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Chapter 22
Gauss’s Law
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22-1 Electric Flux
Flux through a closed surface:
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Think in terms of flux lines:
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22-2 Gauss’s Law
The net number of field lines through the surface
is proportional to the charge enclosed, and also
to the flux, giving Gauss’s law:
This can be used to find the electric field in
situations with a high degree of symmetry.
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22-2 Gauss’s Law
For a point charge,
Spherical
Therefore,
Solving for E gives the result
we expect from Coulomb’s
law:
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22-2 Gauss’s Law
Using Coulomb’s law to
evaluate the integral of
the field of a point charge
over the surface of a
sphere surrounding the
charge gives:
Looking at the arbitrarily shaped surface A2, we
see that the same flux passes through it as
passes through A1. Therefore, this result should
be valid for any closed surface.
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22-2 Gauss’s Law
Finally, if a gaussian surface encloses several
point charges, the superposition principle shows
that:
Therefore, Gauss’s law is valid for any charge
distribution. Note, however, that it only refers to
the field due to charges within the gaussian
surface – charges outside the surface will also
create fields.
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22-2 Gauss’s Law
Conceptual Example 22-2: Flux from Gauss’s law.
Consider the two gaussian surfaces, A1 and A2, as
shown. The only charge present is the charge Q at
the center of surface A1. What is the net flux through
each surface, A1 and A2?
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22-3 Applications of Gauss’s Law
Example 22-3: Spherical
conductor.
A thin spherical shell of
radius r0 possesses a
total net charge Q that is
uniformly distributed on it.
Determine the electric
field at points (a) outside
the shell, and (b) within
the shell. (c) What if the
conductor were a solid
sphere?
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Spherical
22-3 Applications of Gauss’s Law
Example 22-4: Solid
sphere of charge.
An electric charge Q
is distributed uniformly
throughout a
nonconducting sphere
of radius r0.
Determine the electric
field (a) outside the
sphere
(r > r0) and (b) inside
the sphere (r < r0).
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Spherical/radial
22-3 Applications of Gauss’s Law
Example 22-5:
Nonuniformly charged
solid sphere.
Suppose the charge
density of a solid sphere is
given by ρE = αr2, where α
is a constant. (a) Find α in
terms of the total charge Q
on the sphere and its
radius r0. (b) Find the
electric field as a function
of r inside the sphere.
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22-3 Applications of Gauss’s Law
Example 22-6: Long uniform line of charge.
A very long straight wire possesses a uniform
positive charge per unit length, λ. Calculate the
electric field at points near (but outside) the
wire, far from the ends.
Linear,
Cylindrical
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22-3 Applications of Gauss’s Law
Example 22-7: Infinite
plane of charge.
Charge is distributed
uniformly, with a surface
charge density σ (σ =
charge per unit area =
dQ/dA) over a very large
but very thin nonconducting
flat plane surface.
Determine the electric field
at points near the plane.
Planar & cylindrical
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22-3 Applications of Gauss’s Law
Example 22-8: Electric
field near any
conducting surface.
Show that the electric
field just outside the
surface of any good
conductor of arbitrary
shape is given by
E = σ/ε0
where σ is the surface
charge density on the conductor’s
surface at that point.
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22-3 Applications of Gauss’s Law
The difference between the electric field outside
a conducting plane of charge and outside a
nonconducting plane of charge can be thought
of in two ways:
1. The field inside the conductor is zero, so the
flux is all through one end of the cylinder.
2. The nonconducting plane has a total charge
density σ, whereas the conducting plane has a
charge density σ on each side, effectively giving
it twice the charge density.
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22-3 Applications of Gauss’s Law
Conceptual Example
22-9: Conductor with
charge inside a cavity.
Suppose a conductor
carries a net charge
+Q and contains a
cavity, inside of
which resides a point
charge +q. What can
you say about the
charges on the inner
and outer surfaces of
the conductor?
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22-3 Applications of Gauss’s Law
Procedure for Gauss’s law problems:
1. Identify the symmetry, and choose a gaussian
surface that takes advantage of it (with surfaces
along surfaces of constant field).
2. Draw the surface.
3. Use the symmetry to find the direction of E
E.
4. Evaluate the flux by integrating.
5. Calculate the enclosed charge.
6. Solve for the field.
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22-4 Experimental Basis of Gauss’s
and Coulomb’s Laws
In the experiment shown,
Gauss’s law predicts that the
charge on the ball flows onto
the surface of the cylinder when
they are in contact. This can be
tested by measuring the charge
on the ball after it is removed –
it should be zero.
Copyright © 2009 Pearson Education, Inc.
Summary of Chapter 22
• Electric flux:
• Gauss’s law:
• Gauss’s law can be used to calculate the field in
situations with a high degree of symmetry.
• Gauss’s law applies in all situations, and
therefore is more general than Coulomb’s law.
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Chapter 23
Electric Potential
Copyright © 2009 Pearson Education, Inc.
Units of Chapter 23
• Electric Potential Energy and Potential
Difference
• Relation between Electric Potential and
Electric Field
• Electric Potential Due to Point Charges
• Potential Due to Any Charge Distribution
• Equipotential Surfaces
• Electric Dipole Potential
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Units of Chapter 23
• E Determined from V
• Electrostatic Potential Energy; the Electron
Volt
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23-1 Electrostatic Potential Energy
and Potential Difference
The electrostatic force is
conservative – potential
energy can be defined.
Change in electric potential
energy is negative of work
done by electric force:
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ConcepTest 23.1b Electric Potential Energy II
A proton and an electron are in
a constant electric field created
by oppositely charged plates.
You release the proton from
the positive side and the
electron from the negative side.
Which has the larger
acceleration?
1) proton
2) electron
3) both feel the same acceleration
4) neither – there is no acceleration
5) they feel the same magnitude
acceleration but opposite
direction
Electron
electron
-
+

E
Proton
proton
ConcepTest 23.1b Electric Potential Energy II
A proton and an electron are in
a constant electric field created
by oppositely charged plates.
You release the proton from
the positive side and the
electron from the negative side.
Which has the larger
acceleration?
1) proton
2) electron
3) both feel the same acceleration
4) neither – there is no acceleration
5) they feel the same magnitude
acceleration but opposite
direction
Since F = ma and the electron is much less
Electron
electron
-
massive than the proton, the electron
experiences the larger acceleration.
+

E
Proton
proton
23-1 Electrostatic Potential Energy
and Potential Difference
Electric potential is defined as potential
energy per unit charge:
Unit of electric potential: the volt (V):
1 V = 1 J/C.
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23-1 Electrostatic Potential Energy
and Potential Difference
Only changes in potential can be measured,
allowing free assignment of V = 0:
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23-1 Electrostatic Potential Energy
and Potential Difference
Analogy between gravitational and electrical
potential energy:
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23-1 Electrostatic Potential Energy
and Potential Difference
Conceptual Example 23-1: A negative charge.
Suppose a negative charge, such as an electron, is
placed near the negative plate at point b, as shown
here. If the electron is free to move, will its electric
potential energy increase or decrease? How will
the electric potential change?
Copyright © 2009 Pearson Education, Inc.
23-1 Electrostatic Potential Energy
and Potential Difference
Electrical sources
such as batteries and
generators supply a
constant potential
difference. Here are
some typical potential
differences, both
natural and
manufactured:
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23-2 Relation between Electric
Potential and Electric Field
The general relationship
between a conservative force
and potential energy:
Substituting the
potential difference
and the electric field:
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23-2 Relation between Electric
Potential and Electric Field
The simplest case is a uniform field:
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23-2 Relation between Electric
Potential and Electric Field
Example 23-3: Electric field
obtained from voltage.
Two parallel plates are
charged to produce a potential
difference of 50 V. If the
separation between the plates
is 0.050 m, calculate the
magnitude of the electric field
in the space between the
plates.
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23-2 Relation between Electric
Potential and Electric Field
Example 23-4:
Charged conducting
sphere.
Determine the
potential at a distance
r from the center of a
uniformly charged
conducting sphere of
radius r0 for (a) r > r0,
(b) r = r0, (c) r < r0. The
total charge on the
sphere is Q.
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23-2 Relation between Electric
Potential and Electric Field
The previous example
gives the electric
potential as a function of
distance from the surface
of a charged conducting
sphere, which is plotted
here, and compared with
the electric field:
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23-3 Electric Potential Due to Point
Charges
To find the electric potential due to a point
charge, we integrate the field along a field line:
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23-3 Electric Potential Due to Point
Charges
Setting the potential to zero at r = ∞ gives the
general form of the potential due to a point
charge:
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23-3 Electric Potential Due to Point
Charges
Example: Work required to bring two positive
charges close together.
What minimum work must be done by an
external force to bring a proton q = 1.60×10-19
C from a great distance away (take r = ∞) to a
point 1.00×10-15 m from another proton?
Copyright © 2009 Pearson Education, Inc.
23-3 Electric Potential Due to Point
Charges
Example: Work required to bring two positive
charges close together.
What minimum work must be done by an
external force to bring a proton q = 1.60×10-19
C from a great distance away (take r = ∞) to a
point 1.00×10-15 m from another proton?
W = ke2/r = (9×109) (1.6×10-19)2/ (1×10-15)
= 2.3×10-13 J = 2.3×10-13 J
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ConcepTest 23.3 Electric Potential
1) V > 0
What is the electric
potential at point B?
2) V = 0
3) V < 0
A
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B
ConcepTest 23.3 Electric Potential
1) V > 0
What is the electric
potential at point B?
2) V = 0
3) V < 0
Since Q2 and Q1 are equidistant
from point B, and since they have
equal and opposite charges, the
total potential is zero.
Follow-up: What is the potential
at the origin of the x y axes?
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A
B
23-3 Electric Potential Due to Point
Charges
Example 23-7: Potential above two charges.
Calculate the electric potential (a) at point A in
the figure due to the two charges shown.
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23-4 Potential Due to Any Charge
Distribution
The potential due to an arbitrary charge
distribution can be expressed as a sum or
integral (if the distribution is continuous):
or
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23-4 Potential Due to Any Charge
Distribution
Example 23-8:
Potential due to a
ring of charge.
A thin circular ring of
radius R has a
uniformly distributed
charge Q. Determine
the electric potential
at a point P on the
axis of the ring a
distance x from its
center.
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23-4 Potential Due to Any Charge
Distribution
Example 23-9:
Potential due to a
charged disk.
A thin flat disk, of
radius R0, has a
uniformly
distributed charge
Q. Determine the
potential at a point
P on the axis of the
disk, a distance x
from its center.
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ConcepTest 23.5 Equipotential Surfaces
Which of these configurations gives V = 0 at
+2mC
+2mC
+2mC
+1mC
-2mC
+1mC
all
points
on the
x axis?
x
-1mC
-2mC
x
1)
+1mC
-1mC
2)
4) all of the above
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-2mC
-1mC
x
3)
5) none of the above
ConcepTest 23.5 Equipotential Surfaces
Which of these configurations gives V = 0 at
+2mC
+2mC
+2mC
+1mC
-2mC
+1mC
all
points
on the
x axis?
x
-1mC
-2mC
x
-2mC
-1mC
1)
x
2)
4) all of the above
+1mC
-1mC
3)
5) none of the above
Only in case (1), where opposite charges lie
directly across the x axis from each other, do
the potentials from the two charges above the
x axis cancel the ones below the x axis.
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23-5 Equipotential Surfaces
An equipotential is a line
or surface over which the
potential is constant.
Electric field lines are
perpendicular to
equipotentials.
The surface of a conductor
is an equipotential
(E// =0 → ∂V/∂s// = 0)
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23-5 Equipotential Surfaces
Equipotential surfaces are always
perpendicular to field lines; they are
always closed surfaces (unlike field lines,
which begin and end on charges).
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23-5 Equipotential Surfaces
A gravitational analogy to equipotential surfaces
is the topographical map – the lines connect
points of equal gravitational potential (altitude).
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23-6 Electric Dipole Potential
The potential due to an
electric dipole is just the sum
of the potentials due to each
charge, and can be calculated
exactly. For distances large
compared to the charge
separation:
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23-7 E Determined from V
If we know the field, we can determine the
potential by integrating. Inverting this
process, if we know the potential, we can
find the field by differentiating:
This is a vector differential equation;
here it is in component form:
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23-7 E Determined from V
Example 23-11: E for ring and disk.
Use electric potential to determine the electric
field at point P on the axis of (a) a circular ring
of charge and (b) a uniformly charged disk.
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Homework Assignment # 3
Chapter 22 – 10, 22, 34
Chapter 23 – 14, 30, 36
Tentative HW # 4:
Chapter 24 – 6, 16, 28, 46, 60, 82
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