Physics 122B
Electricity and Magnetism
Lecture 1
Charge and Coulomb’s Law
March 26, 2007
Martin Savage
This week
week
date
Lecture Text
tutorial
topic
reading
lab
1
Mar 26
Charge and
Coulombs
Law
No Labs
Mar 28
Electric Field 25.5 to 26.1
Mar 30
Field Lines
and Charge
Distributions
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25.1 to 25.4
Charge
26.2 to 26.3
Physics 122B - Lecture 1
2
The History of Electricity
Electric  elektron = Greek word for amber
“Rub amber with wool, and it will pick up
bits of wood, feathers, straw …”
Thales of Miletus
(640-546 BC)
About 1736, Charles Francois du Fay (1698-1739) learned that rubbing glass
and rubbing resinous substances (e.g., amber) seemed to produce charges of
different kinds. He found that two charges of the same kind repelled each other,
while two of unlike kinds attracted. He suggested that electricity might exist as
two distinctly different types, which he named “vitreous” and “resinous”
electricity.
William Watson (1715-1790) suggested in 1746 that electricity was one “fluid”.
One of the two kind of electricity proposed by Du Fay could be an excess (+) of this
fluid and the other a difficiency of it (-). Flow from + to - could account for
electrical discharge. Benjamin Franklin (1706-1790) adopted and popularized
Watson's “one fluid” theory and chose vitreous electricity to be the positive type
(thereby giving electrons a negative charge). Franklin’s great reputation won
universal acceptance for this view.
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Physics 122B - Lecture 1
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Charging Experiments (1)
1
Observations:
2
1. No force is observed
between un-rubbed rods.
2. Plastic rods rubbed with
wool repel each other.
3
3. Glass rod rubbed with silk
attracts plastic rod
rubbed with wool.
4
4. At increased distance,
forces are decreased.
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Charging Experiments (2)
1
2
3
More Observations:
1. A (charged) plastic rod rubbed with wool
attracts small pieces of paper.
2. A (charged) plastic rods rubbed with wool
is attracted to an un-rubbed (neutral)
plastic rod.
3. A plastic rod rubbed with wood is
attracted to the wool, repelled by the silk.
4. No charged (rubbed) object attracts both
the charged plastic rod and the charged
glass rod.
4
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Charging Experiments (3)
1
3
4
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2
Even More Observations:
1. The charge from a plastic rod rubbed with
wool can be transferred to a metal sphere.
2. After the transfer, the plastic rod does
not attract paper.
3. A 2nd metal sphere connected by a metal
rod acquires the rod’s charge.
4. A 2nd metal sphere connected by a plastic
rod does not acquire the rod’s charge.
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Electric Charge and Atoms
An atom with atomic number Z consists
of a small but massive nucleus of charge
+Ze, surrounded by a cloud of Z electrons,
each with much less mass and with a
charge of -e.
The atom has a diameter of about 10-10
m (0.1 nm), while the atomic nucleus has a
diameter of about 5x10-15 m (5 fm). All
atoms, heavy or light, are about the same
size, but the mass varies from 1 to ~250.
The nucleus contains Z protons with
charge +e and N neutrons with charge 0.
The atomic mass number of the atom is A
= Z+N.
Charge: q = Npe-Nee = (Np-Ne)e ;
for neutral atoms (q=0), Np=Ne=Z
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Positive and Negative Ions
Atoms can have a net charge if
NpNe, where Np=Z is the number of
protons and Ne is the number of
electrons in the atom.
If a neutral atom loses one electron,
it will have a net positive charge of q=+e
because Np-Ne=+1.
If a neutral atom gains one electron,
it will have a net negative charge of q=-e
because Np-Ne=-1.
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Charging by Friction
Friction causes the breaking of molecular
bonds, and can result in a separation of
charges in a formerly neutral molecule,
creating a positive and a negative molecular
ion.
Rubbing a plastic rod with wool or a glass
rod with silk produces such charge separation
effects.
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Conservation of Charge
Electrical charge can be neither created or destroyed. It can
be separated and moved around, but the net charge of an isolated
system must remain constant. qinitial = qfinal
Example: A plastic rod is rubbed with wool, each initially
neutral. Then qwool=-qrod.
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Insulators and Conductors (1)
If a conductor is charged,
all of the charge must reside
on the outer surface (and
none in the interior.)
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If an insulator is charged,
the charge may (or may not)
reside in the interior.
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Insulators and Conductors (2)
In insulators, the electrons are tightly bound
in the atoms and are not free to move around.
When insulators are charged, e.g. by friction,
patches of molecular ions are created on the
surface, but these patches are immobile.
In solid metal conductors, the outer (valence)
electrons of the atoms are only weakly bound and
are free to move around in the solid. The
conductor as a whole may be electrically neutral,
but the electrons are rather like an electrically
charged liquid, a “sea” of electrons within the
material. Electrons are the “charge carriers”.
There are other forms of conduction (in ionic liquids, etc.) in
which the charge carriers are not electrons.
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Charging an Insulator
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Charging a Conductor
Conductors cannot be charged by
friction.
However, charge can be transferred to
a conductor by contact with a charged
object.
The charges arriving at the conductor
stay on the outer surface and distribute
themselves over that surface so that they
are as far away as possible from the
repulsive forces of the other charges.
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The Electroscope
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Charging an Electroscope
The angle of deflection of the leaves provides a rough
indication of the amount of charge that has been deposited
on the electroscope.
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Discharging a Charged Object
The human body, composed mainly of salty
water, is a moderately good conductor.
Therefore, a person touching a charged
object will normally discharge the object.
Where the charge goes next depends on
the degree to which the person is insulated
from ground (e.g., by rubber shoe soles).
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Charge Polarization
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Induced Charge
By using charge
polarization, it is possible to
induce charge on an
electrically neutral object.
Example: Bring a charged
rod near (but not touching) an
electroscope and observe the
effect on the leaves.
-
-
+
+
+
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-
+
+
+
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Charges and Forces
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The Electric Dipole
Experiment: Bring a positive charge near a neutral atom.
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Dipoles and Forces (1)
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Dipoles and Forces (2)
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Charging by Induction
Induction: Charging an object with only neutral contact.
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Coulomb’s Law
Like charges repel.
Charles Augustine
de Coulomb
(1736-1806).
Opposite charges attract.
Coulomb’s Law:
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F1 on 2 = F2 on 1 = K
Physics 122B - Lecture 1
q1 q2
r
2
( Magnitude of force )
25
Units of Charge
F=K
q1 q2
r
2
=
1
q1 q2
4 0
r2
Coulomb’s Law,
written two ways.
K = 8.99 109 N m2 /C2  9.0 109 N m2 /C2
C = coulomb = SI unit of charge;
1.0 nC = 1.0 10-9 C
(Note that in Newton’s law of gravitation, G,
which plays a role similar to K, has the value
G = 6.67  10-11 N m2/kg2.)
1
0 =
= 8.85 10-12 C2 /(N m 2 ) = permittivity constant
4 K
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Using Coulomb’s Law
1. Coulomb’s Law applies only to point charges.
(This is particularly important because charge are
free to move around on conductors.)
1. Strictly speaking, Coulomb’s Law applies only to
electrostatics (non-moving charges).
(However, it is usually OK provided v<<c).
1. Electrostatic forces can be superposed.
Linear superposition !!!!
Fnet = F1 on j + F2 on j + F3 on j +
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Example: Sum of Two Forces
Two +10 nC charged particles are
2 cm apart on the x axis.
(1) What is the net force on a +1.0
nC particle midway between
them?
(2) What is the net force if the
charged particle on the right is
replaced by a -10 nC charge?
F( ++ ) net = F1 on 3 + F2 on 3 = F iˆ - F iˆ = 0
F=K
q1q2
r2
= (9.0  109 N m 2 /C2 )
(1.0 10-8 C )(1.0 10-8 C )
(1.0 10-2 m) 2
= 9.0  10-4 N
F( +- ) net = F1 on 3 + F2 on 3 = F iˆ + F iˆ
= 2 F iˆ = 1.8  10-3 N
-8
-98
-9
q1q2
9
2
2 (1.0  10 C )(1.0  10 C )
F = K 2 = (9.0  10 N m /C )
r
(1.0 10-2 m) 2
= 9.0  10-4 N
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Example: Point of Zero Force
Two positively charged
particles q1 and q2=3q1 are
placed 10. 0 cm apart. Where
(other than infinity) could
another charge q3 be placed so
as to experience no net force?
Need force vectors to be co-linear, so location must be on x axis.
Need force vectors to be in opposite directions, so location must
be between 0 and d.
Need force vectors equal in magnitude, so F=Kq1q3/x2=Kq2q3/(d-x)2.
Therefore, (d-x)2=3x2 or x=d/(1±√3) = 10 cm/(1±1.732);
x+=10.0 cm/2.732 = 3.66 cm;
x-=10.0 cm/-0.732=-13.66 cm, which does not satisfy the 2nd criterion.
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Example: Three Charges (1)
Three charges with q1 = -50 nC,
q2 = +50 nC, and q3 = +30 nC, are
placed at the corners of a 10 cm x
5 cm rectangle as shown.
What is the net force on q3 due
to the other two charges?
F1 on 3 = Kq1q3 / r132
= (9.0  109 N m 2 /C2 )(5.0 10-8 C)(3.0 10-8 C) /(0.10 m) 2
= 1.35  10-3 N
F2 on 3 = Kq2 q3 / r232
= (9.0 109 N m 2 /C 2 )(5.0 10-8 C)(3.0 10-8 C) /[(0.10 m)2 +(0.05 m) 2 ]
= 1.08 10-3 N
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Example: Three Charges (2)
ˆ
ˆ
Sin  ˆ
Fnet = -F2 on Cos
3 Sin i - F1 on 3 j + F2 on 3Cos j
Fx = -(1.08 10-3 N)
5
= -4.83 10-4 N
125
Fy = -(1.35 10-3 N) + (1.08 10-3 N)
10
125
= -3.84 10-4 N
Fnet = Fx 2 + Fy 2 = 6.17 10-4 N
 = ArcTan
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Fy
Fx
= 38.5
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Example: Lifting a Glass Bead
A small plastic sphere is charged to
-10 nC. It is held 1.0 cm above a small
glass bead that rests on a table. The
bead has a mass of 15 mg and a charge
of +10 nC. Will the glass bead “leap up”
to the plastic sphere?
F1 on 2 = Kq1q2 / r12 2
= (9.0 109 N m 2 /C 2 )(1.0 10-8 C)(1.0 10-8 C) /(0.01 m) 2
= 9.0 10-3 N
w = m2 g = (1.5 10-5 kg)(9.80 N/kg) = 1.5 10-4 N
Therefore, F1 on 2 exceeds w by a factor of 60.
Therefore, the sphere should indeed leap upward.
(Note that we have neglected electrical forces between
the bead and table, which could be significant.)
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End of Lecture 1
 Before the next lecture, read
Knight, Chapters 25.5 through 26.1
 Lecture Homework #1 will be
placed on the Tycho system soon.