Transcript q 3

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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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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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  910 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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40
q1 q 2
r2
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• 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
• The charges are described by
– q1 = 0.15 C
– q2 = 0.35 C
x1
x = 0.0 m
x = 0.40 m
x2
• Where do we need to put a third
charge for that charge to be at an equilibrium point?
x
At the equilibrium point, the forces from
the two charges will cancel.
Here the forces from q1 and q2 can balance.
q1q3
q2 q3
k
k
2
( x)
(0.4  x ) 2
x  0.16m
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0.4  x
x
q3
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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
k
q1q3
q2 q3

k
( x) 2
(10  x ) 2
1 10 6
3  10 6

2
( x)
( x  10) 2
x  13.7cm
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Example:
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?
Step 1: Three forces act on each ball:
Coulomb force, gravity and the tension of the
string.
Ball on left :
x
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!
kq2
T sin   2
d
T cos   mg
T sin  kq2 / d 2

T cos 
mg
d=2 l sin 
kq2
mg  2
d tan 
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
r2
Fgravity
m1m2
G
r2
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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Example:
An electron is released above the surface of the Earth. A second electron
directly below it exerts an electrostatic force on the first electron just great
enough to cancel out the gravitational force on it. How far below the first
electron is the second?
FE  mg
q1q2
q1q2
k 2  mg  r  k
r
mg
19 2
(
1
.
6
x
10
)
9
r  (9 10 )
5.1 m
31
(9.11x10 )(9.8)
Fe
e
mg
r=?
e
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The electron and proton of a hydrogen atom are separated (on the average) by a
distance of approximately 5.3 x10-11 m. Find the magnitudes of the electric force
and the gravitational force between the two particles.
Compare the electrostatic and gravitational the forces


2
19
e2 
C
9 N.m  1.60  10

Fe 
. 2   9 10
2
2 
11
40 r
C
5
.
3

10
m


1


2
 8.2 108 N
Fg  G
me m p
r2

 
2
31

kg  1.67 1027 kg
11 N.m  9.11 10

  6.7 10
2
2 
11
k
g
5.3 10 m





 3.6 1047 N
Fe/Fg = 2 x 1039  The force of gravity is much weaker than the electrostatic force
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Electric Forces and Vectors
Electric Fields and Forces are ALL vectors, thus all rules
applying to vectors must be followed.
Consider three point charges, q1 = 6.00 x10-9 C (located at the origin),q3 =
5.00x10-9 C, and q2 = -2.00x10-9 C, located at the corners of a RIGHT triangle.
q2 is located at y= 3 m while q3 is located 4m to the right of q2. Find the
resultant force on q3.
4m
q2
3m
q1

q3
Which way does q2 push q3?
Which way does q1 push q3?
Fon 3 due to 1
5m
Fon 3 due to 2
q3
 = 37
= tan-1(3/4)
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4m
q2
3m
q1

q3
Fon 3 due to 1
5m
= tan-1(3/4)
Fon 3 due to 2
q3
F3,1sin37
 = 37
F3,1cos37
F3, 2
F3, 2
(5.0 x10 9 )( 2 x10 9 )
 (8.99 x10 )
42
 5.6 x10-9 N
9
(6 x10 9 )(5 x10 9 )
F3,1  (8.99 x10 )
52
F3,1  1.1x10-8 N
9
F
F
F
x
 F3,1 cos(37)  F3, 2
x
 3.18 x10 9 N
y
 F3,1 sin( 37)  6.62 x10 9 N
Fresultant  ( Fx ) 2  ( Fy ) 2
Fres  7.34x10-9 N
Direction    tan
1
F

(
F
y
)
x
64.3 0 above the +x
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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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tan  
L
1
L
so   450
3x
L
1
so   450
L
 F 3 cos   1.62  107  0.707  1.145  107 N
3y

tan  
F
F
F
F
F
3
sin   1.62  107  0.707  1.145  107 N
x
 3.24  107  1.145  107  4.385  107 N
y
 3.24  107  1.145  107  4.385  107 N
F 
 


2
2
7
 6.2  107 N
 4.385  10
4.385  107
  tan 1
F
F
y
 450
x
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= 6.20 ‫ ء‬107 N away from the center of the square.
.
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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?
Answer:
F (on q4) = 0.0916 N
… and the direction?
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HOMEWORK :
23-7; Three point charges are located at the corners of an equilateral
triangle. Calculate the net electric force on the 7.00 uC charge.
23-8: Two small beads having positive charges 3q and q are fixed at the
opposite ends of a horizontal insulating rod extending from the origin to the
point x =d. a third small charged bead is free to slide on the rod. At what
position is the third bead in equilibrium? Can it be in stable equilibrium?
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23-12; An object having a net charge of 24.0 C is placed in a uniform electric
field of 610 N/C that is directed vertically. What is the mass of this object if it
“floats” in the field?
3-18; Two 2.00uC point charges are located on the x axis. One is at x = 1.00
m, and the other is at x =- 1.00 m. (a) Determine the electric field on the y
axis at y =0.500 m. (b) Calculate the electric force on a - 3.00uC charge
placed on the y axis at y = 0.500 m.
23-41; An electron and a proton are each placed at rest in an electric field of
520 N/C. Calculate the speed of each particle 48.0 ns after being released.
23-44; The electrons in a particle beam each have a kinetic energy of 1.60 x
10-17 J. What are the magnitude and direction of the electric field that stops
these electrons in a distance of 10.0 cm?
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