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ENE 325
Electromagnetic Fields
and Waves
Lecture 5 Conductor, Semiconductor,
Dielectric and Boundary Conditions
1
Review (1)
The electric potential difference Vba is a work done by an
external force to move a charge from point a to point b in
an electric field divided by the amount of charge moved.

b
W
Vba     E d L
Q
a
The electric potential is the same no matter which
routes are used. Only displacement distance (shortest
route) matters

2
Review (2)

Conductors and Ohm’s law
 Current, I is defined as the amount of charge that passes
through a reference plane in a given amount of time.
dQ
I
dt




Ampere.
Current density, J is defined as the amount of current per unit
I  J S A/m2
area
the relationship between I and J, I   J  S
s
convection current J  v v
conduction current
J  E
3
Outline


Conductor and boundary conditions
Semiconductor and insulator
4
Conductors and boundary conditions



charge density is zero inside a conductor.
Surface charge density DS is on the conductor
surface.
An electric field inside a conductor is zero.
+
+
+
+
+
Ds=0
E=0
+
+
+
+
+
outside charges
cause an electric field.
+
5
Tangential and normal fields.

The electric field on the surface can be divided into two
components.
 tangential electric field, Et, = 0 for an equipotential surface
 normal electric field, En
En
E
Et
6
Boundary conditions (1)

Consider a conductor-free space boundary
 E dl  0
From
a
h
b
w
d
c
7
Boundary conditions (3)

Consider Gauss’s law
 D dS  Q
From
s
Dn
D
Dt
8
Boundary conditions (3)

For conductor-free space boundary conditions (B.Cs.)
Dt = Et = 0
Dn = 0En s
9
Ex1 Let V  100e5 x sin(3 y ) cos(4 z ) V and let a
point P(0.1, /12, /24) locate at the
conductor-free space boundary. At point P,
determine
a)
V
b)
E
10
c) En
d) Et
e) S
11
Semiconductors





Electron and hole currents
conductivity   ee  hh
mobility is 10-100 times higher than conductor.
electron and hole density depend on temperature.
Doping is the process of adding impurities to a
semiconductor to alter the polarity.
12
Dielectric or insulator
+
+
+ -
-
-
+
-
-
+
-
+
+

-

+

no free charge
microscopic electric dipoles
energy stored capability
polar and non-polar molecules
+

polar molecules
+ - +
+ +
+ +
- +
non-polar molecules
13
Alignment of dipoles with E field
E
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
-
-
-
-
-
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Polarization
Each dipole has its dipole moment, p
p  Qd
C m
where Q is the positive one of the two bound charges
d is the vector from the negative to the positive charge.
nv
ptotal   pi
i 1
where n = number of dipoles per volume.
Polarization is dipole moment per unit volume,
1 nv
P  lim
 pi
v 0 v i 1
C / m2
15
Equivalent polarization
 The movement of bound charges when induced by an
electric field causes changes in surface and volume charge
densities.
1. Equivalent polarization surface charge density, s
s  P a n
C / m2
2. Equivalent polarization volume charge density, v
an
+
+
+
+
-
+
+
+
+
+
-
+
+
+
+
+
+
-
+
-
+
+
+
+
+
+
-
+
+
+
+
+
+
+
+
-
E
16
Electric flux density in a dielectric
material (1)

D can be calculated from free and bound charges.
Net bound charges flowing out of the closed surfaces,
Qb    P  d S
C
S
Let
QT  Q  Qb
where QT = total charge
Q = free charge
+
+
+
-
-
17
Electric flux density in a dielectric
material (2)
QT   0 E d S
From
s
then
Q  QT  Qb   (0 E  P)  d S .
s
Let
D  0 E  P
we can write
Q   D dS
s
where D is an electric flux density in a dielectric material.
18
Equivalent divergence relationships
From
Qb    P a n d S ,
s
use a divergence theorem, we have   P d S    Pdv.
s
Since
then
Qb   v dv
 P  v  b .
We can also show that
therefore
v
 0 E  T ,
 D  T  b  v .
19
Electric flux density in dielectric medium
(1)
 If the dielectric material is linear and isotropic,
the polarization P is proportional to the electric
field E .
P   0 e E
where e is an electric susceptibility.
Then
D  0 E  e0 E.
Let
 r  e  1
where r is a relative permittivity or a dielectric constant.
20
Electric flux density in dielectric medium
(2)
So we can write
D   r 0 E
or
where
D  E
.
  r 0 F/m.
21
Ex2 A dielectric material has an electric
susceptibility e =0.12 and has a uniform
electric flux density D = 1.6 nC/m2, determine
a) E
b) P
.
c) Average dipole moment if there are 2x1019
dipoles/m3
22
Boundary conditions for dielectric
materials: tangential fields

It is useful to determine the electric field in a dielectric medium.
Dielectric-Dielectric
1
2
and
w
Dt1 1

Dt 2  2
h
Et1
Et1 =Et2
Et2
23
Boundary conditions for dielectric
materials: normal fields
Dielectric-Dielectric
D n1
a21  ( D1  D2 )  S
s
For a free of charge boundary,
D n2
Dn1 = Dn2
and
a 21 is the unit vector pointing from
medium 2 to medium 1.
1En1 2En2.
24
Use B.C.s to determine magnitude of D
and
E
D n1
q1
1
2
D2
D1
q2
D t1
D n2
D t2
2

D2  D1 cos2 q1  22 sin 2 q1
1
12
E2  E1 sin q1  2 cos2 q1
2
2
25
Ex3 The isotropic dielectric medium with r1 = 3 and
r2 = 2 is connected as shown. Given
E1  a x  5a y  4a z V/m, determine E 2 and its
magnitude, D2 and its magnitude, q1, and q2.
Z
r13
x-y
r22
-Z
26
Boundary conditions for dielectricconductor
En
Et
w
dielectric
conductor
h
Dt = Et = 0
Dn = En = s
27
Ex4 Between a dielectric-conductor interface has a
surface charge density of s = 2x10-9 C/m2. Given
E1  30a x  50a y  70a z V/m, determine E 2 .
28