Gauss’ Law
Class Objectives
• Introduce the idea of the Gauss’ law as
another method to calculate the electric
field.
• Understand that the previous method of
calculating the electric field strength does
not consider symmetry.
• Consider the different types of symmetry.
Class Objectives
•
•
•
•
Introduce the idea of the Gaussian surface.
Define the properties of Gaussian surfaces.
Show how to choose a Gaussian surface.
Show how Gaussian surfaces can be used to
take advantage of symmetry.
Class Objectives
• Show that the results of Gauss’s law are the
same as the standard results but quicker and
easier!
Student Objectives
• Be able to use Gauss’ law to calculate the
electric field for various objects.
• Be able to determine the type of symmetry
in the problem and hence the type of
Gaussian surface to be used.
Gauss’ Law
• Symmetry in problems arise naturally.
Gauss’ Law
• Symmetry in problems arise naturally.
• Gauss’ law is an alternate method to
coulomb’s law where there is symmetry.
Gauss’ Law
• Outline of Gauss’ Law
1. Central idea is a hypothetical closed
surface called a Gaussian surface.
2. The surface can be chosen any shape to
complement the symmetry. Eg cylinder or
sphere.
3. Gauss’ law relates the electric field at a
point on the closed surface to the net
charge enclosed by the surface.
Gauss’ Law
• The amount of charge enclosed is measured
by the amount of flux passing through the
surface.
Gauss’ Law
• Example of a
spherical Gaussian
closed surface.
Electric flux
• The electric flux is a measure of the number
of field lines passing an area.
• The flux is represented by the symbol  .
Electric flux
• The electric flux is a measure of the number
of field lines passing an area.
• The flux is represented by the symbol  .
• Assuming an arbitrary Gaussian surface in a
field.
Electric flux
• If we divide the surface into small squares
of area dA then the curvature of the surface
can be ignored if the squares are taken small
enough.
Electric flux
• If we divide the surface into small squares
of area dA then the curvature of the surface
can be ignored if the squares are taken small
enough.

• By convention we assign an area vector dA .
• The direction of which is perpendicular to
the surface of the element directed away
from its interior.
Electric flux

• Because dA can be made arbitrarily small,E
may be taken as constant for a given
square.
• The flux for a Gaussian surface is given as
 
   E  dA
Electric flux

• Because dA can be made arbitrarily small,E
may be taken as constant for a given
square.
• The flux for a Gaussian surface is given as
 
   E  dA
• The circle indicates that the integral is to be
taken over the entire closed surface.
Electric flux
• The follow diagrams gives three cases for the
relative orientation of the field and dA. The angle
measured is the smallest angle between the
vectors.

Electric flux
• Note that all of the elements on the surface
do not contribution to the flux.
Electric flux
• Note that all of the elements on the surface
do not contribution to the flux.

Gauss’ Law
• Guass’ Law is often written as
  qenc
 E  dA 
0
• Where qenc is the sum of the enclosed
charges.
Gauss’ Law
Choosing a Gaussian surface
Gauss’ Law
• Choosing a Gaussian surface is not hard but
subtle.
Gauss’ Law
•
•
•
Choosing a Gaussian surface is not hard
but subtle.
A simple surface must be chosen to take
advantage of symmetry.
Consider the following rules as a guide.
Gauss’ Law
1. Choose the surface perpendicular to the
field so the E and dA are parallel.
Gauss’ Law
1. Choose the surface perpendicular to the
field so the E and dA are parallel.
2. Choose so that points on the surface are
equal distance away from the charge so
the E doesn’t vary.
Gauss’ Law
1. Choose the surface perpendicular to the
field so the E and dA are parallel.
2. Choose so that points on the surface are
equal distance away from the charge so
the E doesn’t vary.
3. If this is not possible (1 and 2) choose a
surface such that the dot product is zero.
Ie. They are perpendicular.
Gauss’ Law and Coulomb’s Law
• Consider a point charge q.
Gauss’ Law and Coulomb’s Law
• Consider a point
charge q.
• For symmetry we use
a spherical Gaussian
surface.
Gauss’ Law and Coulomb’s Law
• the chosen Gaussian
surface holds true for
the first two
guidelines.
Gauss’ Law and Coulomb’s Law
dA
•
the chosen Gaussian
surface holds true for
the first two
guidelines.
1.
2.
E and dA are parallel.
Points on the surface are
equal distance away
from the charge so the E
doesn’t vary.
Gauss’ Law
• From Gauss’ Law
  qenc
 E  dA 
0
Gauss’ Law
• From Gauss’ Law
  qenc
qenc
 E  dA    E dA 
0
0
• Since the two vectors are parallel.
Gauss’ Law
• From Gauss’ Law
  qenc
qenc
 E  dA    E dA 
0
0
• Since the two vectors are parallel.
• Also we note that the enclosed charge is
simply q. So we can write that,
E  dA 
q
0
Gauss’ Law
• The area of the closed surface is the area of
a circle. So that


E 4r 
2
q
0
Gauss’ Law
• The area of the closed surface is the area of
a circle. So that


E 4r 
2
• Therefore
E
q
4 0 r 2
q
0
Conductor in an Electric Field
Conductor in an Electric Field
• For the cases to investigated, we will
consider the cases where the conductor is in
equilibrium (electrostatics).
Conductor in an Electric Field
• For the cases to investigated, we will
consider the cases where the conductor is in
equilibrium (electrostatics).
• For a conductor in equilibrium we have
following conditions:
Conductor in an Electric Field
1. The charge exists entirely on the surface
of conductor(no charge is found within the
body of the conductor).
Conductor in an Electric Field
1. The charge exists entirely on the surface
of conductor(no charge is found within the
body of the conductor).
2. The electric field within the conductor is
zero.
Conductor in an Electric Field
1. The charge exists entirely on the surface
of conductor(no charge is found within the
body of the conductor).
2. The electric field within the conductor is
zero. The charge distributes itself so as to
get as far from each other as possible.
Conductor in an Electric Field
3. The external electric field is perpendicular
to the surface of the conductor.
Conductor in an Electric Field
3. The external electric field is perpendicular
to the surface of the conductor. If not it
would cause the charges to move along
the surface of the conductor.
Conductor in an Electric Field
3. The external electric field is perpendicular
to the surface of the conductor. If not it
would cause the charges to move along
the surface of the conductor.
 Note: unless the conductor is spherical,
the charge does not distribute itself
uniformly.
Conductor in an Electric Field
• When a conductor is place in an external
field, the mobile electrons experience a
force pushing them in the opposite direction
to the field
E
e
Conductor in an Electric Field
• Thus making the top of the conductor
positive and bottom negative.
E
++++++++++
e
___________
Conductor in an Electric Field
• Thus making the top of the conductor
positive and bottom negative. This sets up
an internal electric which grows in strength
as more electrons move.
E
++++++++++
Eint
e
___________
Electric field caused
by pd between the top
and bottom of the
conductor
Conductor in an Electric Field
• Electrons move until the two fields have the
same magnitude. Hence the net electric field
E=0.
E
++++++++++
Eint
e
___________
Conductor in an Electric Field
• Electrons move until the two fields have the
same magnitude. Hence the net electric field
E=0. The magnitude of the charge on the
top & bottom of the conductor are the same.
E
++++++++++
Eint
e
___________
E
Eint
Conductor in an Electric Field
• Let us now consider the electric field just
outside the surface of a conductor.
Conductor in an Electric Field
• Let us now consider the electric field just
outside the surface of a conductor.
E
Conductor in an Electric Field
• Let us now consider the electric field just
outside the surface of a conductor. We use a
cylindrical Gaussian surface.
A
E
Conductor in an Electric Field
• Assume the conductor has charge per unit
area and total charge Q with uniform
charge distribution.
Conductor in an Electric Field
• Recall Gauss’ Law:
  qenc
 E  dA 
0
Conductor in an Electric Field
• Recall Gauss’ Law:
  qenc
 E  dA 
0
• We note that qenc  A
Conductor in an Electric Field
• Recall Gauss’ Law:
  qenc
 E  dA 
0
• We note that qenc  A
  A
• We can therefore write that:  E  dA  
0
Conductor in an Electric Field
• Integrating as before we get that:
A
EA 
0
Conductor in an Electric Field
• Integrating as before we get that:
A
EA 
0

• Hence E 
0
Conductor in an Electric Field
• Integrating as before we get that:
A
EA 
0

• Hence E 
0
• The electric field.
Conductor in an Electric Field
Cylindrical rod
Conductor in an Electric Field
• Consider an infinitely long cylindrical rod with
uniform linear charge density λ. What is the
electric field a distance r from its axis.
r
+ + + + + + + + + + + + + +
+ + + +
+
Conductor in an Electric Field
• We choose a cylindrical Gaussian surface.
E
E
r
+ + + + + + + + + + + + + +
h
+ + + +
+
• Writing Gauss’ law:
  qenc
 E  dA 
0
• Writing Gauss’ law:
  qenc
 E  dA 
0
• Hence, EA 
q enc
0
• Writing Gauss’ law:
  qenc
 E  dA 
0
• Hence,
EA 
q enc
0
 EA 
h
0
• Writing Gauss’ law:
  qenc
 E  dA 
0
• Hence,
EA 
q enc
0
 EA 
h
0
• The area enclosed is that of a cylinder.
 E 2rh 
h
0
• Writing Gauss’ law:
  qenc
 E  dA 
0
• Hence,
EA 
q enc
0
 EA 
h
0
• The area enclosed is that of a cylinder.
 E 2rh 
h

E
0
2 0 r
• The electric field due to a line of charge is

E
2 0 r
Charged sphere
• We previously looked at the properties of a
conductor.
• For a spherical conductor of charge +Q with
uniform charge density, what is the electric
field outside and inside the conductor?
+Q
• We know the charge resides on the surface.
• Consider a Gaussian surface inside the
sphere.
• We know the charge resides on the surface.
• Consider a Gaussian surface inside the
sphere.
• No charge is enclosed by the surface.
• We know the charge resides on the surface.
• Consider a Gaussian surface inside the
sphere.
• No charge is enclosed by the surface.
• Therefore E = 0 as expected.
• Now consider a Gaussian surface outside
the sphere.
• Now consider a Gaussian surface outside
the sphere.
• The total charge enclosed is Q.
• Now consider a Gaussian surface outside
the sphere.
• The total charge enclosed is Q.
• Therefore from Gauss’ Law,
  qenc
  Q
 E  dA    E  dA 
0
0
• Now consider a Gaussian surface outside
the sphere.
• The total charge enclosed is Q.
• Therefore from Gauss’ Law,
  qenc
  Q
 E  dA    E  dA 
0
 EA 
Q
0


 E 4r 
2
0
Q
0
• Hence, E 
Q
4 0 r 2
• As expected!!
Uniform Charge non-conducting
Sphere
• Consider a solid non-conducting sphere of
radius R with uniform charge distribution ρ.
What is the electric field outside and inside
the sphere?
r
R
• The total charge is given
 4r 3 
by: Q   3  


• The total charge is given
 4r 3 
by: Q   3  


• Case r > R (outside the sphere)
• The total charge is given
 4r 3 
by: Q   3  


• Case r > R (outside the sphere)
  qenc
• Gauss’ Law:  E  dA 
0
• The total charge is given
 4r 3 
by: Q   3  


• Case r > R (outside the sphere)
  qenc
• Gauss’ Law:  E  dA 
0
• The charge enclosed is the total charge.


  Q
E  dA 
0
• The total charge is given
 4r 3 
by: Q   3  


• Case r > R (outside the sphere)
  qenc
• Gauss’ Law:  E  dA 
0
• The charge enclosed is the total charge.


  Q
E  dA 
0

Q
• E is constant on the surface, therefore E dA 
0
• Integrating over the surface we have, E 4r  
2
Q
0
• Integrating over the surface we have, E 4r  
2
Q
E
4r 2 0
Q
0
• Integrating over the surface we have, E 4r  
2
Q
E
4r 2 0
• That is the sphere acts like a point charge (as
expected).
Q
0
• Case r < R (inside the sphere)
r
R
• Case r < R (inside the sphere)
• Again we use Gauss’ Law.
• However the total charge is not enclosed.
r
R
•
•
•
•
Case r < R (inside the sphere)
Again we use Gauss’ Law.
However the total charge is not enclosed.
 4r 3 

The total charge enclosed is Qenc  

 3 
r
R
•
•
•
•
•
Case r < R (inside the sphere)
Again we use Gauss’ Law.
However the total charge is not enclosed.
 4r 3 

The total charge enclosed is Qenc  

3


Therefore,

   4r 3 
 0
E  dA  

3


r
R
•
•
•
•
•
Case r < R (inside the sphere)
Again we use Gauss’ Law.
However the total charge is not enclosed.
 4r 3 

The total charge enclosed is Qenc  

3


Therefore,

   4r 3 
 0  E
E  dA  

3


3


4

r
2
   0
4r  
 3 


r
R
•
•
•
•
•
Case r < R (inside the sphere)
Again we use Gauss’ Law.
However the total charge is not enclosed.
 4r 3 

The total charge enclosed is Qenc  

3


Therefore,

   4r 3 
 0  E
E  dA  

3


E
r
3 0
3


4

r
2
   0
4r  
 3 


r
R