1
Chapter 3. Gauss’ law, Divergence
EMLAB
Displacement flux : Faraday’s Experiment
2
charged sphere
(+Q)
+
+
+
+
•
•
•
•
metal
insulator
Two concentric conducting spheres are separated by an insulating
material.
The inner sphere is charged to +Q. The outer sphere is initially
uncharged.
The outer sphere is grounded momentarily.
The charge on the outer sphere is found to be -Q.
EMLAB
3
• Faraday concluded there was a “displacement” from the charge
on the inner sphere through the inner sphere through the insulator
to the outer sphere.
• The electric displacement (or electric flux) is equal in magnitude
to the charge that produces it, independent of the insulating
material and the size of the spheres.
+Q
-Q
+Q
EMLAB
Gauss’ law
4
D : electric displacement flux density. The electric field lines come out of positive
charges and induce another charge distribution.
D  E
D(r ) 
Q1
4 r  r
2
Rˆ
r
R̂
S
Q1
r
Q1
n̂
S
n̂
Q1 (r  S )
S D(r)  da   0 (r  S )
Gauss’ law : The integral of electric flux density over a closed surface is equal to the
sum of charges inside the surface.
EMLAB
5
Integration surface deformability
If the integrand follows an inverse square law, the integration surface is
deformable as long as the position of the charge remains inside or outside the
surface during deformation. The result of the integration does not change after
deformation.
D(r ) 
Q1
4 r  r
2
Rˆ
r
R̂
S1
r
S2

S1
D(r)  da   D(r)  da  Q1
S2
EMLAB
Proof : Gaussian surface contains a charge
6
The contribution of the
integrand on da1 is equal to
that on da2
da1
S1
da2
S2
Q1 da cos
ˆ
S D(r)  da  S1 4 r  r 2 R  nˆ da  4 S1 R2
Q1
EMLAB
7
n̂
S1 
1
2  0
R2
r
1
R̂
S2
R1
da2 cos2
da1 cos1
da1
Because S1 and S2 are similar, the ratio of the area S1 to that of
S2 is
2
where
S1 R1
 2
S2 R2
S1 S2
 2
2
R1 R2
S1  da1 cos 1
S2  da2 cos 2

da2 cos  2 da1 cos 1

2
R2
R12
EMLAB
8
n̂
S2
da cos 0 Q1 2  r 2 sin  d d

Q1
2
2



S
0
0
4 2 R
4
r
Q1
EMLAB
9
Proof : Gaussian surface contains no charge
An integration surface is deformable as long as the position of the charge
remains outside the surface S. The result of the surface integral remains the
same.
S
Sb
S
n̂
n̂
Sa
n̂
 D  nˆda   D  nˆda   D  nˆda
  D  nˆda   D  ( nˆ )da  0
S
Sa
Sa
Sb
Sa
EMLAB
Gauss’ law
10
Gauss’ law holds for the case with multiple charges.

D

d
a
Q

S
EMLAB
11
Gauss law usefulness
Gaussian surface
n̂
The integration in Gauss’ law becomes trivial when
there exists a symmetry in the problem geometry.
2
ˆ
ˆ
D

d
a

D
r

r
da

D
da

D
da

4

r
Dr  Q1
r

 r
 r
S1
S1

 D  da  Q
S1
S1
S1
‘Dr’ can be taken out of the integral thanks to the
symmetry.
S
EMLAB
Example 1: electric field due to spherically symmetric
volume charges
12
ẑ
Gaussian surface
r
D(r )  
V'

a
x̂
ŷ
(r' ) R
4 r  r '
3
da '
 r 
1 4r 3 
ˆ

 
 r 
2
3
4

r
3


 3 
3
rˆ a    1 4a  
 3r 2  4r 2 3 
Charge enclosed in the
Surface integral
 2   r
4r 3
2
rˆ  rˆ r sin dd 


  0 0 3
3
  D  da  2   3
3
S
 a  rˆ  rˆ r 2 sin dd  4a 
 0 0 3r 2
3
(r<a)
(r>a)
EMLAB
Example 2: E-field due to an axially symmetric charge distribution
ẑ
13
 L 2
 D  da    D ρˆ  ρˆ ddz 2LD  2a S L
S
0 0
 D  ρˆ
a S


a
ŷ
 L 2
 D  da    Dρˆ  ρˆ ddz 2RLD   0
x̂
S

a
•D is determined by the
charges inside the Gaussian
surface regardless of the
charges outside the surface.
x̂
0 0
D  0
ŷ
• Usually, the total charge on the outer
surface is equal to the negative of that
on the inner surface.
EMLAB
14
Gauss’ law in differential form
S
V
d

 D  da   q i    d
S
i
V

D

d
a
 D1a1  D2 a2  D3a3  

S
•
An integral form of Gauss’ law only states that a surface
integral of D over a closed surface is equal to the total
charge inside. We cannot obtain the behavior of D at one
specific point.
•
To observe D at a point, an infinitesimally small integration
surface is chosen for the Gauss’ law.
EMLAB
Gauss’ law on an infinitesimally small surface
  D  lim

D

d
a

S
V 0
ẑ

D

d
a


d
‘Divergence’ D
V

D

d
a


x축에 수직인 면
15

D

d
a


y축에 수직인 면

D

d
a

z축에 수직인 면
dx
dx
 



  Dx  x  , y , z   Dx  x  , y , z  dydz
2
2



 
dy 
dy 
 

  D y  x, y  , z   D y  x, y  , z  dzdx
2 
2 

 
dz
n̂
dx
ŷ
dy
x̂
dz 
dz 
 

  Dz  x, y , z    Dz  x, y , z   dxdy
2
2 

 
 D D D 
  x  y  z  dxdydz   ( x, y , z ) dxdydz
y
z 
 x
 D D D 
   D   x  y  z    ( x, y , z )
y
z 
 x
The coordinate of the
center of the cube is (x,y,z)
The differential form of Gauss’ law states that the
sum of partial derivatives of Dx, Dy, Dz with respect
to x, y, z is equal to the charge density at that point.
EMLAB
Divergence in cylindrical coordinate
16
 Dx D y Dz 


D

d
a

D
dydz

D
dzdx

D
dxdy




 dxdydz
x
y
z
S
S
y
z 
 x
  ( D  ) D  ( Dz  ) 
  ( D  )d dz  D dzd  ( Dz  )d d   


 d d dz

z 
 
S
  (  ,  , z )  d d dz
  D 
  ( D  ) D  ( Dz  ) 
1



d d dz
 d d dz  

z 

1  ( D ) 1 D Dz


  (  , , z)
 
 
z
Divergence in spherical coordinate
 D  da   (D r

r
S
2
sin )dd  (D  r sin )drd  (D  r )drd

S
  (D r r 2 sin )  (D  r sin )  (D  r ) 
 drdd
 


r

 

 (, , z)r 2 sin drdd
  D 
  (D r r 2 sin )  (D  r sin )  (D  r ) 
1

drdd


r 2 sin drdd 
r

 

1  (r 2 D r )
1  (sin D  )
1 D 


 (r, , )
2
r
r
r sin 

r sin  
EMLAB
Boundary Conditions on the Electric Field at the Surface
of a Metallic Conductor
n̂
Et  0
Dn  n̂  D   s
-
 En  nˆ  E 
17
s
0
- - -
E=0
+
+
++ +
EMLAB
Del operator : rectangular coordinate
18
 del operator
  xˆ



 yˆ  zˆ
x
y
z
 


   D   xˆ  yˆ  zˆ   xˆ Dx  yˆ D y  zˆ Dz 
y
z 
 x
D D D
 x y z
x
y
z
Divergence theorem

D

d
a
 Q    d     D d

S
V

    D d   D  da
V
V
S

   D d   D  da
V
S
EMLAB
19
Example 3.5
D  2 xyxˆ  x 2 yˆ [C/m 2 ]
ẑ
3
 D  da  
3 2

0 0
D( x  0, y, z )  ( xˆ dydz )  
3 2
 D( x  1, y, z )  (xˆ dydz )
0 0
S
3 1
3 1
0 0
0 0


2
ŷ
x̂
1
 D( x, y  0, z )  (yˆ dxdz)    D( x, y  2, z )  (yˆ dxdz)
3 2

0 0
Dx ( x  1, y, z )dydz  
3 2

0 0
3
2 ydydz   4dz  12
0
Dx Dy Dz (2 xy) ( x 2 )
D 




 2y
x
y
z
x
y
3 2 1
   D d     2ydxdydz  12
0 0 0
V

    D d   D  da
V
S
EMLAB
How to obtain D using a differential form Gauss’ law
20
x̂
D 
s
Dx Dy Dz Dz



0
x
y
z
z
 Dz  C
R
 Dz   S
ẑ
Due to
boundary
condition
ŷ
r
D 
D  C

a
1 ( D ) 1 D Dz 1 ( D )



0
 
  z  
ŷ
D (   a) 
C
D 

 C  a S
x̂
 D (  ) 
C
 S
a
a S
ρ̂

EMLAB