Transcript Lecture 5 - City University of New York

Lecture 5
Electric Flux Density and
Dielectric Constant
Boundary Conditions
Electromagnetics
Prof. Viviana Vladutescu
Electric Flux Density
 C 
D    3 
m 


Ddv


dv


v
v

 E  0
D

ds

Q
(
C
)

s
Gauss’s Law: The total outward flux of the
dielectric displacement (or simply the
outward flux) over any closed surface is
equal to the total free charge enclosed in
the surface
D   0 E  P   0 E   0 e E 
 C 
  0 (1   e ) E   0 r E   E  2 
m 
Where- ε is the absolute permittivity (F/m)
-εr is the relative permittivity or the dielectric constant
of the medium
-ε0 is the permittivity of free space
-χe is the electric susceptibility (dimensionless)
Material Dielectric Constants
Vacuum 1
Glass 5-10
Mica 3-6
Mylar 3.1
Neoprene 6.70
Plexiglas 3.40
Polyethylene 2.25
Polyvinyl chloride 3.18
Teflon 2.1
Germanium 16
Strontiun titanate 310
Titanium dioxide (rutile) 173 perp
86 para
Water 80.4
Glycerin 42.5
Liquid ammonia(-78°C 25
Benzene 2.284
Air(1 atm) 1.00059
Air(100 atm) 1.0548
Homogeneous
-εr independent of position
Anisotropic
εr is different for different of the electric field
 Dx   11  12
 D   

y
21
22
  
 Dz   31  32
 13   E x 



 23   E y 
 33   E z 
Biaxial, Uniaxial and Isotropic
Medium
-biaxial
 Dx   1 0
D    0 
2
 y 
 Dz   0 0
-uniaxial
1   2
-isotropic
1   2   3
0   Ex 



0 Ey 
 3   E z 
KDP ADP crystals
– Electric field along optic axis
Uniaxial crystal becomes biaxial with applied field!
GaAs CdTe crystals
– Electric field along optic axis
Isotropic crystal become biaxial with applied field!
Dielectric Strength
The maximum electric field intensity that a
dielectric material can stand without breakdown
Material
Dielectric Strength (V/m)
Air
3e6
Bakelite
24e6
Neoprene rubber
12e6
Nylon
14e6
Paper
16e6
Polystyrene
24e6
Pyrex glass
14e6
Quartz
8e6
Silicone oil
15e6
Strontium titanate
8e6
Teflon
60e6
Boundary Conditions
Medium 1&2 are dielectrics
CONDITION I (tangential components)
E

d
l

0

abcda
b
c
d
a
a
b
c
d
E

d
l

E

d
l

E

d
l

E

d
l

0




w
E

d
l


0

0
 h / 2
h / 2
0
E

d
l


E

d
l


0
h / 2
 h / 2
0
0
E

d
l


w
 E  dl   E  dl  0
w
E
T1
0
0
aT dlaT 
E
N1
an dlan 
h / 2
  ET2 aT dlaT 
w
 h / 2
0
E
N2
an dlan
0
h / 2
0
E
 h / 2
N1
an dlan 
E
N2
an dlan  0
0
h
h
ET1 w  E N1
 EN2
 ET2 w
2
2
h
h
 E N1
 EN2
0
2
2
Condition I (tangential components)
ET1  ET1
DT1
1

DT2
2
CONDITION II (normal components)
D

d
s

Q

 Dds   Dds   Dds   Dds
s
top
side
D

d
s

D
a
ds
a

N
n
n

 1
s
top
bottom
D
a
ds
(

a
)

N
n
n
 1
bottom
 DN1 s  DN 2 s
Qencl    s ds   s s
s
Condition II (normal components)
 C 
a21 ( DN1  DN 2 )   s  2 
m 
The normal component of D field is discontinuous across
an interface where a surface charge exists, the amount of
discontinuity being equal to the surface charge density
Example:
Two dielectric media with permittivity ε1 and
respectively ε2 , are separated by a charge free boundary. The
electric field intensity in medium 1 at point P1 has a magnitude
E1 and makes an angle α1. Determine the magnitude and
direction of the electric field intensity at point P2 in medium 2.
P1
α2
α1
P2
ET1  ET1
DN1  DN 2
since a12 ( DN1  DN 2 )   s  0
E1 sin 1  E2 sin  2
1E1 cos1   2 E2 cos 2
tan1

tan 2
1
2
tan1  1

tan 2  2
The magnitude of E2 :
E2  E  E
2
T1

2
N2
E2 sin  2   E2 cos 2 

2
E1 sin 1 
E2  E1
2
 1

  E1 cos1 
 2

sin 1 
2
2
2
 1

  E1 cos1 
 2

2

Boundary conditions at a
Dielectric/Conductor Interface
-inside
a good conductor
E=0
ET=0
D=0
Dn=ρs
Practice Problems
Divergence Theorem and Gauss’s Law
Suppose D = 6rcosf af C/m2. (a) Determine the charge density at
the point (3m, 90, -2m). Find the total flux through the surface
of a quartered-cylinder defined by 0 ≤ r ≤ 4m, 0 ≤ f ≤ 90, and
-4m ≤ z ≤ 0 by evaluating (b) the left side of the divergence
theorem and (c) the right side of the divergence theorem.
(a)
1 D 1   6  cos  

 6sin .
  Dcylinder 
  

C
 v  3,90 , 2   6 3 .
m
(b)
 D dS  
 0

   
 90
top
bottom


outside
,
note that the top, bottom and outside integrals yield zero since
there is no component of D in the these dS directions.

 90

 0
So,
(c)
  6  cos 
  6  cos 
 90
 0
a  d  dza   0
a   d  dza   192C
 D dS  192C.
 D  6sin  , dv   d  d dz
90
4
0
0
0
4
  Ddv  6  sin  d   d   dz  192C.
Electric Potential
The potential field in a material with εr= 10.2
is V = 12 xy2 (V). Find E, P and D.
E  V  
 12 xy 2 
x
ax 
 12 xy 2 
y
V
a y  12 y a x  24 xya y
m
2
D   r  o E  -1.1y a x  2.2 xya y
2
nC
2
m
e   r 1  9.2
P   e o E   9.2   8.854 x10
12
 E = -9.8 y a
2
x
 2.00 xya y
nC
m2
Boundary Conditions
For z ≤ 0, r1 = 9.0 and for z > 0, r2 = 4.0. If E1 makes
a 30 angle with a normal to the surface, what angle
does E2 make with a normal to the surface?
ET 1  E1 sin 1 , ET 2  E2 sin 2 , and ET 1  ET 2
also
ET 1 ET 2

,
DN 1 DN 2
Therefore
DN1   r1o E1 cos1, DN 2   r 2o E2 cos2 , and DN1  DN 2 since s  0
and after routine math we find
 r2

 2  tan 
tan 1 
  r1

1
Using this formula we obtain for this problem 2 = 14°.