Transcript Document
23.1 Properties of Electric Charges
23.2 Charging Objects By Induction
23.3 Coulomb’s Law
23.4 The Electric Field
23.6 Electric Field Lines
23.7 Motion of Charged Particles in a Uniform Electric Field
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King Saud university
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Electric Charge
• Types:
– Positive
• Glass rubbed with silk
• Missing electrons
– Negative
• Rubber/Plastic rubbed with fur
• Extra electrons
• Arbitrary choice
– convention attributed to ?
• Units: amount of charge is measured in
[Coulombs]
• Empirical Observations:
– Like charges repel
– Unlike charges attract
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Charge in the Atom
•
•
•
•
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Protons (+)
Electrons (-)
Ions
Polar Molecules
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23.1 Properties of Electric Charges
• Conservation
electricity is the implication that electric charge is always conserved.
• That is, when one object is rubbed against another, charge is not
created in the process. The electrified state is due to a transfer of
charge from one object to the other.
• One object gains some amount of negative charge while the other
gains an equal amount of positive charge.
• Quantization
– The smallest unit of charge is that on an electron or proton. (e
= 1.6 x 10-19 C)
• It is impossible to have less charge than this
• It is possible to have integer multiples of this charge
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Q Ne
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23.2 Charging Objects By Induction
Conductors and Insulators
• Conductor
transfers charge on contact
• Insulator
does not transfer charge on contact
• Semiconductor might transfer charge on contact
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Charge Transfer Processes
• Conduction
• Polarization
• Induction
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23-3 Coulomb’s Law
• Empirical Observations
F q1q 2
1
F 2
r
Direction of the force is along the line joining the two charges
• Formal Statement
kq1q 2
F12
rˆ21
2
r21
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• Consider two electric charges: q1 and q2
kq1q2
• The electric force F between these two
F 2
charges separated by a distance r is given by
r
Coulomb’s Law
• The constant k is called Coulomb’s constant
9
2
2
k 910 Nm /C
and is given by
• The coulomb constant is also written as
k
1
4 0
where 0 8.85 10
12
C2
Nm 2
• 0 is the “electric permittivity of vacuum”
– A fundamental constant of nature
F
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40
q1 q 2
r2
8
• Double one of the charges
– force doubles
• Change sign of one of the charges
– force changes direction
• Change sign of both charges
– force stays the same
• Double the distance between charges
– force four times weaker
• Double both charges
– force four times stronger
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Example:
What is the force between two charges of 1 C separated by 1 meter?
Answer: 8.99 x 109 N,
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Coulomb’s Law Example
• What is the magnitude of the electric force of
attraction between an iron nucleus (q=+26e) and
its innermost electron if the distance between
them is 1.5 x 10-12 m
• The magnitude of the Coulomb force is
•
F = kQ1Q2/r2
• = (9.0 x 109 N · m2/C2)(26)(1.60 x 10–19 C)(1.60 x
10–19 C)/(1.5x10–12 m)2
• =
2.7 x 10–3 N.
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Example - The Helium Nucleus
Part 1: The nucleus of a helium atom has two protons and two
neutrons. What is the magnitude of the electric force between the
two protons in the helium nucleus?
Answer: 58 N
Part 2: What if the distance is doubled; how will the force
change?
Answer: 14.5 N
Inverse square law: If the distance is doubled then the force is
reduced by a factor of 4.
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Example - Equilibrium Position
• Consider two charges located on the x axis
x1
x2
• The charges are described by
– q1 = 0.15 C
– q2 = 0.35 C
x = 0.0 m
x = 0.40 m
• Where do we need to put a third charge for that
charge to be at an equilibrium point?
– At the equilibrium point, the forces from the two charges will
cancel.
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Example - Equilibrium Position (2)
x1
x3
x2
• The equilibrium point must be along the x-axis.
• Three regions along the x-axis where we might
place our third charge
x3 < x 1
x1 < x 3 < x 2
x3 > x 2
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Example - Equilibrium Position (3)
x1
x2
• x3<x1
– Here the forces from q1 and q2 will always point in the
same direction (to the left for a positive test charge)
• No equilibrium
• x2<x3
– Here the forces from q1 and q2 will always point in the
same direction (to the right for a positive test charge)
• No equilibrium
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Example - Equilibrium Position (4)
0.4 x
x
q3
x 1 < x3 < x 2
x balance.
Here the forces from q1 and q2 can
q1q3
q2 q3
k
k
2
2
( x)
(10 x )
Answer: x3 = 0.16 m
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Zero Resultant Force, Example
Two fixed charges, 1mC and -3mC are separated by 10cm as shown in the figure (a) where may a
third charge be located so that no force acts on it?
– The magnitudes of the individual forces will be
equal
– Directions will be opposite
– Will result in a quadratic
– Choose the root that gives the forces in
opposite directions
q1q3
q2 q3
k
k
2
( x)
(10 x ) 2
X= -13.7 cm
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two charges are located on the positive x-axis of a coordinate system, as shown in
the figure. Charge q1=2nC is 2cm from the origin, and charge q2=-3nC is 4cm
from the origin. What is the total force exerted by these two charges on a charge
q3=5nC located at the origin?
The total force on q3 is the vector sum of the forces due to q1 and q2 individually.
The total force is directed to the left, with magnitude 1.41x10-4N.
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Example - Charged Pendulums
• Consider two identical charged balls hanging
from the ceiling by strings of equal length 1.5
m (in equilibrium). Each ball has a charge of
25 C. The balls hang at an angle = 25 with y
respect to the vertical. What is the mass of
the balls?
x
Step 1: Three forces act on each ball:
Coulomb force, gravity and the tension of the Ball on left :
string.
kq2
Fx T sin 2
d
Fy T cos mg
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Example - Charged Pendulums (2)
Step 2: The balls are in equilibrium
positions. That means the sum of all
forces acting on the ball is zero!
T sin kq 2 / d 2
T cos
mg
kq
mg 2
d tan
2
d=2 l sin
Answer: m = 0.76 kg
A similar analysis applies to the ball on the right.
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Electric Force and Gravitational Force
• Coulomb’s Law that describes the electric force and
Newton’s gravitational law have a similar functional
form
Felectric
q1q2
k 2
r
Fgravity
m1m2
G 2
r
• Both forces vary as the inverse square of the
distance between the objects.
• Gravitation is always attractive.
• k and G give the strength of the force.
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The Hydrogen Atom
Compare the electrostatic and gravitational the forces
(The average separation of the electron and proton
r = 5.3 x 10-11 m
e
r
F
+Ze
2
19
e2
C
9 N.m 1.60 10
Fe
. 2 9 10
2
2
11
40 r
C
5
.
3
10
m
1
v
2
8.2 108 N
Fg G
me m p
r2
2
31
kg 1.67 1027 kg
11 N.m 9.11 10
6.7 10
2
2
11
k
g
5.3 10 m
3.6 1047 N
Fe/Fg = 2 x 1039 The force of gravity is much weaker than the electrostatic force
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King Saud university
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Coulomb’s Law Example
y
+
Q
F1
L
F3
+
Q F2
L
Q
x
• Q = 6.0 mC
• L = 0.10 m
• What is the magnitude and
direction of the net force on one of
the charges?
Q
+
+
We find the magnitudes of the individual forces on the charge at the upper
right corner:
F1= F2 = kQQ/L2 = kQ2/L2
= (9 x109 N · m2/C2)(6 x10–3 C)2/(0.100 m)2 = 3.24 x107 N.
F3= kQQ/(L√2)2 = kQ2/2L2
= (9 x109 N · m2/C2)(6 x10–3 C)2 /2(0.100 m)2
= 1.62 x107 N.
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King Saud university
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The directions of the forces are determined from the signs
of the charges and are indicated on the diagram. For the
forces on the upper-right charge, we see that the net force
will be along the diagonal. For the net force, we have
F = F1 cos 45° + F2 cos 45° + F3
= 2(3.24x107 N) cos 45° + 1.62x107 N
= 6.20 x107 N along the diagonal, or away from the center
of the square.
From the symmetry, each of the other forces will have the
same magnitude and a direction away from the center: The net
force on each charge is
6.20x107 N away from the center of the
square.
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King Saud university
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Example - Four Charges
Consider four charges
placed at the corners of a
square with sides of length
1.25 m as shown on the
right. What is the
magnitude of the electric
force on q4 resulting from
the electric force from the
remaining three charges?
Set up an xy-coordinate system with its origin at q2.
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King Saud university
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Example - Four Charges (2)
Answer:
F (on q4) = 0.0916 N
… and the direction?
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