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Chapter 16Section 1 Electric Charge
Properties of Electric Charge
• electric charge is + or -.
– like charges repel
– unlike charges attract
• Electric charge is conserved.
• Atomic Charges
– Protons (+)charged particles.
– neutronsUncharged particles.
– Electrons (-) charged particles.
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Chapter 16
Electric Charge
Section 1 Electric Charge
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Chapter 16Section 1 Electric Charge
Properties of Electric Charge, continued
• Charge is measured in coulombs (C).
• fundamental unit of charge= e, charge
of a single electron or proton.
e = -1.602 176 x 10–19 C
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Chapter 16
The Milikan Experiment
Section 1 Electric Charge
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Chapter 16
Milikan’s Oil Drop Experiment
Section 1 Electric Charge
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Chapter 16Section 1 Electric Charge
Transfer of Electric Charge
• An electrical conductor is a material in which
charges can move freely.
• An electrical insulator is a material in which charges
cannot move freely. Charge will remain where it was
placed.
• Ground- often the ground reservoir of electrons, can
donate or absorb.
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Chapter 16Section 1 Electric Charge
Transfer of Electric Charge, continued
• Insulators and conductors can be charged by contact.
• Conductors can be charged by induction.
• Induction charging a conductor by bringing it near
another charged object and grounding the conductor.
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Chapter 16
Charging by Induction
Section 1 Electric Charge
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Chapter 16Section 1 Electric Charge
Transfer of Electric Charge, continued
• A surface charge can be
induced on insulators by
polarization.
• With polarization, the
charges within individual
molecules are realigned
such that the molecule
has a slight charge
separation.
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Section 2 Electric Force
Chapter 16
Objectives
• Calculate electric force using Coulomb’s law.
• Compare electric force with gravitational force.
• Apply the superposition principle to find the resultant
force on a charge and to find the position at which the
net force on a charge is zero.
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Coulomb’s Law, continued
• The electrical force is a field force.
• A field force = force exerted by one object on another
with no physical contact between them
Chapter 16Section 2 Electric Force
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Chapter 16
Coulomb’s Law
• Two charges near one another exert a force on one
another called the electric force.
• Kc= 8.99 E 9 n m2/C2
• Coulomb’s law
Section 2 Electric Force
1 2
2
2
charge 1 charge 2electric force = Coulomb constant
distance
electric C
q qF k
r
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Chapter 16
Coulomb’s Law, continued
• The resultant force on a charge is the vector sum of
the individual forces on that charge.
• equilibrium, the net external force acting on that body
is zero.
Section 2 Electric Force
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Chapter 16
Superposition Principle
Section 2 Electric Force
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Chapter 16
Sample Problem
The Superposition Principle
Consider three point
charges at the corners of a
triangle, as shown at right,
where q1 = 6.00 10–9 C,
q2 = –2.00 10–9 C, and
q3 = 5.00 10–9 C. Find
the magnitude and
direction of the resultant
force on q3.
Section 2 Electric Force
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Chapter 16Section 2 Electric Force
Sample Problem, continued
The Superposition Principle
1. Define the problem, and identify the known variables.
Given:
q1 = +6.00 10–9 C r2,1 = 3.00 m
q2 = –2.00 10–9 C r3,2 = 4.00 m
q3 = +5.00 10–9 C r3,1 = 5.00 m
q = 37.0º
Unknown: F3,tot = ? Diagram:
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Chapter 16Section 2 Electric Force
Sample Problem, continued
The Superposition Principle
3. Calculate the magnitudes of the forces with Coulomb’s law.
–9 –9293 1
3,1 22 2
–8
3,1
–9 –9293 2
3,2 22 2
–9
3,1
5.00 10 C 6.00 10 CN m 8.99 10
( 3,1) C 5.00 m
1.08 10 N
5.00 10 C 2.00 10 CN m8.99 10
( 3,2) C 4.00m
5.62 10 N
C
C
q qF k
r
F
q qF k
r
F
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Chapter 16Section 2 Electric Force
Sample Problem, continued
The Superposition Principle
4. Find the x and y components of each force.
At this point, the direction each component must be
taken into account.
F3,1: Fx = (F3,1)(cos 37.0º) = (1.08 10–8 N)(cos 37.0º)
Fx = 8.63 10–9 N
Fy = (F3,1)(sin 37.0º) = (1.08 10–8 N)(sin 37.0º)
Fy = 6.50 10–9 N
F3,2: Fx = –F3,2 = –5.62 10–9 N
Fy = 0 N
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Chapter 16Section 2 Electric Force
Sample Problem, continued
The Superposition Principle
5. Calculate the magnitude of the total force acting
in both directions.
Fx,tot = 8.63 10–9 N – 5.62 10–9 N = 3.01 10–9 N
Fy,tot = 6.50 10–9 N + 0 N = 6.50 10–9 N
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Chapter 16Section 2 Electric Force
Sample Problem, continued
The Superposition Principle
6. Use the Pythagorean theorem to find the magni-
tude of the resultant force.
2 2 9 2 9 2
3, , ,
–9
3,
( ) ( ) (3.01 10 N) (6.50 10 N)
7.16 10 N
tot x tot y tot
tot
F F F
F
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Chapter 16Section 2 Electric Force
Sample Problem, continued
The Superposition Principle
7. Use a suitable trigonometric function to find the
direction of the resultant force.
In this case, you can use the inverse tangent function:
–9,
–9
,
6.50 10 Ntan
3.01 10 N
65.2º
y tot
x tot
F
F
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Section 3 The Electric Field
Chapter 16
Objectives
• Calculate electric field strength.
• Draw and interpret electric field lines.
• Identify the four properties associated with a
conductor in electrostatic equilibrium.
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Chapter 16Section 3 The Electric Field
Electric Field Strength
• An electric field = region where an electric force on
a test charge would be detected.
• Test charge= + charge with no magnitude
• units of electric field, E, (N/C).
• direction of electric field vector= E, the direction the
electric force exerted on a small positive test charge.
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Chapter 16
Electric Fields and Test Charges
Section 3 The Electric Field
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Chapter 16Section 3 The Electric Field
Electric Field Strength, continued
• Electric field strength
• Electric Field Strength Due to a Point Charge
• Or E=F/q1(test charge)
2
2
charge producing the fieldelectric field strength = Coulomb constant
distance
C
qE k
r
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Chapter 16Section 3 The Electric Field
Electric Field Diagrams, continued
• Use arrows to represent lines.
• Originate on +
• Terminate on –
• Never cross
• Number represents relative charge
ratios
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Chapter 16
Rules for Drawing Electric Field Lines
Section 3 The Electric Field
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Chapter 16
Rules for Sketching Fields Created by Several
Charges
Section 3 The Electric Field
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Chapter 16
Calculating Net Electric Field
Section 3 The Electric Field
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Chapter 16Section 3 The Electric Field
Sample Problem
Electric Field Strength
A charge q1 = +7.00 µC is
at the origin, and a charge
q2 = –5.00 µC is on the x-
axis 0.300 m from the
origin, as shown at right.
Find the electric field
strength at point P,which is
on the y-axis 0.400 m from
the origin.
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Chapter 16
Sample Problem, continued
Electric Field Strength
1. Define the problem, and identify the knownvariables.
Given:
q1 = +7.00 µC = 7.00 10–6 C r1 = 0.400 m
q2 = –5.00 µC = –5.00 10–6 C r2 = 0.500 m
q = 53.1º
Unknown:
E at P (y = 0.400 m)
Tip: Apply the principle of superposition. You must first calculate the electric field produced by each charge individually at point P and then add these fields together as vectors.
Section 3 The Electric Field
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Chapter 16
Sample Problem, continued
Electric Field Strength
2. Calculate the electric field strength produced by each charge. Because we are finding the magnitude of the electric field, we can neglect the sign of each charge.
–69 2 2 51
1 2 2
1
–69 2 2 52
2 2 2
2
7.00 10 C8.99 10 N m /C 3.93 10 N/C
(0.400 m)
5.00 10 C8.99 10 N m /C 1.80 10 N/C
(0.500 m)
C
C
qE k
r
qE k
r
Section 3 The Electric Field
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Chapter 16
Sample Problem, continued
Electric Field Strength
3. Analyze the signs of the charges.
The field vector E1 at P due to q1 is directed vertically upward, as shown in the figure, because q1 is positive. Likewise, the field vector E2 at P due to q2 is directed toward q2 because q2 is negative.
Section 3 The Electric Field
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Chapter 16
Sample Problem, continued
Electric Field Strength
4. Find the x and y components of each electric field vector.
For E1: Ex,1 = 0 N/C
Ey,1 = 3.93 105 N/C
For E2: Ex,2= (1.80 105 N/C)(cos 53.1º) = 1.08 105 N/C
Ey,1= (1.80 105 N/C)(sin 53.1º)= –1.44 105 N/C
Section 3 The Electric Field
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Chapter 16
Sample Problem, continued
Electric Field Strength
5. Calculate the total electric field strength in both directions.
Ex,tot = Ex,1 + Ex,2 = 0 N/C + 1.08 105 N/C
= 1.08 105 N/C
Ey,tot = Ey,1 + Ey,2 = 3.93 105 N/C – 1.44 105 N/C
= 2.49 105 N/C
Section 3 The Electric Field
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Chapter 16
Sample Problem, continued
Electric Field Strength
6. Use the Pythagorean theorem to find the magnitude of the resultant electric field strength vector.
22
, ,
2 25 5
5
1.08 10 N/C 2.49 10 N/C
2.71 10 N/C
tot x tot y tot
tot
tot
E E E
E
E
Section 3 The Electric Field
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Chapter 16
Sample Problem, continued
Electric Field Strength
7. Use a suitable trigonometric function to find the direction of the resultant electric field strength vector.
In this case, you can use the inverse tangent function:
5,
5
,
2.49 10 N/Ctan
1.08 10 N/C
66.0
y tot
x tot
E
E
Section 3 The Electric Field
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Chapter 16Section 3 The Electric Field
Conductors in Electrostatic Equilibrium
• Any charge on an isolated conductor is on the conductor’s outer surface only. (Faradays cage)
• electric field outside a charged conductor isperpendicular to the conductor’s surface.
• On an irregularly shaped conductor, charge accumulate at sharp points. St Elmo’s fire
• http://www.youtube.com/watch?feature=player_detailpage&v=ufKMRSkFrds
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Multiple Choice
1. In which way is the electric force similar to the
gravitational force?
A. Electric force is proportional to the mass of the
object.
B. Electric force is similar in strength to gravitational
force.
C. Electric force is both attractive and repulsive.
D. Electric force decreases in strength as the
distance between the charges increases.
Standardized Test PrepChapter 16
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Multiple Choice, continued
1. In which way is the electric force similar to the
gravitational force?
A. Electric force is proportional to the mass of the
object.
B. Electric force is similar in strength to gravitational
force.
C. Electric force is both attractive and repulsive.
D. Electric force decreases in strength as the
distance between the charges increases.
Standardized Test PrepChapter 16
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Multiple Choice, continued
2. What must the charges be for
A and B in the figure so that
they produce the electric
field lines shown?
F. A and B must both be
positive.
G. A and B must both be
negative.
H. A must be negative, and
B must be positive.
J. A must be positive, and B
must be negative.
Standardized Test PrepChapter 16
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Multiple Choice, continued
2. What must the charges be for
A and B in the figure so that
they produce the electric
field lines shown?
F. A and B must both be
positive.
G. A and B must both be
negative.
H. A must be negative, and
B must be positive.
J. A must be positive, and B
must be negative.
Standardized Test PrepChapter 16
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Multiple Choice, continued
3. Which activity does not produce the same results as
the other three?
A. sliding over a plastic-covered automobile seat
B. walking across a woolen carpet
C. scraping food from a metal bowl with a metal
spoon
D. brushing dry hair with a plastic comb
Standardized Test PrepChapter 16
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Multiple Choice, continued
3. Which activity does not produce the same results as
the other three?
A. sliding over a plastic-covered automobile seat
B. walking across a woolen carpet
C. scraping food from a metal bowl with a metal
spoon
D. brushing dry hair with a plastic comb
Standardized Test PrepChapter 16
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Multiple Choice, continued
4. By how much does the electric force between two
charges change when the distance between them is
doubled?
Standardized Test PrepChapter 16
4
2
1 2
1 4
F.
G.
H.
J.
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Multiple Choice, continued
4. By how much does the electric force between two
charges change when the distance between them is
doubled?
Standardized Test PrepChapter 16
4
2
1 4
1
2
F.
G.
H.
J.
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Multiple Choice, continued
Use the passage below to answer questions 5–6.
A negatively charged object is brought close to the
surface of a conductor, whose opposite side is then
grounded.
5. What is this process of charging called?
A. charging by contact
B. charging by induction
C. charging by conduction
D. charging by polarization
Standardized Test PrepChapter 16
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Multiple Choice, continued
Use the passage below to answer questions 5–6.
A negatively charged object is brought close to the
surface of a conductor, whose opposite side is then
grounded.
5. What is this process of charging called?
A. charging by contact
B. charging by induction
C. charging by conduction
D. charging by polarization
Standardized Test PrepChapter 16
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Multiple Choice, continued
Use the passage below to answer questions 5–6.
A negatively charged object is brought close to the
surface of a conductor, whose opposite side is then
grounded.
6. What kind of charge is left on the conductor’s
surface?
F. neutral
G. negative
H. positive
J. both positive and negative
Standardized Test PrepChapter 16
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Multiple Choice, continued
Use the passage below to answer questions 5–6.
A negatively charged object is brought close to the
surface of a conductor, whose opposite side is then
grounded.
6. What kind of charge is left on the conductor’s
surface?
F. neutral
G. negative
H. positive
J. both positive and negative
Standardized Test PrepChapter 16
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Multiple Choice, continued
7. What is the electric field
strength 2.0 m from the
center of the conducting
sphere?
A. 0 N/C
B. 5.0 102 N/C
C. 5.0 103 N/C
D. 7.2 103 N/C
Standardized Test PrepChapter 16
Use the graph below to answer
questions 7–10. The graph shows
the electric field strength at different
distances from the center of the
charged conducting sphere of a Van
de Graaff generator.
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Multiple Choice, continued
7. What is the electric field
strength 2.0 m from the
center of the conducting
sphere?
A. 0 N/C
B. 5.0 102 N/C
C. 5.0 103 N/C
D. 7.2 103 N/C
Standardized Test PrepChapter 16
Use the graph below to answer
questions 7–10. The graph shows
the electric field strength at different
distances from the center of the
charged conducting sphere of a Van
de Graaff generator.
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Multiple Choice, continued
8. What is the strength of
the electric field at the
surface of the
conducting sphere?
F. 0 N/C
G. 1.5 102 N/C
H. 2.0 102 N/C
J. 7.2 103 N/C
Standardized Test PrepChapter 16
Use the graph below to answer
questions 7–10. The graph shows
the electric field strength at different
distances from the center of the
charged conducting sphere of a Van
de Graaff generator.
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ResourcesChapter menu
Multiple Choice, continued
8. What is the strength of
the electric field at the
surface of the
conducting sphere?
F. 0 N/C
G. 1.5 102 N/C
H. 2.0 102 N/C
J. 7.2 103 N/C
Standardized Test PrepChapter 16
Use the graph below to answer
questions 7–10. The graph shows
the electric field strength at different
distances from the center of the
charged conducting sphere of a Van
de Graaff generator.
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ResourcesChapter menu
Multiple Choice, continued
9. What is the strength of
the electric field inside
the conducting sphere?
A. 0 N/C
B. 1.5 102 N/C
C. 2.0 102 N/C
D. 7.2 103 N/C
Standardized Test PrepChapter 16
Use the graph below to answer
questions 7–10. The graph shows
the electric field strength at different
distances from the center of the
charged conducting sphere of a Van
de Graaff generator.
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ResourcesChapter menu
Multiple Choice, continued
9. What is the strength of
the electric field inside
the conducting sphere?
A. 0 N/C
B. 1.5 102 N/C
C. 2.0 102 N/C
D. 7.2 103 N/C
Standardized Test PrepChapter 16
Use the graph below to answer
questions 7–10. The graph shows
the electric field strength at different
distances from the center of the
charged conducting sphere of a Van
de Graaff generator.
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Multiple Choice, continued
10. What is the radius of
the conducting sphere?
F. 0.5 m
G. 1.0 m
H. 1.5 m
J. 2.0 m
Standardized Test PrepChapter 16
Use the graph below to answer
questions 7–10. The graph shows
the electric field strength at different
distances from the center of the
charged conducting sphere of a Van
de Graaff generator.
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ResourcesChapter menu
Multiple Choice, continued
10. What is the radius of
the conducting sphere?
F. 0.5 m
G. 1.0 m
H. 1.5 m
J. 2.0 m
Standardized Test PrepChapter 16
Use the graph below to answer
questions 7–10. The graph shows
the electric field strength at different
distances from the center of the
charged conducting sphere of a Van
de Graaff generator.
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Short Response
11. Three identical
charges (q = +5.0 mC)
are along a circle with a
radius of 2.0 m at angles
of 30°, 150°, and 270°,
as shown in the
figure.What is the
resultant electric field at
the center?
Standardized Test PrepChapter 16
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Short Response, continued
11. Three identical
charges (q = +5.0 mC)
are along a circle with a
radius of 2.0 m at angles
of 30°, 150°, and 270°,
as shown in the
figure.What is the
resultant electric field at
the center?
Answer: 0.0 N/C
Standardized Test PrepChapter 16
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Standardized Test PrepChapter 16
Short Response, continued
12. If a suspended object is attracted to another object
that is charged, can you conclude that the suspended
object is charged? Briefly explain your answer.
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Standardized Test PrepChapter 16
Short Response, continued
12. If a suspended object is attracted to another object
that is charged, can you conclude that the suspended
object is charged? Briefly explain your answer.
Answer: not necessarily; The suspended object might
have a charge induced on it, but its overall charge
could be neutral.
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Standardized Test PrepChapter 16
Short Response, continued
13. One gram of hydrogen contains 6.02 1023 atoms,
each with one electron and one proton. Suppose that
1.00 g of hydrogen is separated into protons and
electrons, that the protons are placed at Earth’s north
pole, and that the electrons are placed at Earth’s
south pole. Assuming the radius of Earth to be 6.38
106 m, what is the magnitude of the resulting
compressional force on Earth?
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Standardized Test PrepChapter 16
Short Response, continued
13. One gram of hydrogen contains 6.02 1023 atoms,
each with one electron and one proton. Suppose that
1.00 g of hydrogen is separated into protons and
electrons, that the protons are placed at Earth’s north
pole, and that the electrons are placed at Earth’s
south pole. Assuming the radius of Earth to be 6.38
106 m, what is the magnitude of the resulting
compressional force on Earth?
Answer: 5.12 105 N
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Standardized Test PrepChapter 16
Short Response, continued
14. Air becomes a conductor when the electric field
strength exceeds 3.0 106 N/C. Determine the
maximum amount of charge that can be carried by a
metal sphere 2.0 m in radius.
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Standardized Test PrepChapter 16
Short Response, continued
14. Air becomes a conductor when the electric field
strength exceeds 3.0 106 N/C. Determine the
maximum amount of charge that can be carried by a
metal sphere 2.0 m in radius.
Answer: 1.3 10–3 C
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Standardized Test PrepChapter 16
Extended Response
Use the information
below to answer
questions 15–18.
A proton, which has a
mass of 1.673 10–27 kg,
accelerates from rest in a
uniform electric field of
640 N/C. At some time
later, its speed is 1.2 106
m/s.
15. What is the magnitude
of the acceleration of
the proton?
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Standardized Test PrepChapter 16
Extended Response, continued
Use the information
below to answer
questions 15–18.
A proton, which has a
mass of 1.673 10–27 kg,
accelerates from rest in a
uniform electric field of
640 N/C. At some time
later, its speed is 1.2 106
m/s.
15. What is the magnitude
of the acceleration of
the proton?
Answer: 6.1 1010 m/s2
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ResourcesChapter menu
Standardized Test PrepChapter 16
Extended Response, continued
Use the information
below to answer
questions 15–18.
A proton, which has a
mass of 1.673 10–27 kg,
accelerates from rest in a
uniform electric field of
640 N/C. At some time
later, its speed is 1.2 106
m/s.
16. How long does it take
the proton to reach this
speed?
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ResourcesChapter menu
Standardized Test PrepChapter 16
Extended Response, continued
Use the information
below to answer
questions 15–18.
A proton, which has a
mass of 1.673 10–27 kg,
accelerates from rest in a
uniform electric field of
640 N/C. At some time
later, its speed is 1.2 106
m/s.
16. How long does it take
the proton to reach this
speed?
Answer: 2.0 10–5 s
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ResourcesChapter menu
Standardized Test PrepChapter 16
Extended Response, continued
Use the information
below to answer
questions 15–18.
A proton, which has a
mass of 1.673 10–27 kg,
accelerates from rest in a
uniform electric field of
640 N/C. At some time
later, its speed is 1.2 106
m/s.
17. How far has it moved
in this time interval?
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ResourcesChapter menu
Standardized Test PrepChapter 16
Extended Response, continued
Use the information
below to answer
questions 15–18.
A proton, which has a
mass of 1.673 10–27 kg,
accelerates from rest in a
uniform electric field of
640 N/C. At some time
later, its speed is 1.2 106
m/s.
17. How far has it moved
in this time interval?
Answer: 12 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Standardized Test PrepChapter 16
Extended Response, continued
Use the information
below to answer
questions 15–18.
A proton, which has a
mass of 1.673 10–27 kg,
accelerates from rest in a
uniform electric field of
640 N/C. At some time
later, its speed is 1.2 106
m/s.
18. What is its kinetic
energy at the later
time?
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Standardized Test PrepChapter 16
Extended Response, continued
Use the information
below to answer
questions 15–18.
A proton, which has a
mass of 1.673 10–27 kg,
accelerates from rest in a
uniform electric field of
640 N/C. At some time
later, its speed is 1.2 106
m/s.
18. What is its kinetic
energy at the later
time?
Answer: 1.2 10–15 J
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Standardized Test PrepChapter 16
Extended Response, continued
19. A student standing on a piece of insulating material
places her hand on a Van de Graaff generator. She
then turns on the generator. Shortly thereafter, her
hairs stand on end. Explain how charge is or is not
transferred in this situation, why the student is not
shocked, and what causes her hairs to stand up after
the generator is started.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Standardized Test PrepChapter 16
Extended Response, continued
19. (See previous slide for question.)
Answer: The charge on the sphere of the Van de Graaff generator is transferred to the student by means of conduction. This charge remains on the student because she is insulated from the ground. As there is no path between the student and the generator and the student and the ground by which charge can escape, the student is not shocked. The accumulation of charges of the same sign on the strands of the student’s hair causes the strands to repel each other and so stand on end.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Chapter 16Section 1 Electric Charge
Charging By Induction
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Chapter 16Section 1 Electric Charge
Transfer of Electric Charge