Fundamentals of Physics
Chapter 23 Gauss Law
1.
2.
3.
4.
5.
6.
7.
8.
9.
A New Look at Coulomb s Law
Flux
Flux of an Electric Field
Gauss Law
Gauss Law & Coulomb s Law
A Charged Isolated Conductor
Applying Gauss Law: Cylindrical Symmetry
Applying Gauss Law: Planar Symmetry
Applying Gauss Law: Spherical Symmetry
Review & Summary
Questions
Exercises & Problems
2006
Physics 2112
Fundamentals of Physics
Chapter 23
1
A New Look at Coulomb s Law
Electrostatics
Coulomb s Law
Electric field (force) due to point charges
(difficult integrations)
E
k
dq
r2
Gauss Law
Electric field due to distributed charges.
(easy to calculate for highly symmetric configurations)
Gauss law relates the electric fields at points on a closed surface to the net
charge enclosed by the surface.
Gaussian Surface
Coulomb's Law and Gauss Law are equivalent
for static charges.
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Fundamentals of Physics
Chapter 23
2
Flux of an Electric Field
The flux of an electric field through a surface:
EA
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Physics 2112
E A cos
Fundamentals of Physics
Chapter 23
3
Flux of an Electric Field
The flux of an electric field through a surface:
E Ai cos
i
E
Ai
i
2
N
m
Unit of flux :
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Physics 2112
Fundamentals of Physics
C
Chapter 23
4
Flux of an Electric Field
The flux of an electric field through a
Gaussian surface (N m2 / C)
E
oE
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A
dA
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Fundamentals of Physics
Chapter 23
5
Example 1
Gaussian surface in the form of a cylinder
E parallel to axis of cylinder
Flux through closed surface?
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Fundamentals of Physics
Chapter 23
6
Checkpoint 1
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Physics 2112
Fundamentals of Physics
Chapter 23
7
Flux through the net = ???
E dA
a2 E
a2 E
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Fundamentals of Physics
Chapter 23
8
Flux of an Electric Field
The flux of an electric field through a Gaussian surface depends on the net
charge inside the surface.
E
qnet
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0
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A
qnet
Fundamentals of Physics
q
Chapter 23
9
Gauss Law
The net flux F of an electric field through a closed
surface is related to the net charge qenc enclosed by
that surface:
Gauss Law:
0
0
oE
dA
q
qenc
Consider the Gaussian surfaces in the figure:
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Physics 2112
Fundamentals of Physics
Chapter 23
10
Checkpoint 2
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Fundamentals of Physics
Chapter 23
11
Example 3
What is the net electric flux through the
surface if q1 = q4 =3.1 nC, q2 = q5 = -5.9 nC
and q3 = -3.1 nC?
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Fundamentals of Physics
Chapter 23
12
Gauss Law & Coulomb s Law
Gauss Law is equivalent to Coulomb s Law.
Consider the special case of a point charge enclosed by a concentric
Gaussian surface:
spherical
Gauss s Law:
E dA
qenc
E dA
qenc
E dA
qenc
2
E
4
r
0
qenc
0
0
0
Coulomb s Law!
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Physics 2112
Fundamentals of Physics
Chapter 23
13
Flux through the surfaces?
q enc
Gauss Law:
0
q
q
S1 :
q
q
S2 :
0
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Physics 2112
S3 :
?
q
S4 :
?
0
S5 :
?
Fundamentals of Physics
Chapter 23
14
Flux through the right-hand face?
L = 1.40 m
Uniform E
E
E
L
6.00 i
E
2.00 j
E
3.00 i
L
A
4.00k
A
L2 j
E A
1.40
2
j
L
E
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6.00 i
E
2.00 j
E
3.00 i
A
A
4.00k
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2
6.00 i
1.40 j
2.00 j
A
2
1.40 j
0
3.92
N m2
C
0
Fundamentals of Physics
Chapter 23
15
Checkpoint 3
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Fundamentals of Physics
Chapter 23
16
A Charged Isolated Conductor
Consider a Gaussian surface just inside the surface of a charged lump of copper hung by a
thread:
No electric field inside.
Charges in the conductor would move until the field was zero.
No flux through surface.
E = 0
No net charge inside the Gaussian surface.
By Gauss Law
0
E dA
qenc
Conclusion: If an excess charge is placed on an isolated conductor, that
amount of charge will move entirely to the surface of the conductor.
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Fundamentals of Physics
Chapter 23
17
A Charged Isolated Conductor with a Cavity
Consider a Gaussian surface around the
cavity inside the lump of copper hung by a
thread:
No electric field inside conductor.
No flux through surface.
No net charge inside.
If an excess charge is placed on an isolated conductor, that amount of charge will
move entirely to the surface of the conductor.
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Fundamentals of Physics
Chapter 23
18
Conductor Completely Removed!
Consider the case where the hollow cavity and
the surrounding Gaussian surface grow until
they are out to the edge of the conductor:
Still no electric field inside!
No flux through surface.
No net charge inside.
The conductor just serves as a medium to allow
the electrons to distribute themselves.
Conclusion: If an excess charge is placed on an isolated conducting shell, the
electric field inside the shell will be zero.
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Fundamentals of Physics
Chapter 23
19
The External Electric Field
Electric Field Outside a Charged Conductor
Surface Charge Density ( )
varies over the surface
Except for a spherical conductor.
Gaussian Surface
Right circular cylinder with cross-section A.
Side normal to the conductor surface (flux = 0)
One end-cap inside the conductor (E=0)
One end-cap outside the conductor (flux = E A)
E A = qenc =
A
E
0
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Fundamentals of Physics
Chapter 23
20
Example 4
Spherical metal shell of inner radius R
If the shell is electrically neutral, what are the
induced charges on it inner and outer radius?
Are the charges uniformly distributed?
What are the field patterns inside and outside the
shell?
No electric field inside a conductor.
No flux through the Gaussian surface.
Gauss Law
no net charge inside
qenc
0 E dA
A positive charge induced on the inner surface of the sphere that cancels
the negative charge inside.
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Fundamentals of Physics
Chapter 23
21
Sample Problem
Positive Point Charge Inside a Neutral Spherical Metal Shell
The position of the charge inside the conducting
sphere does not matter!
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Fundamentals of Physics
Chapter 23
22
Checkpoint 4
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Fundamentals of Physics
Chapter 23
23
Gauss s Law: Cylindrical Symmetry
Gauss s Law:
Symmetry
0
E dA
qenc
E is radial outward
E dA
E A cos 00
E 2 rh 1
qenc
E
h
2
0
r
Approximates the electric field due to a finite line of charge at points not too near
the ends relative to the distance from the line of charge.
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Fundamentals of Physics
Chapter 23
24
a long, thin-walled, charged metal cylinder
Cylindrical Gaussian Surface:
E(r)
= 2.0 x 10-8 C/m
Maximum E?
E dA
E dA
E 2 rL
Gauss Law:
r > R:
qenc
L
0
qenc
Emax
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Physics 2112
0
E R
E
2 rLE
L
2
0
r
0
1.2 10 4
Fundamentals of Physics
qenc
0
E
r < R:
r2
2 0
N
C
Chapter 23
25
Electric Field due to a Charged Disk
Plastic Disk of + Surface Charge:
= charge density
coulomb / meter2
z dq
dE
4
dq
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z
r
dA
z
E
E
0
2
4
2
1
0
3
2
2 r dr
R
0
2
0
2 r dr
z2
r2
3
2
z
z2
R2
Physics 2112
Fundamentals of Physics
Chapter 23
26
Electric Field due to an Infinite Plane of Charge
Consider a very large disk:
E
2
R
E
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1
0
z
z2
R2
z
2
0
Physics 2112
Fundamentals of Physics
Chapter 23
27
Gauss Law: Planar Symmetry
Positive Charge Density on an Infinite Non
Conducting Sheet
0
0
E dA
EA
qenc
EA
E
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A
2
0
Fundamentals of Physics
Chapter 23
28
Example 6
parallel non-conducting sheets
(+) = 6.8 mC/m2
(-) = 4.3 mC/m2
Find E to left, between, and to the
right of the sheets
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Fundamentals of Physics
Chapter 23
29
Electric Field?
Two parallel non-conducting sheets
E above, below, between?
E
0
E
E
0
2
2
0
0
Infinite Non-Conducting Sheet:
E
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Physics 2112
2
Fundamentals of Physics
0
Chapter 23
30
Gauss Law: Planar Symmetry
Two Conducting Plates:
E
1
0
E
1
0
Between 2 Plates:
E
2
Capacitor
1
0
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Fundamentals of Physics
Chapter 23
31
= ?
m = 1.0 mg
q = 2.0 x 10-8 C
= 300
=?
E
2
T
Fq
0
qE
T sin
T cos
Fq
0
mg
0
mg
2
0
m g tan
q
5.0 10
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Physics 2112
Fundamentals of Physics
9C
m2
Chapter 23
32
Gauss Law: Spherical symmetry
Spherical Shell of Total Charge q and Radius R:
S2: Spherical Shell r > R
E 4 r2 = qenc
E = q / (4
r2 )
S1: Spherical Shell r < R
E = 0
A shell of uniform charge attracts or repels a charged particle that is outside the shell as if
all the shell s charge were concentrated at the center of the shell. (Coulomb s Law)
A shell of uniform charge exerts no electrostatic force on a charged particle that is located
inside the shell.
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Fundamentals of Physics
Chapter 23
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Gauss Law: Spherical symmetry
Spherical Distribution of Charge:
Spherical Charge r > R
E 4 r2 = qenc
r2 )
E = q / (4
Spherical Charge r < R
E = q / (4
r2 )
uniform r: q / q = r3 / R3
E = [q / (4
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R3)] r
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Fundamentals of Physics
Chapter 23
34
Uniformly Charged Sphere
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Chapter 23
35
Problem
Point charge inside a cavity radius R
embedded in a metal block
E at point P1 at R/2 (still in cavity)
E at point P2 inside block
Einside
Ein shell
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1
4
0
q
r12
0
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conductor
Fundamentals of Physics
Chapter 23
36
Problem
A charge Q is inside a cavity in a conducting sphere.
Qon outside of shell
Qnet
14 C
10 C
Charge on inner cavity wall and inside cavity?
Qcavity wall
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4 C
Q
Qcavity wall
Q
4 C
Physics 2112
Fundamentals of Physics
Chapter 23
37
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