Part-I: Electro-Static
Lecture 5
Dr. Ahmed Said Eltrass
Electrical Engineering Department
Alexandria University, Alexandria, Egypt
Fall 2020
Office hours:
Wednesday 12:30 p.m. to 01:30 p.m.
4th floor, Electrical Engineering Building
Chapter 3 (Continued)
Electric Flux Density, Gauss’s
Law, and Divergence
Application of Gauss’s Law: Differential Element Volume
• Apply Gauss’s law to a problem that does not possess any
symmetry
• Without symmetry, a Gaussian surface cannot be chosen such
that the normal component of D is constant or zero
everywhere on the surface
• Choose a very small closed surface (Δν ) where D is almost
constant over the surface
• In this procedure, we will not obtain a solution for D
• Instead we will obtain some valuable information about D
Gauss’ Law in the Point Form (Differential Form)
The surface element is very small, D is approximately constant




 D • ds = Dfront • Sfront = Dx, front ax • (yz )ax
front
 D • ds = D
x, front
yz
front
Front face is at Δx/ 2 from P, and hence
Δx
Dx, front = Dx0 +
 rate of change of Dx with x
2
Δx Dx
Dx, front = Dx0 +
2 x
Δx Dx 

D • ds = Dx0 +
yz

2 x 

front
Consider now the integral over the back surface




(
)
D
•
ds
=
D
•

S
=
D
a
•

y

z
(
−
a
back
back
x, back x
x)

back
 D • ds = − D
x, back
yz
back
Back face is at Δx/ 2 from P, and hence

Δx Dx
Dx, back = Dx0 −
2 x
Δx Dx 

D
•
ds
=
−
D
+

yz
x0

2 x 

back
Combine these two integrals
Δx Dx 
Δx Dx 


D
•
ds
=
D
+

y

z
+
−
D
+



yz
x0
x0


2 x 
2 x 


front
back
Dx
D
•
ds
+
D
•
ds
=
Δxyz


x
front
back
D • ds +
Following the same procedure
D y
D • ds +  D • ds =
Δxyz

y
right
left
Dz
topD • ds + bottom
 D • ds = z Δxyz
Combining the three parts

 Dx D y Dz 
S D • dS = Q =  x + y + z  Δxyz

 Dx D y Dz 
S D • dS = Q =  x + y + z  Δv
The expression is an approximation which becomes better as
Δν becomes smaller
Example:
7 - Find an approximate value for the total charge enclosed
in an incremental volume of 10-9 m 3 located at the origin.




−x
−x
D = e sin ya x − e cos ya y + 2 za z C/m 2

 Dx D y Dz 
S D • dS = Q =  x + y + z  Δv
Allowing the volume element Δν to shrink to zero, an exact
relationship can be derived
 Dx D y Dz 
Q 


+
+
= lim 
 = v

y
z  Δv →0  Δv 
 x
Q 
Divergence of D = divD = lim 
 = v
Δv →0  Δv 
The divergence of the vector flux density D is the outflow of flux
from a small closed surface per unit volume as the volume shrinks

to zero.
div D =  v
Maxwell’s First Equation
Gauss’ Law in the Point Form
Maxwell’s First Equation

div D =  v
• The divergence refers to a flux that diverges (or converges) as
determined by the charge distribution.
• In the absence of ρv, the divergence of the electric flux is zero,
which indicates that the flux does not diverge (or converge).
• The presence of a charge distribution at the location (ρv ≠0) will
cause the flux to diverge (positive ρv) or converge (negative ρv
causes negative divergence).
• A positive divergence for any vector quantity indicates a source
of that vector quantity at that point. Similarly, a negative
divergence indicates a sink.
Maxwell’s First Equation

div D =  v
• In different coordinate systems
Will be given in exams
Divergence Theorem
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.


 D • dS =   • Ddv
S
v
The total flux crossing the closed surface is equal to the integral of
the divergence of the flux density throughout the closed volume
Proof

− Starting from Gauss law :  D • dS = Qenclosed
S
where Qenclosed =   v dv
v

 D • dS =  v dv
S
v
S
v

- From Maxwell’s first equation :  • D =  v


 D • dS =   • Ddv
Example:

8 - Evaluate both sides of the divergence theorem for the field D
Over the rectangular parallelepiped formed by the planes
x = 0 and 1, y = 0 and 2, and z = 0 and 3


2
D = 2 xya x + x a y C/m 2