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Chapter 19
Electric Forces
and
Electric Fields
Electricity and
Magnetism, Some History

Many applications


Chinese


Macroscopic and microscopic
Documents suggests that magnetism was
observed as early as 2000 BC
Greeks


Electrical and magnetic phenomena as early as
700 BC
Experiments with amber and magnetite
2
Electricity and
Magnetism, Some History, 2

1600



William Gilbert showed electrification
effects were not confined to just amber
The electrification effects were a general
phenomena
1785

Charles Coulomb confirmed inverse
square law form for electric forces
3
Electricity and
Magnetism, Some History, 3

1820


Hans Oersted found a compass needle
deflected when near a wire carrying an
electric current
1831

Michael Faraday and Joseph Henry
showed that when a wire is moved near a
magnet, an electric current is produced in
the wire
4
Electricity and
Magnetism, Some History, 4

1873

James Clerk Maxwell used observations
and other experimental facts as a basis for
formulating the laws of electromagnetism


Unified electricity and magnetism
1888


Heinrich Hertz verified Maxwell’s
predictions
He produced electromagnetic waves
5
Electric Charges

There are two kinds of electric charges

Called positive and negative



Negative charges are the type possessed by
electrons
Positive charges are the type possessed by
protons
Charges of the same sign repel one
another and charges with opposite
signs attract one another
6
Electric Charges, 2



The rubber rod is
negatively charged
The glass rod is
positively charged
The two rods will
attract
7
Electric Charges, 3



The rubber rod is
negatively charged
The second rubber
rod is also
negatively charged
The two rods will
repel
8
More About Electric Charges

The net charge in an isolated system is
always conserved


For example, charge is not created in the
process of rubbing two objects together
The electrification is due to a transfer of
electrons from one object to another
9
Quantization of
Electric Charges

The electric charge, q, is said to be quantized



q is the standard symbol used for charge as a
variable
Electric charge exists as discrete packets
q=Ne





N is an integer
e is the fundamental unit of charge
|e| = 1.6 x 10-19 C
Electron: q = -e
Proton: q = +e
10
Conservation and Quantization
of Electric Charges, Example





A glass rod is rubbed
with silk
Electrons are transferred
from the glass to the silk
Each electron adds
a negative charge to
the silk
An equal positive charge
is left on the rod
The charges on the two
objects are ±e, or ±2e, …
11
Three objects are brought close to one another, two at a
time. When objects A and B are brought together, they
repel. When objects B and C are brought together, they
also repel. Which of the following statements are true?
a.
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20% 20% 20% 20% 20%
Objects A and C possess
charges of the same sign.
Objects A and C possess
charges of opposite sign.
All three objects possess
charges of the same sign.
One of the objects is neutral.
We need to perform
additional experiments to
determine the signs of the
charges.
O
1.
12
Conductors

Electrical conductors are materials in which
some of the electrons move relatively freely




Free electrons are not bound to the atoms
These electrons can move relatively freely through
the material
Examples of good conductors include copper,
aluminum and silver
When a good conductor is charged in a small
region, the charge readily distributes itself over the
entire surface of the material
13
Insulators

Electrical insulators are materials in
which electric charges do not move
freely


Examples of good insulators include glass,
rubber and wood
When a good insulator is charged in a
small region, the charge is unable to move
to other regions of the material
14
Semiconductors



The electrical properties of semiconductors
are somewhere between those of insulators
and conductors
Examples of semiconductor materials include
silicon and germanium
The electrical properties of semiconductors
can be changed over many orders of
magnitude by adding controlled amounts of
foreign atoms to the materials
15
Charging by Induction


Charging by induction
requires no contact with
the object inducing the
charge
Assume we start with a
neutral metallic sphere

The sphere has the same
number of positive and
negative charges
16
Charging by Induction, 2

A negatively charged
rubber rod is placed near
the sphere


It does not touch the sphere
The electrons in the
neutral sphere are
redistributed

The migration of electrons
leaves the side near the rod
with an effective positive
charge
17
Charging by Induction, 3

The sphere is
grounded


Grounded means the
conductor is connected
to an infinite reservoir
for electrons, such as
the Earth
Some electrons can
leave the sphere
through the ground
wire
18
Charging by Induction, 4



The ground wire is
removed
There will now be
more positive
charges in the
sphere
The positive charge
has been induced
in the sphere
19
Charging by Induction, 5

The rod is removed



The rod has lost none of
its charge during this
process
The electrons remaining
on the sphere
redistribute themselves
There is still a net
positive charge on the
sphere
20
Charge Rearrangement
in Insulators



A process similar to
induction can take
place in insulators
The charges within
the molecules of the
material are
rearranged
The effect is called
polarization
21
Charles Coulomb



1736 – 1806
Major contributions in
the fields of
electrostatics and
magnetism
Also investigated



Strengths of materials
Structural mechanics
Ergonomics

How people and
animals can best do
work
22
Coulomb’s Law


Charles Coulomb
measured the
magnitudes of electric
forces between two
small charged spheres
He found the force
depended on the
charges and the
distance between them
Torsion balance
23
Coulomb’s Law, 2



The electrical force between two stationary
charged particles is given by Coulomb’s Law
The force is inversely proportional to the
square of the separation r between the
particles and directed along the line
joining them
The force is proportional to the product of the
charges, q1 and q2, on the two particles
24
Point Charge

The term point charge refers to a
particle of zero size that carries an
electric charge

The electrical behavior of electrons and
protons is well described by modeling them
as point charges
25
Coulomb’s Law, Equation

Mathematically,
Fe  ke


q1 q2
r2
The SI unit of charge is the Coulomb, C
ke is called the Coulomb Constant



ke = 8.9875 x 109 N.m2/C2 = 1/(4peo)
eo is the permittivity free space
eo = 8.8542 x 10-12 C2 / N.m2
26
Coulomb's Law, Notes

Remember the charges need to be in
Coulombs

e is the smallest unit of charge





Except quarks
e = 1.6 x 10-19 C
So 1 C needs 6.24 x 1018 electrons or protons
Typical charges can be in the µC range
Remember that force is a vector quantity
27
Vector Nature of
Electric Forces



In vector form,
q1 q2
F12  ke 2 rˆ12
r
r̂12 is a unit vector
directed from q1 to
q2
The like charges
produce a repulsive
force between them
28
Vector Nature of
Electrical Forces, 2


Electrical forces obey Newton’s Third Law
The force on q1 is equal in magnitude and
opposite in direction to the force on q2



F12  F21
With like signs for the charges, the product
q1q2 is positive and the force is repulsive
With opposite signs for the charges, the
product q1q2 is negative and the force is
attractive
29
Vector Nature of
Electrical Forces, 3



Two point charges are
separated by a distance r
The unlike charges
produce an attractive
force between them
With unlike signs for
the charges, the
product q1q2 is
negative and the
force is attractive
30
A Final Note About Directions


The sign of the product of q1q2 gives the
relative direction of the force between
q1 and q2
The absolute direction is determined by
the actual location of the charges
31
Hydrogen Atom Example

The electrical force between the
electron and proton is found from
Coulomb’s Law


Fe = keq1q2 / r2 = 8.2 x 10-8 N
This can be compared to the
gravitational force between the electron
and the proton

Fg = Gmemp / r2 = 3.6 x 10-47 N
32
The Superposition Principle

The resultant force on any one particle
equals the vector sum of the individual
forces due to all the other individual
particles


Remember to add the forces as vectors
The resultant force on q1 is the vector
sum of all the forces exerted on it by
other charges: F1  F21  F31  F41
33
Problem 19.5.

Three point charges
are located at the
corners of an
equilateral triangle
as shown in Figure
P19.5. Calculate the
resultant electric
force on the 7.00-μC
charge.
34
Zero Resultant Force,
Superposition Example

Where is the resultant
force equal to zero?




The magnitudes of the
individual forces will be
equal
Directions will be
opposite
Will result in a quadratic
Choose the root that
gives the forces in
opposite directions
35
Electric Field – Test Particle




The electric field is defined in terms of a test
particle, qo
By convention, the test particle is always a
positive electric charge
The test particle is used to detect the
existence of the field
It is also used to evaluate the strength of the
field
The test charge is assumed to
be small enough not to disturb the
charge distribution responsible for
the field

36
Electric Field – Definition

An electric field is said to exist in the region
of space around a charged object

This charged object is the source particle

When another charged object, the test
charge, enters this electric field, an electric
force acts on it

Analogy to gravity: gravity field of the Earth exists around it
independent of the objects; however if an object of test mass m
is placed in the Earth’s gravity field, force of gravity equal to mg
will act on it.
37
Electric Field – Definition, cont


The electric field is defined as the
electric force on the test charge per unit
charge
The electric field vector, E, at a point in
space is defined as the electric force,Fe ,
acting on a positive test charge, qo
placed at that point divided by the test
charge: E  Fe / qo
38
Electric Field,
Notes


E is the field produced
by some charge or charge
distribution, separate from
the test charge
The existence of an electric field is a property of the
source charge



The presence of the test charge is not necessary for the field
to exist
The test charge serves as a detector of the field and
its strength
Test charge does not disturb the charge distribution
responsible for the electric field
39
Relationship Between F and E

Fe  q E





This is valid for a point charge only
One of zero size
For larger objects, the field may vary over the size
of the object
If q is positive, F and E are in the same
direction
If q is negative, F and E are in opposite
directions
40
Electric Field Notes, Final



The direction of E is that of the force on
a positive test charge
The SI units of E are N/C
We can also say that an electric field
exists at a point if a test charge at that
point experiences an electric force
41
Electric Field, Vector Form

Remember Coulomb’s Law, between
the source and test charges, can be
expressed as
qqo
Fe  ke 2 rˆ
r

Then, the electric field will be
Fe
q
E
 ke 2 rˆ
qo
r
42
More About Electric
Field Direction




a) q is positive, the force
is directed away from q
b) The direction of the
field is also away from
the positive source
charge
c) q is negative, the force
is directed toward q
d) The field is also
toward the negative
source charge
43
A test charge of +3 μC is at a point P where an external
electric field is directed to the right and has a magnitude
of 4 × 106 N/C. If the test charge is replaced with
another charge of −3 μC, the external electric field at P
10
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reverses direction
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33% 33% 33%
44
Superposition with
Electric Fields

At any point P, the total electric field due to a
group of source charges equals the vector
sum of electric fields at that point due to all
the particles
qi
E  ke  2 rˆi
i ri

Superposition principle is applied in the same
manner as for addition of electric forces
45
Problem 19.12.

Two point charges are located on the x
axis. The first is a charge +Q at x = –a. The
second is an unknown charge located at x
= +3a. The net electric field these charges
produce at the origin has a magnitude of
2keQ/a2. What are the two possible values
of the unknown charge?
46
Electric Dipole Example



Find the electric field
due to q1, E1
Find the electric field
due to q2, E2
E  E1  E2


Remember, the fields
add as vectors
The direction of the
individual fields is
the direction of the
force on a positive
test charge
Electric dipole
47
Electric Field – Continuous
Charge Distribution



The distances between charges in a group of
charges may be much smaller than the
distance between the group and a point of
interest
In this situation, the system of charges can be
modeled as continuous
The system of closely spaced charges is
equivalent to a total charge that is
continuously distributed along some line, over
some surface, or throughout some volume
48
Electric Field – Continuous
Charge Distribution, cont

Procedure:



Divide the charge
distribution into small
elements, each of which
contains Dq
Calculate the electric
field due to one of these
elements at point P
Evaluate the total field by
summing the
contributions of all the
charge elements
49
Electric Field – Continuous
Charge Distribution, equations

For the individual charge elements
Dqi
DEi  ke 2 rˆi
ri

Because the charge distribution is
continuous
Dqi
dq
lim
E  Dqi 0 ke  2 rˆi  ke  2 rˆ
ri
r
i
50
Charge Densities

Volume charge density – when a charge is
distributed evenly throughout a volume


Surface charge density – when a charge is
distributed evenly over a surface area


r=Q/V
s=Q/A
Linear charge density – when a charge is
distributed along a line

l=Q/l
51
Example 19.4.


The electric field due to a charged rod
A rod of length l has a uniform linear density
l and a total charge Q. Calculate the
electric field at a point P along the axis of
the rod, a distance a from one end.
52
Problem Solving Strategy

Conceptualize


Imagine the type of electric field that would
be created by the charges or charge
distribution
Categorize


Analyzing a group of individual charges or
a continuous charge distribution?
Think about symmetry
53
Problem Solving Hints, cont

Analyze

Group of individual charges: use the
superposition principle


The resultant field is the vector sum of the individual
fields
Continuous charge distributions: the vector
sums for evaluating the total electric field at some
point must be replaced with vector integrals

Divide the charge distribution into infinitesimal pieces,
calculate the vector sum by integrating over the entire
charge distribution
54
Problem Solving Hints, final

Analyze, cont


Symmetry: take advantage of any
symmetry in the system
Finalize


Check to see if your field is consistent with
the mental representation and reflects any
symmetry
Imagine varying parameters to see if the
result changes in a reasonable way
55
Electric Field Lines


Field lines give us a means
of representing the electric
field pictorially
The electric field vector E is tangent to the electric
field line at each point


The line has a direction that is the same as that of the
electric field vector
The number of lines per unit area through a surface
perpendicular to the lines is proportional to the
magnitude of the electric field in that region
56
Electric Field Lines, General



The density of lines
through surface A is
greater than through
surface B
The magnitude of the
electric field is greater
on surface A than B
The lines at different
locations point in
different directions

This indicates the field is
non-uniform
57
Electric Field Lines,
Positive Point Charge

The field lines radiate
outward in all
directions


In three dimensions, the
distribution is spherical
The lines are directed
away from the source
charge

A positive test charge
would be repelled away
from the positive source
charge
58
Electric Field Lines,
Negative Point Charge


The field lines radiate
inward in all directions
The lines are directed
toward the source
charge

A positive test charge
would be attracted
toward the negative
source charge
59
Electric Field Lines –
Rules for Drawing

The lines must begin on a positive charge
and terminate on a negative charge



In the case of an excess of one type of charge,
some lines will begin or end infinitely far away
The number of lines drawn leaving a positive
charge or approaching a negative charge is
proportional to the magnitude of the charge
Field lines cannot intersect
60
Motion of Charged Particles



When a charged particle is placed in an
electric field, it experiences an electrical
force
If this is the only force on the particle, it
must be the net force
The net force will cause the particle to
accelerate according to Newton’s
Second Law
61
Motion of Particles, cont





Fe  q E  m a
If E is uniform, then a is constant
If the particle has a positive charge, its
acceleration is in the direction of the field
If the particle has a negative charge, its
acceleration is in the direction opposite the
electric field
Since the acceleration is constant, the
kinematic equations can be used
62
Electron in a
Uniform Field, Example


The electron is
projected horizontally
into a uniform electric
field
The electron undergoes
a downward
acceleration


The charge is negative,
so the acceleration is
opposite the field
Its motion is parabolic
while between the
plates
63
Cathode Ray Tube (CRT)


A CRT is commonly used to obtain a
visual display of electronic information
in oscilloscopes, radar systems,
televisions, etc
The CRT is a vacuum tube in which a
beam of electrons is accelerated and
deflected under the influence of electric
or magnetic fields
64
CRT, cont


The electrons are
deflected in various
directions by two sets
of plates
The placing of charge
on the plates creates
the electric field
between the plates
and allows the beam
to be steered
65
Problem 19.54.

A small, 2.00-g
plastic ball is
suspended by a
20.0-cm-long string
in a uniform electric
field as shown in the
figure. If the ball is in
equilibrium when the
string makes a 15.0°
angle with the
vertical, what is the
net charge on the
ball?
66
Electric Flux


Electric flux is the
product of the
magnitude of the
electric field and the
surface area, A,
perpendicular to the
field
FE = E A
67
Electric Flux, General Area



The electric flux is
proportional to the
number of electric field
lines penetrating some
surface
The field lines may
make some angle q with
the perpendicular to the
surface
Then FE = E A cos q
68
Electric Flux,
Interpreting the Equation



The flux is a maximum when the
surface is perpendicular to the field
The flux is zero when the surface is
parallel to the field
If the field varies over the surface, F = E
A cos q is valid for only a small element
of the area
69
Electric Flux, General
In the more general
case, look at a small
area element
DFE  Ei Ai cosqi  Ei DAi


In general, this
becomes
lim
FE 
 E  DA
i
DAi 0
i


E  dA
surface
70
Electric Flux, final



The surface integral means the integral
must be evaluated over the surface in
question
In general, the value of the flux will
depend both on the field pattern and on
the surface
The units of electric flux will be N.m2/C2
71
Electric Flux, Closed Surface


Assume a closed
surface
The vectors DA i
point in different
directions


At each point, they
are perpendicular to
the surface
By convention, they
point outward
72
Flux Through Closed
Surface, cont



At (1), the field lines are crossing the surface from the
inside to the outside; q <90o, F is positive
At (2), the field lines graze the surface; q =90o, F = 0
At (3), the field lines are crossing the surface from the
outside to the inside;180o > q >90o, F is negative
73
Flux Through Closed
Surface, final

The net flux through the surface is
proportional to the net number of lines
leaving the surface


This net number of lines is the number of
lines leaving the volume surrounding the
surface minus the number entering the
volume
If En is the component of E
perpendicular to the surface, then
F E   E  dA   En  dA
74
Gauss’ Law, Introduction

Gauss’ Law is an expression of the
general relationship between the net
electric flux through a closed surface
and the charge enclosed by the surface


The closed surface is often called a
Gaussian surface
Gauss’ Law is of fundamental
importance in the study of electric fields
75
Gauss’ Law – General


A positive point
charge, q, is located
at the center of a
sphere of radius r
The magnitude of
the electric field
everywhere on the
surface of the
sphere is
E = ke q / r2
76
Gauss’ Law – General, cont.

The field lines are directed radially outward
and are perpendicular to the surface at
every point
FE 


 E dA  E  dA  EA
n
This will be the net flux through the
Gaussian surface, the sphere of radius r
We know E = kq/r2 and Asphere = 4pr2, so
q
Does not depend on r!
F E  4 pk e q 
eo
77
Gauss’ Law –
General, notes



The net flux through any closed
surface surrounding a point charge, q,
is given by q/eo and is independent of
the shape of that surface
The net electric flux through a closed
surface that surrounds no charge is
zero
Since the electric field due to many
charges is the vector sum of the
electric fields produced by the
individual charges, the flux through any
closed surface can be expressed as
 E  dA   E
1
 E2 
  dA
78
Gauss’ Law – Final

Gauss’ Law states
FE   E  dA 
qin
eo
qin is the net charge inside the surface
 E represents the total electric field at
any point on the surface


The total electric field may have contributions
from charges both inside and outside of the
surface
79
For a Gaussian surface through which the net flux is
zero, the following four statements could be true. Which
of the statements must be true?
1.
2.
3.
4.
No charges are inside
the surface.
The net charge inside
the surface is zero.
The electric field is zero
everywhere on the
surface.
The number of electric
field lines entering the
surface equals the
number leaving the
surface.
80
Consider the charge distribution shown in Active Figure
19.31. (i) What are the charges contributing to the total
electric flux through surface S ’?
10
ar
ge
s
s
no
ne
of
th
e
ch
ha
rg
e
q3
lf
ou
rc
an
d
al
q2
5.
on
ly
4.
q4
3.
on
ly
2.
20% 20% 20% 20% 20%
q1 only
q4 only
q2 and q3
all four charges
none of the charges
q1
1.
81
Consider the charge distribution shown in Active Figure
19.31. (ii) What are the charges contributing to the total
electric field at a chosen point on the surface S ’?
ar
ge
s
s
10
no
ne
of
th
e
ch
ha
rg
e
q3
lf
ou
rc
an
d
al
q2
5.
on
ly
4.
q4
3.
on
ly
2.
q1 only
q4 only
q2 and q3
all four charges
none of the
charges
q1
1.
20% 20% 20% 20% 20%
82
Applying Gauss’ Law


Gauss’ Law is valid for the electric field of any
system of charges or continuous distribution
of charge.
Although Gauss’ Law can, in theory, be
solved to find E for any charge configuration,
in practice it is limited to symmetric situations


Particularly spherical, cylindrical, or plane
symmetry
Remember, the Gaussian surface is a surface
you choose, it does not have to coincide with
a real surface
83
Conditions for a
Gaussian Surface

Try to choose a surface that satisfies one or
more of these conditions:




The value of the electric field can be argued from
symmetry to be constant over the surface
The dot product can be expressed as a simple
algebraic produce E dA because E and A are
parallel
The dot product is 0 because E and A are
perpendicular
The field can be argued to be zero everywhere
over the surface
84
Problem 19.37.

A cylindrical shell of radius 7.00 cm and
length 240 cm has its charge uniformly
distributed on its curved surface. The
magnitude of the electric field at a point 19.0
cm radially outward from its axis (measured
from the midpoint of the shell) is 36.0 kN/C.
Find (a) the net charge on the shell and (b)
the electric field at a point 4.00 cm from the
axis, measured radially outward from the
midpoint of the shell.
85
Example 19.9: The Electric Field
Due to a Point Charge

Choose a sphere as
the Gaussian
surface

E is parallel to dA at
each point on the
surface
FE 
qin
 E  dA   EdA  e
o
 E  dA  E (4p r 2 )
q
q
E
 ke 2
2
4pe o r
r
86
Example 19.10: Field Due to a
Spherically Symmetric Charge
Distribution


Select a sphere as the
Gaussian surface
For r> a
qin
F E   E  dA   EdA 
eo
Q
Q
E
 ke 2
2
4 pe or
r
87
Spherically Symmetric, cont



Select a sphere as
the Gaussian
surface, r < a
qin < Q
qin = r (4/3pr3)
FE 
E
qin
 E  dA   EdA  e
qin
4pe o r 2

o
keQ
r
3
a
88
Spherically Symmetric
Distribution, final

Inside the sphere, E
varies linearly with r


E  0 as r  0
The field outside the
sphere is equivalent
to that of a point
charge located at
the center of the
sphere
89
Example 19.11: Field at a Distance
from a Line of Charge

Select a cylindrical
charge distribution


The cylinder has a radius
of r and a length of l
E is constant in
magnitude and
perpendicular to the
surface at every point
on the curved part of
the surface
90
Field Due to a
Line of Charge, cont


The end view
confirms the field is
perpendicular to the
curved surface
The field through the
ends of the cylinder
is 0 since the field is
parallel to these
surfaces
91
Field Due to a
Line of Charge, final

Use Gauss’ Law to find the field
FE 
qin
 E  dA   EdA  e
o
l
E  2p r  
eo
l
l
E
 2ke
2pe o r
r
92
Example 19.12: Field Due to a
Plane of Charge


E must be perpendicular
to the plane and must
have the same
magnitude at all points
equidistance from the
plane
Choose a small cylinder
whose axis is
perpendicular to the
plane for the Gaussian
surface
93
Field Due to a
Plane of Charge, cont
E is parallel to the curved surface and
there is no contribution to the surface
area from this curved part of the
cylinder
 The flux through each end of the
cylinder is EA and so the total flux is
2EA

94
Field Due to a
Plane of Charge, final


The total charge in the surface is sA
Applying Gauss’ Law
sA
s
F E  2EA 
and E 
eo
2e o


Note, this does not depend on r
Therefore, the field is uniform
everywhere
95