Download Chapter 28. Gauss`s Law

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
high degree of symmetry,
simply from the shape of the
charge distribution.
Chapter Goal: To
understand and apply
Gauss’s law.
Chapter 28. Gauss’s Law
Topics:
• Symmetry
• The Concept of Flux
• Calculating Electric Flux
• Gauss’s Law
• Using Gauss’s Law
• Conductors in Electrostatic Equilibrium
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The amount of electric field
passing through a surface is
called
Chapter 28. Reading Quizzes
A. Electric flux.
B. Gauss’s Law.
C. Electricity.
D. Charge surface density.
E. None of the above.
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Gauss’s law is useful for calculating
electric fields that are
The amount of electric field
passing through a surface is
called
A. symmetric.
B. uniform.
C. due to point charges.
D. due to continuous charges.
A. Electric flux.
B. Gauss’s Law.
C. Electricity.
D. Charge surface density.
E. None of the above.
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Gauss’s law is useful for calculating
electric fields that are
Gauss’s law applies to
A. lines.
B. flat surfaces.
C. spheres only.
D. closed surfaces.
A. symmetric.
B. uniform.
C. due to point charges.
D. due to continuous charges.
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The electric field inside a
conductor in electrostatic
equilibrium is
Gauss’s law applies to
A. lines.
B. flat surfaces.
C. spheres only.
D. closed surfaces.
A. uniform.
B. zero.
C. radial.
D. symmetric.
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The electric field inside a
conductor in electrostatic
equilibrium is
Chapter 28. Basic Content and Examples
A. uniform.
B. zero.
C. radial.
D. symmetric.
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Symmetry and Electric Field
The Concept of Flux
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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An electric field passing through a
surface
The Concept of Flux
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The electric Flux of a constant electric
field
The electric Flux of a Nonuniform
Electric Field
r
r
Φ e = ∑ δ Φ i = ∑ Ei ⋅ (δ A)i
i
Φe =
i
r r
E
∫ ⋅ dA
surface
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Electric Flux
The Flux Through a Curved Surface
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
• Then ΦE = EA cos θ
r
E
normal
θ
θ
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r
E
Φ = EA cos θ
E
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Φ E = EA
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Electric Flux
Electric Flux
Definition:
• Electric flux is the scalar product
r of electric
A
field and the vector
rr
• Φ = EA
• In the more general case, look at a
small flat area element
r
A
θ
r
r
∆Φ E = Ei ∆Ai cos θi = Ei ⋅ ∆Ai
r
E
• In general, this becomes
θ
r
Φ E = lim
r
rr
Φ E = EA = EA cos θ > 0
or
r
E
r
θ
A
∆Ai →0
θ
rr
Φ E = EA = − EA cos θ < 0
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∑E
i
r
⋅ ∆Ai =
r r
E ⋅ dA
∫
surface
• The surface integral means the
integral must be evaluated over the
surface in question
• The units of electric flux will be
N.m2/C
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Electric Flux: Closed Surface
Electric Flux: Closed Surface
r
E
• 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
r
has the same direction as A at every point.
E = ke
Spherical
surface
q
r2
Then
i
E = keq / r2
i
= EA0 = E 4π r 2 = 4π r 2 ke
• Electric field is perpendicular to the
surface at every point, so
= 4π ke q =
r
r
A
has
the
same
direction
as
E
at every point.
Spherical
surface
r r
Φ = ∑ Ei dAi =E ∑ dAi =
q
ε0
q
=
r2
Gauss’s Law
Φ does not depend on r
BECAUSE
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1
2
24 r
E∝
Electric Flux: Closed Surface
r
E
EXAMPLE 28.1 The electric flux inside a
parallel-plate capacitor
r
and A have opposite directions at every point.
E = ke
|q|
r2
QUESTION:
Spherical
surface
Then
r r
Φ = ∑ Ei dAi = − E ∑ dAi =
i
i
= − EA0 = − E 4π r 2 = −4π r 2 ke
= −4π ke | q |=
q
ε0
−
|q|
=
r2
Gauss’s Law
Φ does not depend on r
ONLY BECAUSE
E ∝25
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EXAMPLE 28.1 The electric flux inside a
parallel-plate capacitor
1
r2
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EXAMPLE 28.1 The electric flux inside a
parallel-plate capacitor
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Gauss’s Law
EXAMPLE 28.1 The electric flux inside a
parallel-plate capacitor
Gauss’s law states
ΦE =
r r q
E
∫ ⋅ dA = εino
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
r
Ai
r
E
q
r
E
r
Ai
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ΦE =
q
Φ=
ε0
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q
ε 0 30
Gauss’s Law:
What is the electric flux here?
Gauss’s law states
Φ=
q
r r q
E
∫ ⋅ dA = εino
r
Gauss’s law states
ΦE =
r
qin
∫ E ⋅ dA = ε
o
qin is the net charge inside the surface
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
q5
q2
E is the total electric field and may have contributions
from charges both inside and outside of the surface
q1
q5
q2
Φ=0
q
Φ=0
q3
−q
q4
q6
q7
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q4
q6
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q7
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Chapter 28
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
choose, it does not have to coincide with a real surface
Gauss’s Law: Applications
r
Φ=
r
qin
∫ E ⋅ dA = ε
o
q5
Φ=
q1 + q2 + q3 + q4
q6
q2
ε0
q1
q3
q4
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Gauss’s Law: Point Charge
Gauss’s Law: Applications
r
E
r
E - direction - along the radius
r
E - depends only on radius, r
ε0
– The dot product of E.dA can be expressed as a simple algebraic product
EdA because E and dA are parallel
Gaussian Surface – Sphere
Only in this case the magnitude of
electric field is constant on the
Gaussian surface and the flux can be
easily evaluated
– The dot product is 0 because E and dA are perpendicular
– The field can be argued to be zero over the surface
correct Gaussian surface
wrong Gaussian surface
- Gauss’s Law
r r
Φ = ∑ Ei dAi =E ∑ dAi =EA0 = E 4π r 2
i
- definition of the Flux
+
i
q
r2
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Then
qin
– The value of the electric field can be argued from symmetry to be
constant over the surface
SYMMETRY:
q
r
∫ E ⋅ dA = ε
• Try to choose a surface that satisfies one or more of these conditions:
+
q
Φ=
r
Φ=
q
0
= 4π r 2 E
E = ke
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o
Gauss’s Law: Applications
Gauss’s Law: Applications
Spherically Symmetric Charge Distribution
The total charge is Q
SYMMETRY:
r
∆A
r
E - direction - along the radius
r
E - depends only on radius, r
r
E
Spherically Symmetric Charge Distribution
E=
Q
Q
= ke 2
2
4πεo r
r
The electric field is the same as
for the point charge Q !!!!!
Q
a
• Select a sphere as the gaussian
surface
• For r >a
ΦE =
r r
E
∫ ⋅ dA =
∫ EdA = 4πr
2
E=
For r > a
≡
Q
qin Q
=
εo ε o
For r > a
Q
Q
= ke 2
E=
2
4πεo r
r
≡
The electric field is the same as
for the point charge Q
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Gauss’s Law: Applications
Gauss’s Law: Applications
Spherically Symmetric Charge Distribution
r
∆A
SYMMETRY:
r
E - direction - along the radius
r
E - depends only on radius, r
Spherically Symmetric Charge Distribution
r
E
• Inside the sphere, E varies
linearly with r
E → 0 as r → 0
• Select a sphere as the gaussian
surface, r < a
qin =
Q
4 3
πa
3
• The field outside the sphere is
equivalent to that of a point
charge located at the center of
the sphere
4 3
r3
πr = Q 3 < Q
a
3
ΦE =
E=
r r
E
∫ ⋅ dA =
∫ EdA = 4πr
2
E=
qin
Qr 3 1
Q
=
= ke 3 r
k
e
2
3
2
a r
a
4πεo r
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qin
εo
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Gauss’s Law: Applications
Gauss’s Law: Applications
Field due to a thin spherical shell
Field due to a thin spherical shell
• When r < a, the charge inside the surface is 0 and E = 0
• 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
∆A1
r1
r
∆E 2
the same
solid angle
∆q2 = σ∆A2
∆A1 = r12 ∆Ω
∆A2 = r22 ∆Ω
∆E1 = ke
∆q1
σ∆ A
σ r 2 ∆Ω
= ke 2 1 = ke 1 2 = keσ∆Ω
2
r1
r1
r1
∆E 2 = k e
∆q2
σ∆ A
σ r 2 ∆Ω
= ke 2 2 = ke 2 2 = keσ∆Ω
2
r2
r2
r2
r
∆E1
r2
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∆q1 = σ∆A1
∆A
∆E1 = ∆E2
Only because in Coulomb law
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Gauss’s Law: Applications
Gauss’s Law: Applications
Field from a line of charge
r
r
∆E1 + ∆E2 = 0
E∝
1
42
r2
Field from a line of charge
r
dA
• Select a cylindrical Gaussian surface
The flux through this surface is 0
– The cylinder has a radius of r and a
length of ℓ
The flux through this surface:
• Symmetry:
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
ΦE =
r r
E
∫ ⋅ dA =
qin
∫ EdA = E ( 2πr l ) = ε
qin = λl
E ( 2πr l ) =
E=
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The end view
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2008view
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λl
εo
λ
λ
= 2ke
2πεo r
r
44
Gauss’s Law: Applications
Gauss’s Law: Applications
Field due to a plane of charge
r
E2
• Symmetry:
r
A4
E must be perpendicular to the plane and
must have the same magnitude at all
Field due to a plane of charge
r
A2
h
r
A5
h
r
A6
r
dA
points equidistant from the plane
r
A3
• Choose a small cylinder whose axis is
perpendicular to the plane for the
gaussian surface
r
E1
The flux through this surface is 0
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r
dA
r
A1
E1 = E2 = E
A1 = A2 = A
r r
r r
Φ = E1 A1 + E2 A2 = EA + EA = 2 EA
The flux through this surface is 0
σA
σA
σ
=
2 EA =
E=
does not depend on h
ε
ε
ε
2
ε0
0
0 as Pearson Addison-Wesley.
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Φ=
qin
Gauss’s Law: Applications
46
Chapter 28
Conductors in Electric Field
E=
E = 2ke
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σ
2ε 0
λ
r
47
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Electric Charges: Conductors and Isolators
Electrostatic Equilibrium
Electrical conductors are materials in which
some of the electrons are free electrons
Definition:
when there is no net motion of charge
within a conductor, the conductor is said to
be in electrostatic equilibrium
These electrons can move relatively freely
through the material
Examples of good conductors include copper,
aluminum and silver
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
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
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49
r
If electric field inside the conductor is not 0, E ≠ 0 then
r
r
there is an electric force F = qE and, from the second
Newton’s
law,
there
is aInc.,
motion
freeAddison-Wesley.
electrons.
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Education,
publishingof
as Pearson
r
r
F = qE
50
Conductor in Electrostatic Equilibrium
Conductor in Electrostatic Equilibrium
• The electric field is zero everywhere inside the
conductor
• If an isolated conductor carries a charge, the charge resides
on its surface
Electric filed is 0,
• Before the external field is applied, free
electrons are distributed throughout the
conductor
so the net flux through
Gaussian surface is 0
• When the external field is applied, the
electrons redistribute until the magnitude
of the internal field equals the magnitude
of the external field
Φ=
Then
• There is a net field of zero inside the
conductor
r r q
E
∫ ⋅ dA = εino
qin = 0
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Conductor in Electrostatic Equilibrium
Conductor in Electrostatic Equilibrium
• The electric field just outside a charged conductor is
perpendicular to the surface and has a magnitude of σ/εo
E=
σ >0
σ
εo
σ <0
• Choose a cylinder as the gaussian surface
• The field must be perpendicular to the surface
E =0
– If there were a parallel component to E,
charges would experience a force and
accelerate along the surface and it would not be
in equilibrium
σ >0
σ <0
• 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:
σA
σ
and E =
ε
εo
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Φ E = EA =
σ <0
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General Principles
Chapter 28. Summary Slides
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General Principles
Important Concepts
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Important Concepts
Important Concepts
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Applications
Chapter 28. Questions
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This box contains
This box contains
A. a net positive charge.
B. a net negative charge.
C. a negative charge.
D. a positive charge.
E. no net charge.
A. a net positive charge.
B. a net negative charge.
C. a negative charge.
D. a positive charge.
E. no net charge.
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The total electric flux through this box is
The total electric flux through this box is
Nm2/C.
A. 6 Nm2/C.
B. 4 Nm2/C.
C. 2 Nm2/C.
D. 1 Nm2/C.
E. 0 Nm2/C.
A. 6
B. 4 Nm2/C.
C. 2 Nm2/C.
D. 1 Nm2/C.
E. 0 Nm2/C.
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