Transcript FirstLecturesPHY242-244

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Lecture 1
 Electric Charge
 Structure of Matter
 Conductors and Insulators
 Charging Mechanisms
 Coulomb’s law
 Superposition principle
08/30/2010
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Electric charge - experiments
Plexiglass
Plexiglass
Plexiglass
Repulsion
Plastic
Attraction
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Electric charge
Plexiglass
Plastic
Plexiglass
Repels
Attracts
Plastic
Attracts
Repels
 New physical property of matter: electric charge
 Charged bodies interact through: electrical forces
 Need two flavors to explain the existence of repulsive as well as
attractive forces
positive (+) and negative (-)
Microscopic view of matter
 Elementary particles:
Number of electrons = Number of protons
The atom is neutral
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Electric charges
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 Two types of charges: positive (+) and negative (-)
 Like charges repel; opposite charges attract.
Examples of charges: electrons (negative); protons (positive)
 Type of materials: insulators, conductors (Ex metals), semiconductors
Electric charge properties
 Like charges repel; opposite charges attract.
 Charge is discrete (quantized)
- The smallest charge possible is = 1.602 x 10-19 C (Coulombs)
- Any charge is a integer multiple of the elementary unit of
charge
 Charge is conserved - charge can be
exchanged between different parts of a
closed system, but the total charge of
the system cannot change
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Three pithballs are suspended from thin threads. It is found that
pithballs 1 and 2 repel each other and that pithballs 2 and 3
repel each other. From this we can conclude that:
1.
2.
3.
4.
5.
1 and 3 carry charges of opposite sign.
1 and 3 carry charges of equal size.
all three carry charges of the same sign.
one of the objects carries no charge.
we need to do more experiments
Types of materials
 Conductors - charges move freely (Ex. metals). One consequence
of this property is that the charge that is transferred to a
conductor will spread out uniformly on its surface.
 Insulators - charges do not move freely (Ex. glass, plastic).
 Semiconductors - intermediate case - charges can move freely
under in some special cases (higher temperature, applied voltage)
(silicon, germanium).
 Superconductors – extremely good conductors = zero resistance.
Restricted to very low temperatures. (Ex. niobium, BISCO)
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Electroscope and Van de Graaff generator
 Electroscope: used to measured
charge by measuring the deflection
of charged metal foils
 Van de Graaff generator.
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Charging mechanisms
 Objects can become charged when elementary charged particles
(most probably electrons) are transferred from one object to the
other.
- When a glass rod and fur are rubbed together some electrons are
transferred to the fur (triboelectricity)
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Charging by contact
- Electrons are transferred from the plastic rod to the metal ball
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When a neutral metal sphere is charged by contact with a
positively charged glass rod, the sphere:
1.
2.
3.
4.
loses electrons
gains electrons
loses protons
gains protons
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Charging by induction
- The copper rod is attracted to the charged glass rod even if
initially is uncharged and does not contact the glass rod at any
time.
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The diagram below shows a neutral metal sphere on a isolating
pedestal. Which of the other diagrams shown, describes best
the charge distribution on the sphere when a negatively charged
rod is brought in its vicinity?
1.
2.
3.
4.
A
B
C
D
++++ +
++ + +
A
-
-+
+
+
-+
B
C
++ + +-
D
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The diagram below shows three identical neutral metal spheres
on a isolating pedestals, which are in contact. Which of the other
diagrams shown, describes best the charge distribution on the
spheres when a negatively charged rod is brought in its vicinity?
1.
2.
3.
4.
A
B
C
D
A
B
C
D
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A negatively charged plastic rod is brought in the vicinity of a
neutral metal ball placed on a isolating pedestal. If the opposite
side of the sphere is briefly connected to the ground and then
the plastic rod is removed, what will the final charge on the ball
be?
1. The metal ball will be
neutral as initially.
2. The metal ball will be
positively charged
3. The metal ball will be
negatively charged.
Solution
- Due to the interaction with the plastic rod, positive and negative
charges will be separated on the ball. Through the grounding loop
the electrons are transferred to the Earth which acts as an infinite
sink or source of electrons, leaving the sphere with a deficit of
electrons (positively charged)
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Three metal balls are suspended from thin threads. It is found
that balls 1 and 2 attract each other and that balls 2 and 3 repel
each other. From this we can definitely conclude that:
1.
2.
3.
4.
5.
1 and 3 carry charges of opposite sign.
1 and 3 carry charges of the same sign.
all three carry charges of the same sign.
one of the objects carries no charge.
we need to do more experiments
Coulomb's Law
 Explored by Charles Augustin de Coulomb

q1q 2
F  k 2 r̂
r
2
N

m
k
 8.99  109
40
C2
1
2
C
 0  8.85 10 -12
N  m2
 0 - permitivit y constant
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Coulomb's law properties

q1q 2
F  k 2 r̂
r
 Inverse square law
 Attractive for unlike charges and
repulsive for like charges
 Direction: along the line joining
the charges
 Newton’s 3rd Law action reaction pair:


F12  - F21
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Two uniformly charged spheres are firmly fastened to and
electrically insulated from frictionless pucks on an air table. The
charge on sphere 2 is three times the charge on sphere 1. Which
force diagram correctly shows the magnitude and direction of the
electrostatic forces?
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3.
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5.
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Which graph best represents the magnitude of the interaction
force between two positively charged ions as a function of the
distance separating them?
1.
2.
3.
4.
B
A
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Coulomb's law vs. Newton law of gravitation
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2
N

m
Coulomb' s law : F  k 2 , k  8.99 109
r
C2
2
m1m 2
N

m
Newton' s law of gravitatio n : F  G 2 , G  6.6742 10-10
r
kg 2
q1q 2
 Inverse square laws - force acting along the line joining the particles.
 There are two kinds of charges, but only one type of mass.
 Gravity is always attractive.
 The electrostatic force, (e.g., between an electron and proton), is
enormously stronger (~ 1035 times stronger)
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A hydrogen atom is composed of a nucleus containing a single
proton, about which a single electron orbits. The electric force
between the two particles is 2.3 x 1039 greater than the
gravitational force! If we can adjust the distance between the two
particles, can we find a separation at which the electric and
gravitational forces are equal?
1.
2.
3.
Yes, we must move the particles further apart.
Yes, we must move the particles closer, together.
No, at any distance.
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Superposition principle
 When two or more charges each exert a force on a charge, the
total force on that charge is the vector sum of the forces exerted
by the individual charges.
  

F  F1  F2  ...Fn
Example: - Linear distribution
What is the net force acting on q3?
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Solution:
Step 1 : - Calculate the force produce by each charge
F1on3  k
q1q3
F2 on3  k
q2 q3
r13
2
r23
2
, where r13  2.0cm
, where r23  4.0cm
Step 2 : - Add all the forces using vector summation rules
Ftotal on 3  F2on3 - F1on3  k
q2 q3
r23
2
-k
q1q3
r13
2
- the magnitude of the total force is given by the difference between
the individual forces since they are opposite.
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Example 2: - Planar distribution (vector nature of Coulomb's law)
Find the magnitude and direction of the force on Q.
1
1
3
2
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Step 1 : - find the force exerted by each charge
on Q
F1onQ  k
q1Q
r(q21Q)
Nm2 (2 10 -6 C)(4.0 10 -6 C)
 9.0 10

C2
(0.5m) 2
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 0.29 N

The components on the x and y axis of F1onQ are:
 F1onQ  x  F1onQ  cos  360 - α 
with

 F1onQ  y  F1onQ  sin  360 - α 




0.3
=0.6
0.5
α=sin -1 (0.6)  36.86o
sin  α  
 F1onQ  x  F1onQ  cos 323.13o  0.232 N


o
F

F

sin
323.13
 -0.17 N



1onQ
y
1onQ

Obtain the F2onQ components
in a similar way
F2onQ x  F2onQ  cosα   0.232 N

F2onQ y  -F2onQ  sin α   0.17 N
Step 2 : - Add the two forces (addition using components)
Ftotal x  F1onQ x  F2onQ x  0.23 N  0.23 N  0.46 N
Ftotal y  F1onQ y  F2onQ y  -0.17 N  0.17 N  0
- the total force on charge Q acts only along x
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Problem
Two free point charges +q and +4q are located a distance
L, apart. A third charge is placed so that the whole system is in
equilibrium. Find the location, magnitude, and sign of the third charge.
L
+q
+4q
Extra credit (5 points) – due Wednesday, September 7
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Review – dealing with vector summation
Question: How do we express the sum vector parameters (F, ), if
we know them for the individual vectors, i.e. (F1, 1) and (F2, 2.
F1x  F1 cos 1

F1 y  F1 sin 1
Fx  F1x  F2 x  F1 cos 1  F2 cos  2

Fy  F1 y  F2 y  F1 sin 1  F2 sin  2
F1y
1
F2 x  F2 cos  2

F2 y  F2 sin  2
For the sum vector:

F1
y
2
F1x
F2x
x

F2
F2y
- the  angles are always measured
clockwise towardsthe positive x axis
y
F1

F
Fy

 F  F 2  F 2 - magnitude
x
y


Fy
F
-1  y 

   tan   - angle
tan  
Fx

 Fx 

F2
Fx
x
Extracredit Problem – Assigned on 09/03/2010
Two free point charges +q and +4q are located a distance L, apart.
A third charge is placed so that the whole system is in equilibrium.
Find the location, magnitude, and sign of the third charge.
Solution:
Step 1: - determine the condition for the charge qo to be at equilibrium.
Call the new charge, q0, and let it be distance
x from +q. The free-body diagram shows
relationship of the two forces on the new
charge:
k
q q0
x2
k
4 q q0
( L - x) 2
1
4

x 2 ( L - x) 2
( L - x) 2  4 x 2  L - x  2 x 
L
x
3
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Step 2: - determine the condition for the charge q or 4q to be at equilibrium.
k
q q0
q0
x
2
k
4q
4q q
( L)
2
where
L
x
3
4
4

 9 q0  4 q  q0  q  q
2
2
( L)
9
9
L
9
4
q0  - q
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Problem (Chapter 21, Problem 4 – Page 574)
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Identical isolated conducting spheres 1 and 2 have equal charges and are
separated by a distance that is large compared with their diameters
(Figure a). The electrostatic force acting on sphere 2 due to sphere 1 is
F. Suppose now that a third identical sphere 3, having an insulating
handle and initially neutral, is touched first to sphere 1 (Figure b), then to
sphere 2 (Figure c), and finally removed (Figure d). The electrostatic
force that now acts on sphere 2 has magnitude F’. What is the ratio F’/F?
1
2
3
F12
q
q/2
q/2
q
q
3q/4
0
q/2
3q/4
F
F/2
3F/8
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In the following diagram what is the direction of the electrostatic
force on the negative charge -q?
Q
1.
2.
-q
3.
4.
5. none of the above
Q
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In the figure a charged particle of charge (- q) is surrounded
by two circular rings of charged particles. What is the net
force on the central particle due to the other particles?
2q 2
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4 o r
2.
2q 2
- 4
3.
ĵ
2
ĵ
2q 2
4 o R
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5.
or
2
2
ĵ
2
ĵ
2q 2
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
oR
2q 2
4 o R
q2
 4
2
or
2
 ĵ
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Two small charged objects attract each other with a force F
when separated by a distance d. If the charge on each
object is reduced to one-fourth of its original value and the
distance between them is reduced to d/2 the force
becomes:
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F/16
F/8
F/4
F/2
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Electric Field
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 Why do we need this concept?
- Coulomb's law describes action at distance (like gravitation).
- Any charge changes the properties of the space around it.
- We need a field to describe this change.
 The notion of field is not a new one: - A field is a quantity that
is assigned a value at each point in a region of space
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Vector field
© http://maps.wunderground.com/
Wind speed - the absolute value and direction is known.
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Back to the electric field

 F
E
q0
- Units: N/C
- E does not depend on the
q0  0

F

F
q0  0
test charge q0.
 To calculate the electric field of a charge distribution use
superposition:

 

E total  E1  E 2  E3  ...
Note: The rest of this chapter and chapter 23 are mostly dedicated to
how to determine the magnitude and orientation of different electric.
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Force on a charge in an electric field


F  qE
The direction of the force acting on a
point charge in an electric field
depends on both the direction of the
electric field and the sign of the
charge.
Positive charge q0 placed in an
electric field
-
Negative charge q0 placed in an
electric field
Electric field due to a point charge

q q0
F  k 2 r̂
r

 F
E
q0

E
1
q
r̂
2
4  0 r
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