Physics 122B
Electricity and Magnetism
Lecture 6
Flux and Gauss’s Law
April 06, 2007
Martin Savage
Lecture 6 Announcements
Lecture Homework #2 has been posted on the Tycho
system. It is due at 10 PM on Wednesday, April 11.
 On Friday, April 13, we will have Midterm 1, covering
Chapters 25-27. You may bring one page of notes. Also
bring a Scantron sheet and a scientific calculator with
good batteries.
A discussion forum for this class now exists…use it as
you please…discussions related to physics only !!
Chapter 26 Summary (1)
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Chapter 26 Summary (2)
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Types of Symmetry (again)
Symmetry Principle: The electric field must have the
same symmetries as the charge distribution that produced it.
Translational symmetry:
translation along a line
does not change object.
Full rotational symmetry:
any rotation about any
axis does not change
object: Sphere
Cylindrical symmetry:
any rotation about one
axis does not change
object: Cylinder
Reflectional symmetry:
mirror image reflection
about some plane is same
as object.
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A
(yes)
A
E
(no)
E
Partial rotational symmetry:
720 rotation about one axis
does not change object: Star
720
5
Symmetry of a Long Cylinder
z
A long cylinder has the following symmetries:
1. Translational symmetry along the z axis;
2. Full rotational symmetry about the z axis;
3. Reflectional symmetry in any plane containing
the z axis;
4. Reflectional symmetry in any plane ^ the z axis.
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Rejected Possibilities (1)
Question: Could the E field have z-axis components?
Symmetry Answer: Consider the 4th symmetry:
(reflection in a plane ^ z)
The field changes on this reflection, but the charged object does not.
Therefore, the E-field of the long cylinder cannot have z-axis components
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Rejected Possibilities (2)
Question: Could the E-field have
q-axis components?
Symmetry Answer: Consider the 3rd
symmetry: (reflection plane contains z)
r
z
q
The field changes on this
reflection, but the charged
object does not. Therefore,
the E-field cannot have q-axis
components.
r
Conclusion: In cylindrical coordinates, the
electric field must be radial (r), with no
components in the q or z directions.
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What Good is Symmetry?
 Symmetry allows us to rule out many conceivable field
shapes that are found to be incompatible with the symmetries
of the charge distribution.
 Knowing what cannot happen is often as useful as knowing
what must happen.
 By using symmetry arguments, we can avoid lengthy
mathematical calculations that will ultimately give a zero
result.
 Reasoning on the basis of symmetry can be a powerful tool
in physics, chemistry, and engineering. It is a tool worth
mastering.
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Symmetries of Three
Geometries
1.
2.
3.
Translation along plane
Full rotation about ^ axis.
Reflection about charged
plane or about ^ plane
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1.
2.
3.
Translation along z
Full rotation about z
Reflection about plane
including or ^ z.
Physics 122B - Lecture 6
1.
2.
Full rotation about any
axis through center.
Reflection about any
plane including center.
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Charge Conjugation Symmetry
Charge Conjugation
There is an additional symmetry associated with electric charges.
If we change all positive charges to negative and all negative charges to
positive without changing the magnitudes of the charges, the field must
reverse direction, but must not change in any other way.
Charge Conjugation
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Question 2
A uniformly charged rod has positive charge Q and finite length L. The
rod is symmetric under rotations about the z axis and under reflection in
any plane containing the z axis. It does not have translational symmetry or
symmetry on reflection about any plane ^ z, except the plane through the
center of the rod.
Which of the field shapes below matches the symmetries of the rod?
(a)
(c)
(b)
(d)
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The Porcupine Theory of Flux
Imagine that we have some porcupines
with very long quills. Careful examination
shows each porcupine has exactly 30 quills.
+
+
+
+
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Now suppose that we hide a number of
these porcupines in a loose-weave burlap
bag. Assuming that the quills of one beast
can completely penetrate another beast (i.e
they are transparent to quills, or ghosts) …..
How many porcupines are in the bag?
Answer: Count the number of quills
coming out through the cloth. Then, by
dividing the number of quills by 30, we can
tell how many porcupines are in the bag.
This is a good analogy to Gauss’s Law. The quill-count is the
electric flux, the porcupines are electric charges, and the bag is a
“Gaussian surface” surrounding the charges. Each charge Q has a
“quill count” of Q/e0.
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The Concept of Flux
The basic idea here is that the qualitative idea of field lines can be quantified by
defining electric flux as a conserved quantity that is “delivered” by electric charge.
Charge creates flux, and observing flux allows one to deduce the presence of
charge.
If an electric field is observed to be coming out of all walls of an opaque box as in
(a), we can conclude that the box contains a net positive charge.
If an electric field is observed to be going into all walls of an opaque box as in (b),
we can conclude that the box contains a net negative charge.
If equal amounts of electric field are observed to be entering and exiting the box
as in (c), we can conclude that the box contains no net charge.
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Gaussian Surfaces
A “Gaussian surface” is a closed virtual surface (you get to choose it as
you wish) that surrounds a region of space that may contain a quantity of
charge.
Gaussian surfaces are most useful when the symmetry of the Gaussian
surface matches the symmetry of the electric field that passes through the
surface (the same symmetry as the charge distribution that produces it),
and when the field lines are perpendicular to the surfaces that they cross.
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Matching the Field
with the Gaussian Surface
We would like to choose a closed
Gaussian surface to assess the
flux so that:
1. The electric field & surface are ^.
2. The electric field is constant or
zero over each surface element.
To do this, we must match
the symmetries of the field and
the closed surface.
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Question 1
The box contains:
(a) A single positive charge; (b) A single negative charge; (c) No charge;
(d) A net positive charge; (e) A net negative charge.
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The Basic Definition of Flux
Imagine that you are holding a loop of wire of
area A in front of a fan, and you want to know
the volume per second, F, of air with velocity v
flowing through the loop.
When the loop is perpendicular to the air
flow (a), F will be a maximum, and when the plane
of the loop is parallel to the air flow (b), F=0.
Then F = vA cosq, where q is the angle
^ normal to the
between v and the unit vector n
plane of the loop, i.e., F = v^A.
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By analogy, the electric
flux is:
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The Area Vector
We can make the expression for the flux a vector equation by
defining an area vector:
Here, ^n is the unit vector
perpendicular to the plane of the area. If the surface is closed, ^
n
points away from the interior. We can use this definition of ^n for
curved surfaces as well as flat ones by defining ^
n as perpendicular to
the local surface.
With this definition, the electric flux Fe for a constant E field is:
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The Electric Flux of a
Nonuniform Electric Field
 Fe  E   A
F e    F i   Ei   Ai
i
Fe 
i

surface
d Fe 

E  dA
surface
For a non-uniform field, the flux can
be calculated by breaking up the area A
of interest into small surface elements dA and integrating the E·dA flux
increments over the surface.
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Flux through a Curved Surface
For a non-uniform field plus a curved
surface, the same approach works. Break
up the area A into small surface elements
dA and integrate the E·dA flux increments
over the surface.
When E is ^ to dA the flux
contributions are zero.
When E is || to dA the flux
contributions are maximum.
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Flux Surface Integral
For closed Gaussian surfaces (i.e. encloses a non-zero volume)
we write the flux integral as:
F e   E  dA
Here, the “loop” on the integral sign indicates a surface
integral over a closed surface. Since a simple non-intersecting
closed surface has a definite inside and outside, we take the
vector direction of dA as always pointing out of the volume (and
normal to the surface of course).
This integral is related to the amount of charge enclosed by
the surface !!
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Example:
Flux through a Closed Cylinder
A cylindrically symmetric charge distribution has
created an electric field E(r)=E0(r/r0)2^
r, where E0
and r0 are constants and the unit vector ^r lies in the
xy plane.
Calculate the electric flux through a closed
cylinder of length L and radius R that is centered
along the z axis.
The electric field is directed radially outward
from the z axis, so it will be perpendicular to the
area vectors for the “end caps” of the cylinder. It
will be parallel to the normal vectors of the curved
sides of the cylinder and constant in magnitude
there. Therefore:
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Question 2
The total electric flux through the box is:
(a) 0 Nm2/C; (b) 1 Nm2/C; (c) 2 Nm2/C; (d) 4 Nm2/C; (e) 6 Nm2/C.
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End of Lecture 6
Lecture Homework #2 has been posted on the Tycho
system. It is due at 10 PM on Wednesday, April 11.
 On Friday, April 13, we will have Midterm 1, covering
Chapters 25-27. You may bring one page of notes. Also
bring a Scantron sheet and a scientific calculator with
good batteries.
A discussion forum for this class now exists…use it as
you please…discussions related to physics only !!