cchacon_Aplicaciones ley de Gauss_teoria.pdf

Lecture 4
97.315 Basic E&M and Power Engineering Topic: Using Gauss’ Law
TITLE
USING GAUSS’
LAW

Lecture 4
OUTLINE
• Theory
• How to apply Gauss’ law
• Examples of calculation
• Spherical charge distribution
• Infinite flat surface
• Parallel plate capacitor
• Infinite line of charge
• Infinite cylindrical charge distribution
• Problem solving strategy
• Assignment
• References
• Summary
Gauss, Johann Karl Friedrick (1777-1855)

Lecture 4
TEXT
 
 

S o
q
A
d
E

surface
by
enclosed


GAUSS’ LAW
Gauss’ law is true for any surface enclosing any
charge distribution. When the charge distribution
has sufficient symmetry we can chose a surface ---
Gaussian Surface --- over which the evaluation
of the flux integral becomes simple. Gauss’ law
allows us to calculate the field far more easily than
we could using Coulomb’s law and superposition.

Lecture 4
THEORY
Step 1: Study the symmetry to see if you
can construct a Gaussian surface on which
the field magnitude and its direction
relative to the surface are constant.
Using Gauss’ Law
 


S
E A
d
E


Through step 1 the dot product in the integral can
be replaced by:
 

cos
EdA
Then Ecos() is constant and can be removed from
inside the integral:
 


S
E dA
E 
cos

Lecture 4
THEORY
Step 2: Evaluate the flux. This should be
easy because of your choice of the
Gaussian surface.
Using Gauss’ Law
  
Area
Gaussian
cos 
E
A
d
E
S
E 


 



Lecture 4
THEORY
Step 3: Evaluate the enclosed charge. This
is not the same as the total charge if the
Gaussian surface lies within the charge
distribution.
Using Gauss’ Law
 
surface
Gaussian
by
enclosed
q
The Gaussian surface is an imaginary surface that you place in
the charge distribution. Choosing the Gauusian surface shape
and location depends on the symmetry of the charge
distribution AND on what exactly you want to calculate.

Lecture 4
THEORY
Step 4: Equate the flux
and solve for E. The direction of the
electric field vector should be evident for
the symmetry and polarity of the charges.
Using Gauss’ Law
 
o
E
q

enclosed


In step 1 you reasoned the direction of E. Now go back and use that information.

Lecture 4
TEXT
 
 

S o
q
A
d
E

surface
by
enclosed


GAUSS’ LAW
There are only a dozen or so problems in which
Gauss’ law can be easily applied. These problems
require a symmetric charge distribution (surface,
line, spherical, cylindrical, …).

Lecture 4
EXAMPLE
ELECTRIC FIELD FROM A SPHERICAL CHARGE DISTRIBUTION.
Charged sphere V
In this example we will follow the steps in
applying Gauss’ law in order to obtain an
expression for the electric field at a point
external to the uniform spherical charge
distribution.
Finding the electric field using Coulomb’s law
and superposition is lengthy and involves messy
math. Using Gauss’ law makes the task easy,
but, we must assume the field is radial.
R
Example (Question)

Lecture 4
EXAMPLE
Charged sphere V
P
Step 1: Study the symmetry to see if you can construct a Gaussian surface
on which the field magnitude and its direction relative to the surface are
constant.
Imaginary Gaussian surface (sphere of radius r)
Assume a radial
electric field
E

normal to A
d
 r
Example (Solution)
E

q
 r


Lecture 4
EXAMPLE
Step 2: Evaluate the flux. This should be easy because of your choice of
the Gaussian surface. The integral is just the area of the Gaussian
surface.
P
r

E

n

dA


 

 dA
E
EdA
dA
n
E


Example (Solution)

Lecture 4
EXAMPLE
Step 3: Evaluate the enclosed charge. This is not the same as the total
charge if the Gaussian surface lies within the charge distribution.
P
r
R
Charged sphere V
Imaginary Gaussian surface
All of the spherical charge distribution is
contained inside the spherical Gaussian
surface.


V
V dV
q 
Total charge enclosed
Example (Solution)

Lecture 4
EXAMPLE
P
r
R
o
V
V
E
dV





electric field vector should be
evident for the symmetry and
polarity of the charges.
 
o
E
q

enclosed


 
r
E

Example (Solution)

Lecture 4
EXAMPLE
Combining results and solving for the electric field
P E

 
o
S
q
A
d
E

surface
Gaussian
by
enclosed
 



o
V
V
o
dV
q








S
E dA
E



S
o
V
V
dA
dV
E

 Direction of E from
symmetry of charge
distribution
Example (Solution)

Lecture 4
EXAMPLE
P E




S
o
V
V
dA
dV
E

 3
3
4
R
dV V
V
V 

 

2
4 r
dA o
S
o 
 

2
3
3 r
R
E
o
V



R
r
Integral over charge distribution
Integral over Gaussian surface
SOLVING FOR THE ELECTRIC FIELD E
Example (Solution)

Lecture 4
EXAMPLE
Writing electric field expression in terms of total
charge q inside Gaussian surface
P E




S
o
V
V
dA
dV
E


q
dV
V
V 

2
4 r
dA o
S
o 
 

2
4 r
q
E
o


R
r
Integral over charge distribution
2
r
q
k
E 
END
Example (Solution)

Lecture 4
EXAMPLE
Charged sphere V
internal to the uniform spherical charge
distribution.
Finding the electric field using Coulomb’s law
and superposition is lengthy and involves messy
math. Using Gauss’ law makes the task easy,
but, we must assume the field is radial.
R
Example (Question)

Lecture 4
EXAMPLE
Charged sphere V
Step 1: Study the symmetry to see if you can construct a Gaussian surface
on which the field magnitude and its direction relative to the surface are
constant.
Imaginary Gaussian surface (sphere of radius r)
Assume a radial
electric field
E

normal to A
d

Example (Solution)
P
R
r
E

q
 r


Lecture 4
EXAMPLE
Step 2: Evaluate the flux. This should be easy because of your choice of
the Gaussian surface. The integral is just the area of the Gaussian
surface.
r

E

n

dA


 

 dA
E
EdA
dA
n
E


Example (Solution)

Lecture 4
EXAMPLE
Charged sphere V
Only the charge inside the Gaussian surface is
enclosed by the Gaussian surface. All charge
external to the Gaussian surface is not included.


V
V dV
q 
Total charge
enclosed
Example (Solution)
P
R
r
Volume enclosed by
Gaussian surface
3
3
4
r


Lecture 4
EXAMPLE
Only the charge inside the Gaussian surface is
enclosed by the Gaussian surface. All charge
external to the Gaussian surface is not included.
Example (Solution)
P
R
r External charge does not contribute to flux
External charge
Gaussian surface
E field lines intersect surface twice
contribution to flux cancel

Lecture 4
EXAMPLE
P
r
R
o
V
V
E
dV





electric field vector should be
evident for the symmetry and
polarity of the charges.
 
o
E
q

enclosed


 
r
E

Example (Solution)
V = volume enclosed by Gaussian surface

Lecture 4
EXAMPLE
P
E

 
o
S
q
A
d
E

surface
Gaussian
by
enclosed
 



o
V
V
o
dV
q








S
E dA
E



S
o
V
V
dA
dV
E

 Direction of E from
symmetry of charge
distribution
Example (Solution)

Lecture 4
EXAMPLE



S
o
V
V
dA
dV
E

 3
3
4
r
dV V
V
V 

 

2
4 r
dA o
S
o 
 

o
V r
E


3

Integral over charge
distribution contained inside
Gaussian surface
Example (Solution)
P
E

R
r
END

Lecture 4
SUMMARY
P
2
3
3 r
R
E
o
V



P
r
R
o
V r
E


3

Electric field expression at a point outside and inside the
spherical charge distribution
OUTSIDE INSIDE
Direction of E is radial
R
r

Lecture 4
EXAMPLE
ELECTRIC FIELD FROM A INFINITE FLAT CHARGE DISTRIBUTION.
Charged surface S
above a uniformly charged infinite flat
surface.
Finding the electric field using Coulomb’s
law and superposition is lengthy and
involves messy math. Using Gauss’ law
makes the task easy, but, we must assume
the field is uniform and normal to the flat
surface.
Example (Question)
x
y
z

Lecture 4
EXAMPLE
above a uniformly charged infinite flat
surface.
Example (Question)
View of the flat
surface with
electric field
lines drawn.
Field is uniform
and normal to
the flat surface.
E

E


Step 1: Study the symmetry to see if you can construct a Gaussian
surface on which the field magnitude and its direction relative to the
surface are constant.
Example (Solution)
One possibility is to use
a cylindrical surface
with the top and bottom
surfaces parallel to the
charged plane. The E
field will be normal to
the top and bottom
surfaces.
The E field is parallel to the side of the cylinder and as such no field lines
(flux) pass through it.
This type of surface is often called a Gaussian pill box.
Lecture 4 EXAMPLE
E

E

A1 A2
A3

Step 2: Evaluate the flux. This should be easy because of your choice
of the Gaussian surface.
Example (Solution)
Surface S contains
three parts (Top
A3, Bottom A1 and
Side A2)
 


S
E A
d
E





 








2
1
3 A
A
A
S
E A
d
E
A
d
E
A
d
E
A
d
E








Top Bottom Side
Lecture 4 EXAMPLE
E

E

A1 A2
A3

Example (Solution)
   


3
A
E A
d
E
Top


Top flux
A
d
E


parallel
  3
EA
Top
E 

Uniform over A3
Outward pointing
Lecture 4 EXAMPLE
E

E

A1 A2
A3

Example (Solution)
   


1
A
E A
d
E
Bottom


Bottom flux
A
d
E


parallel
  1
EA
Bottom
E 

Uniform over A1
Outward pointing
Lecture 4 EXAMPLE
E

E

A1 A2
A3

Example (Solution)
   


2
A
E A
d
E
Side


Side flux
A
d
E


perpendicular
  0

 Side
E
Contained in side surface
Outward pointing
Lecture 4 EXAMPLE
E

E

A1 A2
A3

Example (Solution)



 








2
1
3 A
A
A
S
E A
d
E
A
d
E
A
d
E
A
d
E








Top Bottom Side
 
EA
A
A
E
EA
EA
E
E
E
2
1
3
1
3








Total flux through Gaussian
surface is equal to flux through
top an bottom surfaces.
Lecture 4 EXAMPLE
E

E

A1
A2
A3

Example (Solution)
Step 3: Evaluate the enclosed charge. This is not the same as the
total charge if the Gaussian surface lies within the charge
distribution.
The total charge
enclosed consists of only
the charge on the disk
contained within the
Gaussian Surface.
Area A=A1=A3
Disk of charged contained inside Gaussian
surface
  A
q S


enclosed
Lecture 4 EXAMPLE
E

E

A1 A2
A3

Example (Solution)
o
S
E
A




and solve for E. The direction of the electric field vector should be
evident for the symmetry and polarity of the charges.
 
o
E
q

enclosed


Lecture 4 EXAMPLE
E

E

A1 A2
A3

Example (Solution)
o
S
E
A




Solve for electric field
Lecture 4 EXAMPLE
 
 

S o
q
A
d
E

surface
by
enclosed


EA
E 2


Direction of E determined
from symmetry
o
S
E


2

END
E

E

A1 A2
A3

Lecture 4
EXAMPLE
ELECTRIC FIELD FOR UNIFORMLY CHARGED PARALLEL PLATE
CAPACITOR.
In this example we will make use of the
result obtained from the infinite charged
flat surface in order to determine the
electric field inside and outside a parallel
plate capacitor.
The plane of the plates are
taken to extend to infinity.
This is usually a good
approximation when the plate
separation is small compared
to their area.
Example (Question)
Charged surface
S

 S


+q -q
V Battery redistributes charge between
plates but system remains overall
neutral
Area of plates A

Lecture 4
EXAMPLE
CAPACITOR.
Example (Solution)
S


+q
S


-q
V
P
1
E

2
E

PLATE (1) (2) For a point P between plates
By superposition
2
1 E
E
E





A
q
A
q
E
o
o 
 2
2



A
q
E
o



o
s
E




A
q s



Lecture 4
EXAMPLE
CAPACITOR.
Example (Solution)
S


+q
S


-q
V
P
1
E

2
E

By superposition
2
1 E
E
E





A
q
A
q
E
o
o 
 2
2




0

E

PLATE (1) (2)
On Left side
For a point P to the left of both plates
A
q s



Lecture 4
EXAMPLE
CAPACITOR.
Example (Solution)
For a point P to the right of both plates
A
q s


S


+q
S


-q
V
P
1
E

2
E

By superposition
2
1 E
E
E





A
q
A
q
E
o
o 
 2
2



0

E

PLATE (1) (2)
On Right side
END

Lecture 4
SUMMARY
A
q
E
o



o
s
E




S


+q
S


-q
V
P
1
E

2
E

PLATE (1) (2)
0

E

0

E

P
1
E

2
E

P
1
E

2
E

CAPACITOR.

Lecture 4
SUMMARY
CAPACITOR.
Ideal infinite parallel plate capacitor Real finite parallel plate capacitor
+
+
+
+
+
+
+
+
-
-
-
-
-
-
-
-
P
1
E

2
E

P
1
E

2
E

P
1
E

2
E

+
+
+
+
+
+
+
+
-
-
-
-
-
-
-
-

Lecture 4
EXAMPLE
ELECTRIC FIELD FROM A LONG STRAIGHT LINE OF CHARGE.
In one of the examples of lecture 2 we calculated the
magnitude and direction of the electric field from a long
straight line of charge using Coulomb’s law and superposition.
Now we will show how Gauss’ law can be used to obtain the
same result, in a few simple steps.
Example (Question)
Uniform charge
density 

E

L

Lecture 4
EXAMPLE
Step 1: Study the symmetry to see if you can construct a Gaussian
surface on which the field magnitude and its direction relative to the
surface are constant.
Chose a cylinder of length and radius R

E normal and constant to surface A2
E parallel and to surfaces A1 and A3
Example (Solution)

1
A 2
A
3
A
R

Lecture 4
EXAMPLE
Surface S contains
three parts (Front
A1, Back A3 and
Side A2)



 








2
3
1 A
A
A
S
E A
d
E
A
d
E
A
d
E
A
d
E








Front Back Side
Example (Solution)

1
A 2
A
3
A
R

Lecture 4
EXAMPLE
Flux front  


1
)
(
A
E A
d
E
Front


A
d
E


perpendicular
Contained in front surface
Outward pointing
0
)
( 
 Front
E
Example (Solution)


1
A 2
A
3
A
R
E

L

Lecture 4
EXAMPLE
Flux back  


3
)
(
A
E A
d
E
Back


A
d
E


perpendicular
Contained in back surface
Outward pointing
0
)
( 
 Back
E
Example (Solution)
E

L

1
A 2
A
3
A
R

Lecture 4
EXAMPLE
Flux side  


2
)
(
A
E A
d
E
Side


A
d
E


parallel
Normal to side surface (constant on surface)
Outward pointing
2
)
( EA
Side
E 

Example (Solution)

1
A 2
A
3
A
R
E

L

Lecture 4
EXAMPLE
Step 3: Evaluate the enclosed charge. This is not the same as the
total charge if the Gaussian surface lies within the charge
distribution.
Example (Solution)
Uniform charge
density 

The total enclosed by the Gaussian surface is only the
segment of the line within the bounds of the front and back
surfaces.
  



enclosed
q

1
A 2
A
3
A
R

Lecture 4
EXAMPLE
Example (Solution)
Uniform charge
density 

o
E

 



and solve for E. The direction of the electric field vector should be
evident for the symmetry and polarity of the charges.
 
o
E
q

enclosed



1
A 2
A
3
A
R

Lecture 4
EXAMPLE
Step 4: Equate the flux and solve for E.
2
)
( EA
Side
E 

Example (Solution)
o
E

 



 
 

S o
q
A
d
E

surface
by
enclosed


2
A
E
o

 



1
A 2
A
3
A
R
E

L

Lecture 4
EXAMPLE
Step 4: Equate the flux and solve for E. The direction of the electric
field vector should be evident for the symmetry and polarity of the
charges.
Example (Solution)
2
A
E
o

 


Surface area of a cylinder 
R
A 
2
2 
R
E
o


2

 Note R radial distance from
charged line
END

1
A 2
A
3
A
R

Lecture 4
EXAMPLE
Example (Question)
An infinitely long cylindrical wire of radius R carries a
uniform positive surface charge S on its outer surface. Using
Gauss’ law evaluate the electric field vector for a) points external
to the wire (r > R) and b) for points internal to the wire (r < R) .
C) What is the field just above the surface of the wire.
R


S

Lecture 4
EXAMPLE
Example (Solution)
An infinitely long cylindrical wire of radius R carries a uniform positive
surface charge S on its outer surface. Using Gauss’ law evaluate the electric
field vector for a) points external to the wire (r > R) and b) for points internal to
the wire (r < R) . C) What is the field just above the surface of the wire.
R


S
Cylindrical symmetry exists for
the charge distribution thus a
Gaussian surface in the shape of a
cylinder may be used.
r
The Gaussian cylinder has
radius r and length L
L
E


Lecture 4
EXAMPLE
Example (Solution)
a) Electric field at points external to the wire (r > R)
R


S
r
L
We expect the electric field to be
radially outward.
There is no flux
through the ends
of the Gaussian
cylinder.
E

L

Lecture 4
EXAMPLE
Example (Solution)
R


S
r
L
   




Side
E
E A
d
E
Side


Flux through side of cylinder
side
E EA


rL
E
E 
2


A
d
E


parallel

Lecture 4
EXAMPLE
Example (Solution)
R
S
r
L
Total charge enclosed by Gaussian cylinder
The total charge enclosed
by the Gaussian cylinder
is only the charge that lies
within the end of the
cylinder. This charge is
distributed over a
cylindrical surface of
radius R
  

S
S dA
q 
enclosed
  RL
q S 
 2
enclosed 

Lecture 4
EXAMPLE
Example (Solution)
R
S
r
L
Flux and enclosed charge
o
S
E
RL


 2


We can now equate the two flux expressions and obtain an expression for the
electric field.
o
S
E
RL


 2


rL
E
E 
2

 And Gives
r
R
E
o
S



External electric field

Lecture 4
EXAMPLE
Example (Solution)
R
S
r
L
r
R
E
o
S



External electric field
The direction of the electric field can be obtained from the sign
of the surface charge distribution. Since S is positive here,
then the electric field is directed radially outward.
r
r
R
E
o
S
ˆ




r̂

Lecture 4
EXAMPLE
Example (Solution)
b) Electric field at points internal to the wire (r < R).
R
S
r
The Gaussian cylinder has
radius r and length L
L
The Gaussian surface
retains the symmetry of
the charge distribution and
encloses no charge when
(r < R).

Lecture 4
EXAMPLE
Example (Solution)
R
S
r
L
The Gaussian surface
retains the symmetry of
the charge distribution and
encloses no charge when
(r < R).
Thus
  0
enclosed
o




q
E

Lecture 4
EXAMPLE
Example (Solution)
R
S
r
L
If an electric field exists
inside the cylindrical
charge distribution it must
be radial.
Thus
side
S
E EA
A
d
E 


 


rL
E
E 
2



Lecture 4
EXAMPLE
Example (Solution)
R
S
r
L
We can now equate the
two flux expressions and
solve for the electric field
inside the charged
structure.
  0
enclosed
o




q
E
 
o
S
E
q
A
d
E

enclosed



 


rL
E
E 
2


0

E
For this symmetric charge distribution
the internal electric field is zero.

Lecture 4
EXAMPLE
Example (Solution)
C) What is the field just above the surface of the wire.
R
S
E

In part a) we obtained an expression for the
electric field for all points external to the
charge distribution.
r
r
R
E
o
S
ˆ




The field just above the surface
can be obtained by setting r = R.
r
E
o
S
ˆ



 Electric field at the surface of
the charge distribution.

Lecture 4
ASSIGNMENT
These questions are straight forward. Plug in the numbers and get your answer. Being able to
solve this type of question ensures you of at least a grade of 25% on a quiz or final exam
containing questions related to this lecture.
These questions require a few manipulations of equations or numbers before the answer can be
obtained. Being able to solve this type of question ensures you of at least a grade of 50% on a
quiz or final exam containing questions related to this lecture.
These questions are the most difficult and require a thorough understanding of the topic material
and also pull in topics from other lectures and disciplines. Being able to solve this type of
question ensures you an A grade on a quiz or final exam containing questions related to this
lecture.
These question are quite involved and requires a thorough understanding of the topic material.
Being able to solve this type of question ensures you of at least a grade of 75% on a quiz or final
exam containing questions related to this lecture.
25
50
75
100
75 100 These form excellent review questions when preparing for the quiz and final exam.
25 50 75 100
SELL EVALUATION SCALE

Lecture 4
TEXT
PROBLEM SOLVING STRATEGY
1. The first step is to select the surface (which we call the
Gaussian surface) that you are going to use with Gauss’ law. If
you are trying to find the field at a particular point, then that point
must lie on your Gaussian surface.
2. The Gaussian surface does not have to be a real physical
surface, such as the surface of a solid body. Often the
appropriate surface is an imaginary geometric surface; it may be
in empty space, embedded in a solid body, or partly both.

Lecture 4
TEXT
3. The Gaussian surface and the charge distribution must have
symmetry property, so that it is possible to actually evaluate the
integral in Gauss’ law. If the charge distribution has cylindrical
or spherical symmetry, the Gaussain surface will usually be a
coaxial cylinder or a concentric sphere, respectively.
4. Often you can think of the closed Gaussian surface as being
made up of several separate areas, such as the sides and ends of a
cylinder. The integral:
over the entire Gaussian surface is always equal to the sum of the
integrals over all the separate areas.
 
S
A
d
E



Lecture 4
TEXT
5. If the electric field E is perpendicular (normal) to the Gaussian
surface with an area A at every point, and if it also has the same
magnitude at every point on the Gaussian surface, then E is a
constant and the integral:
becomes:
 
S
A
d
E


EA

Lecture 4
TEXT
6. If the electric field E is tangent to the Gaussian surface at every
point, then the dot product:
and the integral:
0



S
A
d
E


0

 A
d
E


7. If the electric field E is zero everywhere on the Gaussian
surface then the integral is also zero.

Lecture 4
TEXT
8. At all times it is important to have a clear picture of the charge
distribution. Always draw a figure if one is not provided.
9. The application of Gauss’ law requires a high degree of
symmetry. For spherical and cylindrical symmetries the
Gaussian surface can be characterized by a radial coordinate r.
That is a sphere of radius r OR a cylinder of radius r. Then when
you solve for the electric field it is usually a function of r.

Lecture 4
ASSIGNMENT
Obtain an expression for the electric field vector at a
point external to a uniform spherical charge distribution
of radius R and charge density V. Here rework the
example given above making as few references to your
notes as possible.
25

Lecture 4
ASSIGNMENT
point internal to a uniform spherical charge distribution
of radius R and charge density V. Here rework the
example given above making as few references to your
notes as possible.
25

Lecture 4
ASSIGNMENT
point external to a uniformly charged line of charge.
Here rework the example given above making as few
references to your notes as possible.
25


Use as the linear charge distribution

Lecture 4
ASSIGNMENT
A infinite flat sheet of charge is contained in the (x,y)
plane and has charge density S. Obtain an expression
for the electric field vector on each side of the plane.
Here rework the example given above making as few
references to your notes as possible. Recall that electric
field is a vector so indicate its direction.
25
x
y
z

Lecture 4
ASSIGNMENT
A uniform spherical charge distribution of radius R and
charge density V is centered on the origin of a
coordinate system. Using the expressions for the electric
field internal and external to the charge distribution make
a plot of the magnitude of E versus the radial coordinate.
Hint: If you wish to use a computer to plot the field try
putting in numbers for V and R. The curve obtained will
have the same overall shape.
50
:
ans

Lecture 4
ASSIGNMENT
point external to a spherical charge distribution of radius
R. The volume charge density is a function of the radial
coordinate r and varies as: V = Mr where M is a
constant.
50
2
2
4
:
r
MR
ans
o


Lecture 4
ASSIGNMENT
point external to a spherical shell of radius R. The
surface charge density S is constant over the entire
surface.
50
2
2
ˆ
:
r
r
R
ans
o
s



Lecture 4
ASSIGNMENT
The charge distribution consist of a uniformly volume
charged straight infinite cylindrical conductor. The
volume charge density is V. A) Obtain an expression
for the electric field vector at points external to the
charged cylinder. B) Obtain an expression for the
electric field at points inside the charged cylinder. Hint:
Be sure to include only the charge enclosed by the
Gaussian surface.
50
o
v R
A
ans


2
:
)
(
2
See TA for 

Lecture 4
ASSIGNMENT
If you “painted” positive charge on the floor, what
surface charge density would be necessary in order to
suspend a 15 C, 5.0 g particle above the floor.
50
2
8
/
10
7846
.
5
: m
C
ans 


Lecture 4
ASSIGNMENT
The parallel plate capacitor shown in the figure has plates
which extend to infinity and separated by 1.0 cm. They
are charged such that they have identical positive surface
charge densities. Calculate the magnitude and direction
of the electric field at A) point P (1 cm to the left of plate
1); B) point R (midway between the plate); C) point T (1
cm to the right of plate 2).
50
S

 S


+q +q
P R T
1 2
x
P
ans
o
s
ˆ
:
)
(




Lecture 4
ASSIGNMENT
A slab of charge extends infinitely in two dimensions and
has thickness d in the third dimension as shown in the
figure. The slab carries a uniform volume charge density
V. Find expressions for the electric field strength a)
inside and b) outside the slab, as a function of the
distance x from the center plane.
50
d
o
v
x
a
ans


:
)
(

Lecture 4
ASSIGNMENT
A point charge +Q is located at the origin of a coordinate
system. This charge is also located at the center of a
spherical shell of total surface charge -Q and radius R.
A) Obtain an expression for the electric field vector at
points external to a spherical shell. B) obtain an
expression for the electric field inside the shell. (exclude
the origin).
75
0
:
)
(A
ans

Lecture 4
ASSIGNMENT
point internal to a spherical charge distribution of radius
R. The volume charge density is a function of the radial
coordinate r and varies as: V = Mr where M is a
constant. Hint: Be careful to include only the charge
contained inside of the Gaussian surface.
75
o
r
Mr
ans

4
ˆ
:
2

Lecture 4
ASSIGNMENT
A uniformly charged spherical volume of radius R1 and
charge density V is centered on a coordinate system. It
carries a total charge +Q. This volume charge is also
located at the center of a spherical shell of total surface
charge -Q and radius R2. A) Obtain an expression for the
electric field vector at points external to a spherical shell.
B) obtain an expression for the electric field between the
sphere and shell. C) Obtain an expression for the electric
field inside the charged spherical volume.
75
o
vr
C
ans


3
:
)
(

Lecture 4
ASSIGNMENT
A spherical charge distribution centered on the origin has
its volume charge density vary as r-2 ( V = M r-2, M a
constant) A) Obtain an expression for the electric field
throughout the volume charge distribution. Does your
result make sense. B) What is the electric field
expression if the volume charge density varies as r-1 ( V
= M r-1, M a constant).
75
o
M
B
ans

2
)
(

Lecture 4
ASSIGNMENT
A square non-conducting plate measures 4.5 m on a side
and carries charge spread uniformly over its surface. The
electric field 10 cm from the plate and not near an edge
has magnitude 430 N/C and points towards the plate.
Find A) the surface charge density on the plate and B) the
total charge on the plate. C) What is the electric field
strength 20 cm from the plate.
75
2
81
.
3
:
)
(
m
nC
a
ans 

Lecture 4
ASSIGNMENT
A point charge +Q = 1 C is located at the origin of a
coordinate system. This charge is surrounded by a thick shell
of inner radius R1 = 1 cm and outer radius R2 = 2 cm. If you
are told that the electric field inside the shell is zero evaluate
the following:
A) The surface charge density on the inner surface of the shell.
B) The surface charge density on the outer surface of the shell.
C) The electric field strength at r = 0.5 cm, r = 2.5 cm.)
100
Point charge
Spherical shell

Lecture 4
ASSIGNMENT
A long coax cable consists of an inner cylindrical conductor of
radius a and an outer coaxial cylinder with inner radius b and outer
radius c. The outer cylinder is mounted on insulating supports and
has no net charge. The inner cylinder has a uniform positive charge
density per unit length. Calculate the electric field A) at any
point between the cylinders, at a distance r from the axis; B) at any
point outside the outer cylinder. C) Sketch a graph of the magnitude
of the electric field as a function of distance r from the axis of the
cable. From r = 0 to r = 2c. D) Find the charge per unit length on
the inner surface and on the outer surface of the outer conductor.
100


insulation
Inner conductor

Lecture 4
ASSIGNMENT
A very long conducting tube (hollow cylinder) has inner
radius a and outer radius b. It carries charge per unit
length +, where  is a positive constant with units of
C/m. A line of charge lies along the axis of the the tube
and also has charge per unit length +. A) Calculate the
electric field in terms of  and the distance r from the
axis of the tube for I) r < a; II) a < r < b; III) r > b. Show
your results in a sketch of E® as a function of r. B) What
is the charge per unit length on I) the inner surface of the
tube; II) the outer surface of the tube.
100

Lecture 4
ASSIGNMENT
A non-uniform but spherically symmetric distribution
charge has charge density  given by:
100
R
r
R
r
R
r
o











for
0
for
1



3
3
R
Q
o

 
A) Show that the total charge contained in the charge
distribution is Q. B) Show that, for the region defined
by r  R, the electric field is identical to that produced
by a point charge Q. C) Obtain an expression for the
electric field in the region r  R. D) Compare your
results in parts B) and C) for r = R.

Lecture 4
ASSIGNMENT
Three large parallel insulating sheets have surface charge densities
+0.0200 C/m2, +0.0100 C/m2, and -0.0200 C/m2, respectively.
Adjacent sheets are a distance 0.300 m from each other. Calculate
the resulting electric field (magnitude and direction) due to all three
sheets at a) point P (0.150 m to the left of sheet I); b) point R
(midway between sheets I and II); c) point S (midway between
sheets II and III); d) point T ( 0.150 m to the right of sheet III).
100
I II III
P R S T
2
/
02
.
0 m
C

2
/
02
.
0 m
C

m
15
.
0 m
15
.
0 m
15
.
0 m
15
.
0 m
15
.
0 m
15
.
0

Lecture 4
REFERENCES
REFERENCES
(0) Textbook: U. S. Inan, A. S. Inan
“Engineering Electromagnetics”
(1) J.D. Kraus, K. R. Carver “Electromagnetics” 2nd
(2) Reitz, Milford, Christy “Foundations of Electromagnetic
theory” 4th
(3) M. Plonus “Applied Electromagnetics”
(4) R. P. Winch “Electricity and Magnetism”
(5) P. Lorrain, D. Corson “Electromagnetic fields and Waves”
2nd
(6) Duckworth “Electricity and Magnetism”
(7) J.D. Jackson “Classical Electrodynamics” 2nd
(8) F. Ulaby, “Fundamentals of applied Electromagnetics”
(0) Inan p. 288 - 310
(1) Kraus p. 43 - 45, 39 - 48, 93 -99
(2) Reitz p. 12 - 14, 37 - 43
(3) Plonus p. 22 - 31, 35 - 38
(4) Winch p. 271 - 289
(5) Lorrain p. 13 - 16, 47 - 51
(6) Duckworth p. 16 - 25
(7) Jackson p. 30 - 38
(8) Ulaby p. 148 - 150, 232

Lecture 4
SUMMARY
 
 

S o
q
A
d
E

surface
by
enclosed


GAUSS’ LAW
Gauss’ law is true for any surface enclosing any charge distribution.
When the charge distribution has sufficient symmetry we can chose a
surface --- Gaussian Surface --- over which the evaluation of the flux
integral becomes simple. Gauss’ law allows us to calculate the field far
more easily than we could using Coulomb’s law and superposition.
E

E

A1 A2
A3

1
A 2
A
3
A
R

Lecture 4
END
END LECTURE 4

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