Transcript 25 - wbm

Coulomb’s Law and Electric
Field
Chapter 24: all
Chapter 25: all
1
Electric charge


Able to attract other objects
Two kinds





Positive – glass rod rubbed with silk
Negative – plastic rod rubbed with fur
Like charges repel
Opposite charge attract
Charge is not created, it is merely transferred
from one material to another
2
Elementary particles


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
Proton – positively charged
Electron – negatively charged
Neutron – no charge
Nucleus – in center of atom, contains
protons and neutrons
Quarks – fundamental particles – make
up protons and neutrons, have
fractional charge
3
ions



Positive ions – have lost one or more
electrons
Negative ions – have gained one or
more electrons
Only electrons are lost or gained under
normal conditions
4
Conservation of charge

The algebraic sum of all the electric
charges in any closed system is
constant.
5
Electrical interactions

Responsible for many things

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The forces that hold molecules and crystals
together
Surface tension
Adhesives
Friction
6
Conductors
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
Permit the movement of charge through
them
Electrons can move freely
Most metals are good conductors
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Insulators
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Do not permit the movement of charge
through them
Most nonmetals are good insulators
Electrons cannot move freely
8
Charging by induction

See pictures on pages 539-540
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Coulomb’s Law



Point charge – has essentially no
volume
The electrical force between two
objects gets smaller as they get farther
apart.
The electrical force between two
objects gets larger as the amount of
charge increases
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Coulomb’s Law
F k



q1q2
r
2
r is the distance between the charges
q1 and q2 are the magnitudes of the
charges
k is a constant

8.99 x 109 N∙m2/C2
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Coulombs


SI unit of charge, abbreviated C
Defined in terms of current – we will
talk about this later
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Coulomb’s law constant

k is defined in terms of the speed of
light
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k = 10-7c
k = 1/4pe0
e0 is another constant that will be more
useful later
e0 = 8.85 x 10-12 C2/N∙m2
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The coulomb

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Very large amount of charge
Charge on 6 x 1018 electrons
Most charges we encounter are
between 10-9 and 10-6 C
1 mC = 10-6 C
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Examples

See pages 543 - 546
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Electric Field
• A field is a region in space where a force
can be experienced.
• Or: a region in space where a quantity has
a definite value at every point.
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Electric Field
• Produced by a charged particle.
• The force felt by another charged particle
is caused by the electric field.
• We can check for an electric field with a
test charge, qt. If it experiences a force,
there is an electric field.
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Electric field
• The definite quantity is a ratio of the
electric force experienced by a charge to
the amount of the charge.
• Vector quantity measured in N/C.
F
E
qt
F  qt E
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Electric field
• To determine the field from a point charge,
Q, we place a test charge, qt, at some
position and determine the force acting
on it.
Q
F
qt
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Direction of E
• If the test charge is positive, E has the same
direction as F.
• If the test charge is negative, E has the
opposite direction as F.
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Electric Field - Point Charge
F
E
qt
Ek
Fk
qt Q
r
2
Q
r
2
E
1
Q
4peo r
2
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Electric Field
• The field is there, independent of a test
charge or anything else!
• The electric field vector points in the
direction a positive charge would be
forced.
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Example 1
• Two charges, Q1 = +2 x 10-8 C and Q2 = +3 x
10-8 C are 50 mm apart as shown below.
• What is the electric field halfway between
them?
Q1
E2
E1
50 mm
Q2
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Example 1
• At the halfway point, r1 = r2 = 25 mm.
• Magnitudes of fields:
kQ1
E1  2 
r1
kQ2
E2  2 
r2
(9 x 10
9
N•m
C
2
2
8
)(2 x 10 C)
2
(2.5 x 10 m)
(9 x 10
9
N•m
C
2
2
2
8
)(3 x 10 C)
2
(2.5 x 10 m)
2
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Example 1
•
•
•
•
•
•
E1 = 2.9 x 105 N/C
E2 = 4.3 x 105 N/C
E1 is to the right and E2 is to the left.
E1 = 2.9 x 105 N/C
E2 = - 4.3 x 105 N/C
E = E1 + E2 = - 1.4 x 105 N/C
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Example 2
• For the charges in Example 1, where is the
electric field equal to zero?
• Since the fields are in opposite directions
between the charges, the point where the
field is zero must be between them.
Q1
E2
E1
Q2
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Example 2
E1  E 2
kQ1 kQ 2
2 
2
r1
r2
Q1 Q 2
2  2
r1
r2
r1 + r2 = s, so r2 = s – r1
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Example 2
Q1 Q 2
2  2
r1
r2
Q1
Q2
2 
2
r1 (s  r1 )
2
(s  r1 )
Q2

2
r1
Q1
s  r1
Q2

r1
Q1
r1 
s
Q2
1
Q1
r1  23 mm
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Field Diagrams
• To represent an electric field we use lines
of force or field lines.
• These represent the sum of the electric
field vectors.
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Field Diagrams
30
Field Diagrams
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Field Diagrams
• At any point on the field lines, the electric
field vector is along a line tangent to the
field line.
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Field Diagrams
33
Field Diagrams
• Lines leave positive charges and enter
negative charges.
• Lines are drawn in the direction of the force
on a positive test charge.
• Lines never cross each other.
• The spacing of the lines represents the
strength or magnitude of the electric field.
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Point Charges
• Lines leave or enter the charges in a
symmetric pattern.
• The number of lines around the charge is
proportional to the magnitude of the
charge.
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Point Charges
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Point Charges
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Gauss’s Law
• Electric flux through a closed surface is
proportional to the total number of field
lines crossing the surface in the outward
direction minus the number crossing in the
inward direction.
  EA 
Q
e0
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Example 25-9 (see page 563)
Field of a charged sphere is the same as if it
were a point charge
1
q
E
4pe0 r 2
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Example 25-10 (see page 564)
Field of a infinite line of charge is
E
1

2pe0 r
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Other scenarios
• See table on page 567
41
Example 3
• Two parallel metal plates are 2 cm apart.
• An electric field of 500 N/C is placed between
them.
• An electron is projected at 107 m/s halfway
between the plates and parallel to them.
• How far will the electron travel before it strikes
the positive plate?
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Example 3
• Two charged parallel plates create a
uniform electric field in the space between
them.
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Example 3
vo
E
This is just like a projectile problem except
that the acceleration is not a given value.
44
Example 3
a=
a=
F
F = qE = eE
m
eE
m
=
(1. 6 x 10
–19
C )(500 N /C )
9. 1 x 10
–31
kg
= 8.8 x 1013 m/s2
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Example 3
• 8.8 x 1013 m/s2 is the vertical acceleration
of the electron.
• Horizontally, the acceleration is zero.
• x = vt
• v = 1 x 107 m/s & t = ?
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Example 3
• Back to vertical direction:
• y = yo + vot + 1/2at2
• y = 1/2at2
2(0.01 m)
2y 
13
2
t
8.8x10
m
/
s
a
= 1.5 x 10-8 s
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Example 3
• Back to horizontal direction:
• x = vt
• x = (1 x 107 m/s)(1.5 x 10–8 s)
• x = 0.15 m = 15 cm
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Dipoles
• A pair of charges with equal and opposite
sign.
• Induced dipoles, molecular dipoles, etc.…
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