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Review: Properties of Electric Charges
„
„
„
„
There are two kinds of charges in nature:
positive and negative;
Charges of the same sign repel each other;
and those of opposite sign attract one
another;
Total charge in an isolated system is
conserved (positive = negative);
Charge is quantized.
Review Questions
1.
2.
3.
How many types of charges in nature?
A glass rod rubbed with silk will attract or
repel with a rubber rubbed with fur?
After a glass rod is rubbed with silk, it
have a total charge of 1000 protons. How
much charge does the silk have?
New Concept
Conductors
„
Electrical conductors are materials in which
some of the electrons are free electrons
„
„
„
Free electrons are not bound to the atoms and
can move relatively freely
Examples : 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
New Concept
Insulators
„
Electrical insulators are materials in which all
of the electrons are bound to atoms
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„
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These electrons can not move relatively freely
through the material
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
New Concept
Semiconductors
„
„
The electrical properties of semiconductors
are somewhere between those of insulators
and conductors
Examples of semiconductor materials
include silicon and germanium
New Physical Process
Charging by Induction
Charging by Induction
„
Charging by induction
requires no contact with
the object inducing the
charge
„
The electrons in the
neutral sphere are
redistributed as the
negatively charged
rod is approaching
Charging by Induction, 2
„
„
„
Some electrons will
leave the sphere through
the ground wire
There will now be more
positive charges after the
ground wire is removed
The positive charge has
been induced in the
sphere
“Ground” and “Negative Polarity”
Pos
Neg
Charging by Induction, 5
„
The electrons remaining
on the sphere
redistribute themselves
„
There is still a net
positive charge on the
sphere after rod is
removed
Charge Rearrangement in
Insulators
„
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In neutral molecules,
the center of positive
charge coincides with
that of negative
charge
The charges within the
molecules of the
material are
rearranged Æa layer
of charge facing the
charged object
Questions:
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What is insulator and is it useful?
What is conductor and is it useful?
What is semi-conductor and is it useful?
Why a rubber rod rubbed with fur can
absorb bits of paper?
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
Proton, electron, and atomic mass
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Proton:
„ proton is a particle smaller than an atom (called subatomic
particle) with an electric charge of one positive fundamental unit
(+1.602 × 10−19 C) and a mass of 1.6726 × 10−27 kg
Electron:
„ electron is a fundamental particle smaller than an atom (called
subatomic particle) which carries a negative fundamental unit
(-1.602 × 10−19 C) and a mass of 9.11x10-31 kg.
18 electrons or protons
„ 1 C needs 6.24 x 10
-6
„ Typical charges can be in the µC range: 1 µC =10 C
Unit of atomic mass:
-27 kg
„ 1u = 1.66x10
Problem solving
Example 1:
The plasma is an ionized gas. The plasma in
a hot star contains quadruply ionized
nitrogen atoms, N+4. Find the charge and
mass of a quadruply ionized nitrogen atoms.
Problem solving
Example 1:
The plasma is an ionized gas. The plasma in a hot star
contains quadruply ionized nitrogen atoms, N+4. Find the
charge and mass of a quadruply ionized nitrogen atoms.
Solution:
1) Charge = 4 electrons x 1.602x10-19C
=6.408x10-19C
2) Mass: mN = mN − 4 × me
+4
= 14.007u − 4me
= 14.007 × 1.66 × 10 27 kg − 4 × 9.11×10 −31 kg
= 2.3248 ×10 − 26 kg
Problem solving
Example 2:
Calculate the number of electrons in 1g of copper.
Problem solving
Example 2:
Calculate the number of electrons in 1g of copper.
Solution:
One atom of copper has 29 electron.
One atom of copper weighs 63.546u, i.e. its molar
mass is 63.546 g/mol.
Within 1 g of copper, the number of electrons is
thus
⎛
⎞⎛
1.0 g
atoms ⎞⎛ electrons ⎞
⎟⎟⎜ 6.02 ×10 23
n = ⎜⎜
⎟⎜ 29
mol ⎠⎝
⎝ 63.546 g / mol ⎠⎝
= 2.747 ×10 23
atom
⎟
⎠
Coulomb’s Law
„
„
Charles Coulomb measured the electric
forces between two small charged
spheres
He found that:
„ The force is inversely proportional to
the square of the separation r
between the particles and directed
along the line joining them. Æ Fe ∝
1/r2.
„ The force is proportional to the
product of the charges, q1 and q2, on
the two particles. Æ Fe ∝ q1q2.
Fe ∝ (q1q2)/r2
„ The force is a conservative force
(work done just depends on
positions not on path).
Torsion balance
Coulomb’s Law, Equation
„
Mathematically,
q1 q2
Fe = ke
2
r
„
The SI unit of charge is the coulomb (C)
ke is called the Coulomb constant
9 . 2 2
„ ke = 8.9875 x 10 N m /C = 1/(4πεo)
„ εo is the permittivity of free space
-12 C2 / N.m2
„ εo = 8.8542 x 10
„
Vector Nature of Electric Forces
„
In vector form,
q1q2
F12 = ke 2 rˆ
r
„
r̂ is a unit vector directed
„
from q1 to q2
The like charges produce
a repulsive force between
them
Vector Nature of Electrical
Forces, 2
„
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Electrical forces obey
Newton’s Third Law
The force on q1 is equal in
magnitude and opposite in
direction to the force on q2
„
„
F21 = -F12
With like signs for the
charges, the product q1q2 is
positive and the force is
repulsive
Active Figure 23.7
(SLIDESHOW MODE ONLY)
A Final Note about Directions
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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
The Superposition Principle
„
The resultant force on any one
charge equals the vector sum
of the forces exerted by the
other individual charges that
are present
„
„
Remember to add the forces as
vectors
The resultant force on q3 is
the vector sum of all the forces
exerted on it by other charges:
F3 = F13 + F23
Problem solving
Example 4: Three point
charges located at the
corners of a right triangle
as shown. q1 = q3 = 10
μC, q2= -5 μC. a = 0.1 m.
Find the total force F3 on
q3 by the other two
charges.
F3
Problem solving
Solution:
The force directions and the
distance between charges
are shown in the figure.
q q
(10 ×10 C )(10 ×10 C ) = 44 N
F =k
= (8.99 × 10 Nm / C )
1
13
e
F23 = ke
3
9
( 2a )
2
−6
2
2(0.1m )
2
q2 q3
a
2
−6
2
(
= 8.99 × 10 Nm / C
9
2
2
) (5 ×10 C )(10 ×10 C ) = 45N
−6
(0.1m )2
F3 x = F13 x + F23 x = F13 cos 45° − F23 = −13.89 N
r
F3 = (− 13.89 )iˆ + 31.11 ˆj
r
2
2
F3 = (− 13.89 ) + (31.11) = 34.07 N
−6
F3 y = F13 y + F23 y = F13 sin 45° − 0 = 31.11N
Problem solving
Example 4:
(Think about)
Case A: q1 and q2 are
fixed (not movable).
The force on q is zero.
If pull it along x axis,
what happen? If it is
pulled along y axis,
then what happen?
Case B: If q is not
negative but positive,
repeat Case A.
y
q1
q2
+
+
q
x
Case A
y
+
+
q1
q
Case B
+
q2
x
Electric Field – Introduction
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Gravitational force is a field force
Field forces can act through space
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„
The effect is produced even with no
physical contact between objects
An electric field is said to exist in the
region of space around a charged
object
„
This charged object is the source charge
Electric Field – Definition, cont
„
The electric
r field
vector, E , at a point in
space is defined
r as the
electric force F acting
on a positive test
charge, qo placed at
that point divided by
the test charge:
r
r F
E=
q0
Relationship Between F and E
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Fe = qE
„
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„
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F = mg
This is valid for a point charge only
For larger objects, the field may vary over the
size of the object
If q is positive, F and E are in the same
direction
If q is negative, F and E are in opposite
directions
Electric Field Notes, Final
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The direction of E is that
of the force on a positive
test charge
SI units: N/C
Electric Field by Point Charge,
Vector Form
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The force between the
source and test charges is
determined by Coulomb’s
law
qqo
Fe = ke 2 rˆ
r
q
Then, the electric field will
be
Fe
q
E=
= ke 2 rˆ
qo
r
r̂
is a unit vector from source to test point
r̂
Electric Field Direction
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„
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a) q is positive, F is directed
away from q
b) The direction of E is also away
from the positive source charge
c) q is negative, F is directed
toward q
d) E is also toward the negative
source charge
Active Figure 23.13
(SLIDESHOW MODE ONLY)
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 of all the charges
qi
E = ke ∑ 2 rˆi
i ri
q1
+
+
q3
q4
Problem solving
Example 5.
Three point charges are
arranged as shown.
(a)Find the vector electric field
that the 6.00-nC and –3.00-nC
charges together create at the
origin.
(b)Find the vector force on the
5.00-nC charge.
Problem solving
Example 5.
Three point charges are arranged as shown.
(a)Find the vector electric field that the 6.00nC and –3.00-nC charges together
create at the origin.
(b)Find the vector force on the 5.00-nC
charge.
Solution:
( ) (
)(
)( ) (
)( ) (
)
9
2
2
−9
r ke q1
×
Nm
C
×
C
8
.
99
10
/
3
.
00
10
ˆj = − 2.70 × 103 N / C ˆj
E1 = 2 − ˆj =
−
2
r1
(
0.100m )
−9
9
2
2
r
k e q2
8
.
99
10
/
6
.
00
10
×
Nm
C
×
C
ˆ = − 5.99 × 10 2 N / C iˆ
E2 = 2 − iˆ =
−
i
2
r
(
)
0
.
300
m
2
r r r
E = E1 + E2 = − 5.99 × 10 2 N / C iˆ − 2.70 ×103 N / C ˆj
r
r
F = qE = 5.00 ×10 −9 C − 599iˆ − 2700 ˆj N / C = − 3.00 ×10 −6 iˆ − 13.50 × 10 −6 ˆj N
= − 3.00iˆ − 13.5 ˆj μN
( ) (
(
(
)(
(
)
)(
) (
)
(
)
)
)