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Chapter 20
Electric Forces and Fields
Topics:
• Electric charge
• Forces between charged
•
•
objects
The field model and the
electric field
Forces and torques on
charged objects in electric
fields
Sample question:
In electrophoresis, what force causes DNA fragments to migrate
through the gel? How can an investigator adjust the migration rate?
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Slide 20-1
Late Paper Homework Policy and Solution Posting
A. Late papers must be handed in by 5 PM 2 days after
assignment is due
• Late penalty 10% per day
• Solutions will post shortly after 5 PM – no late work
accepted after solutions are posted
B. Late papers must be handed in by 5 PM 4 days after
assignment is due
• Late penalty 10% per day
• Solutions will post shortly after 5 PM – no late work
accepted after solutions are posted
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Slide 20-13
Discuss Einstein Video and the Nature of Science
What make science different from other ways
people study and learn about the world around us?
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Slide 20-13
Checking Understanding
Two spheres are touching each other. A charged rod is brought
near. The spheres are then separated, and the rod is taken away. In
the first case, the spheres are aligned with the rod.
After the charged rod is removed, which of the spheres (A & B) is:
i) Positive
ii) Negative
iii) Neutral
A => Sphere A is + and sphere B is –
B => Sphere A is – and Sphere B is +
C => Spheres A and B are both +
D => Spheres A and B are both –
E => Spheres A and B are both neutral
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Slide 20-13
Checking Understanding
Two spheres are touching each other. A charged rod is brought
near. The spheres are then separated, and the rod is taken away. In
the second case, they are perpendicular. After the charged rod is
removed,
which of the spheres (C & D) is:
i) Positive
ii) Negative
iii) Neutral
A => Sphere C is + and sphere D is –
B => Sphere C is – and Sphere D is +
C => Spheres C and D are both +
D => Spheres C and D are both –
E => Spheres C and D are both neutral
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Slide 20-13
Warm-up 1
Charging 2 spheres
You have two conducting spheres. How can you charge them with
opposite charges without touching either one with a charged object?
(Anything else is fair game)
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Slide 20-3
Warm-up 2: Charged Spheres & Forces
Two identical metal spheres are firmly fastened to and electrically insulated from
frictionless plastic air pucks that ride on an air table as shown below. The pucks
are held in place as a charge of 2.0 x 10-8 C is placed on sphere A on the left and
a charge of 6.0 x 10-6 C is placed on sphere B on the right. The pucks are then
released so that the pucks with the spheres attached are now free to move without
across the table.
A. Draw Free-Body Diagrams for the pucks and spheres
(Treat each puck and sphere system as a single object)
B. How do the Coulomb forces acting on spheres A & B
compare? (Use a ratio)
C. Which sphere has the greater acceleration?
How would your answer change if the mass of the puck under
sphere A was reduced by 50%?
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Slide 20-3
Charged Spheres & Forces
Two identical metal spheres are firmly fastened to and electrically insulated from
frictionless plastic air pucks that ride on an air table as shown below. The pucks
are held in place as a charge of 2.0 x 10-8 C is placed on sphere A on the left and a
charge of 6.0 x 10-6 C is placed on sphere B on the right. The pucks are then
released so that the pucks with the spheres attached are now free to move without
across the table.
D. As the two spheres get farther away from one another, how would (if at all)
the following quantities change?
1) Force
2) Speed
3) Acceleration
Choices:
a) Increase
b) Decrease
c) Stay the same
d) Can’t tell
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Slide 20-3
Coulomb’s Law
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Slide 20-15
Example Problem 1
Two 0.10 g honeybees each acquire a charge of +23 pC as they fly
back to their hive. As they approach the hive entrance, they are 1.0
cm apart. What is the magnitude of the repulsive force between the
two bees? How does this force compare with their weight?
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Slide 20-33
Example Problem 1
Two 0.10 g honeybees each acquire a charge of +23 pC as they fly
back to their hive. As they approach the hive entrance, they are 1.0
cm apart. What is the magnitude of the repulsive force between the
two bees? How does this force compare with their weight?
Fe, B1=>B2 = K |q1| |q2| / (r12)2 = (9e9 Nm2/C2)(23e-12 C)2 / (0.010 m)2
Fe, B1=>B2 = 4.8 x 10-8 N
Fg, Earth=>B1 = mg = 0.00010 kg x 9.8 m/s2 = 9.8 x 10-4 N
The electric force between the bees is much smaller than the weight
of the bees on Earth.
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Slide 20-33
The Force of Gravity
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Slide 6-35
Example Problem 1 – take
2 0.10 g honeybees each acquire a charge of +23 pC as they fly
Two
back to their hive. As they approach the hive entrance, they are 1.0
cm apart. What is the magnitude of the repulsive force between the
two bees? How does this force compare with their weight?
Fe, B1=>B2 = K |q1| |q2| / (r12)2 = (9e9 Nm2/C2)(23e-12 C)2 / (0.010 m)2
Fe, B1=>B2 = 4.8 x 10-8 N
Fg, B1=>B2 = G m1 m2 / (r12)2
Fg, B1=>B2 = (6.67e-11 Nm2/kg2)(0.0001 kg)2 / (0.010 m)2
Fg, B1=>B2 = 6.7 x 10-15 N
The electric force between the bees is much larger than the
gravitational forces between them.
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Slide 20-33
Conceptual Example Problem
All charges in the diagrams below are equal magnitude. In each
case, a small positive charge is placed at the blank dot. In which
cases is the force on this charge:
• to the right?
• to the left?
• zero?
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Slide 20-28
Checking Understanding
All charges in the diagrams below are of equal magnitude. In each
case, a small, positive charge is placed at the black dot.
In which case is the force on the small, positive charge the largest?
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Slide 20-25
Introduction to Electric Field
1. Find the Electric force from a 9.00e-6 C charge on the
following charges at 1m, 2m, and 3 m.
a) 3.00e-6 C
b) -4.00e-6 C
c) -10.0e-6 C
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Introduction to Electric Field
1. Find the Electric force from a 9.00e-6 C charge on the
following charges at 1m, 2m, and 3 m.
a) 3.00e-6 C
b) -4.00e-6 C
c) -10.0e-6 C
2. Find the force per charge from a 9.00e-6 C charge on the
charges above at 1m, 2m, and 3 m.
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Nature of Electric Field
E-field Applet 1
http://physics.weber.edu/schroeder/software/EField/
Back-up
applet:https://phet.colorado.edu/en/simulation/charges-andfields
What observations can we make about E-fields?
Source Charges and Test Charges?
Superposition of E-fields?
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Nature of Electric Field
• Test charge is a small positive charge to sample the E-Field
• Charge of test charge is small compared to source charges
(source charges are the charges that generate the field)
• E-field vectors
• E-field is the force per charge
• E-field vectors points away from + charges
• E-field vectors point towards - charges
• E-field for point charges gets weaker as distance from
source point charges increases
• For a point charge E = Fe / q = [k Q q / r2] / q = k Q / r2
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The Electric Field
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Slide 20-34
The Electric Field of a Point Charge
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Slide 20-35
Nature of Electric Field Vectors
• Test charge is a small positive charge to sample the E-Field
• Charge of test charge is small compared to source charges
(source charges are the charges that generate the E-field)
• E-field vectors
• E-field is the force per charge
• E-field vectors points away from + charges
• E-field vectors point towards - charges
• E-field for point charges gets weaker as distance from source
point charges increases
• E-fields add as vectors, at a point in space Enet,x = E1x + E2x + …
• For a point charge E = Fe / |q| = [k |Q| |qt| / r2] / |qt| = k |Q| / r2
• Electric Force
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