Download Section 3 The Electric Field

Chapter 16
Chapter 16: Electric Forces and Fields
16.1 Electric Charge
16.2 Electric Force
16.3 The Electric Field
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Chapter 16
Section 1 Electric Charge
Objectives for Section 1 Electric Charge
• Understand the basic properties of electric charge.
• Differentiate between conductors and insulators.
• Distinguish between charging by contact, charging
by induction, and charging by polarization.
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Chapter 16
Section 1 Electric Charge
Properties of Electric Charge
•
•
•
Atoms are the source of electric charge.
– Positively charged particles are called protons.
– Uncharged particles are called neutrons.
– Negatively charged particles are called electrons.
Friction Rods
• Electrons from animal fur are transferred to
atoms in ebonite (hard rubber). Ebonite
acquires a net excess of electrons.
• Electrons from a glass rod will transfer to a silk
cloth and give rise to an excess of electrons on
the silk.
• Your hair becomes positively charged when you
rub a balloon across it.
Electric charge is conserved. The amount of positive
charge acquired by your hair equals the amount of
negative charge acquired by the balloon.
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Chapter 16
Section 1 Electric Charge
Electric Charge
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Chapter 16
Section 1 Electric Charge
Properties of Electric Charge
• There are two kinds of electric
charge.
– like charges repel
– unlike charges attract
• The magnitude of electrical
forces between two charged
bodies often exceeds the
gravitational attraction between
the bodies.
• Electric charge is conserved.
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Chapter 16
Section 1 Electric Charge
Properties of Electric Charge
• Electric charge is quantized. That is,
when an object is charged, its charge
is always a multiple of a fundamental
unit of charge. +e, +2e, +3e, …
• The fundamental unit of charge, e, is
the magnitude of the charge of a
single electron or proton.
e = 1.602 176 x 10–19 C
• Charge is measured in coulombs (C).
• -1.0 C contains 6.2 x1018 electrons.
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Chapter 16
Section 1 Electric Charge
The Milikan Experiment (1909)
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Chapter 16
Section 1 Electric Charge
Milikan’s Oil Drop Experiment
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Chapter 16
Section 1 Electric Charge
Transfer of Electric Charge
• An electrical conductor is a material in which
charges can move freely. Most metals are
conductors.
• An electrical insulator, or nonconductor, is a
material in which charges cannot move freely.
(Electrons are tightly bound to the atom.)
Nonmetallic materials, such as glass, rubber, and
wood are good insulators.
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Chapter 16
Section 1 Electric Charge
Transfer of Electric Charge
• Insulators and conductors can be
charged by contact.
• Conductors can be charged by
induction.
• Induction is a process of
charging a conductor by bringing
it near another charged object and
grounding the conductor.
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Chapter 16
Section 1 Electric Charge
Charging by Induction
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Chapter 16
Section 1 Electric Charge
Transfer of Electric Charge
• 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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Chapter 16
Section 2 Electric Force
Section 2 Electric Force 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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Chapter 16
Section 2 Electric Force
Coulomb’s Law (Charles Coulomb, 1780’s)
• Two charges near one another exert a
force on one another called the electric
force. Coulomb’s law describes this
force.
• Coulomb’s law states that the electric
force is proportional to the magnitude of
each charge and inversely proportional
to the square of the distance between
them.
 q1q2 
Felectric  kC  2 
 r 
electric force = Coulomb constant 
 charge 1 charge 2 
2
 distance 
kc = 8.99 x 109 N.m2/C2 when F in N, r in m, and q in C
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Chapter 16
Section 2 Electric Force
Coulomb’s Law
• Electric force is a vector. When the charges (q1 and q2) are
alike they repel each other, and when they are opposite they
attract each other.
•
The resultant force on a charge is the vector sum of the
individual forces on that charge. Adding forces this way is
an example of the principle of superposition.
• When a body is in equilibrium, the net external force acting
on that body is zero. A charged particle can be positioned
such that the net electric force on the charge is zero. Set
Coulomb’s Law forces equal and solve for the distance
between either charge and the equilibrium position.
(Practice C)
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Chapter 16
Section 2 Electric Force
Superposition Principle
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Chapter 16
Section 2 Electric Force
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.
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Chapter 16
Section 2 Electric Force
Fel = kC q12q2
r
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 16
Section 2 Electric Force
Sample Problem, continued
The Superposition Principle
Tip: According to the superposition principle, the resultant
force on the charge q3 is the vector sum of the forces
exerted by q1 and q2 on q3. First, find the force exerted on
q3 by each, and then add these two forces together
vectorially to get the resultant force on q3.
2. Determine the direction of the forces by analyzing
the charges.
The force F3,1 is repulsive because q1 and q3 have
the same sign.
The force F3,2 is attractive because q2 and q3 have
opposite signs.
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Chapter 16
Section 2 Electric Force
Sample Problem
3. Calculate the magnitudes
of the forces with
Coulomb’s law.


2  5.00  10 –9 C 6.00  10 –9 C


q3q1
N

m
9
F3,1  kC
  8.99  10
 
2
2
2
(r 3,1)
C
 5.00 m 


 


F3,1  1.08  10 –8 N
F3,2


2  5.00  10 –9 C 2.00  10 –9 C

q3q2
9 Nm 
 kC
  8.99  10
 
2
2
2
(r 3,2)
C 
4.00m




F3,12  5.62  10 –9 N
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 


Chapter 16
Section 2 Electric Force
Sample Problem,
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 16
Section 2 Electric Force
Sample Problem
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 16
Section 2 Electric Force
Sample Problem
6. Use the Pythagorean theorem to find the magnitude of the resultant force.
F3,tot  (Fx ,tot )2  (Fy ,tot )2  (3.01 109 N)2  (6.50  109 N)2
F3,tot  7.16  10 –9 N
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Chapter 16
Section 2 Electric Force
Sample Problem
7. Use a suitable trigonometric function to find the
direction of the resultant force.
In this case, you can use the inverse tangent function:
tan  
Fy ,tot
Fx ,tot
6.50  10 –9 N

3.01 10 –9 N
  65.2º
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Chapter 16
Section 2 Electric Force
Coulomb’s Law
• The Coulomb force is a field force.
Fel = kC q12q2
r
• A field force is a force that is exerted by
one object on another even though there
is no physical contact between the two
objects.
• Gravitational attraction is also a field
force.
FG = G
m1m2
r2
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Chapter 16
Section 3 The Electric Field
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 16
Section 3 The Electric Field
Electric Field Strength (Intensity)
• An electric field is a region where an
electric force on a test charge (small
positive charge) can be detected. Q exerts
an electric field on q0.
Felec
E=
q0
• The SI units of the electric field, E, are
newtons per coulomb (N/C).
• The direction of the electric field vector,
E, is in the direction of the electric force
that would be exerted on a small positive
test charge (the direction a + test charge
accelerates).
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q0
Q
Q in the figure
above is (+)
because it is
exerting a force
of repulsion on
the test charge.
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Chapter 16
Section 3 The Electric Field
Electric Fields and Test Charges
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Chapter 16
Section 3 The Electric Field
Electric Field Strength
• Electric field strength (intensity) depends on charge and distance.
An electric field exists in the region around a charged object.
E=
Felec
q0
and
Felec = kC
• Electric Field Strength Due to a Point Charge
E  kC
qq0
r2
Q
q
0
Q
q
r2
electric field strength = Coulomb constant 
charge producing the field
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 distance 
2
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Chapter 16
Section 3 The Electric Field
Electric Field Strength Direction
• When q is (+) the electric field is directed outward radially from q.
• When q is (-) the electric field is directed toward q.
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Chapter 16
Electric Field Lines
Section 3 The Electric Field
• Electric field lines are used
to analyze electric fields and
show strength and direction.
– The number of electric
field lines is proportional
to the electric field
strength.
– Electric field lines are
tangent to the electric
field vector at any point.
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Chapter 16
Section 3 The Electric Field
Rules for Drawing Electric Field Lines
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Chapter 16
Section 3 The Electric Field
Rules for Sketching Fields Created by Several
Charges
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Chapter 16
Section 3 The Electric Field
4 Properties of Conductors in Electrostatic Equilibrium
(Distribution of Static Charges)
1. The electric field is zero everywhere inside the conductor
(otherwise charges would move and it would not be at equilibrium).
2. Any excess charge on an isolated conductor resides entirely on
the conductor’s outer surface (because the excess charges repel
each other).
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Chapter 16
Section 3 The Electric Field
4 Properties of Conductors in Electrostatic Equilibrium
(Distribution of Static Charges)
3. The electric field just outside a charged conductor is perpendicular
to the conductor’s surface.
4. On an irregularly shaped conductor, charge tends to accumulate
where the radius of curvature of the surface is smallest, that is, at
sharp points.
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Chapter 16
Section 3 The Electric Field
Discharging Effects of Points
•
The charge density is greatest at the point of greatest curvature.
A corona discharge or
brush is a slow
leakage of charge that
occurs when the
electric field intensity is
great enough to
produce ionization at
sharp projections or
corners.
St. Elmo’s Fire
Spark Discharge
Dry air can ionize at atmospheric
pressure when a difference of 30,000
J/C/cm (V/cm) exists between two
charged surfaces.
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Chapter 16
Section 3 The Electric Field
Calculating Net Electric Field
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Chapter 16
Section 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 xaxis 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
Section 3 The Electric Field
Sample Problem, continued
Electric Field Strength
1. Define the problem, and identify the known
variables.
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.
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Chapter 16
Section 3 The Electric Field
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.
–6
q1
9
2
2  7.00  10 C 
5
E1  kC 2  8.99  10 N  m /C 

3.93

10
N/C
2 
r1
 (0.400 m) 


–6
q2
9
2
2  5.00  10 C 
5
E2  kC 2  8.99  10 N  m /C 

1.80

10
N/C
2 
r2
 (0.500 m) 


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Chapter 16
Section 3 The Electric Field
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.
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Chapter 16
Section 3 The Electric Field
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
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Chapter 16
Section 3 The Electric Field
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
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Section 3 The Electric Field
Chapter 16
Sample Problem, continued
Electric Field Strength
6. Use the Pythagorean theorem to find the
magnitude of the resultant electric field strength
vector.
Etot 
E   E 
Etot 
1.08  10 N/C    2.49  10 N/C 
2
x ,tot
2
y ,tot
5
2
5
2
Etot  2.71 105 N/C
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Chapter 16
Section 3 The Electric Field
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:
tan  
E y ,tot
E x ,tot
2.49  105 N/C

1.08  105 N/C
  66.0
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Chapter 16
Section 3 The Electric Field
Sample Problem, continued
Electric Field Strength
8. Evaluate your answer.
The electric field at point P is pointing away from the
charge q1, as expected, because q1 is a positive
charge and is larger than the negative charge q2.
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Chapter 16
Section 1 Electric Charge
Charging By Induction
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Chapter 16
Section 1 Electric Charge
Transfer of Electric Charge
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Chapter 16
Section 3 The Electric Field
Electric Field Lines
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