Download Topics: • Symmetry • The Concept of Flux • Calculating Electric Flux

Topics:
•  Symmetry
•  The Concept of Flux
•  Calculating Electric Flux
•  Gauss’s Law
•  Using Gauss’s Law
•  Conductors in Electrostatic Equilibrium
PHYS 1022: Chap. 28, Pg 2
1
New Topic
PHYS 1022: Chap. 28, Pg 3
For uniform field, if E and A
are perpendicular, define
If E and A are NOT perpendicular,
define scalar product
General (surface integral)
Unit of electric flux: N . m2 / C
 
Electric flux is proportional to the
number of field lines passing
through the area. And to the total
change enclosed
PHYS 1022: Chap. 28, Pg 4
2
+
+
PHYS 1022: Chap. 28, Pg 5
+
+
PHYS 1022: Chap. 28, Pg 6
3
+
r
½r
+
2r
PHYS 1022: Chap. 28, Pg 7
+
r
½r
+
2r
PHYS 1022: Chap. 28, Pg 8
4
A metal block carries a net positive
charge and put in an electric field.
What is the electric field inside?
0
positive
Negative
Cannot be determined
1) 
2) 
3) 
4) 
If there is an electric field inside the
conductor, it will move the charges in the
conductor to where each charge is in
equilibrium. That is, net force = 0 therefore
net field = 0!
PHYS 1022: Chap. 28, Pg 9
A metal block carries a net charge and
put in an electric field. Where does
the net charge reside ?
1) 
2) 
3) 
All inside the conductor
All on the surface
Some inside the conductor,
some on the surface
The net field = 0 inside, and so the net flux
through any surface drawn inside the
conductor = 0. Therefore the net charge
inside = 0!
PHYS 1022: Chap. 28, Pg 10
5
PHYS 1022: Chap. 28, Pg 11
PHYS 1022: Chap. 28, Pg 12
6
PHYS 1022: Chap. 28, Pg 13
PHYS 1022: Chap. 28, Pg 14
7
Group Work
PHYS 1022: Chap. 28, Pg 15
Group Work
PHYS 1022: Chap. 28, Pg 16
8
New Topic
 
PHYS 1022: Chap. 28, Pg 17
Gauss' Law
The total electric flux through a closed surface is
equal to the total charge enclosed by that surface,
divided by ε0.
ε0=8.85x10-12 C2/N.m2
is called permittivity of
free space
The k in Coulomb’s law
F=kq1q2/r2 is related to ε0
by
PHYS 1022: Chap. 28, Pg 18
9
New Topic
 
PHYS 1022: Chap. 28, Pg 19
Gauss’s law can be used to find out the electric fields around
symmetric charge distributions
PHYS 1022: Chap. 28, Pg 20
10
1.  Gauss’s law applies only to a closed surface, called a
Gaussian surface.
2.  A Gaussian surface is not a physical surface. It need not
coincide with the boundary of any physical object
(although it could if we wished). It is an imaginary,
mathematical surface in the space surrounding one or
more charges.
3.  We can’t find the electric field from Gauss’s law alone.
We need to apply Gauss’s law in situations where, from
symmetry and superposition, we already can guess the
shape of the field.
PHYS 1022: Chap. 28, Pg 21
Some charge distributions have translational, rotational, or
reflective symmetry. If this is the case, we can determine
something about the field it produces:
The symmetry of an electric field must match the symmetry
of the charge distribution.
For example, the electric field of a cylindrically symmetric
charge distribution
a)  cannot have a component parallel to the cylinder axis.
b)  cannot have a component tangent to the circular cross
section.
PHYS 1022: Chap. 28, Pg 22
11
PHYS 1022: Chap. 28, Pg 23
PHYS 1022: Chap. 28, Pg 24
12
PHYS 1022: Chap. 28, Pg 25
Whenever you can, make the electric field parallel
to the area vector (normal vector).
PHYS 1022: Chap. 28, Pg 26
13
PHYS 1022: Chap. 28, Pg 27
PHYS 1022: Chap. 28, Pg 28
14
PHYS 1022: Chap. 28, Pg 29
 
We place a positive charge q on a solid conducting sphere with radius R. Find the
electric field at any point inside or outside the sphere.
This problem has spherical symmetry. For a spherical surface
of any radius, the surface integral becomes
+
+
+
Choose the sphere to be inside the
conductor (r<R). Since there is no charge
enclosed: 4π r2 E=0.
So E=0 inside.
+
R
+
+
+
+
Choose the sphere to be outside the
conductor (r>R). Since the entire
charge is enclosed: 4π r2 E=q/ ε0. So
PHYS 1022: Chap. 28, Pg 30
15
 
Positive electric charge Q is distributed uniformly throughout the volume of an
insulating sphere with radius R. Find the E field both inside and outside of the
sphere.
+Q
This problem has spherical symmetry. The E field must be
radially outward. Choose Gaussian surface as a sphere.
R
Inside:
Outside like a
point charge:
PHYS 1022: Chap. 28, Pg 31
 
If the inner shell has a net charge +q, and the outer shell has no net charge,
 how is the charge distributed on the inside and outside wall of each
sphere?
 find the E field everywhere.
r<a,
+q
a<r<b,
b<r<c,
c<r<d,
r>d, E=kq/r2
PHYS 1022: Chap. 28, Pg 32
16
PHYS 1022: Chap. 28, Pg 33
17