Li2

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Li-2

• Coulombs Law

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Review

• Matter is made of particles with positive charge
  (protons) and negative charge (electrons).
• Objects that have an imbalance between these
  particles are charged.
• Charge may be positive or negative.
• Like charges repel, unlike charges attract.

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Polarization

• A charged object can rearrange the charges of a
  nearby neutral object.
• Surprisingly, this occurs for conductors, where
  charges are free to move, and for insulators,
  where charges cannot move.
• Therefore charged objects attract neutral
  objects.
• If objects attract, at least one is charged.
  Little else can be inferred.

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Coulombs Law
The force between charges q1 and q2
is where q1, q2 are the charges measured in
coulombs (C) r is the distance between the
charges (m) K ke is a constant of nature, 8.99
x 109 N-m2/C2
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Coulombs Law, Frepel keq1q2 / r 2
More on ke ke 1/(4p?0) where ?0 8.85 x
10-12 C2/N-m2 is called the permittivity of
free space Thus Coulombs law is often
written Frepel q1q2 / (4p?0r 2)
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Forces are Vectors
The direction of the force of q1 on q2 is along
the line joining the two charges. If the charges
have the same sign, the force is repulsive. If
the charges have the opposite sign, the force is
attractive.
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Uses of Vectors in Mechanics
One set of mathematical rules handles all of
these concepts.
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Charges A and B exert repulsive forces on each
other. qA 4qB. Which statement is true?
1. FA on B gt FB on A 2. FA on B lt FB on A 3.
FA on B FB on A
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Frepel keq1q2 / r 2

• Sample Problem A penny has a net charge of 1.0
  C, and a nickel has a charge of -1.0 C. They are
  held 10 cm away from each other.
• (a) Is the force between them attractive or
  repulsive? What is its magnitude?
• (b) Is this a physically reasonable problem?

Note More reasonable charges are measured in
µC or less.
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Frepel keq1q2 / r 2

• Sample Problem A penny has a net charge of 1.0
  C, and a nickel has a charge of -1.0 C. They are
  held 10 cm away from each other.
• (a) Is the force between them attractive or
  repulsive? What is its magnitude?
• (b) Is this a physically reasonable problem?

Solution. (a) The force is attractive because
opposites attract. Frepel (8.99e9)(1.0)(-1.0)
/(0.10)2 -8.99e11 N which is negative, as it
should be for an attractive force.
(b) 8.99e11 N ? (1 lb / 4.45 N) 2.02e11 pounds
of attractive force, obviously totally
unreasonable and unrealistic!
Note More reasonable charges are measured in
µC or less.
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Unit vector notation
Coulombs law where F12 is the force exerted
by q1 on q2 and r12 is a unit vector pointing
from q1 to q2
Unit vector length is one, used to give a
direction.
r12


q1
q2
F12
Same law, just swap the 1s and 2s, gives
equal and opposite forces as it must.
r21
F21
q1


q2
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But dont get hung up on notation
- same signs, the force is repulsive. -
opposite signs, the force is attractive.
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For opposite charges
?
F12 (keq1q2 / r 2) r12 where F12 is the force
exerted by q1 on q2 and r12 is a unit vector
pointing from q1 to q2
?
?
r12

?
F12
q1
?
q2
?

?
F21
Doesnt matter where we draw the the unit vector.
We like to draw the force with its tail on the
charge that feels that force.
r21
q1
?
q2
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7µC

• F12 (keq1q2 / r 2) r12
• Three point charges are arranged in an
  equilateral triangle. QA 2.00 µC, QB -4.00
  ?C, and QC 7.00 µC. The length of the sides of
  the triangle is 0.500 m. Find the total force on
  QA.

0.5 m
-4µC
2µC
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Solution. First compute force due to QB. The
unit vector from QB to QA is rBA -1,0
keQAQB/r2 (8.99e9)(2e-6)(-4e-6)/0.52 -0.288
N. Therefore FBA (-0.288 N)?-1,0 0.288,
0 N.
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7?C

F12 (keq1q2 / r 2) r12 (from previous slide)
FBA 0.288, 0 N.
The unit vector from QC to QA is rCA -1/2,
-3/ 2
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2
v3
60
90
?4?C
2?C
1
keQAQC/r 2 (8.99e9)(2e-6)(7e-6)/0.52 0.503
N. Therefore FCA .503(-1/2), .503(-3/2)
N -0.252, -0.436 N

rCA
QA
FCA
Finally, add the two vectors to get the total
force on QA Ftotal 0.288-0.252, 0-0.436
0.036, -0.436 N
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• Two identical charges Q are fixed on the x-axis
  at x 1 m and x -1 m. A third charge q will
  experience no net force due to these two charges
  if it is placed
• 1. anywhere on the x-axis between the two other
  charges
• 2. anywhere on the x-axis, not between the two
  other charges
• 3. only at the origin
• 4. anywhere on the plane x 0
• 5 . nowhere there is no place where the force
  on q is zero

z
Q
y
x
Q
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• Opposite charges of equal magnitude Q, -Q are
  fixed on the x-axis at x 1 m and x -1 m. A
  third charge q will experience no net force due
  to these two charges if it is placed
• 1. anywhere on the x-axis between the two other
  charges
• 2. anywhere on the x-axis, not between the two
  other charges
• 3. only at the origin
• 4. anywhere on the plane x 0
• 5. nowhere there is no place where the force on
  q is zero

z
-Q
y
x
Q
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The Electric Field
Our idea up until now Charges directly exert
forces on other charges (attractive or
repulsive), given by Coulombs Law.
Philosophical problems with this - What is
mechanism? - Action at a distance modern
physical theories are local, meaning you can
only directly affect nearby things. -
Instantaneous what if two charges were 106
light-years apart and we moved one of them when
would other know about change? - could use to
send signals faster than light.
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Resolution There is something there that
charges interact with Electric field that fills
all of space. What creates the electric
field? One charge is the source of the field
(makes it nonzero), the field is part of the
charged object and a nonzero field exerts forces
on other charge.
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The idea of a field Math A field is a function
f(x,y,z) that assigns a value at every point
in space Scalar field f(x,y,z) is a single
number Vector field f (x,y,z) carries a
magnitude and direction
- three numbers
Physics - fields satisfy math definition and
also are related to the carriers of forces
electric and magnetic fields
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Example of a Field
The electric field is our first example of a
vector field.
A vector field consists of an infinite set of
vectors, one vector for every point in space. We
can only draw some of them, like this
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The Electric Field, Defined
The ELECTRIC FIELD at a point in space can be
measured by q is a "test charge" placed at
that point (C) F is the force vector on the test
charge (N) E is the electric field vector at that
point (N/C)
Units note Electric fields are usually given in
V/m (volts/meter). But 1 N/C 1 V/m. More on
volts later!
Conversely, the force on a charge q placed in an
electric field E is F qE
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F qE

• An electron is observed to travel in a circular
  orbit, in the plane of the page. This could be
  caused by
• 1. an electric field pointing out of the page
• 2. an electric field pointing into the page
• 3. an electric field that points towards the
  center of the orbit
• 4. an electric field that points away from the
  center of the orbit
• 5. no electric field could cause this

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The Electric Field of a Point Charge
Consider the force on a test charge q when placed
near another charge Q. Coulombs law But
also, F qE from which we deduce
The electric field produced by a point charge Q
is where r is the distance from Q and the
field point, and r is a unit vector from Q to the
field point.
or
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(No Transcript)
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The electric field of a POSITIVE POINT CHARGE
looks like this
F qE
The large red dot at the center is the charge
PRODUCING THE FIELD, E (keQ / r 2) r The small
dots are TEST CHARGESPOSITIVE test charge
(upper left) feels force ALONG THE
FIELD. NEGATIVE test charge (lower left) feels
force OPPOSITE THE FIELD.
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The electric field of a NEGATIVE POINT CHARGE
looks like this
F qE
The large blue dot at the center is the charge
PRODUCING THE FIELD, E (keQ / r 2) r The small
dots are TEST CHARGESPOSITIVE test charge
(upper left) feels force ALONG THE
FIELD. NEGATIVE test charge (lower left) feels
force OPPOSITE THE FIELD
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Another way that the pictures are drawn
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The electric field is the primary object
The field exist at each point in space all the
time - if another charged is placed there it
produces force - charge is interacting with
electric field F q E
Answers philosophical problems - mechanism is
interaction of charge and field - direct
interaction of charge with field local - if
source charge moves, the field would not change
instantaneously change propagates at speed of
light
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An electron is placed at the position marked by
the dot. The force on the electron is
1. to the left. 2. to the right. 3. zero. 4.
Theres not enough information to tell.
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• The electric field 10 cm away from an isolated
  point charge is E 100 V/m. The electric field
  20 cm away from this same charge is
• 1. 200 V/m
• 2. 100 V/m
• 3. 50 V/m
• 4. 25 V/m
• 5. 10 V/m

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• The electric field 10 cm away from an isolated
  point charge is E 500 V/m. At what distance
  from the charge is the field E 5 V/m?
• 1. 20 cm
• 2. 50 cm
• 3. 100 cm
• 4. 1000 cm
• 5. 104 cm

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• A point charge is located at the origin. Along
  the positive x-axis, the electric field due to
  this charge points in the negative x direction.
  Along the negative y-axis, the field points in
• 1. the positive y direction
• 2. the negative negative y direction
• 3. the positive x direction
• 4. the negative x direction
• 5. none of these are correct

z
y
x
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E (keQ / r 2) r
Q2

• Four charges are at the corners of a square of
  side a 1.50 m. Charge q 1.50 ?C is located at
  the origin, and Q2 6q. Find the magnitude and
  direction of the electric field at the location
  of q, due to Q2.

?
r
q
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Rank in order, from largest to smallest, the
electric field strengths E1 to E4 at points 1 to
4.
1. E2 gt E4 gt E1 gt E3 2. E2 gt E1 E4 gt E3 3.
E2 gt E1 gt E4 gt E3 4. E1 E2 gt E3 E4 5. E1 gt
E2 gt E3 gt E4
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Summary
Charge (Q) is measured in coulombs (C). Like
charges repel, opposites attract. Basic charge qe
-1.6 x 10-19 C
The vector force between charges is given by
Coulombs law,
ke 8.99 x 109 N-m2/C2 1/(4p?0) where ?0
8.85 x 10-12 C2/N-m2
The electric field consists of a vector E at each
point in space. F qE