Download Chapter 3. Electric Flux Density, Gauss`s Law, and Divergence

Chapter 3. Electric Flux Density, Gauss’s Law, and
Divergence
3.1 Electric Flux Density
 Michael Faraday’s experiment
1) 내부의 구에는 알고 있는 양(+)의 charge 가 충전됨
2) 내부 및 외부의 구 사이에는 2 cm 두께의 dielectric material 로
채워져 있음
3) 외부의 구는 접지면과 연결함으로써 충분히 방전됨
4) dielectric 봉으로 외부의 구를 접지면에서 떼어놓자 외부의 구에
음(-)의 charge 가 유기됨
5) 외부의 구에 전도된 총 전하량은 내부의 구가 가지고 있는 양
과 같았고 두개의 구 사이에 끼워져 있는 material 과 material
두께와 상관없음
6) Faraday conclusion: There was some sort of “displacement” from the
inner sphere to the outer which was independent of the medium
7) Displacement flux or Electric flux:  [psi]
1
• =Q
(외부 구에 전달된 전하량이 Q 이므로 Electric flux 는 내부 구
의 전하량과 동일)
• At the surface of the inner sphere,  coulombs of electric flux are
produced by the charge Q(=) coulombs distributed uniformly over a
surface having an area of 4πa2 [m2].
•

Electric Flux Density ( D ): The density of the flux at this surface
• Electric flux densities :

D

D
r a
r b

Q 
ar
4a 2
@ surface of inner sphere

Q 
ar
4b 2
@ surface of outer sphere

Q 
D
ar
4r 2
@ a r  b
• Let the inner sphere make smaller and smaller, still retaining a charge of
Q, it becomes a point charge.
• Electric Flux Density

Q 
D
ar
4r 2
(Q lines of flux) are symmetrically directed outward from the point and
pass through an imaginary spherical surface of area 4r2.
2
• Compare

E
Q
4 0 r 2


 D  0 E

ar
@0<rb
(free space only)

 v dv 
E

vol 4 0 R 2 ar
•
(from Ch.2)
• Similar manner

 v dv 
D
ar
vol 4R 2
[Ex.]
z
 L  8 [nC/m]

D (임의의 가상 원통면을 관통하여
밖으로 나가는 electric flux)

y

x

E
 143.8 
8  10 9
L 
a 
a
a  [V/m]
 
2 0 
2  8.854  10 12 




E
47
.
9
a


3
m
@
,
 [ V/m ]
 L 
8  10 9  1.273  10 9 
D
a 
a 
a [C/m 2 ]

2
2


2
@   3m , D  0.424a  [nC/m ]
3
3.2 Gauss’s Law
• “The electric flux passing through only closed surface is equal to the
total charge enclosed by that surface.” (Faraday 의 실험을 구체적인
문구로 표현)
• S : Incremental (surface) element of surface at P


S  Sa N


DS : Angle  with S
(3.1 절에서는 구를 고려하였지만 일반적인 폐 공간을 고려)

  : Flux crossing on S


  DS ,normal S  DS cosS  DS  S
S 의 수직방향성분 고려
Electric flux density 의

S 방향성분
(Tangential 방향 성분의 vector 들의 합은 “0”  무시 가능)
4
• Total flux passing through the closed surface :
   d 

closed
surface


DS  dS
: Closed surface integral
실제로 이루어진 closed surface 의 단위
면적 및 그 normal 벡터 성분
내부로부터 밖(radial 방
향)으로 향하는 flux
where dS  dxdy , ddz , r 2 sin dd , etc.
•


s


DS  dS  charge enclosed  Q
Several point charges case : Q   Qn
: Q
Line charge case
•

S
  L dL
Q    S dS
Surface charge case
:
Volume charge case
: Q
S

vol
v dv


DS  dS   v dv
vol
( Gauss’s Law 에 의하여 폐곡면을 빠져나가는 electric flux
density 와 폐 곡면 내부의 전하량이 같으므로)
5
[Ex.] Check the results of Faraday’s experimental

E
Q

ar
4 0 r 2


Q 
D  0E 
ar
4r 2
At the surface of the sphere,

Q 
DS 
a
2 r
4a
dS  r 2 sin dd  a 2 sin dd



Q 
Q
2
DS  dS  (
a
)

(
a
sin

d

d

a
)

sin  d  d 
r
r
4 a 2
4


  2   Q
sin  d  d 
 Total charge   DS  dS  
  0    0 4

2
0
Q
Q
 cos   0 d  
4
 2
6
2

  Q
0
3.3 Application of Gauss Law: Some Symmetrical Charge
Distributions


Q   S DS  dS

D
(전하분포가 알려져 있을 때 S 를 구하기
위해 Gauss 법칙을 어떻게 적용할 것인가?)
• The solution is easy if

1) DS is everywhere either normal or tangential to the closed surface, so


that DS  dS becomes either DS dS or zero, respectively.


D

d
S
2) On that portion of the closed surface for which
is not zero
S
and DS = constant.
[Ex.]

DS is everywhere normal to the surface.
 
Q   D  dS  
DS dS  DS  dS
s
sphere
 DS 
2
0


0
r 2 sin  d d  Ds 
 4 r 2 DS
 DS 
Q
4 r 2
7
2
0
2r 2 d
•

Since r may have any value and D S is directed radial outward

Q 
D
ar
4r 2

E
Q
4 0 r 2

ar
[Ex.]
•



D  D a  ( a  방향으로만 radiate )


Q   DS  dS  DS  dS  0  dS  0  dS
cyl
sides
 DS  dS  DS 
L

2
z 0  0
sides
top
ddz
 DS 2L
DS  D 
Q
2 L

 L
2
E 
L
2 0 
 Q  LL

 
 D  L a
2

L 
a
 E
2 0 
8
bottom
• Charge distribution the outer surface of the inner conductor:
s
• Total electric flux by coaxial cylindrical conductor which is of length L
and radius ρ, where a < ρ < b :
Q  DS 2L
• Total change on a length L of the inner conductor :
Q
L

2
z 0  0
 DS 
 S addz  2aL S  DS 2L
a S
(임의의 공간면을
지나는 electric

 a 
 D  S a  ( a    b)

flux density)
where s: surface charge density at =a
The previous result might be expressed in terms of line charge per unit
length.
 L  2aL S  2a S  (inner conductor 의 단위길이 당
(L=1m) 전하량)
S 
L
2a
9
 a S 
D
a 
a

L
2a a   L a


2

• Since every line of electric flux starting from the charge on the inner
cylinder must terminate on a negative charge on the inner surface of the
outer cylinder
Qouter cyl  2 aL S .inner cyl
 2 bL S .outer cyl
a
b
 S .outer cyl    S .inner cyl
• At ρ > b,
0  DS 2L (The total charge enclosed would be zero.)
Ds  0
• At ρ < a,
0  D s 2L (The total charge enclosed would be zero.)
Ds  0
[Ex.]L=50 cm, inner=1 mm(=a), outer=4 mm(=b), o (in intermediate space)
Total charge on the inner conductor: 30 nC
 S .inner
30  10 9


 9.55 [μC/m 2 ]
3
2aL
2  10  0.5
Qinner . cyl
10
Internal fields :
D 
a s


10 3  9.55  10 6


9.55

[nC/m 2 ]
9.55  10 9 /  1079
E 


[V/m]
0
8.854  10 12

D
E  D  0
(1    4 mm)
(   1,   4 mm)
3.4 Application of Gauss’s Law: Differential Volume Element
Fig. 3.6 A differential-sized gaussian surface about the point P is

used to investigate the space rate of change of D in the
neighborhood of P




D0  Dx 0a x  D y 0a y  Dz 0a z


Q   S DS  dS

front

back
 
left

right
 
top
bottom
• Since the surface element is very small, D is essentially constant over
this portion of the entire closed surface.
11

front


 D front  S front


 D front  yza x  D x. front yz




( D front  Dx , front a x  D y , front a y  Dz , front a z )
Dx , front  Dxo 
x Dx
2 x
= Dxo 

front
 ( Dxo 
x
 rate of change Dx with x
2
x Dx
)yz
2 x
• Consider the integral over the back surface,

back




 Dback  Sback  Dback  (yzax )
  Dx , back yz
Dx.back  Dxo 

back
x Dx
2 x
 ( Dxo 
x Dx
)yz
2 x
12

front

back

Dx
xyz
x
• By the same processes

right


top


left

bottom
D y

y
xyz
Dz
xyz
z
  Dx Dy Dz
D
)xyz
 dS (


S
x
y
z
D Dy Dz
)v
( x 

x
y
z
Q
• Charge enclosed in volume Δv
(
Dx Dy Dz
)


z  volume v
y
x




D
 e - x sin ya x - e - x cos ya y  2 za z [C/m 2 ]
[Ex.]
Dx
-x
= - e sin y
x
D y
y
-x
= e sin y
13
Dz
=2
z
Charge enclosed in volume Δv=2Δv
3.5 Divergence
Dx D y Dz



x
y
z
 
 D  dS

s
v
Q
v
• As a limit
Dx D y Dz


 lim
v0
x
y
z
 
 D  dS
S
v


Divergence of A  div A  lim
 lim
v0
Q
S
v
v
 
A
  dS
s
 
The divergence of the vector flux density A is the overflow of flux
 0
from a small closed surface per unit volume as the volume shrinks to
zero.
• General form
 D x D y Dz
div D 


x
y
z
(Cartesian)
 1 
1 D Dz

div D 
( D ) 
 
 
z (Cylindrical)
 1  2
1

1 D
div D  2
( r Dr ) 
(sin D ) 
r sin  
r sin  
r r
(Spherical)
14




x
x
D

e
sin
y
a

e
cos
y
a

2
z
a
[Ex.]
x
y
x
div
 D D y D
x
z
D=
+
+
x
y
z
 e  x sin y  e  x sin y  2  2
 
 D  dS

div D = lim
•
s
 0
v
///
: definition of divergence

Dx D y Dz
+
+
div D =
x
y
z
: The result of applying the definition to a differential volume
element in Cartesian coordinates.
 
D
s  dS  Q
 
s D  dS  Q
v
v
•
: Gauss’s law
: Gauss’s law per unit volume
• As the volume shrinks to zero
lim
v  0
 
 D  dS
s
v

Q
 divD   v
v  0 v
 lim
: The first of Maxwell’s four equations
:
The electric flux per unit volume leaving a vanishingly small
volume unit is exactly equal to the volume charge density here.
15
[Ex.]

Q 
D
ar
4r 2
 1  2
1 D
1

(sin D ) 
Since div D  2 (r Dr ) 
,
r r
r sin  
r sin  
div D 
1  Q
( )0
r 2 r 4
 v  0 @r≠0 (everywhere except at the origin where it is infinite)
3.6 The vector operator  and the divergence theorem
• :

“del” operator
 
 
 
ax  a y  az
x
y
z



 
 
 
 



(
)
(




D

a
a
a
D
a
D
a
D
a
x
y
z
x x
y y
z z)

z
y
x



Dx D y Dz




z
x
y
• 





div
D

16
• In other coordinate systems,
 1 
1 D D z
( D  ) 
D 

 
 
z
(Cylindrical coordinates)
 1  2
1
1 D

(r Dr ) 
(sin  D ) 
D  2
r r
r sin  
r sin  
(Spherical coordinates)
•
 
 D  dS  Q
s
Q    v dv
vol

  D  v
 

  D  dS  Q    v dv     D dv
s
vol
vol

 
  D  dS     D dv : Divergence theorem
s
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”
( 체적적분  면적분, 삼중적분  이중적분 )
17



D  2 xya x  x 2 a y [C/m 2 ]
[Ex.]
 


D  dS z 0  0a z  (  dxdya z )  0
z 3
 
D
  dS  
s

back

front
3

0

 
3
0
3
0

3
0
2
0



D
1
0
2
0

1
0
x 0

D

left

right
3 2 

 (dydzax )    D
y 0
0
0
x 1
3 1 

 (dxdza y )    D
0
( Dx ) x 0 dydz  
( Dy ) y 0 dxdz  
3
0
3
0


2
0
1
0
0

 (dydzax )
y 2

 (dxdza y )
( Dx ) x 1 dydz
( Dy ) y  2 dxdz
Since ( D x ) x 0  0 , ( D y ) y 0  ( D y ) y  2 ,

s
 
3 2
3
D  dS    (2 y ) x 1 dydz   [ y 2 ] 02 dz  [4 z ] 30  12
0
0
0
18
Another solution
 

  D  (2 xy )  ( x 2 )  2 y
x
y

   Ddv 
vol

3
2
0 0
3
2
1
   2 ydxdydz
0
0
0
2 ydydz 
3
0 [ y
] dz  [ 4 z ]30  12
2 2
0
 직육면체 안에 12 [C]의 전하존재
19