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Essential University Physics
Richard Wolfson
21
Gauss’s Law
PowerPoint® Lecture prepared by Richard Wolfson
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 21-1
In Chapter 21 you learnt
• To represent electric fields
using field-line diagrams
• To explain Gauss’s law
and how it relates to
Coulomb’s law
• To calculate the electric
fields for symmetric
charge distributions using
Gauss’s law
• To describe the behavior
of charge on conductors in
electrostatic equilibrium
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 21-2
Electric field lines
• Electric field lines provide a convenient and insightful
way to represent electric fields.
• A field line is a curve whose direction at each point is the
direction of the electric field at that point.
• The spacing of field lines describes the magnitude of the field.
• Where lines are closer, the field is stronger.
Vector and field-line diagrams of a point-charge field
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Tracing the field
of an electric dipole
Slide 21-3
Field lines for simple charge distributions
• There are field lines everywhere, so every charge distribution has
infinitely many field lines.
•
In drawing field-line
diagrams, we associate
a certain finite number
of field lines with a
charge of a given
magnitude.
•
In the diagrams shown,
8 lines are associated
with a charge of
magnitude q.
•
Note that field lines of
static charge
distributions always
begin and end on
charges, or extend to
infinity.
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Slide 21-4
Counting field lines
• How many field lines emerge from closed surfaces surrounding
charge?
• Count each field line
crossing going
outward as +1, each
inward crossing as
–1.
• You’ll find that the
number of field lines
crossing any closed
surface is
proportional to the
net charge enclosed.
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Slide 21-5
Electric flux
• Electric flux quantifies the
notion “number of field lines
crossing a surface.”
• The electric flux  through a flat
surface in a uniform electric field
depends on the field strength E, the
surface area A, and the angle 
between the field and the normal to
the surface.
• Mathematically, the flux is given
by   EA cos  E  A.
• Here A is a vector whose magnitude
is the surface area A and whose
orientation is normal to the surface.
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Slide 21-6
Electric flux with curved surfaces and
nonuniform fields
• When the surface is curved or the field is nonuniform, we
calculate the flux by dividing the surface into small
patches dA, so small that each patch is essentially flat and
the field is essentially uniform over each.
• We then sum the fluxes
d   E  dA over each patch.
• In the limit of infinitely many
infinitesimally small patches,
the sum becomes a
surface integral:
   E  dA
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Slide 21-7
Clicker question
•
The flux through side B of the cube in the figure is the
same as the flux through side C. What is a correct
expression for the flux through each of these sides?
A.  s 3 E
B.   s2 E
C.   s3 E cos45
D.   s2 E cos45
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 21-8
Clicker question
•
The flux through side B of the cube in the figure is the
same as the flux through side C. What is a correct
expression for the flux through each of these sides?
A.  s 3 E
B.   s2 E
C.   s3 E cos45
D.   s2 E cos45
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 21-9
Gauss’s law
• Slide 5 showed that the number of field lines emerging from a
closed surface is proportional to the net charge enclosed.
• In the language of electric flux, this statement becomes
The electric flux through any closed surface is proportional to
the charge enclosed.
• The proportionality constant is 4pk, also called 1/0.
qenclosed
• Therefore

OE  dA 
0
where the circle designates any closed surface, and the integral is taken over the
surface. qenclosed is the charge enclosed by that surface.
• This statement is Gauss’s law.
• Gauss’s law is one of the four fundamental
laws of electromagnetism.
• It’s true for any surface and charge
anywhere in the universe.
• Gauss’s law is equivalent to Coulomb’s law;
both describe the inverse square dependence
of the point-charge field.
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Slide 21-10
GOT IT? 21.2
A spherical surface surrounds an isolated positive charge, as
shown. If a second charge is placed outside the surface, which of
the following will be true of the total flux through the surface?
(A) it doesn’t change
(B) it increases
(C) it decreases
(D) it increases or decreases
depending on the sign of the
second charge
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 21-11
GOT IT? 21.2
A spherical surface surrounds an isolated positive charge, as
shown. If a second charge is placed outside the surface, which of
the following will be true of the total flux through the surface?
(A) it doesn’t change
(B) it increases
(C) it decreases
(D) it increases or decreases
depending on the sign of the
second charge
Repeat for the electric field on
the surface at the point
between the charges
D: field increases if charges are
opposite, decreases if same
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Slide 21-12
Using Gauss’s law
• Gauss’s law is always true.
• But it’s useful for calculating the electric field only in
situations with sufficient symmetry:
• Spherical symmetry
• Line symmetry
• Plane symmetry
Gauss’s law is always true, so it holds in both situations
shown. Both surfaces surround the same net charge, so the
flux through each is the same. But only the left-hand
situation has enough symmetry to allow the use of Gauss’s
law to calculate the field. The electric fields differ in the
two situations, even though the flux doesn’t.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 21-13
Clicker question
•
A spherical surface surrounds an isolated positive
charge. We can calculate the electric flux for this
surface. If a second charge is placed outside the
spherical surface, what happens to the magnitude of the
flux?
A. The flux increases proportionally to the magnitude of the
second charge.
B. The flux decreases proportionally to the magnitude of the
second charge.
C. The flux does not change.
D. The answer depends on whether the second charge is positive
or negative.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 21-14
Clicker question
•
A spherical surface surrounds an isolated positive
charge. We can calculate the electric flux for this
surface. If a second charge is placed outside the
spherical surface, what happens to the magnitude of the
flux?
A. The flux increases proportionally to the magnitude of the
second charge.
B. The flux decreases proportionally to the magnitude of the
second charge.
C. The flux does not change.
D. The answer depends on whether the second charge is positive
or negative.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 21-15
 
   E .dA
S
Flux through the surface due to ALL
the charges
This charge contributes ZERO FLUX as
every field line from it that enters the
surface at one point, leaves at another
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Slide 21-16
Gauss’s law: A problem-solving strategy
• INTERPRET: Check that your charge distribution has sufficient
symmetry.
• DEVELOP: Draw a diagram and use symmetry to find the
direction of the electric field. Then draw a gaussian surface on
which you’ll be able to evaluate the surface integral in Gauss’s law.
• EVALUATE:

• Evaluate the flux   O E  dA over your surface. The result contains the
unknown field strength E.
• Evaluate the enclosed charge.
• Equate the flux to qenclosed/0 and solve for E.
• ASSESS: Check that your answer makes sense, especially in
comparison to charge distributions whose fields you know.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 21-17
Example: The field of a uniformly charged
sphere
• INTERPRET: The situation has spherical symmetry.
• DEVELOP: Appropriate gaussian surfaces
are spheres.
• EVALUATE:

• The flux becomes   OE  dA  4p r E.
2
• Outside the sphere, the enclosed charge is
the total charge Q.
2
• Then 4p r E  Q  0 , so E 
Q
4p 0 r 2

kQ
.
2
r
• Thus the field outside a spherical charge distribution is identical to that of a point
charge.
• Inside the sphere, the enclosed charge is proportional to the volume
enclosed: qenclosed = (r3/R3)Q.
• Then 4πr2 E = (r3/R3)Q, so E 
Qr
.
4p 0 R 3
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Slide 21-18
A hollow spherical shell
• Applying Gauss’s law to a hollow spherical shell is
similar to that for a spherical charge, but now the
enclosed charge is zero.
• Therefore 4pr2E = 0, so the field inside the shell is zero.
• This can be understood in
terms of Coulomb’s law
because the inverse square
dependence of the electric
field results in cancellation
at any point in the shell by
the greater but more distant
charge and the lesser but
closer charge.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 21-19
Clicker question
• A spherical shell carries charge Q uniformly distributed
over its surface. If the charge on the shell doubles, what
happens to the electric field strength inside the shell?
A. The electric field strength is zero.
B. The electric field strength quadruples.
C. The electric field strength is halved.
D. The electric field strength doubles.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 21-20
Clicker question
• A spherical shell carries charge Q uniformly distributed
over its surface. If the charge on the shell doubles, what
happens to the electric field strength inside the shell?
A. The electric field strength is zero.
B. The electric field strength quadruples.
C. The electric field strength is halved.
D. The electric field strength doubles.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 21-21
Uses of Gauss’s law: symmetric distributions
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Slide 21-22
Line symmetry
• In line symmetry, the charge density depends only on the
perpendicular distance from a line, the axis of symmetry.
• This requires a charge distribution that is infinitely long.
• However, line symmetry is a good approximation for finite cylindrical
charge distributions with length much greater than diameter, at points close
to the charge.
• Applying Gauss’s law in line symmetry requires the use of a
cylindrical gaussian surface.
• The flux through this gaussian
surface is 2prLE.
• Applying Gauss’s law then shows
that the field outside any charge
distribution with line symmetry
has the 1/r dependence of a line
charge.
• The field inside a hollow charged
cylinder is zero.
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Slide 21-23
Plane symmetry
• In plane symmetry, the charge density depends only on the perpendicular
distance from a plane, the plane of symmetry.
• This requires a charge distribution that extends infinitely in two directions.
• However, plane symmetry is a good approximation for finite slabs of charge whose
thickness is much less than their extent in the other two dimensions, at points close to
the charge.
• Applying Gauss’s law in plane symmetry requires the use of a gaussian surface
that straddles the plane.
• The flux through this gaussian
surface is 2AE.
• Applying Gauss’s law then shows
that the field outside any charge
distribution with plane symmetry is
uniform and given by E =  /20,
where  is the surface charge density.
• This makes sense because the
symmetry precludes the field lines
spreading in any particular direction.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 21-24
Fields of arbitrary charge distributions
• Many real charge distributions can be
approximated by the simple distributions
considered in this chapter.
• In many cases one approximation applies
close to the distribution, another far away.
• Far from any finite-size distribution, the
field approaches that of a point charge.
• Far from any neutral distribution, the field
generally approaches that of a dipole.
• Near a flat, uniformly charged region the
field resembles the uniform field of a plane
charge.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 21-25
Clicker question
•
A square sheet with charge Q uniformly distributed
measures 1 m on each side? Which one of the following
expressions would you use to approximate the electric
field strength at a distance of 1 cm somewhere near the
center of the sheet?
A.
kQ
E
r
C.
E  kQ
r2
D.
E  2 0
E.
E  4 0
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 21-26
Clicker question
•
A square sheet with charge Q uniformly distributed
measures 1 m on each side? Which one of the following
expressions would you use to approximate the electric
field strength at a distance of 1 cm somewhere near the
center of the sheet?
A.
kQ
E
r
C.
E  kQ
r2
D.
E  2 0
E.
E  4 0
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Slide 21-27
Gauss’s law and conductors
• Charges in conductors are
free to move, and they do so
in response to an applied
electric field.
• If a conductor is allowed
to reach electrostatic
equilibrium, a condition in
which there is no net charge
motion, then charges
redistribute themselves to
cancel the applied field inside
the conductor.
• Therefore the electric field is
zero inside a conductor in
electrostatic equilibrium.
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Slide 21-28
Charged conductors
• Gauss’s law requires that
any free charge on a
conductor reside on the
conductor surface.
• When charge resides
inside a hollow, charged
conductor, then there may
be charge on the inside
surface of the conductor.
This charged conductor (shaded) carries a net charge of
1  C. There’s a 2-  C point charge within a hollow
cavity in the conductor. Notice how the charge
redistributes itself to be consistent with Gauss’s law.
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Slide 21-29
Clicker question
•
A conductor carries a net charge of +Q. A cavity within
the conductor contains a point charge of −Q. What is
the charge on the outer surface of the conductor in
electrostatic equilibrium?
A. Q
B. 0
C. Q
D. 2Q
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Slide 21-30
Clicker question
•
A conductor carries a net charge of +Q. A cavity within
the conductor contains a point charge of −Q. What is
the charge on the outer surface of the conductor in
electrostatic equilibrium?
A. Q
B. 0
C. Q
D. 2Q
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Slide 21-31
The field at a conductor surface
• The electric field at the
surface of a charged
conductor in electrostatic
equilibrium is perpendicular
to the surface.
• If it weren’t, charge would
move along the surface until
equilibrium was reached.
• Gauss’s law shows that the
field at the conductor surface
has magnitude E = /0,
where  is the local surface
charge density.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 21-32
Summary
• Gauss’s law is one of the four fundamental laws of
electromagnetism.
• In terms of the field line representation of electric fields, Gauss’s law
expresses the fact that the number of field lines emerging from any closed
surface is proportional to the net charge enclosed.

• Mathematically, Gauss’s law states that the electric flux   OE  dA
through any closed surface is proportional to the net charge enclosed:

O E  dA  qenclosed
0
• Gauss’s law embodies the inverse-square dependence of the point-charge
field, and is equivalent to Coulomb’s law.
• Gauss’s law is always true.
• It can be used to calculate electric fields in situations with sufficient symmetry:
spherical symmetry, line symmetry, or plane symmetry.
• Gauss’s law requires that any net charge on a charged conductor reside on the
conductor surface, and that the electric field at the conductor surface be
perpendicular to the surface.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 21-33
TrueTrue
A or False?
False F
1.
If the net electric flux out of a closed surface is zero, the electric
field must be zero everywhere on the surface.
False
2.
If the net electric flux out of a closed surface is zero, the charge
False
density must be zero everywhere inside the surface.
3.
The electric field is zero everywhere within the material of a
conductor in electrostatic equilibrium.
4.
The tangential component of the electric field is zero at all points
just outside the surface of a conductor in electrostatic
True
equilibrium.
False
The normal component of the electric field is the same at all points
5.
True
just outside the surface of a conductor in electrostatic
equilibrium.
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Slide 21-34
Chapter 21 Problem 55
A long solid rod 4.5 cm in radius carries a uniform volume
charge density. If the electric field strength at the
surface of the rod (not near either end) is 16 kN/C, what
is the volume charge density?
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Slide 21-35