7/6/2015 1 University Physics (PHY 2326) Introduction Syllabus and teaching strategy Electricity and Magnetism • Properties of electric charges • Insulators and conductors • Coulomb’s law Lecture 1. Chapter 21
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7/6/2015 1
University Physics (PHY 2326)
Introduction
Syllabus and teaching strategy
Electricity and Magnetism
• Properties of electric charges
• Insulators and conductors
• Coulomb’s law
Lecture 1.
Chapter 21
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Lecturer: Dr. Sunil Karna, Room 119 Hill Hall,
Phone: 361-593-2624, e-mail: [email protected], Web: http://www.tamuk.edu/artsci/physics_geosci/
Meetings: MTWR 10:30AM – 12:30PM at Hill Hall 119
Office Hours: MTW 2:30-5:00 PM, Hill Hall 208
Grading: Homework 15%
Regular Tests 60%
Final 25%
Attendance maximum 3 points (see details of syllabus)
Class Rules and Regulations: No electronic devices should be used without instructor permission. It is important for you to come to class prepared!
Homework Sessions: Homework is assigned every week and must be submitted on due time.
Final Exam: There will be THREE (2) Hour Final Exam on 08/15 Thursday at 10.30 am – 12.30 pm.
Online Content: Lecture Online will be made available to you as a supplemental reference.
Syllabus and teaching strategy
When A comb passed though hair, it attracts small pieces of paper.
Two kinds of charges Named by Benjamin Franklin (1706-1790) as positive
and negative.
Like charges repel one another and unlike charges attract one another.
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Charge has natural tendency to be transferred between unlike materials.
Electric charge is however always conserved in the process. Charge is not created.
Usually, negative charge is transferred from one object to the other.
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Robert Millikan found, in 1909, that charged objects may only have an integer multiple of a fundamental unit of charge.
Charge is quantized.
An object may have a charge e, or 2e, or 3e, etc but not say 1.5e.
Proton has a charge +1e.
Electron has a charge –1e.
Some particles such a neutron have no (zero) charge.
A neutral atom has as many positive and negative charges.
Units
In SI, electrical charge is measured in coulomb ( C).
The value of |e| = 1.602 19 x 10-19 C.
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Materials/substances may be classified according to their capacity to carry or conduct electric charge
Conductors are material in which electric charges move freely.
Insulator are materials in which electrical charge do not move freely.
Glass, Rubber are good insulators.
Copper, aluminum, and silver are good conductors.
Semiconductors are a third class of materials with electrical properties somewhere between those of insulators and conductors.
Silicon and germanium are semiconductors used widely in the fabrication of electronic devices.
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Consider negatively charge rubber rod brought into contact with a neutral conducting but insulated sphere.
Some electrons located on the rubber move to the sphere.
Remove the rubber rod.
Excess electrons left on the sphere. It is negatively charged.
This process is referred as charging by conduction.
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When a conductor is connected to Earth with a conducting wire or pipe, it is said to be grounded.
Earth provides a quasi infinite reservoir of electrons: can accept or supply an unlimited number of electrons.
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Consider a negatively charged rubber rod brought near a neutral conducting sphere insulated from the ground.
Repulsive force between electrons causes redistribution of charges on the sphere.
Electrons move away from the rod leaving an excess of positive charges near the rod.
Connect a wire between sphere and Earth on the far side of the sphere.
Repulsion between electrons cause electrons to move from sphere to Earth.
Disconnect the wire.
The sphere now has a positive net charge.
This process is referred as charging by induction.
Charging by induction requires no contact with the object inducing the charge.
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Q: How does this mechanism work if we bring a positively charged glass rod instead?
Polarization is realignment of charge within individual molecules.
Produces induced charge on the surface of insulators.
how e.g. rubber or glass can be used to supply electrons.
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A positively charged object hanging from a string is brought near a non conducting object (ball). The ball is seen to be attracted to the object.
1.Explain why it is not possible to determine whether the object is negatively charged or neutral.
2.What additional experiment is needed to reveal the electrical charge state of the object?
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? +
Two possibilities:
Attraction between objects of unlike charges.
Attraction between a charged object and a neutral object subject to polarization.
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- +
+ - - - -
+
+ + +
Two Experiments:
Bring known neutral ball near the object and observe whether there is an attraction.
Bring a known negatively charge object near the first one. If there is an attraction, the object is neutral, and the attraction is achieved by polarization.
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? 0
- + + +
+
+-
-+- +
-
Charles Coulomb discovered in 1785 the fundamental law of electrical force between two stationary charged particles.
An electric force has the following properties:
Inversely proportional to the square of the separation, r, between the particles, and is along a line joining them.
Proportional to the product of the magnitudes of the charges |q1| and |q2| on the two particles.
Attractive if the charges are of opposite sign and repulsive if the charges have the same sign.
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q1 q2
r
ke known as the Coulomb constant.
Value of ke depends on the choice of units.
SI units
Force: the Newton (N)
Charge: the coulomb ( C).
Current: the ampere (A =1 C/s).
Distance: the meter (m).
Experimentally measurement: ke = 8.9875109 Nm2/C2.
Reasonable approximate value: ke = 8.99109 Nm2/C2.
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Charge and Mass of the Electron, Proton
and Neutron.
Particle Charge ( C) Mass (kg)
Electron -1.60 10-19 9.11 10-31
Proton +1.60 10-19 1.67 10-27
Neutron 0 1.67 10-27
1e = -1.60 10-19 c
Takes 1/e=6.6 1018 protons to create a total charge of 1C
Number of free electrons in 1 cm3 copper ~ 1023
Charge obtained in typical electrostatic experiments with rubber or glass 10-6 C = 1 mc
A very small fraction of the total available charge
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The electrostatic force is often called Coulomb force.
It is a force: Has a magnitude And a direction.
Second example of action at a distance.
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Question:
The electron and proton of a hydrogen atom are separated (on the average) by a distance of about 5.3x10-11 m. Find the magnitude of the electric force that each particle exerts on the other.
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Question:
The electron and proton of a hydrogen atom are separated (on the average) by a distance of about 5.3x10-11 m. Find the magnitude of the electric force that each particle exerts on the other.
21 7/6/2015
Observations:
We are interested in finding the magnitude of the force between two
particles of known charge, and a given distance of each other.
The magnitude is given by Coulomb’s law.
q1 =-1.60x10+19 C
q2 =1.60x10+19 C
r = 5.3x10+11 m
Observations:
We are interested in finding the magnitude of the force between two particles of known charge, and a given distance of each other.
The magnitude is given by Coulomb’s law.
q1 =-1.60x10-19 C
q2 =1.60x10-19 C
r = 5.3x10-11 m
Solution:
Attractive force with a magnitude of 8.2x10-8 N.
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2
2
22 19
9 8
2211
1.6 108.99 10 8.2 10
5.3 10
Nme e C
CeF k N
r m
From observations: one finds that whenever multiple charges are present, the net force on a given charge is the vector sum of all forces exerted by other charges.
Electric force obeys a superposition principle.
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Consider three point charges at the corners of a triangle, as shown below. Find the resultant force on q3 if
q1 = 6.00 x 10-9 C
q2 = -2.00 x 10-9 C
q3 = 5.00 x 10-9 C
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+ x
y
- + q2
q1
3.00 m
4.00 m
q3
F32
F31
37.0o
Observations:
The superposition principle tells us that the net force on q3 is the vector sum of the forces F32 and F31.
The magnitude of the forces F32 and F31 can calculated using Coulomb’s law.
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+ x
y
- + q2
q1
3.00 m
4.00 m
q3
F32
F31
37.0o
2
2
2
2
9 9
3 2 9 9
32 22
9 9
3 1 9 8
31 22
9
32 31
9
31
2 2 9
5.00 10 2.00 108.99 10 5.62 10
4.00
5.00 10 6.00 108.99 10 1.08 10
5.00
cos37.0 3.01 10
sin 37.0 6.50 10
7.16 10
Nme C
Nme C
o
x
o
y
x y
C Cq qF k N
r m
C Cq qF k N
r m
F F F N
F F N
F F F N
65.2o
26 7/6/2015
Consider three point charges at the corners of a triangle, as shown
below. Find the resultant force on q3.
5.00 m
Solution:
+ x
y
- + q2
q1
3.00 m
4.00 m q3
F32
F31
37.0o
27
28
29
1. Opposite charges attract
2. “Like” charges repel
30
Newton’s Law of Gravitation
rr
mGmF ˆ
2
21
31
Millikan Oil Drop Experiment
In 1910, Millikan was able to measure the charge of the electron Recall: Atom made up of nucleus and clouds of
electrons outside nucleus
Recall: nucleus: made up of protons and neutrons. Protons have charge equivalent to electrons. Neutrons are neutral
Smallest charge possible is
1.602 x 10-19 Coulombs (C) aka e
32
Abbreviation: C
Amount of charge through a cross-section of wire in 1 second when there is 1 Ampere (A) of current.
(We’ll cover the amp later)
33
Quarks– particles which make up the proton and neutron
The “up” quark has charge of +2/3 e
The “down” quark has charge of -1/3 e
They don’t count because there are no “free” quarks. They always are confined in a particle
Proton- uud Neutron-udd
34
Particle Symbol Charge in units of e
Electron e, e- , b- -1
Proton p +1
Neutron n 0
Anti-electron (positron) b- +1
Anti-proton -1
Anti-neutron 0
Alpha particle a or 4He++ +2
Up quark u +2/3
Down quark d -1/3
Any element of atomic
number, z
ZNX z
n
p
35
Conductors– charges move freely
Insulators—charges cannot move easily
Semiconductors—charges only move freely when certain conditions are met (heat, sufficient voltage, etc)
Superconductors-charges move effortlessly and cannot be stopped once they are moving
36
Energye b
What is X?
HeThU X
4
2
234238
92
37
Charles Augustin de Coulomb used a torsion pendulum to establish “Coulomb’s Law”
rr
qqkF ˆ
2
21
38
k is equal to 1 for electrostatic units
We use SI so in this case k is equal to
8.98 x 109 N·m2/C2 k is actually formed from two other
constants p =3.1415928….
e0 = 8.854 x 10-12 C2/(N·m2) Called the permittivity of free space
2
29
0 C
mN109
4
1
pek
39
If the product, q1q2 ,is negative then the force is attractive
If the product, q1q2 ,is positive then the force is repulsive
Your book uses the absolute value in the case of determining magnitude of force.
40
The force is
directed along the
shortest distance
between two points,
just like gravitation.
In the case to the
right, the force is
directed along lines
from the center of
the spheres.
41
Sometimes difficult problems can be made simple by using the principle of superposition.
Problem: Find the electric field of sphere with a hole in
it.
= -
The E-field of the
whole sphere
The E-field of a
sphere with a
hole in it
The E-field of a
small sphere
The principle of superposition is one of the most
powerful problem solving tools that you have
42
Why do I need this concept? Assume that you have a charge in space: we need a general
expression for when we add another charge, q. What force will be exerted on q?
Have I seen this before? Remember F=mg
Our new expression: F=qE
E is the electric field that is present in the space wherein q was placed. E is usually the result of other charges which previously have been located in the same space.
Since E=F/q then the units are newtons per coulomb (N/C). Another set of units is volts per meter (V/m).
43
Rules for Field Lines
1. Electric field lines point to negative charges
2. Electric field lines extend away from positive charges
3. Equipotential (same voltage) lines are perpendicular to a line tangent of the electric field lines
44
For the rest of this chapter and chapter 22, we will investigate how to calculate the electric field
rr
qkE ˆ
2
This quantity represents an infinite set of
vector quantities, in other words, a vector
field.
45
In order to calculate this quantity, we need to know how the charge creating the electric field is distributed in space
The geometrical distribution of the charge will have the biggest effect on the magnitude and direction of the electric field
rr
dqkEd ˆ
2
46
Point charge: All charge resides at a geometric point so there is no geometrical distribution
r-hat points out from the geometric point
rr
qkE
and
qdq
ˆ2
47
Line charge: All charge resides along a line
A charge density must be created: a mathematical description of the geometrical distribution of the charge
For a line charge, this is called the linear charge density, l (units C/m)
2r
dskdE
and
dsdq
ds
dq
l
l
l
drds
drdds
dzordyordxds
p
2
48
Surface charge: All charge resides on top or under a surface (or area)
surface charge density, s (units C/m2)
2r
dakdE
and
dadq
da
dq
s
s
s
dxdzordydzordxdyda
rdrda
rdrdda
p
2
49
Volume charge: All charge resides in a particular volume
volume charge density, r (units C/m3)
2r
dVkdE
and
dVdq
dV
dq
r
r
r
dzrdrdzdrdrdV
drrdV
dddrrdV
dxdydzdV
p
p
2
4
sin
2
2
50
A pair of charges, one “+” and the other “-” which are separated by a short distance
Electric dipole is represents the electrical distribution of many molecules
Positive and negative are relative concepts: “positive” means less negative charges than “negative”
51
Why is this important? Principle of microwave oven, amongst other applications
Recall: t=r x F If F=qE, then t=qE r sin ( where is the angle between E and
r)
Let d=distance between two charges
Electric Dipole Moment Necessary because the charge and distance between charges are
easy to characterize
p=qd Note: p is a vector in the direction pointing from 1 charge to the other
t=pE sin or t=p x E
52
Recall that DW=-DU
DW=F·r=Fr cos=qEd cos
DU=-qd E cos
U = - p·E which is the potential energy of a dipole in an electric field
53
Two free point charges +q and +4q are located a distance, L, apart. A third charge is placed so that the whole system is in equilibrium
Find the location, magnitude, and sign of the third charge
54
+q +4q L
55
+q +4q L 1 2 3
There are 3 positions available for the new charge (marked in
blue).
Which position do you think is the most likely position and more
importantly why?
Positions 1 & 3 are untenable since by equilibrium, we mean the
force on each particle is zero. If 1, for example, then there will be
a repulsive force on +4q that cannot be overcome without creating
a strong attractive force on +q
56
+q +4q L 2
Call the new charge, q0, and let it be distance
x from +q
The free-body diagram shows relationship of
the two forces, F1 and F2 on the new charge
x
F1 F2
xL
xxL
xxL
xLx
xL
qqk
x
qqk
3
2
4)(
)(
41
)(
4
22
22
2
0
2
0
57
+q +4q L 2
The free-body diagram shows relationship of
the two forces, F1 and F4 on charge +q
x
F1
F4
L
q
L
q
xL
whereL
qqk
x
qqk
9
4
49
)(
4
9
30
)(
4
0
0
22
0
22
0
58
An electron is projected with an initial speed of v0, 1.6 x 106 m/s in the uniform field between the parallel plates shown below. Assume that the plates are uniform and the field is directed vertically downward and the field outside the plates is zero. The electron enters the field at a point midway between the two plates
If the electron just misses the upper plate as it emerges
from the field, find the magnitude of the electric field
e-
2.0 cm
1.0 cm
59
Obviously, the electron is 0.5 cm below the upper plate when entering the field so, the initial position is (0,0.5) and the final position is (2.0, 0.0)
SI: (0,0.005) and (0.02, 0)
Also, v0 is actually in the x-direction so vinitial=
(v0,0)
e-
2.0 cm
1.0 cm
60
This reminds me of a projectile problem except it is kinda upside down.
In this case, there is no acceleration in the x-direction only in the y and since F=ma=eE then a=eE/m
Where e=charge of electron and m is the mass of the electron. Since I can never remember the trajectory equation, then I must
solve for the time it takes the electron to traverse 2 cm and then plug that into the an equation relating acceleration to distance in the y-direction
e-
2.0 cm
1.0 cm
61
EC
N
E
Ee
maor
m
eEa
Ce
and
kgm
364
10602.1
)104.6)(1011.9(
10602.1
1011.9
19
1331
19
31
st
t
xtvx
8
6
00
1025.1
0)106.1(02.0
e-
2.0 cm
1.0 cm
213
28
0
0
2
/104.6
)1025.1(010.0
005.
2
1
sma
sa
myy
yaty
62
What is the magnitude and direction of the electric field at the center of the square if q= 1.0 x 10-8 C and distance, a, is 5 cm
+q -2q
+2q -q
a
a
a
a
63
+q -2q
+2q -q
a
a
a
a
Free body diagram of + test
charge in center
2q
-2q
-q q
64
2
222
2
2222
1
2
12
r
qkE
aaarwhere
r
qkE
q
q
First find the magnitudes of all the
electric fields
65
The line from +2q makes an angle of 1350 with the positive x-axis
So
E2qx=E2q cos (1350)=-E2q
cos (450)
E2qx=-0.707*E2q
E2qy=E2q sin (1350)=E2q
sin (450)
E2qy=0.707*E2q
1350
66
E-2qx=0.707*E2q
E-2qy=0.707*E2q
67
-q makes angle of 2250
E-qx=-0.707*Eq
E-qy=-0.707*Eq
-q makes angle of 3150
Eqx=0.707*Eq
Eqy=-0.707*Eq
Now we add the x and y components, respectively
68
2
22
2
22
22
2
42
2
2
a
qkE
a
qkEE
r
qkE
arwhere
r
qkE
q
q
q
2
22
2
22
22
22
22
22
222
)22(*707.0
)(*707.0
0
)(*707.0
a
qkEEE
a
qkEE
EEEEE
EEEEE
EEEEE
E
EEEEE
EEEEE
yx
qy
qqqqy
qqqqy
qyqyqyqyy
x
qqqqx
qxqxqxqxx
01 90)(tan
tan
undefined
UndefinedE
E
x
y
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