2. Demonstration #1
1. Demonstrate how you can pick up the
tissue without touching it in any way with
your body.
2. What is occurring on the atomic level that
lets you do this?
3. The atom
The atom has positive charge in the nucleus,
located in the protons. The positive charge
cannot move from the atom unless there is a
nuclear reaction.
The atom has negative charge in the electron
cloud on the outside of the atom. Electrons can
move from atom to atom without all that much
difficulty.
5. • Demonstration #2
• 1. Rub the black rod with the fur. Bring the rod
toward the pole of the electroscope. What
happens to the vanes?
• ---
• 2. Come up with an atomic level explanation for
your observations.
6. • Demonstration #3
+++
• 1. Rub the glass rod with the silk. Bring
the rod toward the pole of the
electroscope. What happens to the
vanes?
• 2. Come up with an atomic level
explanation for your observations.
7. • Demonstration #4
• 1. What happens when your touch the
electroscope with the glass rod?
9. Sample Problem
A certain static discharge delivers -0.5
Coulombs of electrical charge.
• How many electrons are in this discharge?
10. Sample Problem
1. How much positive charge resides in two moles of
hydrogen gas (H2)?
2. How much negative charge?
3. How much net charge?
11. Sample Problem (end 1)
The total charge of a system composed of
1800 particles, all of which are protons or
electrons, is 31x10-18 C.
How many protons are in the system?
How many electrons are in the system?
17. Sample Problem
A point charge of positive 12.0 μC experiences
an attractive force of 51 mN when it is placed 15
cm from another point charge. What is the other
charge?
31. Sample Problem end 3
A 400 μg styrofoam bead has 600 excess electrons on
its surface. What is the magnitude and direction of the
electric field that will suspend the bead in midair?
33. Sample Problem
A proton traveling at 440 m/s in the +x direction enters
an electric field of magnitude 5400 N/C directed in the
+y direction. Find the acceleration.
34. Sample Problem
A particle bearing -5.0 μC is placed at -2.0 cm, and a
particle bearing 5.0 μC is placed at 2.0 cm. What is the
field at the origin?
35. Sample Problem
A particle bearing 10.0 mC is placed at the origin, and a
particle bearing 5.0 mC is placed at 1.0 m. Where is the
field zero?
36. Sample Problem end 4
What is the charge on the bead? It’s mass is 32 mg.
55. Sample Problem
A 3.0 μC charge is moved through a potential difference
of 640 V. What is its potential energy change?
56. Electrical Potential in Uniform Electric Fields
The electric potential is related in a simple way to a
uniform electric field.
DV = -Ed
DV: change in electrical potential (V)
E: Constant electric field strength (N/C or V/m)
d: distance moved (m)
57. Sample Problem
An electric field is parallel to the x-axis. What is its
magnitude and direction of the electric field if the
potential difference between x =1.0 m and x = 2.5 m is
found to be +900 V?
58. Sample Problem
What is the voltmeter reading between A and B?
Between A and C? Assume that the electric field has a
magnitude of 400 N/C.
59. Sample Problem end 5
How much work would be done BY THE ELECTRIC
FIELD in moving a 2 mC charge from A to C? From A to
B? from B to C?. How much work would be done by an
external force in each case? End 5
62. Sample Problem
If a proton is accelerated through a potential difference
of -2,000 V, what is its change in potential energy?
How fast will this proton be moving if it started at rest?
63. Sample Problem
A proton at rest is released in a uniform electric field.
How fast is it moving after it travels through a potential
difference of -1200 V? How far has it moved?
69. Sample Problem
Draw field lines for the charge configuration below. The
field is 600 V/m, and the plates are 2 m apart. Label
each plate with its properpotential, and draw and label 3
equipotential surfaces between the plates. You may
ignore edge effects.
-- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
+++++++++++++++++++++++++++++
70. Sample Problem
Draw a negative point charge of -Q and its associated
electric field. Draw 4 equipotential surfaces such that DV
is the same between the surfaces, and draw them at the
correct relative locations. What do you observe about
the spacing between the equipotential surfaces?
71. Objectives: After finishing this
unit, you should be able to:
• Explain and demonstrate the first law of electro-
statics and discuss charging by contact and by
induction.
• Write and apply Coulomb’s
Law and apply it to problems
involving electric forces.
• Define the electron, the
coulomb, and the microcoulomb
as units of electric charge.
72. Objectives: After finishing this
unit you should be able to:
• Define the electric field and explain what
determines its magnitude and direction.
• Write and apply formulas for the
electric field intensity at known
distances from point charges.
• Discuss electric field lines and the
meaning of permittivity of space.
• Write and apply Gauss's law for fields around
surfaces of known charge densities.
73. 16.1 Static Electricity; Electric Charge
and Its Conservation
11.1 When a rubber rod is rubbed against fur, electrons
are removed from the fur and deposited on the rod.
Electrons negative
move from
positive
fur to the -- --
++++
rubber rod.
The rod is said to be negatively charged because of an
excess of electrons. The fur is said to be positively
charged because of a deficiency of electrons.
74. Two Negative Charges Repel
1. Charge the rubber rod by rubbing against fur.
2. Transfer electrons from rod to each pith ball.
The two negative charges repel each other.
The two negative charges repel each other.
75. The Two Types of Charge
Rubber glass
Attraction
fur silk
Note that the negatively charged (green) ball is
attracted to the positively charged (red) ball.
Opposite Charges Attract!
Opposite Charges Attract!
76. 16.2 Electric Charge in the Atom
Atom:
Nucleus (small,
massive, positive
charge)
Electron cloud (large,
very low density,
negative charge)
11.2
77. 16.2 Electric Charge in the Atom
Atom is electrically neutral.
Rubbing charges objects by moving electrons
from one to the other.
78. 16.2 Electric Charge in the Atom
Polar molecule: neutral overall, but charge not
evenly distributed
79. 16.3 Insulators and Conductors
Conductor: Insulator:
Charge flows freely Almost no charge flows
Metals Most other materials
Some materials are semiconductors.
80. 16.4 Induced Charge; the Electroscope
Charging Spheres by Induction
Induction
--- - -
- - Electrons
++
++ -- Repelled
Uncharged Spheres Separation of Charge
--- - -
++
++
-- +
+ +
-
- -
-- + -
Isolation of Spheres Charged by Induction
11.3
81. Induction for a Single Sphere
Induction
--- - -
+ ----
+ --
++ --
Uncharged Sphere Separation of Charge
--- - -
-- - - - -
++ -
++ -
+
+ +
+
Electrons move Charged by Induction
to ground.
82. The Quantity of Charge
The quantity of charge (q) can be defined in
terms of the number of electrons, but the
Coulomb (C) is a better unit for later work. A
temporary definition might be as given below:
The Coulomb: 1 C = 6.25 x 1018 electrons
The Coulomb: 1 C = 6.25 x 1018 electrons
Which means that the charge on a single electron is:
1 electron: e-- = -1.6 x 10-19 C
1 electron: e = -1.6 x 10-19 C
83. Units of Charge
The coulomb (selected for use with electric
currents) is actually a very large unit for static
electricity. Thus, we often encounter a need to
use the metric prefixes.
1 µC = 1 x 10-6 C
1 µC = 1 x 10-6 C 1 nC = 1 x 10-9 C
1 nC = 1 x 10-9 C
1 pC = 1 x 10-12 C
1 pC = 1 x 10-12 C
84. 16.4 Induced Charge; the Electroscope
The electroscope
can be used for
detecting charge:
85. 16.4 Induced Charge; the Electroscope
The electroscope can be charged either by
conduction or by induction.
86. 16.4 Induced Charge; the Electroscope
The charged electroscope can then be used to
determine the sign of an unknown charge.
87. 16.5 Coulomb’s Law
Experiment shows that the electric force
between two charges is proportional to the
product of the charges and inversely
proportional to the distance between them.
88. 11.4
Coulomb’s Law
The force of attraction or repulsion between two
The force of attraction or repulsion between two
point charges is directly proportional to the product
point charges is directly proportional to the product
of the two charges and inversely proportional to the
of the two charges and inversely proportional to the
square of the distance between them.
square of the distance between them.
F
- q q’ +
r qq '
F∝ 2
F F r
q q’
- -
89. 16.5 Coulomb’s Law
The force is along the line connecting the
charges, and is attractive if the charges are
opposite, and repulsive if they are the same.
91. 16.5 Coulomb’s Law
Unit of charge: coulomb, C
The proportionality constant in Coulomb’s
law is then:
Charges produced by rubbing are
typically around a microcoulomb:
92. 16.5 Coulomb’s Law
Charge on the electron:
Electric charge is quantized in units
of the electron charge.
93. 16.5 Coulomb’s Law
The proportionality constant k can also be
written in terms of , the permittivity of free
space:
(16-2)
94. Problem-Solving Strategies
1. Read, draw, and label a sketch showing all
given information in appropriate SI units.
2. Do not confuse sign of charge with sign of
forces. Attraction/Repulsion determines the
direction (or sign) of the force.
3. Resultant force is found by considering force
due to each charge independently. Review
module on vectors, if necessary.
4. For forces in equilibrium: ΣFx = 0 = ΣFy = 0.
97. 16.5 Coulomb’s Law
Coulomb’s law strictly applies only to point charges.
Superposition: for multiple point charges, the forces
on each charge from every other charge can be
calculated and then added as vectors.
102. Summary of Formulas:
Like Charges Repel; Unlike Charges Attract.
Like Charges Repel; Unlike Charges Attract.
kqq ' N ⋅m 2
F= 2 k =9 x 109
r C2
1 µC = 1 x 10-6 C
C
1 µ = 1 x 10-6 C 1 nC = 1 x 10-9 C
1 nC = 1 x 10-9 C
1 pC = 1 x 10-12 C
1 pC = 1 x 10-12 C 1 electron: e-- = -1.6 x 10-19 C
1 electron: e = -1.6 x 10-19 C
103. 16.6 Solving Problems Involving
Coulomb’s Law and Vectors
The net force on a charge is the vector
sum of all the forces acting on it.
107. 16.7 The Electric Field
Force on a point charge in an electric field:
(16-5)
Superposition principle for electric fields:
108. 16.7 The Electric Field
Problem solving in electrostatics: electric
forces and electric fields
1. Draw a diagram; show all charges, with
signs, and electric fields and forces with
directions
2. Calculate forces using Coulomb’s law
3. Add forces vectorially to get result
109. 16.8 Field Lines
The electric field can be represented by field
lines. These lines start on a positive charge
and end on a negative charge.
110. 16.8 Field Lines
The number of field lines starting (ending)
on a positive (negative) charge is
proportional to the magnitude of the charge.
The electric field is stronger where the field
lines are closer together.
111. 16.8 Field Lines
Examples of E-Field Lines
Two equal but Two identical
opposite charges. charges (both +).
Notice that lines leave + charges and enter - charges.
Also, E is strongest where field lines are most dense.
112. 16.8 Field Lines
The electric field between
two closely spaced,
oppositely charged parallel
plates is constant.
113. 16.8 Field Lines
Summary of field lines:
1. Field lines indicate the direction of the
field; the field is tangent to the line.
2. The magnitude of the field is proportional
to the density of the lines.
3. Field lines start on positive charges and
end on negative charges; the number is
proportional to the magnitude of the
charge.
114. 16.9 Electric Fields and Conductors
The static electric field inside a conductor is
zero – if it were not, the charges would move.
The net charge on a conductor is on its
surface.
115. 16.9 Electric Fields and Conductors
The electric field is
perpendicular to the
surface of a conductor –
again, if it were not,
charges would move.
116. 16.10 Gauss’s Law
Electric flux:
11.6
(16-7)
Electric flux through an
area is proportional to
the total number of field
lines crossing the area.
118. 16.10 Gauss’s Law
The net number of field lines through the
surface is proportional to the charge
enclosed, and also to the flux, giving
Gauss’s law:
(16-9)
This can be used to find the electric field
in situations with a high degree of
symmetry.
119. The Density of Field Lines
Gauss’s Law: The field E at any point in space
Gauss’s Law: The field E at any point in space
is proportional to the line density σ at that point.
is proportional to the line density σ at that point.
Gaussian Surface Line density σ ∆N
r
∆A
∆N
σ=
Radius r ∆A
120. Gauss’s Law
Gauss’s Law: The net number of electric field
Gauss’s Law: The net number of electric field
lines crossing any closed surface in an outward
lines crossing any closed surface in an outward
direction is numerically equal to the net total
direction is numerically equal to the net total
charge within that surface.
charge within that surface.
N = Σε 0 EA = Σq
If we represent q as net enclosed q
positive charge, we can write ΣEA =
rewrite Gauss’s law as: ε0