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Syllabus and teaching strategy
General Physics (PHY 2140)
Introduction
Lecturer:
Dr. Alan A. Sebastian, Physics Building
Phone: 313-577-2720 (to leave a message with a secretary)
e-mail: [email protected], Web: http://www.physics.wayne.edu/~alan
Office Hours:
TBD, Physics Building, , or by appointment.
Grading:
Reading Quizzes
Quiz section performance/Homework
Best Hour Exam
Second Best Hour Exam
Final
¾ Syllabus and teaching strategy
¾ Electricity and Magnetism
• Properties of electric charges
• Insulators and conductors
• Coulomb’s law
bonus
10%
25%
25%
40%
Reading Quizzes:
It is important for you to come to class prepared!
Homework and QUIZ Sessions:
The quiz sessions meet once a week; quizzes will count towards your grade.
Hour Exams and Final Exam:
There will be THREE (3) Hour Exams and one Final Exam.
Online Content:
Lectures will be made available to you as a supplemental reference.
Lecture 1. Chapter 15
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15.1 Properties of Electric Charges Discovery
Introduction
Knowledge of electricity dates back to Greek antiquity
(700 BC).
Began with the realization that amber (fossil) when
rubbed with wool, attracts small objects.
This phenomenon is not restricted to amber/wool but may
occur whenever two nonnon-conducting substances are
rubbed together.
Observation of “Static Electricity”
Electricity”
„
„
A comb passed though hair attracts small pieces of paper.
An inflated balloon rubbed with wool.
“Electrically charged”
charged”
„
„
Rub shoes against carpet/car seat to charge your body.
Remove this charge by touching another person/a piece of
metal.
Two kinds of charges
„
Named by Benjamin Franklin (1706(1706-1790) as positive and
negative.
negative.
Like charges repel one another and unlike charges
attract one another.
another.
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1
15.1 Properties of Electric Charges
Nature of Electrical Charge
15.1 Properties of Electric Charges
Quantization
Origin of charge is at the atomic level.
„
„
Robert Millikan found, in 1909, that charged objects may only have
have
an integer multiple of a fundamental unit of charge.
Nucleus : “robust”
robust”, positive.
positive.
Electrons : mobile, negative.
negative.
„
„
Usual state of the atom is neutral.
neutral.
Charge has natural tendency to be transferred between
unlike materials.
Electric charge is however always conserved in the
process.
„
„
„
„
„
„
Units
Charge is not created.
created.
Usually, negative charge is transferred from one object to the
other.
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„
„
5
15.2 Insulators and Conductors –Material
classification
„
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Identify substances or materials that can be classified as
„
Conductors ?
„
Insulators?
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.
„
In SI, electrical charge is measured in coulomb ( C).
The value of |e|
|e| = 1.602 19 x 10-19 C.
Mini-quiz:
Materials/substances may be classified according to their capacity
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.
„
Charge is quantized.
quantized.
An object may have a charge ±e, or ± 2e, or ± 3e, etc. but not ±1.5e.
Proton has a charge +1e.
+1e.
Electron has a charge –1e.
1e.
Some particles such a neutron have no (zero) charge.
A neutral atom has as many positive and negative charges.
Silicon and germanium are semiconductors used widely in the
fabrication of electronic devices.
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2
15.2 Insulators and Conductors – Charging
by Conduction.
15.2 Insulators and Conductors –
Earth/Ground.
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.
charged.
This process is referred as charging by conduction.
conduction.
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15.2 Insulators and Conductors – Charging
by Induction.
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15.2 Insulators and Conductors – Charging
by Induction.
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.
induction.
Charging by induction requires no contact
with the object inducing the charge.
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When a conductor is connected to Earth with a
conducting wire or pipe, it is said to be grounded.
grounded.
Earth provides a quasi infinite reservoir of electrons: can
accept or supply an unlimited number of electrons.
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.
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 use a positively
charged glass rod
instead?
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3
15.2 Insulators and Conductors –
Polarization.
Mini-quiz
A positively charged object hanging from a string is brought near
near a non
conducting object (ball). The ball is seen to be attracted to the
the object.
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.
1.Explain
1.Explain why it is not possible to determine whether the object is
is
negatively charged or neutral.
2.What
2.What additional experiment is needed to reveal the electrical charge
charge
state of the object?
?
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Explain why it is not possible to determine whether
the object is negatively charged or neutral.
Bring a known neutral ball
near the object and observe
whether there is an attraction.
?
+
Attraction between a charged object and a neutral object subject
to polarization.
++ -++ - -
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What additional experiment is needed to reveal the
electrical charge state of the object?
„
Attraction between objects of unlike charges.
„
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Two Experiments:
Two possibilities:
„
+
„
+
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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.
-+++
-- -++
++
0
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4
15.3 Coulomb’s Law - Observation
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,
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.
sign.
15.3 Coulomb’s Law – Mathematical
Formulation
q q
F = ke 1 2 2
r
ke known as the Coulomb constant.
Value of ke depends on the choice of units.
SI units
„
„
„
„
q1
Force: the Newton (N)
Charge: the coulomb ( C).
Current: the ampere (A =1 C/s).
Distance: the meter (m).
Experimentally measurement: ke = 8.9875×109 Nm2/C2.
Reasonable approximate value: ke = 8.99×109 Nm2/C2.
q2
r
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Example
Charge and Mass of the Electron, Proton
and Neutron.
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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 μc
A very small fraction of the total available charge
19
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5
15.3 Coulomb’s Law – Remarks
Mini-Quiz
Name the first action at a distance force you have
encountered in physics so far.
The electrostatic force is often called Coulomb force.
It is a force (thus, a vector):
vector):
„
„
a magnitude
a direction.
r
F 21
+
q1
+
q2
r
F 21
+
q1
F 21
F 21
q2
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.
Example: Electrical Force
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.
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’
Coulomb’s law.
F = ke
q1 q2
r2
q1 =-1.60x10-19 C
q2 =1.60x10-19 C
r = 5.3x10-11 m
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6
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.
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’
Coulomb’s law.
q1 =-1.60x10-19 C
q2 =1.60x10-19 C
r = 5.3x10-11 m
Solution:
Fe = ke
e
2
r2
= 8.99 × 10
9 Nm 2
C2
(1.6 ×10
( 5.3 ×10
−19
−11
)
m)
C
Superposition Principle
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.
principle.
2
2
= 8.2 × 10 −8 N
Attractive force with a magnitude of 8.2x10-8 N.
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Consider three point charges at the corners of a triangle, as shown
shown
below. Find the resultant force on q3.
Example: Using the Superposition Principle
y
Consider three point charges at the corners of a triangle, as shown
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
y
q2
3.00 m
q1 +
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q2
4.00 m
F32
+
q3
F31
-
3.00 m
q1 +
F31
-
26
37.0o
4.00 m
F32
+
q3
37.0o
x
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’
Coulomb’s
law.
x
27
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7
Lightning Review
Consider three point charges at the corners of a triangle, as shown
shown
below. Find the resultant force on q3.
y
q2 3.00 m
F32 = ke
q3 q2
F31 = ke
q3 q1
r
37.0o
+
q3
5.00 m
q1 +
Solution:
F31
F32
4.00 m
2
r2
x
( 5.00 ×10 C )( 2.00 ×10 C ) = 5.62 ×10
−9
= 8.99 ×109
= 8.99 ×109
Nm 2
C2
Nm 2
C2
1. Properties of electric charge
9 two types: positive and negative
9 always conserved and quantized
( 4.00m )
(
2. Insulators and conductors
9 charges move freely in conductors; opposite
is true for insulators
9 conductors can be charged by conduction and
induction; insulators can be polarized
−9
)(
5.00 ×10−9 C 6.00 ×10−9 C
( 5.00m )
−9
2
2
) = 1.08 ×10
−8
N
N
Fx = − F32 + F31 cos 37.0o = 3.01× 10−9 N
Review Problem: OperatingOperating-room personnel must wear special conducting
shoes while working around oxygen. Why? What might
happen if personnel wore ordinary rubber shoes
(sneakers)?
Fy = F31 sin 37.0o = 6.50 × 10−9 N
F = Fx2 + Fy2 = 7.16 × 10−9 N
θ = 65.2o
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Recall that units can be manipulated:
q1 q2
r
2
[ F ] = [ ke ]
[ Newton] = [ ke ]
Electric forces act through space even in the absence of
physical contact.
Suggests the notion of electrical field (first introduced
by Michael Faraday (1791(1791-1867).
An electric field is said to exist in a region of space
surrounding a charged object.
If another charged object enters a region where an
electrical field is present, it will be subject to an electrical
force.
[ q1 ][ q2 ]
2
[r ]
[Coulomb][Coulomb]
2
[ meter ]
[ ke ] = N ⋅ m
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15.4 Electric Field - Discovery
Example: Fun with units
F = ke
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2
C2
31
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8
15.4 Electric Field – Quantitative Definition
15.4 Electric Field – Quantitative Definition (2)
Direction defined as the direction of the electrical force
exerted on a small positive charge placed at that
location.
A field : generally changes with position (location)
A vector quantity : magnitude and direction.
Magnitude at a given location
„
Expressed as a function of the force imparted by the field on
a given test charge.
E =
E
F
+
E
qo
- -
-
-
-
-
+
-
+
-
- -
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15.4 Electric Field – Electric Field of a
Charge “q”
Given
One finds
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F = ke
E = ke
+
+ +
+ +
+ + +
+ +
+
-
+
+
+
+
+
+
+ +
+
+ +
+
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• If q>0, field at a given point is radially outward from q.
q qo
r
r2
+
q
q
qo
E
• If q<0, field at a given point is radially inward from q.
r
r2
q
35
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E
qo
36
9
Problem-Solving Strategy
Example:
An electron moving horizontally passes between two
horizontal planes, the upper plane charged negatively,
and the lower positively. A uniform, upwardupward-directed
electric field exists in this region. This field exerts a force
on the electron. Describe the motion of the electron in
this region.
Electric Forces and Fields
Units:
„
For calculations that use the Coulomb constant, ke, charges must
be in coulombs, and distances in meters.
Conversion are required if quantities are provided in other units.
units.
Applying Coulomb’
Coulomb’s law to point charges.
„
It is important to use the superposition principle properly.
Determine the individual forces first.
Determine the vector sum.
Determine the magnitude and/or the direction as needed.
-
-
37
-
-
- -
-
+
+
+ +
-
„
„
„
vo
+ +
+
-
-
- -
-
+
+ + + + + + + + + + + + + + + + +
+
Ex = 0
+ + + + + + + + + + + + + + + + +
38
vo
No electric field
No force
No acceleration
Constant horizontal velocity
+
+ +
-
- - - - - - - - - - - - - - - - -
vo
+
+ + + + + + + + + + + + + + + + +
Observations:
Vertically:
Fx = 0
„
ax = 0
„
„
vx = vo
„
x = vot
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+
- - - - - - - - - - - - - - - - -
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- - - - - - - - - - - - - - - - -
Observations:
Horizontally:
„
-
-
+
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- -
Constant electric field
Constant force
Constant acceleration
Vertical velocity increase
linearly with time.
E y = Eo
Fy = qo Eo
a y = qo Eo / mo
v y = qo Eo t / mo
y=
39
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1
qo Eot 2 / mo
2
40
10
-
-
- -
-
- - - - - - - - - - - - - - - - -
Example: Electric Field Due to Two Point Charges
Question:
Charge q1=7.00 μC is at the origin, and charge q2=-10.00 μC is on the x
axis, 0.300 m from the origin. Find the electric field at point P, which
has coordinates (0,0.400) m.
-
+
+
+ +
+
y
+ + + + + + + + + + + + + + + + +
E1
Conclusions:
The charge will follow a parabolic path downward.
Motion similar to motion under gravitational field only except the
the
downward acceleration is now larger.
0.400 m
P
q1
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E
E2
0.300 m
x
q2
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42
Question:
Charge q1=7.00 μC is at the origin, and charge q2=-10.00 μC is on the x
axis, 0.300 m from the origin. Find the electric field at point P, which
has coordinates (0,0.400) m.
Question:
Charge q1=7.00 μC is at the origin, and charge q2=-10.00 μC is on the x
axis, 0.300 m from the origin. Find the electric field at point P, which
has coordinates (0,0.400) m.
Observations:
First find the field at point P due to charge q1 and q2.
Field E1 at P due to q1 is vertically upward.
Field E2 at due to q2 is directed towards q2.
The net field at point P is the vector sum of E1 and E2.
The magnitude is obtained with
Solution:
E1 = ke
E2 = ke
q1
r12
q2
r22
( 7.00 ×10 C ) = 3.93 ×10 N / C
−6
= 8.99 ×109
Nm2
C2
( 0.400m )
5
2
(10.00 ×10 C ) = 3.60 ×10 N / C
−6
= 8.99 ×109
Nm2
C2
( 0.500m )
5
2
Ex = 53 E2 = 2.16 ×105 N / C
E y = E1 − E2 sin θ = E1 − 54 E2 = 1.05 ×105 N / C
E = ke
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q
r
E = Ex2 + E y2 = 2.4 ×105 N / C
φ = arctan( E y / Ex ) = 25.9o
2
43
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11
15.5 Electric Field Lines
15.5 Electric Field Lines (2)
A convenient way to visualize field patterns is to draw
lines in the direction of the electric field.
Such lines are called field lines.
lines.
Remarks:
1.
2.
Electric field lines of single positive (a) and (b) negative
charges.
a)
Electric field vector, E, is tangent to the electric field lines at
each point in space.
The number of lines per unit area through a surface
perpendicular to the lines is proportional to the strength of the
the
electric field in a given region.
b)
+ q
- q
E is large when the field lines are close together and small
when far apart.
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15.5 Electric Field Lines (3)
2.
3.
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46
15.5 Electric Field Lines (4)
Electric field lines of a dipole.
dipole.
Rules for drawing electric field lines for any charge
distribution.
1.
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Lines must begin on positive charges (or at infinity) and must
terminate on negative charges or in the case of excess charge
at infinity.
The number of lines drawn leaving a positive charge or
approaching a negative charge is proportional to the magnitude
of the charge.
No two field lines can cross each other.
47
+
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-
48
12
Application: Measurement of the atmospheric electric field
The electric field near the surface of the Earth is about
100 N/C downward. Under a thundercloud, the electric
field can be as large as 20000 N/C.
How can such a (large) field be measured?
A
A
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15.6 Conductors in Electrostatic Equilibrium
50
15.6 Conductors in Electrostatic Equilibrium
Good conductors (e.g. copper, gold) contain charges
(electron) that are not bound to a particular atom, and
are free to move within the material.
When no net motion of these electrons occur the
conductor is said to be in electroelectro-static equilibrium.
equilibrium.
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Properties of an isolated conductor (insulated from the
ground).
1. Electric field is zero everywhere within the conductor.
2. Any excess charge field on an isolated conductor resides
entirely on its surface.
3. The electric field just outside a charged conductor is
perpendicular to the conductor’
conductor’s surface.
4. On an irregular shaped conductor, the charge tends to
accumulate at locations where the radius of curvature of the
surface is smallest – at sharp points.
51
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52
13
1.
Electric field is zero everywhere within the conductor.
2.
This property is a direct result of the 1/r2 repulsion
between like charges.
If an excess of charge is placed within the volume, the
repulsive force pushes them as far apart as they can go.
They thus migrate to the surface.
If this was not true,
true, the field inside would be finite.
Free charge there would move under the influence of the
field.
A current would be induced.
The conductor would not be in an electrostatic state.
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3.
53
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4.
The electric field just outside a charged conductor is
perpendicular to the conductor’
conductor’s surface.
Any excess charge field on an isolated conductor resides entirely
entirely
on its surface.
54
On an irregular shaped conductor, the charge tends to accumulate at
locations where the radius of curvature of the surface is smallest
smallest – at
sharp points.
Consider, for instance, a conductor fairly flat at one end and relatively
relatively pointed at the
other.
Excess of charge move to the surface.
Forces between charges on the flat surface, tend to be parallel to the surface.
Those charges move apart until repulsion from other charges creates
creates an equilibrium.
At the sharp ends, the forces are predominantly directed away from
from the surface.
There is less of tendency for charges located at sharp edges to move away from one
another.
Produces large fields (and force) near sharp edges.
If not true, the field would have components parallel to
the surface of the conductor.
This field component would cause free charges of the
conductor to move.
A current would be created.
There would no longer be a electroelectro-static equilibrium.
-
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-
56
14
Remarks
Faraday’s ice-pail experiment
Property 4 is the basis for the use of lightning rods near
houses and buildings. (Very important application)
„
„
Most of any charge on the house will pass through the sharp
point of the lightning rod.
First developed by B. Franklin.
+++++
+
+
+
+
+
+
+
-
-
+
+++++
-
-
-
+
+
+
+
+
+
+
+
+
+
+
+
+
-
+
+
+
-
-
+
+
+
-
+
+
+
+
+
+
+
Demonstrates that the charge resides on the surface of a conductor.
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57
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58
Mini-quiz
Question:
Suppose a point charge +Q is in empty space. Wearing rubber gloves,
gloves, we sneak up and surround the
charge with a spherical conducting shell. What effect does this have on the field lines of the
charge?
Question:
Suppose a point charge +Q is in empty space. Wearing rubber gloves,
gloves,
we sneak up and surround the charge with a spherical conducting shell.
What effect does this have on the field lines of the charge?
Answer:
Negative charge will build up on the inside of the shell.
Positive charge will build up on the outside of the shell.
There will be no field lines inside the conductor but the field lines will remain outside the shell.
?
+
+
-
+
-
+ q
-
+
-
+
59
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+
-
-
+
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+
-
-
+
+ q
+
+
-
+
+
60
15
Question:
Is it safe to stay inside an automobile during a lightning storm?
storm? Why?
Answer:
Yes. It is. The metal body of the car carries the excess charges on its
external surface. Occupants touching the inner surface are in no
danger.
Mini-Quiz
Question:
Is it safe to stay inside an automobile during a lightning
storm? Why?
SAFE
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62
16