### Electric Force and Field

```Chapter 18. Electric Charges and Forces
The electric force is one of
the fundamental forces of
nature. Controlled electricity
is the cornerstone of our
modern, technological
society.
Chapter Goal: To develop
a basic understanding of
electric phenomena in terms
of charges, forces, and
fields.
Learning objectives
• To become familiar with basic electric
phenomena
• To understand and use Coulomb’s Law for
point charges
• To begin the process of understanding the field
model and the concept of the electric field
• Electric charge is a
fundamental conserved property
of some subatomic particles,
which determines how they
interact (i.e. attraction/repulsion)
• Substantial experimentation
supports the existence of only 2
different types of charge.
• Benjamin Franklin introduced
the modern nomenclature of
positive/negative for the 2 types
of charge. These terms have
nothing to do with greater or less
than (as in math) or direction (as
in vectors).
• Electric charge is a conserved
quantity. If a system becomes
negatively charged, the rest of
the universe acquires a positive
charge of the same magnitude
and vice versa.
• Familiar charged particles
include the proton and the
electron.
• The SI unit of electric charge is
the Coulomb (C).
• A Coulomb is a lot of charge!
The charge of one proton or
electron is:
e  1.601019 C
•Objects become charged when
they have an excess or a deficit of
electrons, which are mobile.
• Like charges (pos/pos or
neg/neg) repel each other
• Unlike charges attract.
• Neutral objects are attracted
to both kinds of charge. That
will be explained later in the
chapter.
Charged particles
Each of three objects carries a
charge. As the drawing shows,
objects A and B attract each other,
and objects C and A also attract
each other. Which one of the
following statements concerning
objects B and C is true?
A. They attract each other
B. They repel each other
Conservation of Charge
The rod carries a charge of -3.0 μC,
while the conducting sphere carries a
charge of +2.0 μC. The two charged
objects touch.
a. What is the final charge state of
both rod and sphere?
b. How many electrons travel from
the rod to the sphere to produce the
final charge state?
Conservation of Charge
The rod carries a charge of -3.0 μC, while the
conducting sphere carries a charge of +2.0 μC. The
two charged objects touch.
a. What is the final charge state of both rod and
sphere? -0.5 µC (-3 + 2)/2
b. How many electrons travel from the rod to the
sphere to produce the final charge state?
(1electron/-1.6x10-19C) (-2.5x 10-6 C) = 1.6 x 1013 electrons
Four spheres-A
• Four identical metal spheres have charges of
qA = −8.0 µC, qB = −2.0 µC, qC = +5.0 µC,
and qD = +12.0 µC. Two of the spheres are
brought together so they touch, and then they
are separated. Which spheres are they, if the
final charge on each of those two is +5.0 µC?
A. AB
B. AC
D. BD
Which diagram best represents the charge distribution on three
neutral conducting spheres when a positively charged rod is
brought near sphere x, but does not touch it?
A.
B.
C.
D.
E.
1
2
3
4
None of the
diagrams
Insulators
• The electrons in the insulator
are all tightly bound to the
positive nuclei and not free to
move around.
• Charging an insulator
by friction leaves patches
of molecular ions on the
surface, but these patches
are immobile.
•Plastic, styrofoam, wool, silk
are examples of insulators.
Conductors
• In conductors, the outer
atomic electrons are only
weakly bound to the nuclei.
• These outer electrons become
detached from their parent
nuclei and are free to wander
• The solid as a whole remains
electrically neutral, but the
electrons act like a negatively
charged liquid permeating an
array of positively charged ion
cores.
• Most conductors are metals.
Insulators and Conductors can both acquire a
charge but….
Charging Objects
Frictional charging: Some insulators
lose or gain surface electrons if rubbed. In
a conductor the charge will “diffuse” over
the entire object instead of staying near the
rubbed area.
Charging by contact: Charge can be
transferred from one charged object to
another. This works for both insulators
and conductors, but in the latter case, the
charge will travel and spread out over the
surface of the conductor.
John Travoltage
Notice how charge transferred from the carpet acts and where
it goes when John rubs his foot.
2. Describe what happens when he touches the door handle.
Explain why this happens.
Charging Objects, cont’d
Charging by Induction: If the negatively-charged rod is
brought close to a conductor, the conductor will become
“polarized”, meaning charge will separate (a). If a wire is
attached between sphere and the earth (ground), electrons
will travel down the wire (b). When first the wire, then the
rod, is removed, a positive charge remains on the conductive
sphere. This only works for conductors.
Stop to Think: Is Order Important?
What would happen if, in situation (b) below, we took the
rod away first, then disconnected the grounding wire?
A. The metal sphere would still have a positive charge,
because the order of events doesn’t matter
B. The metal sphere would remain neutral because with the
rod gone, the electrons that went down the wire would
travel back up the grounding wire
C. The metal sphere would become negative because once
the rod is gone, too many electrons would travel up the
grounding wire.
Polarization: Why neutral objects are attracted by
charged objects
The picture below should explain why a
neutral conductor is attracted to a charged
object.
•The positive charges (electron
holes?) are attracted to the rod while
the electrons are repelled
•The attractive force is stronger than
the repulsive force because the
electron holes are closer than the
electrons.
•If there were no frictional force
between the stand and the floor, the
sphere would slide to the rod.
But insulators are attracted too. What
The Electric Dipole
The Electric Dipole
The air is an insulator, except…..
• Although air is normally an
excellent insulator, when stressed
by a sufficiently large electric
field, air can begin to ionize,
becoming partially conductive.
This culminates in an electrical
spark or arc that bridges the entire
gap.
• This is the same phenomenon that
occurs during a lightning storm.
COULOMB’S LAW
In SI units k = 8.99 × 109 N m2/C2.
k = 1/4πε0, where ε0 is called the permittivity of free
space and equals 8.85x 10-12 C2/(N m2).
Note: Assume the weight force is negligible unless
otherwise specified in problems and diagrams.
Attractive and repulsive forces between charges
Charges A and B exert repulsive forces on
each other. qA = 4qB. Which statement is
true?
A. FA on B > FB on A
B. FA on B < FB on A
C. FA on B = FB on A
18.5 Coulomb’s Law
Example 4 Three Charges on a Line
Determine the magnitude and direction of the net electrostatic force on q1.
18.5 Coulomb’s Law
F12  k
F13  k
q1 q2
r
2
q1 q3
r
2
8.9910

9
8.9910







N  m 2 C2 3.0 106 C 4.0 106 C
0.20m2
9
N  m 2 C2 3.0 106 C 7.0 106 C
0.15m2
 2.7 N
 8.4 N
18.5 Coulomb’s Law
F12  k
F13  k
q1 q2
r
2
q1 q3
r
2
8.9910

9
8.9910







N  m 2 C2 3.0 106 C 4.0 106 C
0.20m2
9
N  m 2 C2 3.0 106 C 7.0 106 C
0.15m2
 

F  F12  F13  2.7 N  8.4N  5.7N
 2.7 N
 8.4 N
Charges in a line
The drawing shows three point charges arranged in three different
ways. The charges are +q, -q, and -q; each has the same
magnitude, with one positive and the other two negative. In
each of the arrangements the distance d is the same. Rank the
arrangements in descending order (largest first) according to
the magnitude of the net electrostatic force that acts on the
positive charge.
A. ABC
B. BCA
C. CAB
D. CBA
18.5 Coulomb’s Law
Bohr model of the Hydrogen Atom
In the Bohr model of the hydrogen atom, the electron is in orbit about the
nuclear proton at a radius of 5.29x10-11m. If me = 9.11 x 10-31 kg, determine
a. the speed of the electron, assuming the orbit to be circular.
b. the kinetic energy of the electron
18.5 Coulomb’s Law
1. Draw a fbd of the electron (ignore weight force,
even though mass is used)
2. Use Newton’s 2nd Law: ΣF = ma. In this case,
there is only the electrical force, and a is a centripetal
acceleration; a = v2/r and
Fe  k
q1 q2
r2
m v2

r
18.5 Coulomb’s Law
v  k q1 q2
rm   2.19 10 6 m s
18.5 Coulomb’s Law
Now calculate kinetic energy: KE = ½ mv2
K = 2.16 x 10-18 J
The Electric Field Model: Newton vs Faraday
• In this illustration,
consider A to be one or
more source charges, in
that it is the source of
the “alteration of
space”.
• Consider B to be the test
charge (q0), a small
visitor to the
neighborhood.
The Electric Field
1. Some charges, which we will call the source charges,
alter the space around them by creating an electric field.
2. A test charge in the electric field experiences a force
exerted by the field.
Suppose test charge q0 experiences an electric force F due
to the source charges. The value of the electric field is the
force experienced by q0 divided by the value of q0.

 F
E
qo
The units of the electric field are N/C. The magnitude E of
the electric field is called the electric field strength.
The Electric Field
The electric field vector points in the same directions as
that of the electrostatic force on a positive test charge
• The electric field vector points in the opposite direction
from the electrostatic force on a negative test charge
• The electric field due to one or more source charges
exists in all space whether there is a test charge or not
An electron in an E field
The electron is placed at the position marked by the
dot. The force on the electron is:
A. Zero
C. To the left
B. To the right
D. Not enough information
Electric field strength due to a point charge q
q qo 1
F
E
k 2
qo
r qo
Note that the electric field strength does not depend on the test charge; it is
divided out
For a point charge q:
Ek
q
r2
Example Problem
Two charges are placed on the x axis. One of the
charges (q1 = +9.1 µC) is at x1 = +2.8 cm and the other
(q2 = -22 µC) is at x2 = +8.7 cm. Find the net electric
field (magnitude and direction) at x = 0 cm. (Use the
In SI units k = 8.99 × 109 N m2/C2.
1 cm = 10-2 m
Example Problem
Two charges are placed on the x axis. One of the
charges (q1 = +9.1 µC) is at x1 = +2.8 cm and the other
(q2 = -22 µC) is at x2 = +8.7 cm. Find the net electric
field (magnitude and direction) at x = 0 cm. (Use the
In SI units k = 8.99 × 109 N m2/C2.
1 cm = 10-2 m
Ans: -7.82 x 107 N/C
Electric field strength
Rank electric field strengths at points a,b,c,d from
greatest to least:
The E field vector
The electric field vector at 1 can be represented by
the arrow shown above. How long are the arrows
at 2 and 4? Explain your reasoning.
c. If the distance between the unseen point charge and the 900
N/C point is r1, and the distance between the unseen point
charge and the 400 N/C is r2, by what factor is r2 > r1 ?
THE PARALLEL PLATE CAPACITOR
q

E

o A o
  8.851012 C2 N  m2 
σ = q/A = charge density
•The capacitor is a chargestorage device used in
many electrical appliances
and circuits.
•The simplest type is the
parallel plate capacitor, two
metal plates of area A.
•Each plate has an equal
amount of charge, with one
being positive and the
other negative.
•The electric field inside the
capacitor is constant and
does not depend on the
distance from the plates.
Electric field of a cell membrane
The membrane surrounding a living
cell consists of an inner and an outer
wall that are separated by a small
space. Assume that the membrane
acts like a parallel plate capacitor in
which the effective charge density
on the inner and outer walls has a
magnitude of 7.1x10-6 C/m2. What
is the magnitude of the electric field
within the cell membrane?
Electric field of a cell membrane
The membrane surrounding a living
cell consists of an inner and an outer
wall that are separated by a small
space. Assume that the membrane
acts like a parallel plate capacitor in
which the effective charge density
on the inner and outer walls has a
magnitude of 7.1 x 10-6 C/m2. What
is the magnitude of the electric field
within the cell membrane?
E = σ/ε0 and
  8.851012 C2 N  m2 
E = σ/ε0
= 7.1 x 10-6 C/m2
8.85 x 10-12 C2 /(N · m2 )
E = 8.02 x 105 N/C
18.7 Electric Field Lines
Electric field lines or lines of force provide a map of the electric
field in the space surrounding electric charges. Figure a shows
electric field vectors.
Figure b shows the electric field lines due to a
positive point charge source.
18.7 Electric Field Lines
•Electric field lines are always directed away from positive
source charges and toward negative source charges.
•The line spacing indicates the strength of the field. More
closely spaced lines indicate a stronger field. Parallel lines
indicate a constant field.
•The number of field lines drawn is proportional to the
strength of the field; twice as many lines would emerge
from a charge of 2q than from a charge of q.
•Electric field lines always begin on a positive charge and
end on a negative charge and do not stop in midspace.
•The direction of electric field (force on a positive test
charge) is tangent to the field line at that place in space.
Electric field lines due to a point charge
Positive
Negative
Electric field lines are always directed away from
positive source charges and toward negative source
charges. Lines are more closely spaced close to the
charge, where the field is stronger.
18.7 Electric Field Lines
Field lines for an electric dipole. Note the direction of the
electric force vector, tangent to the field line.
Field lines for 2 like charges. Where is the
electric field zero?
18.7 Electric Field Lines
The parallel plate capacitor has a
constant electric field in the interior
of the plate, shown by parallel
lines. The field strength is the
same, regardless of position.
q

E

o A o
Which is the correct
field line
configuration?
Why are the other 3
incorrect?
Focus on Concepts
The drawing shows some electric field lines. For
the points indicated, rank the magnitudes of the
electric field in descending order (largest first).
A. BCA C. CAB E. ABC
B. BAC D. ACB
```