Download Welcome to Physics 112N

Welcome to Physics 112N
Professor Charles E. Hyde-Wright
Spring 2005
Navigate from http:www.physics.odu.edu, or
http:www.physics.odu.edu/hyde/Teaching/Spring05/Phys112_2005.htm
Topics to be covered
• Electricity and Magnetism (Chapters 19-24)
• Light and Optics (Chapters 25, 26, 28)
• Modern Physics (Chapter 30)
10 January 2005
Walker, Chapter 19
2
Phys 111: Chapters 1-18
• Description of motion: Kinematics
 Position (in 1-, 2-, 3- dimensions)
 Velocity (rate of change of position)
 Acceleration (rate of change of velocity)
• Relationship between Force and Motion
 Net Force equals mass times acceleration
• Description of motion in terms of Energy
 Kinetic Energy
 Potential Energy (Gravity, Springs…)
 Thermal Energy (non conservative forces)
• Examples of forces:
 Contact forces (friction, “normal” force = force perpendicular to surface)
 Spring Force F = - kx
 Gravity
10 January 2005
Walker, Chapter 19
3
Gravity
•
•
•
•
|F| = G M m / r2
Near surface of earth ( h << R )
|F| = G M m /(R+h)2  m [ GM/R2 ] = mg
Circular, Elliptical, Parabolic, & Hyperbolic orbits of
moons, planets, asteroids, comets, possible visitors from
outer space.
• Potential Energy:
 U = - G M m /r
 Note minus sign, Potential energy decreases as two masses
approach: Conservation of energy means Kinetic Energy increases
as Potential energy decreaces.
10 January 2005
Walker, Chapter 19
4
Chapter 19
Electric Charges, Forces, and Fields
•
•
•
•
Fundamental Forces in Physics
Gravity (gravitons)
Electromagnetism (photons)
Weak Interaction (W and Z bosons)
Strong Interaction (gluons)
All of physics is based on these four forces
All four forces have similar equations.
10 January 2005
Walker, Chapter 19
5
Energy in our World
• Nuclear Fusion in sun E=mc2
 H H H H  He + n + n + Energy
 Thermal Energy at surface converted to visible
light energy
Fusion
• Light Energy  Chemical Energy
(photosynthesis)
• Plants  Fossil Fuels
 Fuel for cars (motion)
 Fuel for power plants
Radiation
(electrical energy  lighting for your Physics HW).
• Plants  Food –> Krebs cycle ADP/ATP
 Energy for thought, motion of muscles  HW
10 January 2005
Walker, Chapter 19
6
Electromagnetism in our World
• Gravity holds us to the earth.
• Electromagnetism dominates every other aspect of
our physical world
 Atoms, Molecules, Solids, & Liquids held together by
electrostatics
 Chemistry
 Light
 Virtually all technology:
• Electronics, Electric motors, Electric Lighting
• Even fire is fundamentally an electromagnetic phenomenon
• Profound insights into the physical nature of life.
10 January 2005
Walker, Chapter 19
7
Electrostatic Phenomena
• Rubbing things makes an electrostatic
charge
 Spark, Hair standing on end
 Thunderclouds rub rising microscopic icecrystals against falling hail  Clouds charge up
• Electrostatic phenomena do not require any
obvious macroscopic change (mass,
material change)
10 January 2005
Walker, Chapter 19
8
Basic Model of Electrostatics
Franklin, Coulomb, 18th Century
• When two dissimilar materials are in contact, microscopic
particles can be transferred from one to the other
 Modern view: electrons e-, or ions Ca++…
• These particles carry a property (like mass) called electriccharge.
 Electric charge can be positive or negative
 Electric charge is a scalar: It is a quantity independent of any direction in
space (compare temperature vs. velocity)
 Positive attracts negative
 Positive repels positive, Negative repels negative
 Charge adds linearly: put together charge q1 and charge q2, they act like
q3 = q1 + q2.
 Charge is conserved: 0 = +q + (-q)
10 January 2005
Walker, Chapter 19
9
Insulators & Conductors
• Charges placed on an insulator (plastic, wood,
ceramic) stay put—in spite of the electric forces on
them:
Fnet=0
Binding force acts like a microscopic spring. As external
electric force pulls on charge on insulator, the binding
force pulls back (up to some limit: spring breaks)
Charges placed on a metal are free to move in response to
electro-magnetic (or other forces): ma = F
10 January 2005
Walker, Chapter 19
10
Electrical Charge
All physical quantities must be measured as multiples of a
standard
Time is measured in multiples of the standard second (now
defined by atomic physics phenomena)
Distance is measured in multiples of the standard meter (now
defined in reference to the speed of light times the second).
Mass is measured in multiples of the standard kilogram,
housed near Paris, France.
The SI unit of electrical charge is the Coulomb
The Coulomb (C ) is defined from magnetic phenomena
(skip till later).
10 January 2005
Walker, Chapter 19
11
The Structure of an Atom
The atom consists of a positively
charged nucleus, orbited by
negatively charged electrons.
The nucleus contains protons
(positive) and neutrons (neutral).
The orbital lines are an accurate
description of the orbits of
electrons in a highly excited
atom
The fuzzy red blob is a better
representation of the electron
wave in the atomic ground state
(see Chap. 30).
10 January 2005
Walker, Chapter 19
12
The Electron
One of the fundamental particles found in nature is the
electron.
• The electron mass is 9.11  10-31 kg.
• The electron charge (-e) is -1.6  10-19 C.
 The symbol e is the magnitude of the electron’s charge
• The electron is part of a family of fundamental
particles known as leptons.
• Electron lifetime > 1023 years (test of charge
conservation). Eur. Phys. J. C 3, 1 (1998)
10 January 2005
Walker, Chapter 19
13
The Proton
The proton is not a fundamental particle. It has a finite size (10-15
m) and a spectrum of excited states. It is understood to consist
of three quarks bound together by a cloud of gluons and
quark—anti-quark pairs.
• The proton mass is 1.67  10-27 kg.
• The proton is 2000 times heavier than the electron, so the
vast majority of an atom’s mass resides in the nucleus.
• The proton charge (+e) is +1.6  10-19 C.
• The proton charge and electron charge are known to be equal
and opposite to very high precision.
• |qp + qe|/e < 10-21
10 January 2005
Eur. Phys. J. C 3, 1 (1998)
Walker, Chapter 19
14
• An object may contain both positive and negative
charges. If the object possesses a net charge it is
said to be charged. If the object possesses no net
charge it is said to be neutral.
• An atom is normally neutral, because it possesses
an equal number of electrons and protons.
However, if one or more electrons are removed
from or added to an atom, an ion is formed, which
is charged.
• Charge is always conserved: charge may be
transferred but it is never created or destroyed.
 However, charges can be created and destroyed in
positive and negative pairs, so that the net charge in the
universe does not change.
10 January 2005
Walker, Chapter 19
15
Electrical Forces
Two charged objects will exert forces on one another.
• Unlike charges attract one another.
+
-
• Like charges repel one another.
+
+
• The force decreases with the square of the distance between the
charges
10 January 2005
Walker, Chapter 19
16
Polarization
An object is polarized
when its charges are
rearranged so that
there is a net charge
separation. Charged
objects can be
attracted to neutral
objects because of
polarization.
10 January 2005
charged
Walker, Chapter 19
neutral & polarized
17
Insulators and Conductors
Materials are classified by how easily charged
particles can “flow” through them.
• If charges flow freely, the material is a
conductor (metals, for example)
• If charges are unable to move freely, the
material is an insulator (glass, for example)
• Some materials have properties in between
insulators and conductors, these are called
semiconductors.
10 January 2005
Walker, Chapter 19
18
Charge Transfer
Charge is usually transferred because electrons move
from one place to another.
But sometimes the flow of both positively or
negatively charged ions (atoms or molecules) is
important (cells, batteries…).
The earth can be viewed as an infinite (conducting)
reservoir of electrons. An object in electrical
contact with the earth is said to be grounded.
What happens when I ground the Van de Graaff generator?
[And why do I do this before touching the generator?]
10 January 2005
Walker, Chapter 19
19
Properties of the Mutual Electrical
Forces Acting on Two Charges
• Each of the two charged object experiences a force
that is equal and opposite to the force experienced
by the other charge (Newton’s Third Law).
• The force is attractive if the charges are unlike and
repulsive if the charges are like.
• The force is inversely proportional to the square of
the separation of the two charges, and is directed
along the line joining them (attractive or repulsive).
• The force is proportional to the product of the
magnitudes of the 2 charges.
Remember force is a vector!
10 January 2005
Walker, Chapter 19
20
Coulomb’s Law
The magnitude of the force between two point* objects
separated by a distance r with charges q1 and q2 is given by
Coulomb’s Law:
F k
q1q2
r
2
q1 and q2 are the values
(+ or -) of the two
charges
where k = 8.99…  109 Nm2/C2 , precision of 10-7 is linked to
measurement of electron charge.
*or spherical charge distributions, or any objects whose size is much less
than the separation distance r
The direction of the force on one charge is either toward
(negative) or away (positive) from the other charge.
10 January 2005
Walker, Chapter 19
21
Force: vector, magnitude, component

| q1 || q2 | • Magnitude (strictly positive)
| F | k
2
r
F k
q1
r
10 January 2005
q1q2
r
2
q2
• Component along direction from q1
to q2 of Force from q1 acting on q2.
• If q1q2> 0, force is repulsive
(pushes q2 away from q1)
• If q1q2< 0, force is attractive
(pulls q2 towards q1)
Walker, Chapter 19
22
Comments on Coulomb’s Law
• 1/r2  Charge is conserved, Gauss’ Law
 Deviations from 1/r2 are measured to be less than 1 part in 1010 over distance
scales from (10-10 m to 1.0 m)
• The force is linear in the value of each charge.
 If an amount of charge 0.2q1 is brought from far away and added to q1,
the force on q2 is increased to
(q1 + 0.2q1 )q2
1.2q1q2
F k
k
2
2
r
r
• Why k? (Why not k=1?)
 In Gaussian (or cgs) units, 1.00 esu is defined such that
 Two charges of 1.00 esu each separated by 1cm exert mutual forces on
each other of 1 dyne = 1 gm cm2/sec2
 k = 8.99  109 Nm2/C2 = 1.00 dyne (cm)2/esu2.
 The value of k depends upon our choice of units for Force, Distance and
Charge.
10 January 2005
Walker, Chapter 19
23
Subscript labels on Force
F k

F12  Force on charge q1 from charge q2

F21  Force on charge q2 from charge q1


F21  -F12 : Newton's Third Law : Action - Reaction
q1
10 January 2005

F12

F21
Walker, Chapter 19
q1q2
2
r
q2
24
Vectors and Scalars
• A scalar is a physical quantity with magnitude, but without
direction in space.
 Temperature
 Mass
 Energy, Time, Charge …
• A Vector is a physical quantity with magnitude and
direction in space.
 Displacement
 Momentum
 Velocity, Force
10 January 2005
Walker, Chapter 19
25
Vector Components
& Unit Vectors
• A vector can be expressed in terms of a
coordinate system.
• Force vector
 F = 1.6 N oriented 110° counterclockwise from x-axis.
• Force Vector
 F = (1.6 N)(cos110°) along +x-axis
plus (1.6N)(sin110°) along +y-axis

F
1.5 Nˆy
q110
x
-0.55 Nˆx
xˆ  displacement of 1.0 along x - axis (no units)
yˆ  displacement of 1.0 along y - axis (no units)

F  (1.6N)(cos110) xˆ + (1.6N)(sin110) yˆ
 (1.6 N )(-0.342) xˆ + (1.6N)(0.940) yˆ
 -0.55 Nxˆ + 1.5 Nyˆ
10 January 2005
Walker, Chapter 19
26
Walker Problem 13, pg. 641
Given that q = +12 mC and d = 16 cm, (a) find the direction
and magnitude of the net electrostatic force exerted on
the point charge q2 in Figure 19-30. (b) How would your
answers to part (a) change if the distance d were tripled?
10 January 2005
Walker, Chapter 19
27
Solution
Draw free body for JUST q2

F21 
k
(12mC )- 24mC 
2
(0.16m)
in direction from 1 to 2
 - 101N xˆ
10 January 2005

F23 
Walker, Chapter 19
k
(-24mC )+ 36mC 
(0.16m) 2
in direction from 3 to 2
 + 303N xˆ
28
Problem 13, Solution, cont’d

F2  (-101N + 303N ) xˆ  202N in + x - direction
B) Tripling the separations decreases all forces by a
factor of 32=9
F2 = 22.4 N, +x direction
10 January 2005
Walker, Chapter 19
29
Relative Strength of Gravity and
Electrostatics
• In the hydrogen atom, the electron and proton are
separated by 0.5·10-10 m
• The ratio of gravitational attraction between the
electron and proton divided by the electrostatic
attraction is
 FG/FQ = 10-39 (see text)
• This ratio is independent of the separation
 Both forces are 1/r2.
• Why is gravity so much more important in the
solar system?
10 January 2005
Walker, Chapter 19
30
Multiple Charges
• If there are more than two
charges present, the net
force on any one charge is
given by the vector sum of
the forces on that charge
from all surrounding
charges. This is an
example of the Principle
of Superposition.
F+
-
F-
+
+
What is the direction of the net
force on each charge (roughly)?
10 January 2005
Walker, Chapter 19
31
Walker (1st edition)
Problem 19, pg. 641
(a) Find the direction and
magnitude of the net
electrostatic force exerted
on the point charge q3 in
the Figure. Let q = +1.8
mC and d = 22 cm. (b)
How would your answers
to part (a) change if the
distance d were doubled?
10 January 2005
Walker, Chapter 19
32
Solution
•
•
•
•
•
•
Force F3,2 on q3 from q2 is repulsive
Force F3,1 on q3 from q1 is attractive
Force F3,4 on q3 from q4 is repulsive
Distance from q2 to q3 is d
Distance from q4 to q3 is d
Distance from q1 to q3 is (2)d
F3, 2  k
q2 q3
d
2
k
- 2.0q - 3.0q
d
2
 6k

q2
d2
2
-6

Nm
1
.
8

10
C
9

Define F  k 2   8.99  10

d
C 2  0.22m 2

q2
2  0.602 N
F3, 2  6 F  3.61N

F3, 2  3.61Nxˆ
10 January 2005
Walker, Chapter 19
33
Solution, cont’d
F3, 4  k
q3 q4
- 3.0q - 4.0q
k
d2
 12 F  7.22 N
d2
 12k
q2
d2
F3, 4

F3, 4  7.22 Nyˆ

q3 q1
- 3.0q q 3 q 2 3
F3,1  k 2  k
 k
 F  0.903 N
d
 2d 2 2 d 2 2
x - component of force F3,1
F3,1x  F3,1 cosq31  (0.903N ) cos(180 + 45)  -0.638 N

FNet  (2.97 N ) xˆ + (6.58 N ) yˆ

FNet  (2.97 N ) 2 + (6.58 N ) 2
 52.15 N 2  7.22 N
Add the force vectors
graphically
y - component of force F3,1
F3,1y  F3,1 sin q31  (0.903N ) sin(180 + 45)  -0.638N

F3,1  -0.638 Nxˆ - 0.638 Nyˆ




FNet  F3,1 + F3, 2 + F3, 4

FNet  -0.638 Nxˆ - 0.638 Nyˆ + 3.61Nxˆ + 7.22 Nyˆ

FNet  (3.61N - 0.638 N ) xˆ + (7.22 N - 0.638 N ) yˆ
F3,4
FNet
F3,2
F3,1
10 January 2005
Walker, Chapter 19
34
Solution, four charges

FNet  (2.97 N ) xˆ + (6.58 N ) yˆ

FNet  7.22 N
• Find angle q from x-axis
• Cosq = [FNet,x]/ |FNet|
• Cosq  2.97N/7.22N
 q  65.7
10 January 2005
Walker, Chapter 19
y
FNet
q
x
35
Spherical Charge Distributions
In general a spherical charge distribution
behaves as if all of its charge were at the
center of the sphere. Use the distance to the
center of the sphere to calculate the
electrostatic force.
q2
q1q2
q1
F k
10 January 2005
r
2
Walker, Chapter 19
r
36
Newton: Action at a Distance
Faraday: Force Fields
• A mass m exerts a gravitational force GmM/r2 on a second
mass M separated by a distance r, and vice versa.
• Coulomb gave us the same picture for electrostatic forces
• Faraday offered a new insight, introducing the Electric Field,
which can be thought of as carrying the force from charge q
to charge Q.
 In physics, a field means a physical variable that has a defined value
at every point in space. Examples:
• Temperature map, Barometric Pressure map (a scalar field)
• Wind velocity map (a vector field)
• Initially just a mathematical trick, with our understanding of
electromagnetic waves and the quantum nature of light,
Electric and Magnetic fields are as real as charge and mass.
10 January 2005
Walker, Chapter 19
37
Electric Field
• If a test charge q0 experiences a force F at a given location r,
the magnitude of the electric field at that location is defined

by
 F
E
q0
• The electric field is a “what if” concept. What would be the
electrostatic force acting on a charge q0 if it were placed at
position r?
• The electric field can also be thought of as a disturbance in
space caused by nearby charges.
• The electrostatic force experienced by a charge is the
interaction between the charge and the electric field at that
position.
• The SI units of electric field are Newtons/Coulomb = N/C
10 January 2005
Walker, Chapter 19
38
Electric Force F(r) from charge Q acting on a test charge q0
at various locations r =(x,y,z): F=kQq0/r2
Electric Field E(r)= F/ q0
q0
Q
10 January 2005
Walker, Chapter 19
39
Electric Field E(r) from charge Q at various locations r:
E=kQ/r2
r
Q
10 January 2005
Walker, Chapter 19
40
Electric Field

E ( x, y, z ) at
• A vector
every point in space that
tells us the magnitude
and direction of the

force Fq  qE ( x, y, z ) a charge q will experience
if the charge q is placed at the position (x,y,z).
• If q<0, then the force F on q is opposite E.
• To measure E = F/q, q must be small enough
that it doesn’t change the distribution of
charges that created the electric field in the
first place.
10 January 2005
Walker, Chapter 19
41
Electric Field Direction
The direction of the electric field is defined to be the
direction of the force that would be experienced if
the test charge is positive. Because the field has a
direction, it must be a vector.
E
q0
q0
E
10 January 2005
+
Walker, Chapter 19
42
Electric Field (cont.)
The electric field is the force per charge at a given
location. If you know the electric field, then the force
on a charge can easily be found using
F = qE
Example: A charge q of 8 mC experiences a uniform
electric field of 1000 N/C to the right. (a) What is the
force on the charge? (b) What would the force be if
the charge were –8 mC?
q
Note: In problems like this we do not need to know
what charges created the electric field.
10 January 2005
Walker, Chapter 19
E = 1000 N/C
43
Electric Field of a Point Charge
From Coulomb’s Law, the magnitude of the force
experienced by a test charge q0 a distance r from a
charge q is
qq
F k
r
0.
2
Since the definition of the electric field is
F
E ,
q0
q0
the magnitude of the electric field from a point charge is
given by
q
Ek
10 January 2005
r
.
2
Walker, Chapter 19
q
44
Walker Problem 28, pg. 642
What is the magnitude of the electric field produced by a
charge of magnitude 10.0 mC at a distance of (a) 1.00
m and (b) 2.00 m?
Q
Ek 2
r
k = 8.99 ·109 N m2/C2
10 January 2005
Walker, Chapter 19
45
Electric Field & Polarization
• What is the magnitude of an electric field
strong enough to polarize the molecules in
the air to the point that electrons are pulled
out of the air (ionization produces a spark)?
 Several Million Newton/Coulomb.
 Several Million Volt/meter
10 January 2005
Walker, Chapter 19
46
Electric Fields in Nuclear Physics
• What is the electric field at the surface of a
proton?





(radius 10-15 m, charge 1.6·10-19 C)
E = (8.99 ·109 N m2/C2)(1.6·10-19 C)/(10-15 m) 2
E=(14.4) 109-19+30 N/C
Q
Ek 2
E=1.44 · 1019 N/C
r
That’s big!
10 January 2005
Walker, Chapter 19
47
Electric Fields in
Atomic/Molecular physics
• What is the electric field from the hydrogen
nucleus (proton) at a distance of one atomic radius
(r=0.5Å=0.5·10-10m)




E = k q / r2
E = (8.99 ·109 N m2/C2)(1.6·10-19 C)/(0.5·10-10 m) 2
E = (58)(109-19+20) ( N/C)
E = 5.8 ·1011 N/C
• Smaller, but still very large.
10 January 2005
Walker, Chapter 19
48
Superposition
Just like with forces, electric fields must be
added as vectors. The electric field from
several charges is the vector sum of the
electric field from each charge.
Example: Consider two identical negative charges
as shown. At which lettered point is the
magnitude of the electric field greatest? Least?
a
10 January 2005
b
-
c
Walker, Chapter 19
-
d
49
Superposition
E2
E1
-
-
E
Q2 = Q 1
Q1<0
E
E2
E1
10 January 2005
Walker, Chapter 19
50
Walker Problem 66, pg. 644
An object of mass m = 3.7 g and charge q = +44 mC is
attached to a string and placed in a uniform electric
field that is inclined at an angle of 30.0° with the
horizontal. The object is in static equilibrium when the
string is horizontal. Find (a) the magnitude of the
electric field and (b) the tension in the string.
10 January 2005
Walker, Chapter 19
51
Walker Problem 66, pg. 644
• Free Body Diagram
• Net force = 0
• S Fx=0: qEsin30o – mg = 0
qE
T
mg
m = 3.7E-3 kg, q = 44.E-6 C
E = mg/(q sin30o) = (3.7E-3 kg)(9.8m/s2)/(0.5*44.E-6 C)
E = 1.65E+3 (kg·m/s2)/C = 1.65E+3 N/C
• S Fy=0: qEcos30o – T = 0
T = (44.E-6 C) (1.65E+3 N/C)0.866 = 6.3E-2 N
10 January 2005
Walker, Chapter 19
52
Electric Field Lines
In order to visualize the electric field in space it is convenient
to draw Electric field-lines (see Fig. 19-13). The field
lines are directional [curved] lines that everywhere point in
the direction of the electric field at that point.
+
-
+
Dipole
10 January 2005
Walker, Chapter 19
53
Field Line Properties
• The electric field is tangent to the field line at any point
in space.
• The strength of the electric field is proportional to the
density of field lines (areal density measured
perpendicular to field line).
• The field lines always begin on positive charges or at
infinity and end on negative charges or at infinity.
• No two field lines can ever cross.
• The number of field lines leaving a positive charge or
approaching a negative charge is proportional to the
magnitude of the charge.
10 January 2005
Walker, Chapter 19
54
Electric Field Lines
Note that twice as many field lines originate from
the +2q charge than the +q or –q charges.
10 January 2005
Walker, Chapter 19
55
Lecture 2, Quiz 1
1. The net charge inside
the green blob is:
a) Positive
b) Zero
c) Negative
Hint: Are there more
Electric Field lines
entering, or leaving
the blob, or is it
equal?
10 January 2005
Walker, Chapter 19
56
Lecture 2, Quiz 2
2. The net charge inside the green
blob is:
a)
Positive
b)
Zero
c)
Negative
Hint: Are there more Electric
Field lines entering, or
leaving the blob, or is it
equal?
10 January 2005
Walker, Chapter 19
57
Lecture 2, Quiz 3
3. The net charge inside the green
blob is:
a)
Positive
b)
Zero
c)
Negative
Hint: Are there more Electric
Field lines entering, or
leaving the blob, or is it
equal?
10 January 2005
Walker, Chapter 19
58
Walker Problem 37, pg. 642
The electric field lines
surrounding three
charges are shown in
the Figure. The center
charge is q2 = -10.0
mC. (a) What are the
signs of q1 and q3? (b)
Find q1. (c) Find q3.
10 January 2005
Walker, Chapter 19
59
Parallel-Plate Capacitor
Two parallel conducting plates
with opposite charge, separated
by a distance d, is known as a
parallel-plate capacitor. The
electric field is uniform
between the plates (except near
the edges, not shown).
Uniform means the electric field
magnitude and direction are the
same everywhere (in gap).
This is because of, not in spite
of Coulombs 1/r2 law!!
10 January 2005
Walker, Chapter 19
60
Electrostatic Equilibrium
Recall that charges within a conductor are free
to move around easily.
If the charges within a conductor are not in
motion, then the system is said to be in
electrostatic equilibrium.
10 January 2005
Walker, Chapter 19
61
Properties of Electrostatic
Equilibrium
• In the presence of electrostatic forces, the charges
on the conductor move around until the following
static conditions are achieved:
 The electric field is zero everywhere inside a conductor.
 The excess charge on a conductor resides entirely on its
surfaces.
 The electric field just outside a charged conductor is
perpendicular to its surface.
• On irregularly shaped objects, the charge
accumulates at sharp points, and the electric field
is most intense at sharp points.
10 January 2005
Walker, Chapter 19
62
Electric Flux
We define electric flux F as
the product of the surface
area A times the
component Ecosq of the
electric field
perpendicular to the
surface.
In general, F = EAcosq.
a F  EA
b F = 0
(c) F = EAcosq
10 January 2005
q is the angle between the electric field and
the line perpendicular to the surface.
Walker, Chapter 19
63
Gauss’s Law
Consider an arbitrary (imaginary) closed surface (called a
Gaussian surface) enclosing a total charge q. The
electric flux through the surface is
q
F
0
 0  41k  8.8510-12 C2 /N  m2
This integral property is a consequence of
the 1/r2 Coulomb Law, and is valid for
any irregular surface, no matter how
complicated the electric field produced
by internal or external charges.
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Walker, Chapter 19
64
Example
Three point charges are arranged as shown. q1 = +4 mC,
q2 = -6 mC and q3 = -4 mC. Find the electric flux
through the three Gaussian surfaces labeled a, b and c.
a
b q
1
c
q3
q2
10 January 2005
Walker, Chapter 19
65
Walker Problem 49, pg. 643
A thin wire of infinite extent has a charge per unit length
of l. Using the cylindrical Gaussian surface shown in
the Figure, show that the electric field produced by this
wire at a radial distance r has a magnitude given by
l
E
2 0r
10 January 2005
Walker, Chapter 19
66
Walker Problem 49, pg. 643
solution
By symmetry, Electric force on a test
charge is directed radially
outward (if l>0).
Closed Gaussian surface consists of
the cylinder and its two end caps.
Electric flux through end caps is
zero because E is parallel to
surface.
Electric flux through cylinder wall:
F=Area · E(r )
F = 2 r L E(r )
Net Flux = 0 + 0 + 2 r L E(r )
= (charge enclosed)/0 = L l /0
10 January 2005
Walker, Chapter 19
E (r ) 
1 l
l
 2k
2 0 r
r
67
Charges on (and in) a conductor
• Charge on a conductor is free to move
under the influence of its mutual
repulsion.
• Are the charges in a) or b) “farther
apart”?
 The quantitative meaning to this
question is “Which configuration
gives the lowest value for the
electrostatic energy?” (See Chap 20.)
• It is a property of the 1/r2 law (not just
repulsion) that all the excess charge on a
conductor ends up on the SURFACE.
 This can be an inside, as well as
outside surface!!
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68
Quiz 1
Jan 10, 2005
• Two charges Q1 and Q2 are separated by a distance of
0.010 m. The Electrostatic force of Q1 on Q2 is 2.0e-5 N.
• At what distance of separation between Q1 and Q2 would
the force be 1.0e-5N?
 a) 0.02 m
 d) 0.007m
10 January 2005
b) 0.014 m
e) 0.005 m
Walker, Chapter 19
c) 0.01 m
69
Quiz 1
12 January 2004
Name……………………
• In the diagram at right, F1 is the
electrostatic force of Q1 acting on
charge q=1.0E-9C .
• Draw a vector with its tail at q to
represent the magnitude and direction
of the electrostatic force F2 of Q2
acting on charge q (the length of your
vector should roughly describe the
relative magnitudes of F2 and F1.
• Draw a vector with its tail at q to
represent the magnitude and direction
of the net force FNet acting on q from
both Q1 and Q2
• Label your vectors F1 and FNet

 
Q2 = -1.0E-6 C
• Note:
F F +F
Net
10 January 2005
1
Q1 = +1.0E-6 C
q
F1
2
Walker, Chapter 19
70
Quiz 2
Sketch the electric field lines
generated by these two charges.
Hint: Consider the electric flux
through the three gaussian
surfaces defined by the three
dashed lines.
2 February 2004
Name……………………
+4mC
10 January 2005
-2mC
Walker, Chapter 19
71
Preparation for Lab 2 (Chapter 21)
• Electric Current in wire equals steady flow of charge
(not equilibrium!).
• Unit of measure is Coulomb per second = Amp
• 1.00 C/s = 1.0 A
• Think of electric current like flow of water in pipe.
• Voltage = Electrostatic Potential difference of power
supply or battery (e.g. AA=1.5 V)
 How hard the current is being forced around circuit.
 Think of difference in height of two ends of a water pipe.
Water flows with greater force when the height difference
is greater.
• Resistance R = measure of how hard you have to
push to obtain current (flow). R = V/I
Pump
 Think of long thin pipe (high resistance to flow) versus
short broad pipe (low resistance to flow).
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72
Equivalence of
Gauss’ Law and Coulomb’s Law
• Coulomb: Electric field at a distance r from
a point charge Q:
E
 E(r) = k Q / = Q / (4 0
 For Q>0, E>0: E points away from Q
 For Q<0, E<0: E points towards Q.
r2
r2)
• Gauss: Electric flux through an imaginary
closed spherical shell a distance r from Q
 Flux = E(r)•(Surface area of shell)= E(r) 4 r2
Q
r
• Outward flux is positive
• Inward flux is negative
 Gauss: Flux = Q/ 0.
 E(r) = Q / (4 0 r2)
10 January 2005
Walker, Chapter 19
73
Gauss’ Law and the Parallel Plate Capacitor
• Consider a rectangular Gaussian surface penetrating into
the metal of a parallel plate capacitor:
 Total Charge on left plate = Q, right plate = -Q
 Total area or each plate = A
 Surface charge density = s = Q/A
 Surface area of face of Gaussian surface parallel to
plate = a.
 Charge enclosed by Gaussian surface: sa
 Flux through portion of Gaussian surface inside metal
= 0 (E=0).
 Flux through top and bottom surfaces outside metal =
0 (Electric field parallel to surface).
 Flux through face of Gaussian surface parallel to
plate (outside) = Ea.
 Gauss’ Law: sa/0 = Ea
+
+
+
+
a
+
+
+
+
+
+
+
-
• Uniform Electric Field in gap: E = s/0
10 January 2005
Walker, Chapter 19
74