Engineering Electromagnetics
Lecture 4
Dr.-Ing. Erwin Sitompul
President University
http://zitompul.wordpress.com
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Erwin Sitompul
EEM 4/1
Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence
Application of Gauss’s Law: Differential Volume Element
 We are now going to apply the methods of Gauss’s law to a
slightly different type of problem: a surface without symmetry.
 We have to choose such a very small closed surface that D is
almost constant over the surface, and the small change in D
may be adequately represented by using the first two terms of
the Taylor’s-series expansion for D.
 The result will become more nearly correct as the volume
enclosed by the gaussian surface decreases. We intend
eventually to allow this volume to approach zero.
President University
Erwin Sitompul
EEM 4/2
Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence
Taylor’s Series Expansion
f ( x0  x)  f ( x)
f ( x0 )
x0
x  x0  x
A point near x0
f ( x)  f ( x0  x)
f ( x0 )
f ( x0 )
f ( x)  f ( x0 ) 
x 
(x) 2 
1!
2!
f n ( x0 )

(x) n
n!
Only the linear terms are
used for the linearization
President University
Erwin Sitompul
EEM 4/3
Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence
Application of Gauss’s Law: Differential Volume Element
 Consider any point P, located by a
rectangular coordinate system.
 The value of D at the point P may be
expressed in rectangular components:
D0  Dx0ax  Dy 0a y  Dz 0az
 We now choose as our closed surface,
the small rectangular box, centered at P,
having sides of lengths Δx, Δy, and Δz,
and apply Gauss’s law:


D  dS  Q
S
S
D  dS  
front
President University

back
 
left
right
 
top
bottom
Erwin Sitompul
EEM 4/4
Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence
Application of Gauss’s Law: Differential Volume Element
 We will now consider the front surface
in detail.
 The surface element is very small, thus
D is essentially constant over this
surface (a portion of the entire closed
surface):

front
Dfront  Sfront
Dfront  yz ax
Dx,front yz
 The front face is at a distance of Δx/2 from P, and therefore:
Dx ,front
x
Dx 0 
 rate of change of Dx with x
2
x Dx
Dx 0 
2 x
President University
Erwin Sitompul
EEM 4/5
Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence
Application of Gauss’s Law: Differential Volume Element
 We have now, for front surface:

front
x Dx 

 Dx 0 
 yz
2 x 

 In the same way, the integral over the back surface can be
found as:

back
Dback  S back
Dback  (yz a x )
Dx,back yz
Dx ,back

back
x Dx
Dx 0 
2 x
x Dx 


D

 x0
 yz
2 x 

President University
Erwin Sitompul
EEM 4/6
Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence
Application of Gauss’s Law: Differential Volume Element
 If we combine the two integrals over the front and back surface,
we have:

front


back
Dx
xyz
x
 Repeating the same process to the remaining surfaces, we find:

right

top



left

bottom
Dy
y
yxz
Dz
zxy
z
 These results may be collected to yield:
 Dx Dy Dz 
 S D  dS  x  y  z  xyz
 Dx Dy Dz 
 S D  dS  Q  x  y  z  v
President University
Erwin Sitompul
EEM 4/7
Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence
Application of Gauss’s Law: Differential Volume Element
 The previous equation is an approximation, which becomes
better as Δv becomes smaller, and in the following section the
volume Δv will be let to approach zero.
 For the moment, we have applied Gauss’s law to the closed
surface surrounding the volume element Δv.
 The result is the approximation stating that:
 Dx Dy Dz 
Charge enclosed in volume v 


  v
y
z 
 x
President University
Erwin Sitompul
EEM 4/8
Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence
Application of Gauss’s Law: Differential Volume Element
 Example
Let D = y2z3 ax + 2xyz3 ay + 3xy2z2 az nC/m2 in free space.
(a) Find the total electric flux passing through the surface x = 3,
0 ≤ y ≤ 2, 0 ≤ z ≤ 1 in a direction away from the origin. (b) Find
|E| at P(3,2,1). (c) Find the total charge contained in an
incremental sphere having a radius of 2 mm centered at
P(3,2,1).
(a) ψ  S DS  dS

1
z 0
 y z a

1
2
2

y 0
0 0
 y
1
3
2 3
3
3
2 2

2
xyz
a

3
xy
z a z    dydz a x 
x
y
x 3
y 2 z 3dydz
2 1
0 4
z
4
1
0
 23 nC
President University
Erwin Sitompul
EEM 4/9
Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence
Application of Gauss’s Law: Differential Volume Element
(b) D = y2 z3ax  2xyz3a y  3xy2 z 2az
DP = (2)2 (1)3 ax  2(3)(2)(1)3 a y  3(3)(2)2 (1)2 az
= 4ax 12a y  36az nC m2
DP = DP  (4) 2  (12) 2  (36) 2
 38.158nC m2
EP 
DP
0
38.158 nC m 2

8.854 1012
 4.31 kV m
President University
Erwin Sitompul
EEM 4/10
Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence
Application of Gauss’s Law: Differential Volume Element
(c)
 Dx Dy Dz 
Q 


 v
y
z 
 x
 Dx Dy Dz 
QP 


 v
y
z  P
 x
 0  2 xz
3
 6 xy 2 z  x 3 nC m3  43  (2 103 )3 m3
y 2
z 1
4
3
2
0

2(3)(1)

6(3)(2)
(1)


 3  (2 103 )3 nC
2.611015 C
President University
Erwin Sitompul
EEM 4/11
Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence
Divergence
 We shall now obtain an exact relationship, by allowing the
volume element Δv to shrink to zero.

 Dx Dy Dz 





x

y

z


D  dS
S
v

 Dx Dy Dz 



  lim
y
z  v0
 x


Q
v
D  dS
S
v
Q
 lim
v 0 v
 The last term is the volume charge density ρv, so that:
 Dx Dy Dz 



  lim
y
z  v0
 x
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
D  dS
S
v
 v
Erwin Sitompul
EEM 4/12
Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence
Divergence
 Let us no consider one information that can be obtained from
the last equation:
 Dx Dy Dz 



  lim
y
z  v0
 x

D  dS
S
v
 This equation is valid not only for electric flux density D, but
also to any vector field A to find the surface integral for a small
closed surface.
 Ax Ay Az 



  lim
y
z  v0
 x
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
A  dS
S
v
Erwin Sitompul
EEM 4/13
Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence
Divergence
 This operation received a descriptive name, divergence. The
divergence of A is defined as:

Divergence of A  div A  lim
v 0
A  dS
S
v
“The divergence of the vector flux density A is the
outflow of flux from a small closed surface per unit
volume as the volume shrinks to zero.”
 A positive divergence of a vector quantity indicates a source of
that vector quantity at that point.
 Similarly, a negative divergence indicates a sink.
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Erwin Sitompul
EEM 4/14
Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence
Divergence
Dx Dy Dz
div D 


x
y
z
Rectangular
1 
1 D Dz
div D 
(  D ) 

 
 
z
Cylindrical
1  2
1

1 D
div D  2 (r Dr ) 
(sin  D ) 
r r
r sin  
r sin  
Spherical
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EEM 4/15
Erwin Sitompul
Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence
Divergence
 Example
If D = e–xsiny ax – e–x cosy ay + 2z az, find div D at the origin
and P(1,2,3).
Dx Dy Dz
 e x sin y  e x sin y  2
div D 


x
y
z
2
Regardless of location the divergence of D equals 2 C/m3.
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Erwin Sitompul
EEM 4/16
Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence
Maxwell’s First Equation (Electrostatics)
 We may now rewrite the expressions developed until now:
div D  lim

D  dS
S
v  0
div D 
v
Dy
Dx
D

 z
x
y
z
div D  v
Maxwell’s First Equation
Point Form of Gauss’s Law
 This first of Maxwell’s four equations applies to electrostatics
and steady magnetic field.
 Physically it states that the electric flux per unit volume leaving
a vanishingly small volume unit is exactly equal to the volume
charge density there.
President University
Erwin Sitompul
EEM 4/17
Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence
The Vector Operator  and The Divergence Theorem
 Divergence is an operation on a vector yielding a scalar, just
like the dot product.
 We define the del operator  as a vector operator:




ax  a y  az
x
y
z
 Then, treating the del operator as an ordinary vector, we can
write:


 
 D   a x  a y  a z   ( Dxa x  Dy a y  Dz a z )
y
z 
 x
Dx Dy Dz
D 


x
y
z
Dx Dy Dz
div D =   D =


x
y
z
President University
Erwin Sitompul
EEM 4/18
Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence
The Vector Operator  and The Divergence Theorem
 The operator does not have a specific form in other
coordinate systems than rectangular coordinate system.
 Nevertheless,
1 
1 D Dz
D 
(  D ) 

 
 
z
Cylindrical
1  2
1

1 D
  D  2 (r Dr ) 
(sin  D ) 
r r
r sin  
r sin  
Spherical
President University
EEM 4/19
Erwin Sitompul
Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence
The Vector Operator  and The Divergence Theorem
 We shall now give name to a theorem that we actually have
obtained, the Divergence Theorem:

S
D  dS  Q   v dv     Ddv
vol
vol
 The first and last terms constitute the divergence theorem:

S
D  dS     D dv
vol
“The integral of the normal
component of any vector field
over a closed surface is equal to
the integral of the divergence of
this vector field throughout the
volume enclosed by the closed
surface.”
President University
Erwin Sitompul
EEM 4/20
Electric Flux Density, Gauss’s Law, and DIvergence
Chapter 3
The Vector Operator  and The Divergence Theorem
 Example
Evaluate both sides of the divergence theorem for the field
D = 2xy ax + x2 ay C/m2 and the rectangular parallelepiped
fomed by the planes x = 0 and 1, y = 0 and 2, and z = 0 and 3.


S
S
D  dS     D dv
Divergence Theorem
vol
3
2
0
0
DS  dS  
 (D)
   (D)
x 0
3
1
0
0
3
2
0
0
 (dydz a x )  
y 0
 (dxdz a y
 (D)
)    (D)
x 1
3
1
0
0
 (dydz a x )
y 2
 (dxdz a y )
But ( Dx ) x0  0, ( Dy ) y0  ( Dy ) y2
D
S
S
 dS  
3
0

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2
0
( Dx ) x1 dydz  
3
0

2
0
2ydydz  12 C
Erwin Sitompul
EEM 4/21
Electric Flux Density, Gauss’s Law, and DIvergence
Chapter 3
The Vector Operator  and The Divergence Theorem
D =

vol


(2 xy)  ( x 2 )  2 y
x
y
 D dv  
3
2
 
1
z 0 y 0 x 0
 x0 y
1
2 2
0
(2 y)dxdydz
3
z0
 12 C
  D  dS     D dv  12 C
S
vol
President University
Erwin Sitompul
EEM 4/22
Chapter 3
Electric Flux Density, Gauss’s Law, and DIvergence
Homework 4
 D3.6.
 D3.7.
 D3.9. All homework problems from Hayt and Buck, 7th Edition.
 Deadline: 8 May 2012, at 08:00.
President University
Erwin Sitompul
EEM 4/23