Chapter 16: Electric Forces and Fields
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Transcript of Chapter 16: Electric Forces and Fields
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Chapter 16: Electric Forces and Fields
16.1 Electric Charge
16.2 Electric Force
16.3 The Electric Field
Chapter 16
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Section 1 Electric ChargeChapter 16
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 16Section 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
Electric Charge
Section 1 Electric Charge
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Chapter 16Section 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 16Section 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
The Milikan Experiment (1909)
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. 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 16Section 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
Charging by Induction
Section 1 Electric Charge
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Chapter 16Section 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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Section 2 Electric ForceChapter 16
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
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.
Section 2 Electric Force
1 22
2
charge 1 charge 2electric force = Coulomb constant
distance
electric Cq qF kr
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
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)
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 PrincipleConsider 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, continuedThe Superposition Principle1. Define the problem, and identify the known
variables.Given:
q1 = +6.00 10–9 C r2,1 = 3.00 mq2 = –2.00 10–9 C r3,2 = 4.00 mq3 = +5.00 10–9 C r3,1
= 5.00 mq = 37.0º
Unknown: F3,tot = ? Diagram:
Fel = kCq1q2
r2
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Chapter 16Section 2 Electric Force
Sample Problem, continuedThe Superposition PrincipleTip: 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 16Section 2 Electric Force
Sample Problem3. Calculate the magnitudes
of the forces with Coulomb’s law.
–9 –9293 1
3,1 22 2
–83,1
–9 –9293 2
3,2 22 2
–93,1
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 kr
F
q qF kr
F
2
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Chapter 16Section 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 NFy = (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 NFy = 0 N
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Chapter 16Section 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 NFy,tot = 6.50 10–9 N + 0 N = 6.50 10–9 N
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Chapter 16Section 2 Electric Force
Sample Problem 6. Use the Pythagorean theorem to find the magni-
tude of the resultant force.
2 2 9 2 9 23, , ,
–93,
( ) ( ) (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 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 Ntan3.01 10 N
65.2º
y tot
x tot
FF
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Coulomb’s Law• The Coulomb force is a field force.• 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
Chapter 16Section 2 Electric Force
m1m2
r2
Fel = kCq1q2
r2
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Section 3 The Electric FieldChapter 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 (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.
• 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).
E = Felec
q0
Q q0
Q in the figure above is (+) because it is exerting a force of repulsion on the 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 • Electric field strength (intensity) depends on charge and distance.
An electric field exists in the region around a charged object.
• Electric Field Strength Due to a Point Charge
2
2
charge producing the fieldelectric field strength = Coulomb constant
distance
CqE kr
Felec = kC
qq0
r2E = Felec
q0 and
Q q0Q
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Chapter 16Section 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 16Section 3 The Electric Field
Electric Field Lines• 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
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 16Section 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 16Section 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 16Section 3 The Electric Field
Discharging Effects of Points• The charge density is greatest at the point of greatest curvature.
St. Elmo’s Fire
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.
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
Calculating Net Electric Field
Section 3 The Electric Field
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Chapter 16Section 3 The Electric Field
Sample Problem
Electric Field StrengthA 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, continuedElectric Field Strength1. 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 mq = 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, continuedElectric Field Strength2. 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 21
–69 2 2 52
2 2 22
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 kr
qE kr
Section 3 The Electric Field
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Chapter 16
Sample Problem, continuedElectric Field Strength3. 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, continuedElectric Field Strength4. 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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Sample Problem, continuedElectric Field Strength5. 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, continuedElectric Field Strength6. 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, continuedElectric Field Strength7. 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/Ctan1.08 10 N/C
66.0
y tot
x tot
EE
Section 3 The Electric Field
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Chapter 16
Sample Problem, continuedElectric Field Strength8. 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.
Section 3 The Electric Field
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Chapter 16Section 1 Electric Charge
Charging By Induction
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Chapter 16Section 1 Electric Charge
Transfer of Electric Charge
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Chapter 16Section 3 The Electric Field
Electric Field Lines