Download Lecture 2 - Purdue Physics

Lecture 2
1
Wednesday, January 15, 2014
iClicker question
You place same two charges a distance r apart. Then you
double each charge and double the distance between
the charges. How does the force between the two
charges change?
a) The new force is twice a large.
b) The new force is half as large.
c) The new force is four times as large.
d) The new force is four times smaller.
e) The new force is the same
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Wednesday, January 15, 2014
Electric Force - Coulomb’s Law
 Consider two electric charges: q1
and q2
 The electric force F between
these two charges separated by
a distance r is given by
Coulomb’s Law
 The constant k is called
Coulomb’s constant and is given
by
 Coulomb’s law in vector form
q1 q2 r12
F=k 2
r12 r12
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Wednesday, January 15, 2014
2
2
2
q1 q2
q
(2q)
q
F =k 2 =k 2 →k
=
k
2
2
r
r
(2r)
r
Force remains the same
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Wednesday, January 15, 2014
Demo: electrostatic charging
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Wednesday, January 15, 2014
+++
---
6
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F
+++
---
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F
+++
+++
---
---
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F
+++
+++
--F
-
---
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Wednesday, January 15, 2014
F
+++
+++
+++ +
--F
-
-----
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Wednesday, January 15, 2014
F
+++
--F
+++
-
---
F
+++ +
---
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Wednesday, January 15, 2014
Charging by induction
--++
++
+q
+
+
extra electrons
+q
+
+
This will not work if no free electrons
+
+
++
+
++-+- ++ -+ - + -
+q
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Wednesday, January 15, 2014
Example - Forces between Electrons
 What is relative strength of the electric force compared
with the force of gravity for two electrons?
Gravity is irrelevant for atomic and subatomic processes –
the electric force is much much stronger.
 But sometimes gravity is most important; e.g., the motion
of the planets.
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Wednesday, January 15, 2014
Example - Forces between Electrons
 What is relative strength of the electric force compared
with the force of gravity for two electrons?
Gravity is irrelevant for atomic and subatomic processes –
the electric force is much much stronger.
 But sometimes gravity is most important; e.g., the motion
of the planets.
Why?
If Coulomb forces are 42 orders of magnitude
stronger than gravity, why we do not “clearly” feel it?
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Wednesday, January 15, 2014
Superposition Principle
 The net force acting on any charge is the
linear vector sum of the forces due to the
remaining charges in the distribution.
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Wednesday, January 15, 2014
Vector addition of force
Electric force in vector form
F23
q1
Ftot=F13+F23
r13
r23
q3
F13
q2
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Wednesday, January 15, 2014
iClicker question
Three charges are arranged on a straight line as
shown in the figure. What is the direction of the
electrostatic force on the middle charge?
Wednesday, January 15, 2014
x
Positive force directed along x, negative along “- x”
F32
1
a
2
a
F32
2
q
= k 2 = F12
a
3
F12
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Wednesday, January 15, 2014
x
Positive force directed along x, negative along “- x”
F32
1
a
2
a
F32
2
q
= k 2 = F12
a
3
F12
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Wednesday, January 15, 2014
iClicker question
Three charges are arranged on a straight line,
separated by “a”. What is the force on the right
charge? (Positive: along x)
x
a
a
2
q
A : 0;
B : −k 2 ;
2a
2
2
q
q
D : −k 2 ;
E: k 2
a
a
Wednesday, January 15, 2014
2
q
C: k 2
2a
iClicker question
1
a
x
a
2
2
3
F23
F13
2
F13
2
(−2)q
q
=k
=
−k
2
2
4a
2a
2
Ftot = F13 + F23
Wednesday, January 15, 2014
q
=k 2
a
2
2
q
q
q
= −k 2 + k 2 = k 2
2a
a
2a
Example - Charged Balls
 Consider two identical charged balls hanging from the
ceiling by strings of equal length ℓ = 1.5 m (in equilibrium).
Each ball has a charge of q=25 µC. The balls hang at an
angle θ = 25° with respect to the vertical. What is the
mass of the balls?
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Wednesday, January 15, 2014
Example - Charged Balls
 Consider two identical charged balls hanging from the
ceiling by strings of equal length ℓ = 1.5 m (in equilibrium).
Each ball has a charge of q=25 µC. The balls hang at an
angle θ = 25° with respect to the vertical. What is the
mass of the balls?
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Wednesday, January 15, 2014
General Problem Solving Strategies
1. Write what is given.
2. Sketch
3. Write equations
4. Count # of equations and unknown (should be equal!)
5. Solve the system (DO NOT PLUG IN NUMBERS!!!)
6. Check dimensions (if dimensions are wrong, the answer is
wrong), Check simple cases
7. Plug in the numbers...
University Physics, Chapter 1
Wednesday, January 15, 2014
29
General Problem Solving Strategies
1. Write what is given.
2. Sketch
3. Write equations
4. Count # of equations and unknown (should be equal!)
5. Solve the system (DO NOT PLUG IN NUMBERS!!!)
6. Check dimensions (if dimensions are wrong, the answer is
wrong), Check simple cases
7. Plug in the numbers...
University Physics, Chapter 1
Wednesday, January 15, 2014
29
Example - Charged Balls
 Consider two identical charged balls hanging from the
ceiling by strings of equal length ℓ = 1.5 m (in equilibrium).
Each ball has a charge of q=25 µC. The balls hang at an
angle θ = 25° with respect to the vertical. What is the
mass of the balls?

45
Wednesday, January 15, 2014
Example - Charged Balls
 Consider two identical charged balls hanging from the
ceiling by strings of equal length ℓ = 1.5 m (in equilibrium).
Each ball has a charge of q=25 µC. The balls hang at an
angle θ = 25° with respect to the vertical. What is the
mass of the balls?

45
Wednesday, January 15, 2014
Example - Charged Balls
 Consider two identical charged balls hanging from the
ceiling by strings of equal length ℓ = 1.5 m (in equilibrium).
Each ball has a charge of q=25 µC. The balls hang at an
angle θ = 25° with respect to the vertical. What is the
mass of the balls?

 Write what is given:
45
Wednesday, January 15, 2014
Example - Charged Balls
 Consider two identical charged balls hanging from the
ceiling by strings of equal length ℓ = 1.5 m (in equilibrium).
Each ball has a charge of q=25 µC. The balls hang at an
angle θ = 25° with respect to the vertical. What is the
mass of the balls?

 Write what is given:
 ℓ = 1.5 m, q1=q2=25 µC, θ = 25°
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Wednesday, January 15, 2014
 Sketch
Three forces act on each ball:
Coulomb force Fc, gravity Fg, and the tension of the string T
The Coulomb force is horizontal and must be repulsive to keep the balls apart
The gravitational force points down
The tension T is in the direction of the string
The balls are in equilibrium, which means the three forces are balanced
Add the three forces to the drawing
Define distance d between the balls
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Wednesday, January 15, 2014
Example - Charged Balls
 Make a separate free body diagram for one of the charged balls
 Define x-y coordinate system
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Wednesday, January 15, 2014
Example - Charged Balls
 Make a separate free body diagram for one of the charged balls
 Define x-y coordinate system
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Wednesday, January 15, 2014
Example - Charged Balls
Write equations
 The condition of static equilibrium tells us
that the sum of the x-components of the
three forces acting on the ball must equal
zero and the sum of y-components of these
forces must equal zero
 The sum of the x-components of the forces is
• T is magnitude of the string tension
• θ is the angle of the string relative to the vertical
• FC is the magnitude of the Coulomb force
 The sum of the y-components of the forces is
 The force of gravity, Fg, is just the weight of the charged ball
• m is the mass of the charged ball
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Wednesday, January 15, 2014
Example - Charged Balls
Write equations
 The condition of static equilibrium tells us
that the sum of the x-components of the
three forces acting on the ball must equal
zero and the sum of y-components of these
forces must equal zero
 The sum of the x-components of the forces is
• T is magnitude of the string tension
• θ is the angle of the string relative to the vertical
• FC is the magnitude of the Coulomb force
 The sum of the y-components of the forces is
 The force of gravity, Fg, is just the weight of the charged ball
• m is the mass of the charged ball
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Wednesday, January 15, 2014
Example - Charged Balls
 The electric force between the two balls is
• d is the distance between the two balls
 We can express d in terms of ℓ and θ
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Wednesday, January 15, 2014
Count # of equations and unknown (should be equal!)
T sin θ = Fc
T cos θ = Fg
Fg = mg
2
q
Fc = k 2
d
d/2
sin θ =
l
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Wednesday, January 15, 2014
Count # of equations and unknown (should be equal!)
T sin θ = Fc
T cos θ = Fg
Fg = mg
2
q
Fc = k 2
d
d/2
sin θ =
l
22
Wednesday, January 15, 2014
Count # of equations and unknown (should be equal!)
T sin θ = Fc
T cos θ = Fg
Fg = mg
2
q
Fc = k 2
d
d/2
sin θ =
l
22
Wednesday, January 15, 2014
Count # of equations and unknown (should be equal!)
T sin θ = Fc
T cos θ = Fg
Fg = mg
2
q
Fc = k 2
d
d/2
sin θ =
l
22
Wednesday, January 15, 2014
Count # of equations and unknown (should be equal!)
T sin θ = Fc
T cos θ = Fg
Fg = mg
2
q
Fc = k 2
d
d/2
sin θ =
l
22
Wednesday, January 15, 2014
Count # of equations and unknown (should be equal!)
T sin θ = Fc
T cos θ = Fg
Fg = mg
2
q
Fc = k 2
d
d/2
sin θ =
l
22
Wednesday, January 15, 2014
Count # of equations and unknown (should be equal!)
T sin θ = Fc
T cos θ = Fg
Fg = mg
2
q
Fc = k 2
d
d/2
sin θ =
l
5=5
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Wednesday, January 15, 2014
Count # of equations and unknown (should be equal!)
T sin θ = Fc
T cos θ = Fg
Fg = mg
5=5
2
q
Fc = k 2
d
d/2
sin θ =
l
}
eliminate d
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Wednesday, January 15, 2014
Count # of equations and unknown (should be equal!)
T sin θ = Fc
T cos θ = Fg
}
Fg = mg
5=5
2
q
Fc = k 2
d
d/2
sin θ =
l
eliminate T
}
eliminate d
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Wednesday, January 15, 2014
Example - Charged Balls
 Solve the system
 We divide the two free-body force equations containing the
string tension
 Thus we eliminate the string tension and get
 Putting in our expressions for the force of gravity and the
electric force, we get
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Wednesday, January 15, 2014
Example - Charged Balls
 Solve the system
 We divide the two free-body force equations containing the
string tension
 Thus we eliminate the string tension and get
 Putting in our expressions for the force of gravity and the
electric force, we get
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Wednesday, January 15, 2014
Check dimensions
 Dimensions
2
2
Nm
C
kg =
2
2
2
C (m/s )m
2
N = kgm/s
2
2
2
(kgm/s )m
C
kg =
2
2
2
C
(m/s )m
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Wednesday, January 15, 2014
Check dimensions
 Dimensions
2
2
Nm
C
kg =
2
2
2
C (m/s )m
2
N = kgm/s
2
2
2
(kgm/s )m
C
kg =
2
2
2
C
(m/s )m
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Wednesday, January 15, 2014
Check dimensions
 Dimensions
2
2
Nm
C
kg =
2
2
2
C (m/s )m
2
N = kgm/s
2
2
2
(kgm/s )m
C
kg =
2
2
2
C
(m/s )m
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Wednesday, January 15, 2014
Check dimensions
 Dimensions
2
2
Nm
C
kg =
2
2
2
C (m/s )m
2
N = kgm/s
2
2
2
(kgm/s )m
C
kg =
2
2
2
C
(m/s )m
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Wednesday, January 15, 2014
Check dimensions
 Dimensions
2
2
Nm
C
kg =
2
2
2
C (m/s )m
2
N = kgm/s
2
2
2
(kgm/s )m
C
kg =
2
2
2
C
(m/s )m
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Wednesday, January 15, 2014
Simple cases?
Large mass -> small angle: OK
Large charge-> large mass (for given angle): OK
Shorter line: large angle: OK
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Wednesday, January 15, 2014
Simple cases?
Large mass -> small angle: OK
Large charge-> large mass (for given angle): OK
Shorter line: large angle: OK
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Wednesday, January 15, 2014
 Calculate
Now, put in numbers...
 Putting in our numerical values we obtain
 Round

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Wednesday, January 15, 2014
 Calculate
Now, put in numbers...
 Putting in our numerical values we obtain
 Round

50
Wednesday, January 15, 2014
 Calculate
Now, put in numbers...
 Putting in our numerical values we obtain
 Round

50
Wednesday, January 15, 2014
 Calculate
Now, put in numbers...
 Putting in our numerical values we obtain
 Round

We report our result to three significant figures
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Wednesday, January 15, 2014
Electric field
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Wednesday, January 15, 2014
The Electric Field
 A charge creates an
electric field around itself
and the other charge feels
that field.
+
Electric field at a given
point in space: place a
positive test charge q at
the point and measure the
electrostatic force that
acts on the test charge;
then
3
Wednesday, January 15, 2014
The Electric Field
 A charge creates an
electric field around itself
and the other charge feels
that field.
+
+
Electric field at a given
point in space: place a
positive test charge q at
the point and measure the
electrostatic force that
acts on the test charge;
then
Test charge q
3
Wednesday, January 15, 2014
The Electric Field
 A charge creates an
electric field around itself
and the other charge feels
that field.
+
+
Electric field at a given
point in space: place a
positive test charge q at
the point and measure the
electrostatic force that
acts on the test charge;
then
Test charge q
3
Wednesday, January 15, 2014
Precise Definition of Electric Field
 We define the electric field in terms of the force it
exerts on a positive point charge:
 Unit of the electric field:
N/C (Newtons per Coulomb)
 Note that the electric force is parallel to the electric
field and is proportional to the charge
• Force alined with field for q>0, counter alined for q< 0
E
5
Wednesday, January 15, 2014
Precise Definition of Electric Field
 We define the electric field in terms of the force it
exerts on a positive point charge:
 Unit of the electric field:
N/C (Newtons per Coulomb)
 Note that the electric force is parallel to the electric
field and is proportional to the charge
• Force alined with field for q>0, counter alined for q< 0
E
q>0, F
5
Wednesday, January 15, 2014
Precise Definition of Electric Field
 We define the electric field in terms of the force it
exerts on a positive point charge:
 Unit of the electric field:
N/C (Newtons per Coulomb)
 Note that the electric force is parallel to the electric
field and is proportional to the charge
• Force alined with field for q>0, counter alined for q< 0
q<0, F
E
q>0, F
5
Wednesday, January 15, 2014
EM fields are real physical substance
 Electromagnetic fields are real, these are not just
mathematical concepts. EM fields is another material
substance: one can assign energy-density to
electromagnetic field. Empty space with EM field has
more energy than just empty space.
 There is no “substance” that “transports” EM forces
(like mechanical forces are transported by the
medium, sound is transported by medium), EM wave
propagate “on their own”, no medium to carry.
 (Actually, gravitational force is not “real”).
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Wednesday, January 15, 2014
Example: Field of a Point Charge
Question: What is the field created by a point charge q?
Answer: Consider a “test charge” q0 at point x.
 Force on q0:
E
Electric field at x
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Wednesday, January 15, 2014
Superposition of Electric Fields
 Suppose we have many charges
 The electric field at any point in space will have
contributions from all the charges.
 The electric field at any point in space is the
superposition of the electric field from n charges is
(vectors!)
 Note that the superposition applies to each
component of the field (x, y, z)
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Wednesday, January 15, 2014
Vector addition of force
Electric force in vector form
F23
q1
Ftot=F13+F23
r13
r23
q3
F13
q2
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Wednesday, January 15, 2014
Vector addition of force
Electric force in vector form
Еtot=Е1+Е23
Е2
q1
r13
r23
q3
Е1
q2
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Wednesday, January 15, 2014
Electric Field Lines
 We can represent the electric field graphically by
electric field lines — i.e., curves that represent the
vector force exerted on a positive test charge.
 Electric field lines will originate on positive charges and
terminate on negative charges.
 Electric field lines do not cross.
•
Why?
 Field lines are not real, just a useful mathematical
concept
8
Wednesday, January 15, 2014
Electric Field Lines
 We can represent the electric field graphically by
electric field lines — i.e., curves that represent the
vector force exerted on a positive test charge.
 Electric field lines will originate on positive charges and
terminate on negative charges.
 Electric field lines do not cross.
•
Why?
 Field lines are not real, just a useful mathematical
concept
8
Wednesday, January 15, 2014
Electric Field Lines
 The electric force at a given point in space is tangent
to the electric field line through that point.
F2
F1
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Wednesday, January 15, 2014
Electric Field Lines
 The electric force at a given point in space is tangent
to the electric field line through that point.
F2
F1
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Wednesday, January 15, 2014
Properties of Field Lines
 The strength of the electric field is represented by the
density of electric field lines
Weak
Strong
The direction of the electric field is tangent to the electric
field lines
9
Wednesday, January 15, 2014
Field Lines from a Point Charge
 The electric field lines from a
point charge extend out
radially
 For a positive point charge,
the field lines point outward
F
• Terminate at infinity
 For a negative charge, the
field lines point inward
• Originate at infinity
q
E = k 2 r̂
r
10
Wednesday, January 15, 2014
Field Lines from a Point Charge
 The electric field lines from a
point charge extend out
radially
 For a positive point charge,
the field lines point outward
F
• Terminate at infinity
 For a negative charge, the
field lines point inward
• Originate at infinity
q
E = k 2 r̂
r
10
Wednesday, January 15, 2014
Electric Field Lines for Two Point Charges
 We can use the superposition principle to calculate the
electric field from two point charges.
 The field lines will originate from the positive charge and
terminate on the negative charge.
3d
2d
11
Wednesday, January 15, 2014
Electric Field Lines (2)
+
-
What is the direction of the E field at points halfway
between two charges of different sign?
Use superposition principle:
12
Wednesday, January 15, 2014
Electric Field Lines
13
Wednesday, January 15, 2014
iClicker
A negative charge -q is placed in a nonuniform
electric field as shown in the figure. What is the
direction of the electric force on this negative
charge?
Wednesday, January 15, 2014
F=q E
C
since q< 0
E-field tangent to field line
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Wednesday, January 15, 2014
End
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Wednesday, January 15, 2014