3913479.ppt

Chapter 19
Electric Forces
and
Electric Fields

2
Electricity and
Magnetism, Some History
 Many applications
 Macroscopic and microscopic
 Chinese
 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

3
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

4
Electricity and
 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

5
Electricity and
 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

6
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

7
Electric Charges, 2
 The rubber rod is
negatively charged
 The glass rod is
positively charged
 The two rods will
attract

8
Electric Charges, 3
 The rubber rod is
negatively charged
 The second rubber
rod is also
negatively charged
 The two rods will
repel

9
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

10
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 = N e
 N is an integer
 e is the fundamental unit of charge
 |e| = 1.6 x 10-19 C
 Electron: q = -e
 Proton: q = +e

11
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, …

12
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?
O
b
j
e
c
t
s
A
a
n
d
C
p
o
s
.
.
.
O
b
j
e
c
t
s
A
a
n
d
C
p
o
s
.
.
.
A
l
l
t
h
r
e
e
o
b
j
e
c
t
s
p
o
s
.
.
.
O
n
e
o
f
t
h
e
o
b
j
e
c
t
s
i
s
.
.
.
W
e
n
e
e
d
t
o
p
e
r
f
o
r
m
a
.
.
.
20% 20% 20%
20%
20%
1. Objects A and C possess
charges of the same sign.
2. Objects A and C possess
charges of opposite sign.
3. All three objects possess
charges of the same sign.
4. One of the objects is neutral.
5. We need to perform
additional experiments to
determine the signs of the
charges.

13
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

14
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

15
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

16
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

17
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

18
 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

19
 The ground wire is
removed
 There will now be
more positive
charges in the
sphere
 The positive charge
has been induced
in the sphere

20
 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

21
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

22
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

23
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

24
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

25
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

26
Coulomb’s Law, Equation
 Mathematically,
 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
1 2
2
e e
q q
F k
r


27
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

28
Vector Nature of
Electric Forces
 In vector form,
 is a unit vector
directed from q1 to
q2
 The like charges
produce a repulsive
force between them
1 2
12 12
2
ˆ
e
q q
k
r

F r
12
r̂

29
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

 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
12 21
 
F F

30
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

31
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

32
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

33
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: 1 21 31 41
  
F F F F

34
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.

35
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

36
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

37
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.

38
Electric Field – Definition, cont
 The electric field is defined as the
electric force on the test charge per unit
charge
 The electric field vector, , at a point in
space is defined as the electric force, ,
acting on a positive test charge, qo
placed at that point divided by the test
charge: /
e o
q

E F
e
F
E

39
Electric Field,
Notes
 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
E

40
Relationship Between F and 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, and are in the same
direction
 If q is negative, and are in opposite
directions
e q

F E
E
F
F E

41
Electric Field Notes, Final
 The direction of is that of the force on
a positive test charge
 The SI units of 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
E
E

42
Electric Field, Vector Form
 Remember Coulomb’s Law, between
the source and test charges, can be
expressed as
 Then, the electric field will be
2
ˆ
o
e e
qq
k
r

F r
2
ˆ
e
e
o
q
k
q r
 
F
E r

43
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

44
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
i
s
u
n
a
f
f
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c
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a
y
t
h
a
.
.
.
33% 33%
33%
1. is unaffected
2. reverses direction
3. changes in a way
that cannot be
determined.
10

45
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
 Superposition principle is applied in the same
manner as for addition of electric forces
2
ˆ
 
E r
i
e i
i i
q
k
r

46
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?

47
Electric Dipole Example
 Find the electric field
due to q1,
 Find the electric field
due to q2,

 Remember, the fields
add as vectors
 The direction of the
individual fields is
the direction of the
force on a positive
test charge
1
E
1 2
 
E E E
2
E
Electric dipole

48
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

49
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

50
Charge Distribution, equations
 For the individual charge elements
 Because the charge distribution is
continuous
2
ˆ
i
i e i
i
q
k
r
D
D 
E r
lim
0 2 2
ˆ ˆ
i
i
q e i e
i i
q dq
k k
r r
D 
D
 
 
E r r

51
Charge Densities
 Volume charge density – when a charge is
distributed evenly throughout a volume
 r = Q / V
 Surface charge density – when a charge is
distributed evenly over a surface area
 s = Q / A
 Linear charge density – when a charge is
distributed along a line
 l = Q / l

52
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.

53
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

54
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

55
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

56
Electric Field Lines
 Field lines give us a means
of representing the electric
field pictorially
 The electric field vector 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
E

57
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

58
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

59
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

60
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

61
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

62
Motion of Particles, cont

 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
e q m
 
F E a

63
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

64
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

65
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

66
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?

67
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

68
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

69
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

70
Electric Flux, General
 In the more general
case, look at a small
area element
 In general, this
becomes
cos
E i i i i i
E A q
DF   D
E A
lim
0
i
E i i
surface
A
d
D 
F   D  
 
E A E A

71
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

72
Electric Flux, Closed Surface
 Assume a closed
surface
 The vectors
point in different
directions
 At each point, they
are perpendicular to
the surface
 By convention, they
point outward
i
DA

73
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

74
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
E n
d E dA
F    
 
E A

75
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

76
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

77
Gauss’ Law – General, cont.
 The field lines are directed radially outward
and are perpendicular to the surface at
every point
 This will be the net flux through the
Gaussian surface, the sphere of radius r
 We know E = kq/r2 and Asphere = 4pr2, so
E n
E dA E dA EA
F   
 
o
e
E
q
q
k
4
e

p

F Does not depend on r!

78
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
 
1 2
d d
    
 
E A E E A

79
Gauss’ Law – Final
 Gauss’ Law states
 qin is the net charge inside the surface
 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
in
E
o
q
d
e
F   
E A
E

80
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. No charges are inside
the surface.
2. The net charge inside
the surface is zero.
3. The electric field is zero
everywhere on the
surface.
4. The number of electric
field lines entering the
surface equals the
number leaving the
surface.

81
Consider the charge distribution shown in Active Figure
19.31. (i) What are the charges contributing to the total
electric flux through surface S ’?
q
1
o
n
l
y
q
4
o
n
l
y
q
2
a
n
d
q
3
a
l
l
f
o
u
r
c
h
a
r
g
e
s
n
o
n
e
o
f
t
h
e
c
h
a
r
g
e
s
20% 20% 20%
20%
20%
1. q1 only
2. q4 only
3. q2 and q3
4. all four charges
5. none of the charges
10

82
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 ’?
q
1
o
n
l
y
q
4
o
n
l
y
q
2
a
n
d
q
3
a
l
l
f
o
u
r
c
h
a
r
g
e
s
n
o
n
e
o
f
t
h
e
c
h
a
r
g
e
s
20% 20% 20%
20%
20%
10
1. q1 only
2. q4 only
3. q2 and q3
4. all four charges
5. none of the
charges

83
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

84
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 and are
parallel
 The dot product is 0 because and are
perpendicular
 The field can be argued to be zero everywhere
over the surface
E A
E A

85
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.

86
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
2
2 2
(4 )
4
in
E
o
e
o
q
d EdA
E dA E r
q q
E k
r r
e
p
pe
F    
 
 
 

E A

87
Example 19.10: Field Due to a
Spherically Symmetric Charge
Distribution
 Select a sphere as the
Gaussian surface
 For r> a
2
e
2
o
o
in
E
r
Q
k
r
4
Q
E
q
EdA
d

pe

 
e




F A
E

88
Spherically Symmetric, cont
 Select a sphere as
the Gaussian
surface, r < a
 qin < Q
 qin = r (4/3pr3)
2 3
4
in
E
o
e
in
o
q
d EdA
k Q
q
E r
r a
e
pe
F    
 
 
E A

89
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

90
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
surface at every point
on the curved part of
the surface

91
Field Due to a
Line of Charge, cont
 The end view
confirms the field is
curved surface
 The field through the
ends of the cylinder
is 0 since the field is
parallel to these
surfaces

92
Field Due to a
Line of Charge, final
 Use Gauss’ Law to find the field
 
2
2
2
in
E
o
o
e
o
q
d EdA
E r
E k
r r
e
l
p
e
l l
pe
F    

 
 
E A

93
Example 19.12: Field Due to a
Plane of Charge
 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
plane for the Gaussian
surface
E

94
Field Due to a
Plane of Charge, cont
 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
E

95
Field Due to a
Plane of Charge, final
 The total charge in the surface is sA
 Applying Gauss’ Law
 Note, this does not depend on r
 Therefore, the field is uniform
everywhere
2
2
E
o o
A
EA and E
s s
e e
F   

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