Transcript ENE 429 Antenna and Transmission Lines

ENE 325
Electromagnetic Fields
and Waves
Lecture 4 Electric potential, Gradient, Current
and Conductor, and Ohm’s law
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Review
Gauss’s law is another approach to evaluate the electric
field and is proper for highly symmetrical configuration.

Qenc   D dS
Dx Dy Dz
 D


.
x
y
z

Divergence is defined by

Point form of Gauss’s law:  D  v
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Outline




Electric potential
Gradient
Current and Conductor
Ohm’s law
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Electric work
 A work done to move one charge from one to point to
another is defined by
b
W   F d L.
a
 A work done by the field in moving the charge from point a
to point b is
b
WE  field  Q  E d L.
a
 A work done by an external force against the field is
b
W  Q  E d L.
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a
4
Ex1 Calculate work required to move a 5 nC
charge from the origin to point P (1, 1, 0)
against the static field given E  5a xV/m.
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Electric potential (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
Vba  Vb  Va
Va and Vb are the absolute potentials measured with
respect to the reference potential at ground plane.
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Electric potential (2)

If a closed path is chosen, no work is done.
 E dL  0

An absolute potential at some finite radius from a point
charge fixed at the origin is
Q
V
40 r

The electric potential resulted from N charges is found
by adding the potential for each charge.
Qi
V
i 1 40 ri
N
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Electric potential (3)

If a collection of charges become a continuous
distribution, the total potential is then
dQ
V 
40 r
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L
Ex2 LetE 
and L is 100 nC/m
aV/m

20
a)
b)
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Find the work done in moving a 10 nC from  = 3 m
to  = 1 m.
Determine the potential difference Vba.
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Ex3 Find a work done in moving a charge Q = 5 nC
from the origin to point P (2, /4, /2) in spherical
coordinates by giving E  5e r / 4 a r  10 a  V/m.
Note: line different element
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r sin 
dl  drar  rd a  r sin .d a
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Gradient
A plot of the electrostatic potential superimposed over the
field lines for a point charge. The electric field can be found by
finding the maximum rate and direction of spatial change of the
potential field.
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A gradient equation
E  V

For a Cartesian coordinate system:
V
V
V
V 
ax 
ay 
az
x
y
z

For a cylindrical coordinate system:
V 

V
1 V
V
a 
a 
az

 
z
For a spherical coordinate system:
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V
1 V
1 V
V 
ar 
a 
a
r
r 
r sin  
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Introduction to electromagnetic material
 The properties of electromagnetic material is
specified by , , .
 Homogeneous material is the material that
possesses the same properties at every point
in the material.
 Isotropic material is the material that its
properties are independent of direction.
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Conductor
The material that electrons move freely.
The conductivity, , depends on charge density
and scattering of electrons by their interactions
with crystal lattice.  decreases with increasing
temperature.
Perfect conductor
The conductor that has an infinite conductivity.
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Current and current density


Current, current density, resistance, and
capacitance can be explained using
electromagnetics.
Current, I is defined as the amount of
charge that passes through a reference
plane in a given amount of time.
dQ
I
dt

Amperes (A)
Current density, J is defined as the
amount of current per unit area
I  J S

A/m2
total current can be expressed as
I   J S
s
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Current density and volume charge
density
x
L
S
x
t=0
t = t
Consider Q  v v  v S L
At time t, charges move for a distance x crossing a reference plane
that is normal to the direction of charge movement.
Since
Q   S x
v
then
I  v Svx
where nx is a charge velocity (m/s)
J  v vx
J  v v
or
12/12/50 The movement of charge creates “Convection current”.
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Continuity of current
The principle of conservation of charge “charges can be
neither created nor destroyed, although equal amounts of
positive and negative charge may be simultaneously created,
obtained by separation, destroyed, or lost by recombination.”
The integral form of the continuity equation,
dQi
I   J S  
,
dt
s
indicates an outward-flowing current where Qi is the charge
inside the closed surface. We can show its point form as
v
 J 
.
t
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Conduction current
Conduction band
Conduction band
Conduction band
Energy
Energy
Energy
Valence band
Valence band
Valence band
Conductor
Insulator
Semiconductor
Conduction current arises from free electrons in a conductor.
Electrons in valence band have high enough energy to get
into the conduction band.
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Drift velocity
vd  e E
where e is the mobility of the electron in the given material (m2/V-s)
aluminum = 0.0012 m2/V-s
copper = 0.0032 m2/V-s
silver = 0.0056 m2/V-s
Then
J  ee E.
where e is the free-electron charge density, a negative
value.
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The point form of Ohm’s law
a
Vab    E dl
b
where   ee.
 The conductivity  is measured in siemens per meter (S/m).
aluminum = 3.82x107 S/m
copper = 5.8x107 S/m
silver = 6.17x107 S/m
 The conductivity depends on the temperature.
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The application of Ohm’s law
Assume J and E are uniform,
I   J d S  JS
s
b
b
a
a
Vab    E dl   E  dl   E Lba
 E Lab  EL
L
V
I.
S
we can write Ohm’s law as
L
where R 
is the resistance with the measured unit
a
of Ohm (). S
  E dl
Vab
R
 b
General form:
I
 E d S
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s
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Ex4 Determine the magnitude of E in
silver when
a) nd = 1 mm/s
b) J = 107 A/m2
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c) A current of 80 A flows through a silver dice of width
= 3 mm and length = 3 mm.
d) A same silver dice with a 0.5 mV drop across a top and
a bottom face.
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Ex5 An aluminum rod with the length of 1000 feet
has a cross section with the diameter of 0.8 inch.
There is 1.2 V potential drop across both ends,
determine
a) J
b) current
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c) power dissipated in the rod
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