Post on 22-Mar-2018
Chapter 28. Gauss’s Law
The nearly spherical shape
of the girl’s head determines
the electric field that causes
her hair to stream outward.
Using Guass’s law, we can
deduce electric fields,
particularly those with a
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particularly those with a
high degree of symmetry,
simply from the shape of the
charge distribution.
Chapter Goal: To
understand and apply
Gauss’s law.1
Topics:
• Symmetry
• The Concept of Flux
• Calculating Electric Flux
Chapter 28. Gauss’s Law
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• Gauss’s Law
• Using Gauss’s Law
• Conductors in Electrostatic Equilibrium
2
Chapter 28. Reading Quizzes
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Chapter 28. Reading Quizzes
3
The amount of electric field
passing through a surface is
called
A. Electric flux.
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A. Electric flux.
B. Gauss’s Law.
C. Electricity.
D. Charge surface density.
E. None of the above.
4
The amount of electric field
passing through a surface is
called
A. Electric flux.
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A. Electric flux.
B. Gauss’s Law.
C. Electricity.
D. Charge surface density.
E. None of the above.
5
Gauss’s law is useful for calculating
electric fields that are
A. symmetric.
B. uniform.
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B. uniform.
C. due to point charges.
D. due to continuous charges.
6
Gauss’s law is useful for calculating
electric fields that are
A. symmetric.
B. uniform.
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B. uniform.
C. due to point charges.
D. due to continuous charges.
7
Gauss’s law applies to
A. lines.
B. flat surfaces.
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B. flat surfaces.
C. spheres only.
D. closed surfaces.
8
Gauss’s law applies to
A. lines.
B. flat surfaces.
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B. flat surfaces.
C. spheres only.
D. closed surfaces.
9
The electric field inside a
conductor in electrostatic
equilibrium is
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A. uniform.
B. zero.
C. radial.
D. symmetric.
10
The electric field inside a
conductor in electrostatic
equilibrium is
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A. uniform.
B. zero.
C. radial.
D. symmetric.
11
Chapter 28. Basic Content and Examples
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Chapter 28. Basic Content and Examples
12
Symmetry and Electric Field
We say that a charge distribution is symmetric if there are a group of geometric transformations that do not cause any physical change.
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The Concept of Flux
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The Concept of Flux
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An electric field passing through a
surface
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16
The electric Flux of a constant electric
field
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17
The electric Flux of a Nonuniform
Electric Field
i
i
ii
i
e AE )(rr
∑∑ ⋅=Φ=Φ δδ
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∫ ⋅=Φsurface
e AdErr
The Flux Through a Curved Surface
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Electric Flux
Definition:
• Electric flux is the product of the
magnitude of the electric field and the
surface area, A, perpendicular to the field
• ΦE = EA
• The field lines may make some angle θ
with the perpendicular to the surface
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20
with the perpendicular to the surface
• Then ΦE = EA cos θ Errrr
EEAΦ =Φ =Φ =Φ =
Errrr
cosE
EA θθθθΦ =Φ =Φ =Φ =
normal
θθθθ
θθθθ
Electric Flux
Definition:
• Electric flux is the scalar product of electric
field and the vector
• Errrr
θθθθ
θθθθ
Arrrr
EAΦ =Φ =Φ =Φ =rrrrrrrr
Arrrr
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21
cos 0E
EA EA θθθθΦ = = >Φ = = >Φ = = >Φ = = >rrrrrrrr
orErrrr
cos 0E
EA EA θθθθΦ = = − <Φ = = − <Φ = = − <Φ = = − <rrrrrrrr
θθθθ
θθθθ
Arrrr
Electric Flux
• In the more general case, look at a
small flat area element
• In general, this becomes
cosE i i i i iE A θ E A∆Φ = ∆ = ⋅ ∆r r
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22
• The surface integral means the
integral must be evaluated over the
surface in question
• The units of electric flux will be
N.m2/C
0surface
limi
E i iA
E A E dA∆ →
Φ = ⋅ ∆ = ⋅∑ ∫r
r r r r
Electric Flux: Closed Surface
• A positive point charge, q, is
located at the center of a sphere of
radius r
• The magnitude of the electric field
everywhere on the surface of the
sphere is
Spherical
surface
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23
sphere is
E = keq / r2
• Electric field is perpendicular to the
surface at every point, so
has the same direction as
at every point.
Errrr
Arrrr
Electric Flux: Closed Surface
has the same direction as at every point.
Spherical
surface
Errrr
Arrrr
2e
qE k
r====
Then
i i i
i i
E dA E dAΦ = = =Φ = = =Φ = = =Φ = = =∑ ∑∑ ∑∑ ∑∑ ∑rrrrrrrr
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24
2 2
0 2
0
4 4
4
i i
e
e
qEA E r r k
r
qk q
π ππ ππ ππ π
ππππεεεε
= = = == = = == = = == = = =
= == == == =
∑ ∑∑ ∑∑ ∑∑ ∑
does not depend on rΦΦΦΦ
Gauss’s Law
BECAUSE2
1E
r∝∝∝∝
Electric Flux: Closed Surface
and have opposite directions at every point.
Spherical
surface
Errrr
Arrrr
2
| |e
qE k
r====
Then
E dA E dAΦ = = − =Φ = = − =Φ = = − =Φ = = − =∑ ∑∑ ∑∑ ∑∑ ∑rrrrrrrr
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25
2 2
0 2
0
| |4 4
4 | |
i i i
i i
e
e
E dA E dA
qEA E r r k
r
qk q
π ππ ππ ππ π
ππππεεεε
Φ = = − =Φ = = − =Φ = = − =Φ = = − =
= − = − = − == − = − = − == − = − = − == − = − = − =
= − == − == − == − =
∑ ∑∑ ∑∑ ∑∑ ∑
does not depend on rΦΦΦΦ
−−−−
Gauss’s Law
ONLY BECAUSE2
1E
r∝∝∝∝
EXAMPLE 28.1 The electric flux inside a
parallel-plate capacitor
QUESTION:
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26
EXAMPLE 28.1 The electric flux inside a
parallel-plate capacitor
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EXAMPLE 28.1 The electric flux inside a
parallel-plate capacitor
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EXAMPLE 28.1 The electric flux inside a
parallel-plate capacitor
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Gauss’s Law
Gauss’s law states
qin is the net charge inside the surface
E is the total electric field and may have contributions
inE
o
qE dA
εΦ = ⋅ =∫
r r
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30
E is the total electric field and may have contributions
from charges both inside and outside of the surface
q q
0
q
εεεεΦ =Φ =Φ =Φ =
0
q
εεεεΦ =Φ =Φ =Φ =
iArrrr
Errrr
Errrr
iArrrr
Gauss’s law states
qin is the net charge inside the surface
E is the total electric field and may have contributions
from charges both inside and outside of the surface
What is the electric flux here?
inE
o
qE dA
εΦ = ⋅ =∫
r r
q
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31
1q
0Φ =Φ =Φ =Φ =
2q
3q
4q
5q
7q6
q
Gauss’s Law:
Gauss’s law states
qin is the net charge inside the surface
E is the total electric field and may have contributions
from charges both inside and outside of the surface
inE
o
qE dA
εΦ = ⋅ =∫
r r
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32
from charges both inside and outside of the surface
q−−−−
0Φ =Φ =Φ =Φ =2q
q
4q
5q
7q6
q
Gauss’s Law: Applications
Chapter 28
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33
Gauss’s Law: Applications
Although Gauss’s law can, in theory, be solved to find
E for any charge configuration, in practice it is limited to
symmetric situations
To use Gauss’s law, you want to choose a Gaussian
surface over which the surface integral can be simplified
and the electric field determined
Take advantage of symmetry
Remember, the gaussian surface is a surface you
in
o
qE dA
εΦ = ⋅ =∫
r r
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34
Remember, the gaussian surface is a surface you
choose, it does not have to coincide with a real surface
1q
1 2 3 4
0
q q q q
εεεε
+ + ++ + ++ + ++ + +Φ =Φ =Φ =Φ =
2q
3q
4q
5q
6q
SYMMETRY:
Gauss’s Law: Point Charge
++++q
Errrr
Errrr
- direction - along the radius
Errrr
Gaussian Surface – Sphere
Only in this case the magnitude of
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35
Errrr
- depends only on radius, r
2e
qE k
r====
0
q
εεεεΦ =Φ =Φ =Φ = - Gauss’s Law
2
04
i i i
i i
E dA E dA EA E rππππΦ = = = =Φ = = = =Φ = = = =Φ = = = =∑ ∑∑ ∑∑ ∑∑ ∑rrrrrrrr
- definition of the Flux
Then 2
0
4q
r Eππππεεεε
====
Only in this case the magnitude of electric field is constant on the
Gaussian surface and the flux can be
easily evaluated
Gauss’s Law: Applications in
o
qE dA
εΦ = ⋅ =∫
r r
• Try to choose a surface that satisfies one or more of these conditions:
– The value of the electric field can be argued from symmetry to be
constant over the surface
– The dot product of E.dA can be expressed as a simple algebraic product
EdA because E and dA are parallel
– The dot product is 0 because E and dA are perpendicular
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– The field can be argued to be zero over the surface
correct Gaussian surface
++++
wrong Gaussian surface
Spherically Symmetric Charge Distribution
Gauss’s Law: Applications
The total charge is Q
• Select a sphere as the gaussian
SYMMETRY:
Errrr
- direction - along the radius
Errrr
- depends only on radius, r
Errrr
A∆∆∆∆rrrr
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37
• Select a sphere as the gaussian
surface
• For r >a
2 in
2 2
4
4
E
o o
e
o
q QE dA EdA πr E
ε ε
Q QE k
πε r r
Φ = ⋅ = = = =
= =
∫ ∫r r
The electric field is the same as for the point charge Q
Spherically Symmetric Charge Distribution
Gauss’s Law: Applications
2 24e
o
Q QE k
πε r r= = The electric field is the same as
for the point charge Q !!!!!
Q
Qa
≡≡≡≡For r > a
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38
≡≡≡≡
≡≡≡≡For r > a
Spherically Symmetric Charge Distribution
Gauss’s Law: Applications
SYMMETRY:
Errrr
- direction - along the radius
Errrr
- depends only on radius, r
Errrr
A∆∆∆∆rrrr
• Select a sphere as the gaussian
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39
• Select a sphere as the gaussian
surface, r < a
33
33
4
4 3
3
in
Q rq r Q Q
aa
ππππ
ππππ
= = <= = <= = <= = <
2 in
3
in
2 3 2 3
4
1
4
E
o
e e
o
qE dA EdA πr E
ε
q Qr QE k k r
πε r a r a
Φ = ⋅ = = =
= = =
∫ ∫r r
Spherically Symmetric Charge Distribution
Gauss’s Law: Applications
• Inside the sphere, E varies
linearly with r
E→ 0 as r→ 0
• The field outside the sphere is
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40
• The field outside the sphere is
equivalent to that of a point
charge located at the center of
the sphere
Field due to a thin spherical shell
Gauss’s Law: Applications
• Use spheres as the gaussian surfaces
• When r > a, the charge inside the surface is Q and
E = keQ / r2
• When r < a, the charge inside the surface is 0 and E = 0
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Field due to a thin spherical shell
Gauss’s Law: Applications
• When r < a, the charge inside the surface is 0 and E = 0
1r
1A∆∆∆∆
1 1q Aσσσσ∆ = ∆∆ = ∆∆ = ∆∆ = ∆
2 2q Aσσσσ∆ = ∆∆ = ∆∆ = ∆∆ = ∆
2
1 1A r∆ = ∆Ω∆ = ∆Ω∆ = ∆Ω∆ = ∆Ω 2
2 2A r∆ = ∆Ω∆ = ∆Ω∆ = ∆Ω∆ = ∆Ω
E∆∆∆∆rrrr
2q A rσ σσ σσ σσ σ∆ ∆ ∆Ω∆ ∆ ∆Ω∆ ∆ ∆Ω∆ ∆ ∆Ω
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422
A∆∆∆∆
2r
the same
solid angle1
E∆∆∆∆rrrr
2E∆∆∆∆rrrr
2
1 1 1
1 2 2 2
1 1 1
e e e e
q A rE k k k k
r r r
σ σσ σσ σσ σσσσσ
∆ ∆ ∆Ω∆ ∆ ∆Ω∆ ∆ ∆Ω∆ ∆ ∆Ω∆ = = = = ∆Ω∆ = = = = ∆Ω∆ = = = = ∆Ω∆ = = = = ∆Ω
2
2 2 2
2 2 2 2
2 2 2
e e e e
q A rE k k k k
r r r
σ σσ σσ σσ σσσσσ
∆ ∆ ∆Ω∆ ∆ ∆Ω∆ ∆ ∆Ω∆ ∆ ∆Ω∆ = = = = ∆Ω∆ = = = = ∆Ω∆ = = = = ∆Ω∆ = = = = ∆Ω
1 2E E∆ = ∆∆ = ∆∆ = ∆∆ = ∆
1 20E E∆ + ∆ =∆ + ∆ =∆ + ∆ =∆ + ∆ =
r rr rr rr r
2
1E
r∝∝∝∝Only because in Coulomb law
Field from a line of charge
Gauss’s Law: Applications
• Select a cylindrical Gaussian surface
– The cylinder has a radius of r and a
length of ℓ
• Symmetry:
E is constant in magnitude (depends only
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E is constant in magnitude (depends only
on radius r) and perpendicular to the
surface at every point on the curved part
of the surface
The end view
Field from a line of charge
Gauss’s Law: Applications
dArrrr The flux through this surface is 0
The flux through this surface:
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44The end view
( ) in2E
o
qE dA EdA E πr
εΦ = ⋅ = = =∫ ∫
r rl
( )2
22
o
e
o
λE πr
ε
λ λE k
πε r r
=
= =
ll
inq λ= l
Field due to a plane of charge
Gauss’s Law: Applications
• Symmetry:
E must be perpendicular to the plane and
must have the same magnitude at all
points equidistant from the plane
• Choose a small cylinder whose axis is
dArrrr
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45
• Choose a small cylinder whose axis is
perpendicular to the plane for the
gaussian surface
The flux through this surface is 0
Field due to a plane of charge
Gauss’s Law: Applications
dArrrr5
Arrrr
6Arrrr
3Arrrr
4Arrrr
2Arrrr
h
h
2Errrr
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46
The flux through this surface is 0
63
A
1Arrrr
1Errrr
1 1 2 22E A E A EA EA EAΦ = + = + =Φ = + = + =Φ = + = + =Φ = + = + =
r rr rr rr rr rr rr rr r
1 2E E E= == == == =
1 2A A A= == == == =
0 0
inq Aσσσσ
ε εε εε εε εΦ = =Φ = =Φ = =Φ = =
0
2A
EAσσσσ
εεεε====
02
Eσσσσ
εεεε==== does not depend on h
Gauss’s Law: Applications
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47
02
Eσσσσ
εεεε====
2e
E kr
λλλλ====
Conductors in Electric Field
Chapter 28
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48
Electric Charges: Conductors and Isolators
Electrical conductors are materials in which
some of the electrons are free electrons
These electrons can move relatively freely
through the material
Examples of good conductors include copper,
aluminum and silver
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49
Electrical insulators are materials in which all
of the electrons are bound to atoms
These electrons can not move relatively freely
through the material
Examples of good insulators include glass, rubber
and wood
Semiconductors are somewhere between
insulators and conductors
Electrostatic Equilibrium
Definition:
when there is no net motion of charge
within a conductor, the conductor is said to
be in electrostatic equilibrium
Because the electrons can move freely through the
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50
Because the electrons can move freely through the
material
no motion means that there are no electric forces
no electric forces means that the electric field
inside the conductor is 0
If electric field inside the conductor is not 0, then
there is an electric force and, from the second
Newton’s law, there is a motion of free electrons.
0E ≠≠≠≠rrrr
F qE====r rr rr rr r
F qE====r rr rr rr r
Conductor in Electrostatic Equilibrium
• The electric field is zero everywhere inside the conductor
• Before the external field is applied, free
electrons are distributed throughout the
conductor
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51
• When the external field is applied, the
electrons redistribute until the magnitude
of the internal field equals the magnitude
of the external field
• There is a net field of zero inside the
conductor
Conductor in Electrostatic Equilibrium
• If an isolated conductor carries a charge, the charge resides on its surface
Electric filed is 0,
so the net flux through
Gaussian surface is 0
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52
in
o
qE dA
εΦ = ⋅ =∫
r r
Then in 0q =
Conductor in Electrostatic Equilibrium
• The electric field just outside a charged conductor is perpendicular to the surface and has a magnitude of σ/εo
• Choose a cylinder as the gaussian surface
• The field must be perpendicular to the surface
– If there were a parallel component to E,
charges would experience a force and
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53
charges would experience a force and
accelerate along the surface and it would not be
in equilibrium
• The net flux through the gaussian surface is
through only the flat face outside the conductor
– The field here is perpendicular to the surface
• Gauss’s law:
E
o o
σA σEA and E
ε εΦ = = =
Conductor in Electrostatic Equilibrium
o
σE
ε=
0E =
0σ >
0σ <
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54
0σ >
0σ <
0σ <
Chapter 28. Summary Slides
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Chapter 28. Summary Slides
55
General Principles
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General Principles
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Important Concepts
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Important Concepts
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Important Concepts
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Applications
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Chapter 28. Questions
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Chapter 28. Questions
62
This box contains
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A. a net positive charge.
B. a net negative charge.
C. a negative charge.
D. a positive charge.
E. no net charge.
63
This box contains
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A. a net positive charge.
B. a net negative charge.
C. a negative charge.
D. a positive charge.
E. no net charge.
64
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The total electric flux through this box isA. 6 Nm2/C.
B. 4 Nm2/C.
C. 2 Nm2/C.
D. 1 Nm2/C.
E. 0 Nm2/C.
65Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
A. 6 Nm2/C.
B. 4 Nm2/C.
C. 2 Nm2/C.
D. 1 Nm2/C.
E. 0 Nm2/C.
The total electric flux through this box is
66