our calculations
is the shape often used.

 Ddsˆ  Q
0
l  constant
 dŝ
D
S
h
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• The total charge inside the Gaussian
volume is:
Q  l h
• The integral is:

2 h
 D  dsˆ    Dr rˆ  rˆrddz
0
o
S
• The right and left surfaces do not contribute
since.
Dz  0
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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 Dr r 
2
0

h
o
ddz
 2hDr r
and
2hDr r  l h
l
Dr 
  o Er
2r
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Two infinite lines of charge.
• Each carrying a charge density, l.
• Each parallel to the z-axis at
• x = 1 and x = -1.
• What is the E-field at any point along the y-axis?
l  constant
x
1
z
l  constant
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
1
38
• For a single line of constant charge
l
Er 
2o r
• Using the principle of superposition of
fields:

 
Etot  E1  E2
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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r1 
0 12   y  02
 1 y2
r2 
rˆ2 
 xˆ  yyˆ
rˆ1 
r1
y
x
r2
-1
0  12   y  02
 1 y2
xˆ  yyˆ
r2

   xˆ  yyˆ xˆ  yyˆ 
Etot  l 

2
2o  1  y
1  y 2 

E (0, y,0)

 

r1
1
x
z
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Only interested in the y-component of the
field

l
Etot 
2o
 yyˆ
yyˆ 
 1 y2  1 y2 




l
Etot 
2o
 

 2 yyˆ 
 1 y2 





l  yyˆ 
Etot 
o  1  y 2 

Dr. Blanton - ENTC 3331 - Gauss’s Theorem

41
• A spherical volume of radius a
contains a uniform

 charge density V.
• Determine D  E for
• R  a and
• Ra
Note: Charge distribution for
an atomic nucleus where
a = 1.210-15 m  A⅓ (A is the
mass number)
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
+
q
dŝ D
42
• Outside the sphere (R  a), use
Gauss’s Law
 D  dsˆ
S
• To take advantage of symmetry, use
the spherical coordinates:
ds  R 2 sin dd
• and

D  Dr rˆ
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Field is always perpendicular for any
sphere around the volume.
• The left hand side of Gauss’s Law is

 2
2
2
ˆ
ˆ
 D  dsˆ   DR R  RR sin dd  DR R   sin dd
0
S
0
S
 4
Q
 4DR R  Q  DR 
4R 2
2
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Recall that

  D  V

   DdV   V dV
V
V

 V dV  V 
0
2
a
0
0
 
R 2 sin dRd d
V

2
a
0
0
0
2

dV


sin

d

d

R
V
 V
  dR
V
 4
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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4V 
a
0
3
a
R 2 dR  4V
Q
3
4V a 3 V a 3
Q
DR 


2
2
4R
4  3R
3R 2
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Inside the sphere (R  a), use
Gauss’s Law

 D  dsˆ   V dV  Q
S
V
4DR R 2
previously calculated
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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 V dV  
R
0

2
0
0

V R 2 sin dRd d
V
R
 V 4  R 2 dR
0

V 4R 3
3
4DR R 2 
V 4R 3
3
V R
DR 
3
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Thin spherical shell
• Find E-field for
• R  a and
• Ra
S  0
a
 S  constant
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Inside (R  a)
• Gauss’s Law

 D  dsˆ  Q  0
S
S  0
a
 S  constant
 
• This is only possible if D  0.
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Outside (R  a)
• Gauss’s Law

 D  dsˆ  Q    S dS
S
S  0
a
 S  constant
S
4DR R 2
previously calculated

  S dS  
0
S

S

2
0
 S a 2 sin dd
dS  4 S a
2
S
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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4DR R 2  4 S a 2
2
a
DR   S 2
R
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• An electric field is given as

1
E x, y, z   xˆ 2 x  y   yˆ 3 x  2 y  V

m
• Determine
• V
• Q in a 2m  2m  2m cube.
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
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• Maxwell’s equation of
Electrostatics


div D  V   div E
y



 div E  2 x  y   3x  2 y 
x
y

 div E  2  2  0  V  0
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
x
z
54

 D  dsˆ  Q
y
S
1
For the surface 1 directed in the x-direction.
  2x  y xˆ  xˆ dydz
2
2
0
0
x

x  y dydz

0 0
2
2
2
z
2

y 
2   x  y  z  dy  4   x  y dy  4 xy  
0
0
2 0

2
2
0
2
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
2
55
y
2
2

y 
4 xy    8 x  8
2 0

1
For the surface 2 directed in the -x-direction.
  2x  y xˆ  xˆdydz
2
2
0
0

2
0
 2x  y dydz  8x  8
2
2
x
z
0
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
56
y
For the surface 3 & 4 directed in the z- & -z
directions.

 D  dsˆ  0
4
3
S
x
z
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
57
y
For the surface 5 directed in the y-direction.
5
  3x  2 y yˆ  yˆ dxdz
2
2
0
0
  3x  2 y dxdz
2
2
0
0
 3xz  2 yz 
2
2
0
0
x
dx   6 x  4 y  dx
2
z
0
2
 6x

 
 4 xy   12  8 y
 2
0
2
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
58
y
For the surface 6 directed in the -y-direction.
  3x  2 y yˆ  yˆ dxdz
2
2
0
0

2
0
6
 3x  2 y dxdz
2
x
0
 12  8 y 
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
z
59
• By superposition

 D  dsˆ  0
S
• Indeed, there is no charge in the cube.
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
60

• Find D in all regions of an infinitely
long cylindrical shell.
• Inner shell( r  1 )
• Cylindrical volume.

 D  dsˆ  Q  0
V  constant
V  0
1
S
3
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
61
• Shell itself ( 1  r  3 )
• Cylindrical coordinates.
V  constant

2 h
 D  dsˆ    Dr rˆ  rˆ rd dz
0
V  0
0
S
1
r
 dŝ
D r
3
h
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
62
2
h
0
0
 
Dr rˆ  rˆ rd dz  
2
0
 Dr r 
2
0

h
0

h
0
Dr rd dz
2
d dz  Dr rh  d 2Dr rh
0
• Top and bottom face
 of cylinder do
not contribute to D .
r
2
h
0
0
 
1
V rdrd dz
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
63
r
2
h
0
0
 
1
r
V rdrd dz  2 V h rdr
1
r
 2 V h
 2
2
r

2
   V h r  1
1



 D  dsˆ   V dV
S
V
2Dr rh   V h(r 2  1)
Dr. Blanton - ENTC 3331 - Gauss’s Theorem
64
2Dr rh   V h(r 2  1)
Dr 
V (r 2  1)
2
r
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65