21_lecture_ElCharge_Field.pdf slideshow for

PowerPoint® Lectures for
University Physics, 14th Edition
– Hugh D. Young and Roger A. Freedman Lectures by Jason Harlow
Electric Charge and
Electric Field
Chapter 21
© 2016 Pearson Education Inc. Modified by J. Tucci

Learning Goals for Chapter 21
Looking forward at …
• how objects become electrically charged, and how we know
that electric charge is conserved.
• how to use Coulomb’s law to calculate the electric force
between charges.
• the distinction between electric force and electric field.
• how to use the idea of electric field lines to visualize and
interpret electric fields.
• how to calculate the properties of electric charge
distributions, including dipoles.
© 2016 Pearson Education Inc.

Introduction
• Water makes life
possible: The cells of
your body could not
function without
water in which to
dissolve essential
biological molecules.
• What electrical properties of the water molecule allow it to be
such a good solvent?
• We now begin our study of electromagnetism, one of the four
fundamental forces.
• We start with electric charge and look at electric fields.

Electric charge
• Plastic rods and fur (real or
fake) are particularly good
for demonstrating
electrostatics, the
interactions between electric
charges that are at rest (or
nearly so).
• After we charge both plastic
rods by rubbing them with
the piece of fur, we find that
the rods repel each other.

Electric charge
• When we rub glass rods with
silk, the glass rods also
become charged and repel
each other.

Electric charge
• A charged plastic rod attracts a
charged glass rod; furthermore,
the plastic rod and the fur
attract each other, and the glass
rod and the silk attract each
other.
• These experiments and many
others like them have shown
that there are exactly two kinds
of electric charge: The kind on
the plastic rod rubbed with fur
(negative) and the kind on the
glass rod rubbed with silk
(positive).
© 2016 Pearson Education Inc. “Likes repel, Opposites attract”

Electric charge and the structure of matter
• The particles of the atom are the
negative electrons (dark blue
spheres in this figure), the
positive protons (red spheres),
and the uncharged neutrons
(gray spheres).
• Protons and neutrons make up
the tiny dense nucleus, which is
surrounded by electrons.

Atoms and ions
• A neutral atom has the same
number of protons as electrons.
• The electron “shells” are a
schematic representation of the
actual electron distribution, a
diffuse cloud many times larger
than the nucleus.
• Note: The neutrons do not affect
the net charge because they are
uncharged.

Atoms and ions
• A positive ion is an atom
with one or more electrons
removed.

Atoms and ions
• A negative ion is an atom
with an excess of electrons.

Conservation of charge
• The proton and electron have the same magnitude charge.
• The magnitude of charge of the electron or proton is a natural
unit of charge. All observable charge is quantized in this unit.
• The universal principle of charge conservation states that
the algebraic sum of all the electric charges in any closed
system is constant.
TERMS
•Electrically neutral: object contains equal amounts of
positive and negative charges
•Net charge: imbalance in charge

Electric Charge
Electric charge is quantized
•Elementary charge:
q = ne, n = 1,2,3,.....
e = 1.60602176462(63) x 10–19 C
Coulomb (C): one coulomb is the amount of charge that is
transferred through the cross section of a wire in 1 second
when there is a current of 1 ampere in the wire.
$ = n
http://scrapetv.com/News/News%20Pages/B
usiness/images/us%20penny.jpg

Charge of Particles
Electron
Positron
Proton
Anti-Proton
Neutron
Photon
Up Quark
Down Quark
0
Particles Charge
Nucleus charge= +Ze, atom with Z electrons is neutral.

e−

e−

e+

e+
2
3
e
−
1
3
e
0
Proton (uud) charge: |e+ | = 1.60 x 10–19 C
Electron charge: |e- | = 1.60 x 10–19 C

Conductors versus Insulators
• Conductors: material in which electric charges can move
around “freely.”
• Insulators: material in which electric charges are “frozen”
in place.
• Semi-conductor: material in which electric charges can
move around but not as freely as in conductors.
• Super-conductor: zero resistance to the movement of
charge.

Mobility of Charge
• Conductors: material in which electric charges can move
around “freely.”
• Negatively charged plastic rod
will attract either end of the
electrically isolated copper rod
• Reason: charges in copper rod
can redistribute themselves.

Interaction of Charges: Insulators
Force of Repulsion Force of Attraction
Charges with opposite
electrical signs attract each
other.
Charges with the same
electrical sign repel each
other

Conductors and insulators
• Copper is a good conductor of electricity; nylon is a good
insulator. The copper wire shown conducts charge between
the metal ball and the charged plastic rod to charge the ball
negatively.

• After it is negatively charged, the metal ball is repelled by a
negatively charged plastic rod.

• After it is negatively charged, the metal ball is attracted by a
positively charged glass rod.

Charging by induction in 4 steps:
Steps 1 and 2
1. Start with an uncharged
metal ball supported by an
insulating stand.
2. When you bring a
negatively charged rod near
it, without actually
touching it, the free
electrons in the metal ball
are repelled by the excess
electrons on the rod, and
they shift toward the right,
away from the rod.

Charging by induction in 4 steps:
Steps 3 and 4
3. While the plastic rod is nearby,
you touch one end of a
conducting wire to the right
surface of the ball and the other
end to the ground.
4. Now disconnect the wire, and
then remove the rod. A net
positive charge is left on the
ball. The earth acquires a
negative charge that is equal in
magnitude to the induced
positive charge remaining on
the ball.

Electric forces on uncharged objects
• A charged body can exert forces even on objects that are not
charged themselves.
• If you rub a balloon on the rug and then hold the balloon
against the ceiling, it sticks, even though the ceiling has no
net electric charge.
• After you electrify a comb by
running it through your hair,
you can pick up uncharged bits
of paper or plastic with it.
• How is this possible?

• The negatively charged plastic comb causes a slight shifting
of charge within the molecules of the neutral insulator, an
effect called polarization.

• Note that a neutral insulator is also attracted to a positively
charged comb.
• A charged object of either sign exerts an attractive force on
an uncharged insulator.

Electrostatic painting
• Induced positive charge on the metal object attracts the
negatively charged paint droplets.

Mobility of Charge
• Demo: Pie Tins
https://www.youtube.com/watch?v=TV4oq_-OXkg

Charge Induction
• Demo: Chimes
Conducting
thread
Insulating
thread
Grounded
Charged

Measuring the electric force between point
charges
• Coulomb studied the interaction
forces of charged particles in
detail in 1784.
• He used a torsion balance similar
to the one used 13 years later by
Cavendish to study the much
weaker gravitational interaction.
• For point charges, Coulomb
found that the magnitude of the
electric force is inversely
proportional to the square of the
distance between the charges.

Coulomb’s Law of Electro-static Force
q1
q2
r
The electro-static force of attraction/repulsion has a
magnitude:
F = k
q1 q2
r2
Coulomb’s Law
9 2 2
1
8.99 10 /
4 o
k Nm C

= = 
where:
and the permittivity constant is
12 2 2
8.55 10 /
o C Nm
 −
= 
Charles-Augustin de
Coulomb
(1736 - 1806)

*Each particle exerts a force of this magnitude on the other
particle.
*The two forces form an action-reaction pair.
1 2
2
1
ˆ
4 o
Q Q
F r
r

=
Force repulsive
1
+
+
2
r
F12
Force attractive
+
-
2
r
F12
1
Force by “1” on “2”

Force exerted by q1 on q2 at a distance r12
1 2
12 1,2
2
1,2
ˆ
kq q
F r
r
=
12 1,2
12 1,2
Same sign charges: is in the direction of .
Opposite sign charges: is in the direction opposite to .
F r
F r
q1, q2 in coulombs (C)
r12 in meters (m)
F12 in newtons (N)
12
F

Example 21.1 Adding Electric Forces
in One Dimension (1 of 5)
Two +10 nC charged particles are 2.0 cm apart on the x-
axis. What is the net force on a +1.0 nC charge midway
between them? What is the net force if the charged particle
on the right is replaced by a +10 nC charge?
STRATEGIZE We will
proceed using the steps of
the Problem-Solving
Approach. We will model
the charged particles as
point charges.
SHOW

Problem Solving Strategies:
• Draw a clear FORCE diagram
• Use consistent units (meter, Coulomb, Newton)
• Remember that the force is a vector
• Look for (possible) symmetry

Question 1.
Two charges q = + 1 µC and Q = +10 µC are placed near each other as
shown below. Which diagram best depicts the electrostatic forces
acting on the charges?
+1 µC +10 µC
A
B
C

Principle of Superposition
• When several point charges are put together, the total force on
any one charge is the vector sum of the each of the separate
forces acting on that charge.
F = F21y + F31y = 2F21y
F = 2  k
Q1Q2
r2
cos300
F =
2  9 109 N  m2
C2
 (10−6
C)2
 0.866
1m
( )2
F = 15.59 10−3
N
F
F21
• Exercise:
Q2
Determine force on Q1
Q1=Q2=Q3=1C
Q3
R=1m
600
Q1
y
x
Determine force on Q1
F31

Coulomb’s Law Analogous to Newton’s Equation of
Gravitation
F = k
q1 q2
r2
F = G
m1m2
r2
* k electro-static constant
* Inverse Square Law
* Charge
*Attractive/repulsive
depending on sign of
charges
*Two kinds of charges
*Dominates on small scale
* G gravitational constant
* Inverse Square Law
* Mass
*Always attractive
*One kind of mass
*Dominates on large scales
DIFFERS
Analogous

Electro-Static Force versus Newton’s Force of
Gravitational Attraction
⚫ Given such strong electrical interactions, atoms tend to remain uncharged.
Matter prefers to be neutral.
⚫ Forces we experience, if not gravitational, are electrical in nature (even
though the net charge may be zero).
2
e
grav
2
2
el
m
F
e
F
r
m
G
r
k
p
=
=
Fel
Fgrav
=
k
G
e2
/ r2
memp / r2
= 1.35 1020 kg2
C2
(1.6 10−19
C)2
(1.7 10−27
Kg)(9.110−31
Kg)
Fel
Fgrav
= 2.3  1039
Fel = 8.2 10-8
N Fgrav = 3.6 10−47
N
⚫ Fel between the proton and the electron in a hydrogen atom in the ground
state. From the Bohr model r=0.53 x 10-10 m.

Electric field: Introduction Slide 1 of 3
• To introduce the concept of electric field, first consider the
mutual repulsion of two positively charged bodies A and B.

• Next consider body A on its own.
• We can say that body A somehow modifies the properties
of the space at point P.

• We can measure the electric field produced by A with a test
charge.

Electric force produced by an electric field

The electric field of a point charge
• Using a unit vector that points away from the origin, we can
write a vector equation that gives both the magnitude and the
direction of the electric field:
Note: only one q here (qsource).
Fcoul has two q’s (qsource and qtest).

Electric field of a positive point charge
• A point charge q produces an
electric field at all points in
space.
• The field strength decreases
with increasing distance.
• The field produced by a
positive point charge points
away from the charge.

Electric field of a negative point charge
• A point charge q produces an
electric field at all points in
space.
• The field strength decreases
with increasing distance.
• The field produced by a
negative point charge points
toward the charge.

Electric Field
0
q
F
E


=








+

= 2
2
2
2
1
2
1
1
0
0
r̂
r
q
r̂
r
q
4
q
F

Observation: The net Coulomb force on a given charge is always
proportional to the strength of that charge.
q0
q1
q2
F1
F
F2
test charge
Define the electric field, which is independent of the test
charge, q0, and depends only on position in space:
q1 and q2 in F are the sources
of the electric field, q0 is the test charge
2
1 F
F
F



+
=

• An particle of charge q injected in a E field will experience a
force given by F = q E. The resulting acceleration can be found
from Newton's second law.
• In a region of uniform E field, the particle will experience a
constant acceleration and the resulting parabolic trajectory.
• The control of electrons by so-called deflection plates is the
principle behind the operation of the cathode-ray tube used in
oscilloscopes and many televisions and computer monitors.
A Point Charge in a Uniform Field
E
m
q
a
a
m
F




=
=

Electric Field due to Multiple Point Charges
To find the resultant field from n point charge:





+
+
+
= 3
2
1 F
F
F
F
so the electric field is, by definition, given by










+
+
+
=
+
+
+
=
=
3
2
1
0
3
0
2
0
1
0
E
E
E
q
F
q
F
q
F
q
F
E
Principle of Superposition!

Two charges q1 and q2 are fixed at points (-a,0) and (a,0) as
shown. Together they produce an electric field at point (0,d)
which is directed along the negative y-axis.
x
y
(−a,0) (a,0)
(0,d)
Which of the following statements is true:
A) Both charges are negative
B) Both charges are positive
C) The charges are opposite
D) There is not enough information to tell how the charges are
related
E
q1 q2
Question

Electric Field From Two Charges
_ _
+ +
_
+

Electric field lines
• An electric field line is an imaginary line or curve whose
tangent at any point is the direction of the electric field vector
at that point.

Electric field lines of a point charge
• Electric field lines show the
direction of the electric field
at each point.
• The spacing of field lines
gives a general idea of the
magnitude of the electric
field at each point.

Electric field lines of a dipole
• Field lines point away from + charges and toward – charges.

Electric field lines of two equal positive
charges
• At any point, the electric field has a unique direction, so field
lines never intersect.

# lines proportional to
the magnitude of
charge
Far from charges,
the field lines are as
if they are due to a
point charge of +2q-
q=+q
Electric Field Lines
Opposite Charges: unequal magnitude

Summary: Electric Field
E =
1
40
qi
ri
2
 r̂i
With this concept, we can “map” the electric field anywhere
in space produced by any arbitrary:
+
+
+ +
+
+
-
-
-
-
-
Bunch of Charges
+
+ +
+ + +
+
+ +
+
Charge Distribution
“Net” E at origin
+
F
These charges or this charge distribution
“source” the electric field throughout space
E =
1
40
dqi
r2
 r̂

1/20/2021 66
Charge Distributions Problems
Step 1:Understand the geometry
Step 2:Choose dq
Step 3:Evaluate dE contribution from the
infinitesimal charge element
Step 4:Exploit symmetry as appropriate
Step 5:Set up the integral
Step 6:Solve the integral
Step 7:The Result!
Step 8:Check Limiting Cases

1/20/2021 67
• How do we represent the charge “Q” on an extended
object?
total charge
Q
small pieces
of charge
dq
Line of charge:
= charge per
unit length [C/m]
dq =  dx
Surface of charge:
 = charge per
unit area [C/m2]
dq =  dA =  dxdy
(Cartesian coordinates)
Volume of Charge:
 = charge per
unit volume [C/m3]
dq = dV =  dx dy dz
(Cartesian coordinates)
Charge Densities

1/20/2021 69
Charge on a ring
Problem: calculate the electric field along
z-axis due to a (circular) ring (of radius R)
of uniform positive charge (with density ).
• Charge of element
• The electric field due to this element
Steps 1-3
dq ds

= 
2
0
1
ˆ
4
dq
dE r
r

=

1/20/2021 70
• The z-component of the field:
Field due to charge on a ring
Step 4:Exploit symmetry as appropriate
• Symmetry: direction of the field at point P
must be along the positive z direction.
dEz = dE cos =
1
40
dq
r2
cos

1/20/2021 71
Field due to charge on a ring
• Integrate over the entire ring with  uniform; z and R
fixed
Steps 5-7
( )
( ) ( )
1/2
2 2 2 2 2
0 0
2
3/2 3/2
2 2 2 2
0
0 0
1 1
cos
4 4
4 4
z z
R
dq ds z
E dE
r z R z R
z qz
ds
z R z R



 

 
= = =
+ +
= =
+ +
  


1/20/2021 72
Limiting cases
E =
qz
40 z2
+ R2
( )
3/2
Step 8
• z>>R: point-like charge → ~q/z2, Coulomb formula
• z<<R: E=0 at the center of the ring, all field elements cancel.

1/20/2021 73
Charged Disk
Problem: calculate the electric field along
z-axis due to a (circular) disk (of radius R) of
uniform positive charge (with density ).
Pick any ring element (of infinitesimal
width dr) of the disk. The charge of the
element is:
The electric field due to the ring element
2
dq rdr
 
= 
( )
3/2
2 2
0
2
4
z
rdr z
dE
z r
 


=
+

1/20/2021 74
Field due to the disk
( )
3/2
2 2
0
0 2
R
z z
rdr z
E dE
z r


= =
+
 
• Integrate over the entire disk
Xm
dX =
Xm+1
m +1

• Use substitution of variables
2 2 2 2
0 0
0
2 2
r R
z
r
z z z
E
z
z r z R
 
 
=
=
 
= − = −
 
 
+ +
 
Let 𝑋 = 𝑧2
+ 𝑅2 𝑚
and
m = -3/2

1/20/2021 75
Field due to the disk
• Check Limiting Cases
z>>R:
point-like charge → ~Q/z2
2 2
0
2
z
z z
E
z z R


 
= −
 
 
+
 
2 2
2
0 0
2 2
2 2 2
0 0 0
1 1
1 1
2 2 1 / (2 )
1 ( / )
1 1
1 1
2 2 4 4
z
E
R z
R z
R R Q
z z z
 
 
 
  
   
=   
   
+
 +   
 
 
 
  − =  = 
 
 
 
 
z>0: upper sign
z<0: lower sign
First-order
series
approximation

1/20/2021 76
There is a
discontinuity
at z = 0.
Ez
z
Important limiting cases for charged disk
R →  : Ez →

20
= 2k for z  0
Ez → −

20
= −2k for z  0
Same results for z  0.
z<<R:
2 2
0
2
z
z z
E
z z R


 
= −
 
 
+
 

1/20/2021 77
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
z
xy-plane
discontinuity in electric field at z = 0
Uniform electric
fields generated on
both sides of sheet.
Infinite sheet of positive charge
𝐸𝑧 = 𝜎/2𝜀0
𝐸𝑧 = −𝜎/2𝜀0

Electric field from a (in)finite line of charge
1/20/2021 79

1/20/2021 80
In this coordinate system
you have to deal with Ex and Ey.
E
In this coordinate system
you have to deal with the
horizontal component of E
only.
uniform charge distribution
total charge = Q
y
x
E
Y
X
900 arc of charge

1/20/2021 81
y
x
dE
 =
Q
2r
4
=
2Q
r
dq = r d
dEx =
k dq
r2
cos =
kr cos d
r2
Ex =
k
r
cos d
1
2

=
k
r
sin2 − sin1
( )
For the 900
arc, 1 = −450
and 2 = 450
, so
Ex =
k
r
1
2
− −
1
2











 =
2k
r
=
2k
r
2Q
r
=
2 2kQ
r2
rd
900 arc of charge (continued)

The water molecule is an electric dipole
• The water molecule as a whole is electrically neutral, but the
chemical bonds within the molecule cause a displacement of
charge.
• The result is a net negative charge on the oxygen end of the
molecule and a net positive charge on the hydrogen end,
forming an electric dipole.

The water molecule is an electric dipole
• When dissolved in water, salt dissociates into a positive
sodium ion and a negative chlorine ion, which tend to be
attracted to the negative and positive ends of water
molecules.
• This holds the ions in solution.
• If water molecules were not
electric dipoles, water would
be a poor solvent, and almost
all of the chemistry that occurs
in aqueous solutions would be
impossible!

Electric Dipole Moment
Dipole moment:
Important: The direction of the dipole moment is taken to
be from the negative to the positive end of the dipole.
L is the separation of the two
charges
p 610-30 C·m0.04 e·nm
p q L
= 
Note: This means L points
opposite direction of E!

Force and torque on a dipole
• When a dipole is placed
in a uniform electric
field, the net force is
always zero, but there
can be a net torque on
the dipole.
 = F
d
2
sin + F
d
2
sin = Fdsin
= qEdsin
= pEsin E
p




=


Potential Energy of an Electric Dipole
Potential energy can be associated with the orientation of
an electric dipole in an electric field.
U is least =0 U=-pE
U is greatest =180 U=pE
U =0 when =90
𝑈 = −𝑊 = − න
0
𝜃
𝜏𝑑𝜃′
= − න
0
𝜃
𝑝𝐸 sin 𝜃′ 𝑑𝜃′
= −𝑝𝐸 cos 𝜃′
𝜃′ = 𝜃
𝜃′ = 0
= −𝑝𝐸 cos 𝜃 + 𝑈0
Then choose U0 = 0, therefore U = -pEcosθ… 𝑈 = − റ
𝑝 ∙ 𝐸

Electric Field due to an Electric Dipole
From symmetry must lie on dipole axis
E
E = E+ − E−
=
1
4o
q
r+
2
−
1
4o
q
r−
2
=
q
4o (z −
d
2
)2
−
q
4o (z +
d
2
)2
+
-
P
Dipole
center
z
r+
r-
E+
E−
q+
q-
d
Use Binomial Expansion & d << z,
the magnitude of the electric field at P:
E =
qd
2oz3

Electric Field due to an Electric Dipole
+
-
P
Dipole
center
z
r+
r-
E+
E−
q+
q-
d
E =
qd
2oz3
The electric dipole moment (C-m):
p qd
=
+
-
p
Rewrite magnitude of electric field at P:
E =
p
2oz3
Direction of p is from
negative to positive

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