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Coulomb’s Law and Electric Field
Chapter 24: allChapter 25: all
Physics chapters 24 - 25 2
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
Physics chapters 24 - 25 3
Elementary particles 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
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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
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Conservation of charge The algebraic sum of all the
electric charges in any closed system is constant.
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Electrical interactions Responsible for many things
The forces that hold molecules and crystals together
Surface tension Adhesives Friction
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Conductors Permit the movement of charge
through them Electrons can move freely Most metals are good conductors
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Insulators Do not permit the movement of
charge through them Most nonmetals are good
insulators Electrons cannot move freely
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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
221
rqq
kF
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 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 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.
EF tqtq
FE
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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 qt
F
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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
2
Q4
1r
Eope
tqFE 2r
Qqk tF
2rQ
kE
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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 Q250 mm
E1E2
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Example 1• At the halfway point, r1 = r2 = 25
mm.• Magnitudes of fields:
E1 kQ1r1
2 (9 x 109 N • m
2
C2 )(2 x 10 8C)
(2.5 x 10 2 m)2
E2 kQ2r2
2 (9 x 109 N • m
2
C2 )(3 x 10 8 C)
(2.5 x 10 2 m)2
Physics chapters 24 - 25 25
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 Q2
E1E2
Physics chapters 24 - 25 27
Example 2E1 E2
kQ1r1
2 kQ2r2
2
Q1r1
2 Q2r2
2
r1 + r2 = s, so r2 = s – r1
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Example 2Q1r1
2 Q2r2
2
Q1r1
2 Q2
(s r1)2
(s r1 )2
r12
Q2Q1
s r1r1
Q2Q1
r1 s
1 Q2
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
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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
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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.
0eQEA
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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
204
1rqE
pe
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Example 25-10 (see page 564)Field of a infinite line of charge is
rE
pe021
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Other scenarios • See table on page 567
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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
Evo
This is just like a projectile problem except that the acceleration is not a given value.
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Example 3
a =Fm
F = qE = eE
a =eEm
=(1.6 x 10–19C)(500 N/C)
9.1 x 10–31kg
= 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
a2yt
2(0.01 m)8.8x1013 m / s2
= 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.…