1
1
Swartz 06/5/13
ECE 303 –
Electromagnetic Fields and Waves
– Fall 2008
Lecture 3:
ECE 3030
Electromagnetic Fields and Waves
Instructor:
Dr. Wesley E. Swartz
Fall 2008 Lecture 3
2008/9/3
Electrostatics
Application of Gauss’ Law
Superposition Principle
Fields of Some Charge Distributions
q
+
q
−
x
E
A
2
Swartz 06/5/13
ECE 303 –
Electromagnetic Fields and Waves
– Fall 2008
Lecture 3:
Announcements
•
We will not be able to “move” any of the Demo or
Workshop sections to accommodate conflicts.
•
If you have only a conflict for the Demo sections
–
there is something blowing in the wind that might help.
–
However, if your conflict results from trying to take ENGRD 2300,
ECE 3100, and ECE 3030, then
•
You should realize you are taking on a huge burden…
–
All three of these courses are major undertakings!
•
I recommend you have a talk with your advisor!
•
If you have only a conflict for the Workshop sections, then
–
You
3
Swartz 06/5/13
ECE 303 –
Electromagnetic Fields and Waves
– Fall 2008
Lecture 3:
Simple Charge Distributions
•
A single charge
•
A Charged plate
•
A ring of charge
•
A line charge
•
Oppositely charged plates
•
Dipoles – a pair of opposite charges
4
Swartz 06/5/13
ECE 303 –
Electromagnetic Fields and Waves
– Fall 2008
Lecture 3:
Field of a Point Charge
•
Consider a point charge of
q
Coulombs
sitting at
•
Surround the charge by a
Gaussian
surface
in the form of a
spherical shell
of
radius
r
–
By symmetry
, the E-field magnitude on
the surface must be uniform and
pointing in the radial direction
0
=
r
q
Using Gauss’ Law:
(
)
q
r
E
r
o
=
2
4
π
ε
r
r
q
E
r
q
E
o
o
r
ˆ
4
or
4
2
2
πε
πε
=
=
r
q
5
Swartz 06/5/13
ECE 303 –
Electromagnetic Fields and Waves
– Fall 2008
Lecture 3:
Field of an Infinite Charged Plane
•
Suppose we have a charged plane.
–
Let the surface charge density be
•
Coulombs/m2
–
Symmetry Argument: The charge distribution
is symmetric w.r.t. +z and –z directions.
•
Therefore, if at any point there is an
E-field
component in the +
z
direction, there must also
be E-field component in the –
z
direction.
•
Since the field cannot have a
z
-component
pointing in both +
z
and –
z
directions at the
same time, there cannot be a
z
-component of
the field.
•
Similarly, there cannot be a
y
-component of
the E-field
•
So the field can only have an
x
-component
x
z
y
6
Swartz 06/5/13
ECE 303 –
Electromagnetic Fields and Waves
– Fall 2008
Lecture 3:
Field of an Infinite Charged Plane
•
Since the field can only have an x-component,
–
Draw a Gaussian surface in the form of a cylinder of
area
A
piercing the charged plane
–
Total flux coming out of the cylinder ends =
–
Total charge enclosed by the surface =
–
By Gauss’ Law:
x
z
y
A
E
x
o
ε
2
A
o
x
x
o
E
A
A
E
ε
ε
2
2
=
=
x
E
A

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