Chapter 16
Section 1 Electric Charge
Properties of Electric Charge
• There are two kinds of electric charge.
– like charges repel
– unlike charges attract
• Electric charge is conserved.
– Positively charged particles are called protons.
– Uncharged particles are called neutrons.
– Negatively charged particles are called electrons.
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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, continued
• Electric charge is quantized. That is, when an object
is charged, its charge is always a multiple of a
fundamental unit of charge.
• Charge is measured in coulombs (C).
• 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
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Chapter 16
Section 1 Electric Charge
The Milikan Experiment
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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.
• An electrical insulator is a material in which charges
cannot move freely.
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Chapter 16
Section 1 Electric Charge
Transfer of Electric Charge, continued
• 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, 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
Coulomb’s Law
• Two charges near one another exert a force on one
another called the electric 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.
qq 
Felectric  kC  1 2 2 
 r 
electric force = Coulomb constant 
 charge 1 charge 2 
2
distance


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Chapter 16
Section 2 Electric Force
Coulomb’s Law, continued
• 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.
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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
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, continued
The Superposition Principle
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

q3 q 2
9 Nm 
 kC
  8.99  10
 
2
2
2
(r 3,2)
C 
4.00m




F3,1  5.62  10 –9 N
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 


Chapter 16
Section 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 16
Section 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 16
Section 2 Electric Force
Sample Problem, continued
The Superposition Principle
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, 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:
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, continued
• 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.
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Chapter 16
Section 3 The Electric Field
Electric Field Strength
• An electric field is a region where an electric force
on a test charge can be detected.
• 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.
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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, continued
• Electric field strength depends on charge and
distance. An electric field exists in the region around
a charged object.
• Electric Field Strength Due to a Point Charge
E  kC
q
r2
electric field strength = Coulomb constant 
charge producing the field
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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 3 The Electric Field
Electric Field Lines
• 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
Conductors in Electrostatic Equilibrium
• The electric field is zero everywhere inside the
conductor.
• Any excess charge on an isolated conductor resides
entirely on the conductor’s outer surface.
• The electric field just outside a charged conductor is
perpendicular to the conductor’s surface.
• 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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