PHYS 192-01
Summer 2017
Critical Information
Instructor: W.A. Parkinson
Classroom: SBA 235
Office: 123 Pursley Hall
email: [email protected]
Office Hours: MTWTH 9:15 – 10:30 A.M.
I. Course Texts
“College Physics Reasoning and
Relationships” by Nicholas Giordano
“College Physics, 4th Edition”
by Nicholas Giambattista, Richardson,
and Richardson
II. Content Tentative schedule of chapters covered:
Giordano
Giambattista
Ch 17
Ch 18
Ch 19
Ch 20
Ch 21
Ch 22
Ch 23
Ch 24
Ch 16
Ch 17
Ch 18
Ch 19
Ch 20
Ch 21
Ch 22
Ch 23
– Electric Forces and Fields
– Electric Potential
– Electric Currents and Circuits
– Magnetic Fields and Forces
– Magnetic Induction
– Alternating-Current Circuits
– Electromagnetic Waves
– Geometric Optics
III. Tests There will be five tests, one after every
fifth lecture. Test dates are given below.
I
Thur
June 8
II
Tues
June 20
III
Thu
June 29
IV
Wed
July 12
V*
Mon
July 22
*Test V takes place during the final examination
period from 2:10 - 3:50 P.M. in the regular classroom
IV. Grading There are five 90 point exams. I will
give 2 points attendance per lecture period.(NOT TEST
DAYS). The final grade is calculated using the scale:
100-90% 89-80%
79-70% 69-60% < 60%
500 - 450 449 - 400 399 - 350 349 - 300 < 300
A
B
C
D
F
NO TESTS WILL BE DROPPED
THERE IS NO EXTRA CREDIT WORK
THERE ARE NO BONUS POINTS
THERE WILL BE NO CURVE. TEST SCORES
ARE TOTALLED, AND THE FINAL GRADE IS
DETERMINED BY WHICH CATEGORY YOUR
TOTAL SCORE FALLS UNDER.
V. Makeup Policy If a student knows in advance that
he or she will be unable to attend one of the scheduled
examination periods, please inform the instructor
beforehand, so that suitable arrangements
can be made. Otherwise if a test is missed, a
comprehensive test covering material from the first
four tests will be taken as its replacement. This
comprehensive test will take place the week before
final exams at a time suitable to both the student and
instructor.
VI. Calculator Policy Each student is responsible
for bringing their own functioning calculator for use
during examinations. Cellular phones may NOT be
used as a calculator. Cellular phones are also to be
stowed away during the examination period. In
addition, a student is NOT permitted to share his or her
calculator with another student during an examination.
Violation of these policies results in a score of zero on
the examination for all students involved.
VII. Attendance Southeastern Louisiana University
policy mandates that the instructor keep accurate
attendance records.
VIII. Students with Disabilities If you are a qualified
student with a disability seeking accommodations
under the Americans with Disabilities Act, you are
required to self-identify with the Office of Student
Life, Room 203, Student Union.
IX. Emergency Contingency Plans In the event that
the university experiences unforeseen closure due to
weather or other emergencies, students should access
Moodle to find information on class-related
announcements, assignments, and schedule changes.
X. Student Decorum Free discussion, inquiry,
and expression are encouraged in this class.
Classroom behavior that interferes with either (a)
the instructor’s ability to conduct class or (b) the
ability of students to benefit from instruction is not
acceptable. Examples may include routinely entering
class late or departing early; use of beepers, cellular
phones, or other electronic devices; repeatedly talking
in class without being recognized, talking while others
are speaking, or arguing in a way which is perceived
as “crossing the civility line.” In the event of a
situation where a student needs to carry a beeper, or
cellular phone, please advise the instructor.
XI. Academic Integrity
Students are expected to maintain the highest
standards of academic integrity. Behavior that
violates these standards is not acceptable.
Guidelines and procedures for disciplinary action
are outlined in the current General Catalogue.
XII. Withdrawals It is the student’s responsibility to
assure that all required course work is completed,
or that they have formally withdrawn. Failure to
attend does not constitute withdrawal. The last day to
withdraw from classes or resign from the school
with the University Registrar is Thursday, July 6,
2017 by 5:30 P.M.
Vector Quantities
J. Willard Gibbs
(1839-1903)
Introduced the concept of the vector and the
mathematical study known as vector analysis.
Vector Quantities
Physical quantities such as forces have two equally
important components.
1. Magnitude (Amount)
2. Direction of Action
of Action
Common Vector Quantities
1. Position (Spatial Coordinates)
2. Velocity
3. Acceleration
4. Momentum
5. Force (Mechanical, Electrostatic, Magnetostatic)
6. Gravitational, Electric, or Magnetic Fields
Scalar Quantities
“A scalar quantity can have magnitude, algebraic sign,
and units, but not a direction in space.”
Common Scalar Quantity Examples
1. Mass
4. Volume
2. Time
5. Energy
3. Temperature
6. Voltage
Scalars can always be added, subtracted multiplied,
or divided using the normal rules of arithmetic.
Vectors have specific rules of mathematical
combination that differ significantly from scalars.
Vector Addition
Multiple vector quantities (e.g. the simultaneous action
of several electrostatic or magnetostatic forces or
several electric or magnetic fields) can be combined,
or resolved, into a single vector known as the net or
resultant vector.
Only vector quantities representing the same physical
phenomenon can be added to each other.
Vectors may be qualitatively added in a graphical
fashion, or quantitatively added using the rules of
trigonometry.
Graphical Representation of Vectors
Vectors are symbolically represented by arrows.
The direction in which the tip of the arrow points
indicates the direction of action of the vector.
The length of the arrow can be drawn proportional to
the magnitude or scalar component of the vector.
Two vectors
acting in the positive
x-direction


F1
F2


In terms of their magnitude: F2  2 F1
Adding Collinear Vectors
Collinear vectors are parallel vectors, or vectors acting
in the same direction.
If they point in the same direction, connect the
individual arrows tail to head (in any order you desire)
The resultant or net vector points from the beginning
of the first to the end of the second, and has a
magnitudewhich is the
sum
of
the
individual
vectors.

F1
F2

Fnet



In terms of their magnitude: Fnet  3 F1  1.5 F2
Adding Antiparallel Collinear Vectors
Collinear vectors that are antiparallel act
in opposite directions.
If they are antiparallel, the magnitude of their sum is
the difference between them, always the larger minus
the smaller because the resultant magnitude is always
positive. The resultant direction of action is in the

direction of the larger vector.
F1
Fnet

In terms of their magnitude: Fnet

F2


 F1  0.5 F2
Collinear Vector Addition Examples

F1  8 N

F2  12 N


F1  20 N F2  30 N

Fnet  0 N

Fnet  20 N


F3  40 N F4  10 N
Adding Non-Collinear Vectors
Vectors that do not act in the same direction can be
added in much the same manner.
Each consecutive vector is placed tail-to-head. The
resultant vector begins at the first vector’s tail and
ends at the final vector’s head.
The magnitude of the resultant vector is proportional
to its length.
The exact magnitude can be quantitatively found only
through the use of geometry.
Non-Collinear Vector Addition

F1

F2

Fnet
The same result is obtained if added in
the opposite order:

Fnet
Non-Collinear Vector Addition

F1


F
3
F2

Fnet

F4
Finding a Change inVector Quantity
Initial direction vector Vi:
Final direction vector Vf:
What is the change in vector direction V?
Vi + Vf
Vf
The vector difference is
found by adding Vf to -Vi.
-Vi
Vi
This is the vector sum,
not the vector difference
V= Vf - Vi
Vf
The Acceleration Vector
Suppose an object initially has
the following velocity vector:
After time t, the velocity
vector has changed to:
What average acceleration vector caused this velocity change?
vf + v i
vf
vi
v =
vf - vi
Averaged over time, this a  Δv
is the acceleration vector
Δt
vf
-vi
5,531
13,472
7,028
21,764
18,842
53,767
47,304
95,749
88,752
Social
Sciences
Psychology
Education
Biology
Visual and
Performing Arts
Journalism
Communication
English
Computer
Sciences
Foreign
Language
Mathematics
Chemistry
Engineering
Physics
Business
Degrees Conferred by Major (2011-2012)
105,785
178,543
366,815
95,849
108,986
Source: National Center for Education Statistics (NCES)
Trigonometry Review 
Hypotenuse
c
Right Triangle:
a

90o angle
b
The sum of all angles in a triangle is 180o:  +  = 90o
Pythagorean Theorem: a2 + b2 = c2
Trigonometric relations: S O H C A H T O A
opposite
sin 
hypotenuse
a
sin θ 
c
b
sin  
c
adjacent
cos 
hypotenuse
b
cos θ 
c
a
cos  
c
opposite
tan 
adjacent
a
tan θ 
b
b
tan  
a
Calculator Keying
cos-1(.22)
77.29096701
18
=?
4
adjacent
4
cos  
  0.22
hypotenuse 18
  cos 1 0.22  77.29o
2nd COS
.22
ENTER
Geometry Review
c=?
5
=?
8
12
a=?
 = 11o
b=?
c  52  82  89  9.4
opposite 5
tan  
  0.625
adjacent 8
  tan 0.625  32
1
o
opposite
sin  
hypotenuse
o
a  12 sin( 11 )  2.3
adjacent
cos  
hypotenuse
b  12 cos(11o )  11.8
Resolving Vectors Into Components
We will focus at this time on two -dimensional (in
the plane of the page) vectors.
Vectors of this type can be resolved into their x- and
y- components.
Using similar techniques, the x- and y- components
of net vectors can be found by addition of their all
individual vector components.
Resolving Vectors

F1  8 N
F1(x) = 8 N
0o angle
with the x-axis

F2  19 N
40o angle
with the x-axis

F2  19 N
F2(x) = 19 cos 40
= 14.6 N
F2(y) = 19 sin 40
= 12.2 N
F1(y) = 0 N

F3  125 N
A vector’s angle may be defined in
other ways but it is then important
to specify directions of each component.
205o - 180o = 25o angle
with the negative x-axis

F3  125 N
F3(x) = 125 cos (205)= -113 N

F3  125 N
F3(y) = 125 sin (25)
= 52.8 N (negative y)
205o angle with
the positive x-axis
When a vector’s angle is measured relative
to the positive x-axis, a vector’s x-component
is the cosine of the angle and a vector’s ycomponent is the sine of the angle.
F3(y) = 125 sin (205)
= -52.8 N
Resolving Vectors
F3(x) = 125 cos (25)
= 113 N (negative x)

F3  125 N
Using Other Coordinate Frames in Vector Addition
F1 has:
negative x component
positive y component

F1  16.0 N
F2 has:
positive x component
negative y component
25o
30o

F2  18.0 N
F1(x) = -16.0 cos(25o) = -14.5 N
F1(y) = +16.0 sin(25o) = +6.8 N
F2(x) = +18.0 sin(30o) = +9.0 N
F2(y) = -18.0 cos(30o) = -15.6 N
Fnet(x) = -5.5 N
Fnet(y) = -8.8 N
Fnet  - 5.52  (-8.8)2  10.4 N

8.8 N
tanθ 
 1.6
5.5 N
 = tan-1(1.6) = 58o
Fnet makes angle
180 + 58 = 238o
with positive x-axis
Finding Magnitude and Direction of Acceleration
Suppose a velocity vector due north at 3 m/s is changed to a
velocity vector of 4 m/s heading due east over a time period of
0.20 seconds. What is the magnitude and direction of the
acceleration vector that causes this change in velocity?
4 m/s
vf = 4 m/s
v1 = 3 m/s
- 3 m/s
Δv  4  ( 3)  5 m/s
2
a
Δv 5 m/s

 25 m/s 2
Δt .20 s
4 m/s

5 m/s
2
 3 m/s 
o
θ  tan 
  37
 4 m/s 
1
- 3 m/s
37o below the positive x-axis or
323o around from the positive x-axis.
Resolving Multiple Vectors
= 45 N
50o
=32 N
15o
F1(x) = 45 cos(50) = 29 N
F1(y) = 45 sin(50) = 35 N
F2(x) = 32 cos(15) = 31 N
F2(y) = 32 sin(15) = 8 N
Fnet(x) = 60 N
Fnet(y) = 43 N

Fnet  74 N
 = 36o
Fnet(y) = 43 N
Resolving Multiple Vectors
Qualitative
vector addition
Fnet(x) = 60 N

Fnet  Fnet ( x) 2  Fnet ( y ) 2  3600  1849  74 N
43
o
Angle of resultant with x-axis:   tan
 36
60
1
Resolving Multiple Vectors
= 100 N
= 80 N
x
=150 N
F1 has direction of 45o from
the positive x-axis
F2 has direction of 190o from
the positive x-axis
F3 has direction of 320o from
the positive x-axis
F1(x) = 100 cos(45) = 71 N
F1(y) = 100 sin(45) = 71 N
F2(x) = -80 cos(10) = -79 N
F2(y) = -80 sin(10) = -14 N
F3(x) = 150 cos(40) = 115 N
F3(y) = -150 sin(40) = -96 N
Fnet(x) = 107 N
Fnet(y) = -39 N
Resolving Multiple Vectors
20o

Fnet  114 N
Qualitative
vector addition
Fnet(y) = -39 N
Fnet(x) = 107 N

2
2
Fnet  Fnet ( x)  Fnet ( y )  11449  1521  114 N
Angle of resultant
(below the x-axis):
39
  tan
 20 o
107
1
Gravitational Force
Newton’s Universal Law of Gravitation “Two bodies exert a gravitational force
upon one another that is directly proportional to their masses and inversely
proportional to the square of the distance between them.”
Isaac Newton
(1642-1747)
Gm1m2 Masses of the objects
F
2
r Distance between objects
Magnitude of
gravitational force
Gravitational constant
G  6.67 10
11
N m
kg 2
2
Electrostatic Force
Coulombs Law “Two charged bodies exert an electrostatic
force upon one another that is directly proportional to their charges and inversely
proportional to the square of the distance between them.”
Charles Augustin Coulomb
(1736-1806)
Magnitude of
electrostatic force
Coulomb constant
F
k q1 q2
r
2
Charges of the objects
Distance between objects
2
N

m
k  8.99  109
C2
Comparing Electrostatics and Gravitation
Gravitation
Electrostatics
F
k q1 q2
r
2
9 N m
k  8.99  10
C2
2
Gm1m2
F
2
r
G  6.67 10
11
N m
2
kg
Gravitational forces are only important for very massive
bodies.
All matter is held together by electrostatic forces.
Electrostatic forces can be attractive or repulsive.
Gravitational Forces are attractive only.
2
Charge
Unit of charge - the Coulomb
Symbol - C
Derived unit with SI form: 1 C = 1 Amp ·sec
Elementary charge - the smallest possible amount
Robert Millikan
of charge
Symbol
-e
(1868 -1953)
e = 1.602176487  10-19 C
All larger quantities of charge
are integer multiples of the
elementary charge.
measured the
elementary charge
Attractive and Repulsive Charges
Benjamin Franklin
(1706 - 1790)
First to designate the two
distinct types of charge as
positive and negative.
Determined that lightning was
electrical in nature.
Subatomic Particles and Charge
Electrons possess the elementary charge with
polarity assigned to be negative:
Electron charge :
- 1.602176487  10-19 C
(or -1)
Protons possess the elementary charge with
polarity assigned to be positive:
Proton charge :
+ 1.602176487  10-19 C (or +1)
Charged Objects
Objects become negatively-charged by adding
electrons to them.
Objects become positively-charged by removing
electrons from them.
How many electrons must be added to an object so
that it acquires a negative charge of 1.00 C?
1 electron
18
1C

6.25

10
electrons
19
1.60 10 C
= 6,250,000,000,000,000,000 electrons!
Attractive Force of Opposite Charges
Opposite charges exert an attractive force drawing
each toward the other
+
-
According to Newton’s Third Law the positive and
negative charges exert a force on one another that
is equal in magnitude and opposite in direction.
The attractive force acts along a straight line
connecting the two charges independent of the angle
that the charges make relative to one another.
Repulsive Force of Like Charges
Like charges exert a repulsive force pushing
each away from the other
+
+
or
-
-
According to Newton’s Third Law, two like charges
exert a force on one another that is equal in
magnitude and opposite in direction.
The repulsive force acts along a straight line
connecting the two charges independent of the angle
that the charges make relative to one another.
Coulomb’s Law Example 1
An electron in a hydrogen atom is separated from the
proton by an average distance of 5.29 10-11 m. What
is the force they exert upon one another?
q1 = 1.60 10-19 C
F12  F21 
k q1 q2
r
2
q2 = -1.60 10-19 C
19
N  m 1.60 10 C
 8.99 10
2
2

11
C
5.29 10 m
2
9

= 8.22 10-8 N
F12
+
2
F21
-
The force exerted by the electron on the
proton (F12) is equal in magnitude and
opposite in direction to the force exerted
by the proton on the electron (F21).

Coulomb’s Law Example 2
How far apart must two charges of 2.00 C be placed
so that they exert a force of 1.00 N upon one another?
2.00 C = 2.00 10-6 C
F
k q1 q2
r2
N m
6
2
(8.99 10
)( 2.0 10 C )
2
k q1 q2
C
r

F
1.00 N
 0.190 m  19.0 cm
2
9
Conservation of Electric Charge
Benjamin Franklin
(1706 - 1790)
“The total charge of the universe
is constant.”
The total charge of any closed
system (a system which does not
allow matter in or out) is constant.
An amber rod does not attract
paper scraps.
The amber rod rubs electrons
off of the fur.
After rubbing the amber rod with Afterward, the rod and fur have
charges which are equal and
fur it does attract paper scraps.
opposite in compliance with
conservation of charge.
That still does not explain
how the rod attracts the paper!
Polarization
The attraction of neutral paper scraps to a charged
amber rod can be explained as polarization.
A charged object placed
near an uncharged object
causes the atoms in the
uncharged object to align
so that their opposites
face the incoming charge.
This causes an attraction
between the two objects.
Polarization does not charge the object. It creates an
uneven distribution of charge in the neutral object.
Insulators and Conductors
A conductor of electricity allow charge to move
freely through it.
Metals (copper, iron, gold…) are good conductors.
An insulator does not allow charge to move freely
through it.
Wood, air, most plastics, water are good insulators.
A general rule of thumb is that a material is both a
conductors of heat and electricity, or an insulator of
heat and electricity.
Charging a Conductor or an Insulator
Either a conductor or an insulator can have
charge transferred to it at some point of contact.
The transferred charge is of the same sign and attempts to
repel. Since it is able to move freely in the conductor, the
charge almost instantaneously spreads to its outer edge.
However the insulator does not support charge mobility so
the charge remains localized at the point of contact.
Conductor
Insulator
Coulomb’s Law for Multiple Charges – Ex. 1
What is the magnitude and direction of the net force
acting on charge q1?
q1 = -2.4 C q2 = -3.6 C
12 cm
q3 = -5.9 C
18 cm
6
6
2
.
4

10
C
3
.
6

10
C
9 N m
F12  8.99 10
 5.39 N
2
2
C
(0.12 m)
F12 direction
2
6
6
2
.
4

10
C
5
.
9

10
C
9 N m
F13  8.99 10
 1.41 N
2
2
C
(0.30 m)
F13 direction
2
Fnet = F12 + F13 = 5.39 + 1.41 = 6.80 N
Fnet direction
Coulomb’s Law for Multiple Charges – Ex. 2
Charges q1 and q2 are separated by 24 cm. How far
from charge q1 should the third charge be placed so
that it experiences no net force?
q1 = +8.4 C q3 = -1.2 C
q2 = +6.1 C
?
24 cm
F31 direction
F32 direction
we require that F31 and F32 are opposite in direction but equal
in magnitude:
F31 = F32
We represent the distance between q1 and q3 as x
Then the distance between q2 and q3 is: 24 - x
Coulomb’s Law for Multiple Charges – Ex. 2
q1 = +8.4 C q3 = -1.2 C
q2 = +6.1 C
?
24 cm
cancel
like
terms:
F31 = F32
k
take square
root of both
sides:
q3 8.4 106 C
x
2
8.4 10
2
x
k
6
q3 6.1106 C
(24  x) 2
6
6.110

2
(24  x)
2.9 10 3 2.5 10 3

x
24  x
Coulomb’s Law for Multiple Charges – Ex. 2
q1 = +8.4 C q3 = -1.2 C
q2 = +6.1 C
?
24 cm
0.0029(24  x)  0.0025 x
0.0696  0.0029x  0.0025x
0.0696  0.0054x
x = 12.9 cm
Coulomb’s Law for Non-Collinear Charges
q1 exerts equal and
opposite forces on
charges q2, q3, and q4.
F31
F21
If all charges are the
same sign, then q1 has
repulsive forces acting
on it from charges
q2, q3, and q4.
F41
The individual force vectors
acting on q1 combine to
give the resultant force
vector q1 experiences
(represented here by F1net).
Coulomb’s Law for Non-Collinear Charges
q1 = -6.0 C
12.0 cm
19.0 cm
q3 = +2.0 C
12.0 cm
q2 = -6.0 C
12.0 cm
12.0 cm
q1 = -6.0 C
r31
19.0 cm
r32
q2 = -6.0 C
F31
F32
q3 = +2.0 C
What magnitude and direction
of net force is exerted on q3
due to q1 and q2?
F31 and F32 act along the straight lines
connecting q3 to q1 and q2 respectively.
For the right triangles: a2 +b2 = c2
so r31 =r32 = (12.02 + 19.02)1/2 = 22.5 cm
2
6
6
N

m
(6.0

10
C)(2.0

10
C)
9
F31  F32  (8.99 10
)
 2.13 N
2
2
C
(0.225m)
Coulomb’s Law for Non-Collinear Charges
F31
12.0 cm


19.0 cm
12.0 cm
 12.0 
o
θ  tan 

32.3

 19.0 
1
F32
F31 x = F31cos  = 2.13 N(cos32.3o) = 1.80 N (to the left)
F31 y
F32 y
F31
F31 x = F32 x
F32
F32 x = F32cos  = 2.13 N(cos32.3o) = 1.80 N (to the left)


Fnet x = 1.80 + 1.80 = 3.60 N (to the left)
Fnet = Fnet x = 3.60 N
left
F31 y = F31sin  = 2.13 N(sin32.3o) = 1.14 N (up)
F32 y = F32sin  = 2.13 N(sin32.3o) = 1.14 N (down)
Fnet y = 1.14 - 1.14 = 0 N
Coulomb’s Law for Multiple Charges - Ex. 2
q1 = +7.3 C
17.0 cm
8.0 cm
25.2o
q2 = -15.6 C
q4 = +6.1 C
18.8 cm
17.0 cm
8.0 cm
q3 = +5.0 C
F43
F42
F41
r42  17.02  8.02  18.8 cm
 8.0 
o
  tan 
  25.2
 17.0 
1
What is the net force
acting on q4?
What angle does this
force make with the
positive x-axis?
Coulomb’s Law for Multiple Charges - Ex. 2
q1 = +7.3 C
q4 = +6.1 C
17.0 cm
8.0 cm
q2 = -15.6 C
8.0 cm
17.0 cm
q3 = +5.0 C
What is the net force
acting on q4?
What angle does this
force make with the
positive x-axis?
6
6
6
.
1

10
C
7
.
3

10
C
9 N m
F41  8.99 10
 13.9 N
2
2
C
0.17 m 
2
6
6
6
.
1

10
C
5
.
0

10
C
9 N m
F43  8.99 10
 42.8 N
2
2
C
0.08 m 
2 6.1 10 6 C 15.6  10 6 C
9 N m
F42  8.99 10
 24.2 N
2
2
C
0.188 m 
2
Coulomb’s Law for Multiple Charges - Ex. 2
F42 =24.2 N
F43 =42.8 N
F41 = 13.9 N
24.2 N
25.2 o
F42(y) =24.2 sin(25.2)
=10.3 N (negative y)
F42(x) = 24.2 cos (25.2) = 21.9 N (negative x)
Fnet (x)
13.9
-21.9
-8.0 N
Fnet (y)
42.8
-10.3
32.5 N
Fnet   8.0 2  32.5 2  33.4 N
Fnet(y)
= +32.5 N
Fnet
= 33.4 N
Fnet(x)
= -8.0 N
Coulomb’s Law for Multiple Charges - Ex. 2
Fnet
= 33.4 N
Fnet(y)
= +32.5 N

Fnet(x)
= -8.0 N
Fnet
= 33.4 N
76.2o
What angle does Fnet make
with the positive x-axis?
 32.5 
o
θ  tan 
  76.2
 8.0 
1
180 - 76.2 = 103.8o angle
with the positive x-axis