Physics II: Electricity & Magnetism Binomial Expansions, Riemann Sums, Sections 21.6 to 21.11

Physics II:Electricity & Magnetism

Physics II:Electricity & Magnetism

Binomial Expansions, Riemann Sums, Sections 21.6

to 21.11

Binomial Expansions, Riemann Sums, Sections 21.6

to 21.11

Thursday (Day 11)Thursday (Day 11)

Binomial ExpansionBinomial Expansion

Warm-UpWarm-Up

Thurs, Feb 5 Calculate the force acting on Q2 at distance of 0.50 m.

Place your homework on my desk:“Foundational Mathematics’ Skills of Physics” Packet (Page 5)Electrostatics Lab #2: Lab Report

Have you complete WebAssign Problems: 21.1 - 21.4?

For future assignments - check online at www.plutonium-239.com

Thurs, Feb 5 Calculate the force acting on Q2 at distance of 0.50 m.

Place your homework on my desk:“Foundational Mathematics’ Skills of Physics” Packet (Page 5)Electrostatics Lab #2: Lab Report

Have you complete WebAssign Problems: 21.1 - 21.4?


Warm-Up ReviewWarm-Up Review

Calculate the force acting on Q2 at distance of 0.50 m.

Calculate the force acting on Q2 at distance of 0.50 m.

Essential Question(s)Essential Question(s)

WHAT PRIOR FOUNDATIONAL MATHEMATICS’ SKILLS ARE NECESSARY IN PHYSICS II?


VocabularyVocabulary

Static ElectricityElectric ChargePositive / NegativeAttraction / RepulsionCharging / DischargingFriction InductionConductionLaw of Conservation of

Electric ChargeNon-polar Molecules

Static ElectricityElectric ChargePositive / NegativeAttraction / RepulsionCharging / DischargingFriction InductionConductionLaw of Conservation of

Electric ChargeNon-polar Molecules

Polar Molecules Ion Ionic CompoundsForceDerivative Integration (Integrals)Test ChargeElectric FieldField LinesElectric DipoleDipole Moment

Polar Molecules Ion Ionic CompoundsForceDerivative Integration (Integrals)Test ChargeElectric FieldField LinesElectric DipoleDipole Moment

Foundational Mathematics Skills in Physics Timeline


Day Pg(s) Day Pg(s) Day Pg(s) Day Pg(s)

11

26 3 11 16 16 21

213

147 4 12 17 17 8

322

238 5 13 18 18 9

424

†129 6 14 19 19 10

5 15 10 7 15 20 20 11


AgendaAgenda Review “Foundational Mathematics’ Skills of Physics”

Packet (Page 5) with answer guide. Begin The Four Circles Graphic Organizer DERIVATIVE PROOF USING BINOMIAL EXPANSION Derivative practice

FRIDAY: INTEGRAL PROOF USING RIEMANN SUMS Integral Practice

MONDAY: Discuss Electric Fields & Gravitational Field Apply Electric Fields Continue with The Four Circles Graphic Organizer

Review “Foundational Mathematics’ Skills of Physics” Packet (Page 5) with answer guide.

Begin The Four Circles Graphic Organizer DERIVATIVE PROOF USING BINOMIAL EXPANSION Derivative practice

FRIDAY: INTEGRAL PROOF USING RIEMANN SUMS Integral Practice

MONDAY: Discuss Electric Fields & Gravitational Field Apply Electric Fields Continue with The Four Circles Graphic Organizer

Topic #1: Determine the slope at point A for f(x)=xn


y = 1/2 x




y = 1/4 x2




The something is determined by using binomial expansion.

Binomial Expansion is defined as:

The something is determined by using binomial expansion.

Binomial Expansion is defined as:

Slope “is defined as” the rise over the run.

Trick #1: Factor out (b-a)

slope ≡limb→ a

f (b)− f (a)b−a

slope ≡limb→ a

f (b)− f (a)b−a

=limb→ a

bn −an

b−a=lim

b→ a

(b−a)("something")b−a( )

bn −an =(b−a)(bn−1a0 +bn−2a1 +bn−3a2…+ b2an−3 +b1an−2 +b0an−1)


Aside #1: Binomial Expansion Practice


Using binomial expansion:

Expand b1-a1:






Expand b2-a2:






Expand b3-a3:






Expand b4-a4:






Expand b5-a5:



Determination of SlopeDetermination of Slope

slope ≡limb→ a

f (b)− f (a)b−a

=limb→ a

bn −an

b−a=lim

b→ a


slope =limb→ a

b1 −a1

b−a=lim

b→ a

(b−a)(1)b−a( )

=limb→ a

(1)

Recall: The definition for slope of a polynomial is now:

Calculate the slope for f(x) = x1 The expansion of b1-a1 = (b-a)(1)


Calculate the slope for f(x) = x1 The expansion of b1-a1 = (b-a)(1)



slope ≡limb→ a

f (b)− f (a)b−a

=limb→ a

bn −an

b−a=lim

b→ a


slope =limb→ a

b2 −a2

b−a=lim

b→ a

(b−a)(b+a)b−a( )

=limb→ a

(b+a)


Calculate the slope for f(x) = x2 The expansion of b2-a2 = (b-a)(b+a)


Calculate the slope for f(x) = x2 The expansion of b2-a2 = (b-a)(b+a)



slope ≡limb→ a

f (b)− f (a)b−a

=limb→ a

bn −an

b−a=lim

b→ a


slope =limb→ a

b3 −a3

b−a=lim

b→ a

(b−a)(b2 +ba+a2 )b−a( )

=limb→ a

(b2 +ba+a2 )


Calculate the slope for f(x) = x3 The expansion of b3-a3 = (b-a)(b2+ba+a2)


Calculate the slope for f(x) = x3 The expansion of b3-a3 = (b-a)(b2+ba+a2)



slope ≡limb→ a

f (b)− f (a)b−a

=limb→ a

bn −an

b−a=lim

b→ a


slope =limb→ a

b4 −a4

b−a=lim

b→ a

(b−a)(b3 +b2a+ba2 +a3)b−a( )

=limb→ a

(b3 +b2a+ba2 +a3)


Calculate the slope for f(x) = x4 The expansion of b4-a4 = (b-a)(b3+b2a+ba2+a3)


Calculate the slope for f(x) = x4 The expansion of b4-a4 = (b-a)(b3+b2a+ba2+a3)



slope ≡limb→ a

f (b)− f (a)b−a

=limb→ a

bn −an

b−a=lim

b→ a


slope =limb→ a

b5 −a5

b−a=lim

b→ a

(b−a)(b4 +b3a+b2a2 +ba3 +a4 )b−a( )

=limb→ a

(b4 +b3a+b2a2 +ba3 +a4 )


Calculate the slope for f(x) = x5 The expansion of b5-a5 = (b-a)(b4+b3a+b2a2+ba3 +a4)


Calculate the slope for f(x) = x5 The expansion of b5-a5 = (b-a)(b4+b3a+b2a2+ba3 +a4)



Now we can show that:

Now this is where it gets fun! It is time to decrease the distance between points a and b to get a more accurate slope at point a. This is called taking the ‘limit’ as “b” approaches “a.” All we have to do is change all “b’s” to “a’s”

Now we can show that:

Now this is where it gets fun! It is time to decrease the distance between points a and b to get a more accurate slope at point a. This is called taking the ‘limit’ as “b” approaches “a.” All we have to do is change all “b’s” to “a’s”

slope =limb→ a

(bn−1a0 +bn−2a1 +bn−3a2 +bn−4a3… )

slope =(an−1a0 +an−2a1 +an−3a2 +an−4a3… )

slope=(an−1+0 +an−2+1 +an−3+2 +an−4+3… )

slope=(an−1 +an−1 +an−1 +an−1… )

slope=n⋅an−1



The slope of any power is now: The slope of any power is now:

slope =n⋅an−1


Method Determination Comparision


Checking the slope for f(a) = an where n = 2 Using the binomial expansion method

Using the slope = n an-1 method

Do they agree?



Do they agree?

slope =limb→ a

(b+a) =(a+a) =2a

slope =n⋅an−1 =2⋅a2−1 =2a






Do they agree?



Do they agree?

slope =limb→ a

(1) =1

slope =n⋅an−1 =1⋅a1−1 =1






Do they agree?



Do they agree?

slope =limb→ a

(b2 +ba+a2 ) =(a2 +aa+a2 ) =3a2

slope =n⋅an−1 =3⋅a3−1 =3a2






Do they agree?



Do they agree?

slope =limb→ a

(b3 +b2a+ba2 +a3)

=(a3 +a2a+aa2 +a3) =4a3

slope =n⋅an−1 =4⋅a4−1 =4a3






Do they agree?



Do they agree?

slope =limb→ a

(b4 +b3a+b2a2 +ba3 +a4 )

=(a4 +a3a+a2a2 +aa3 +a4 ) =5a4

slope =n⋅an−1 =5⋅a5−1 =5a4


Summary of DerivativeSummary of Derivative

After adding the constant back in and changing a to the variable x we get

f(x) =C xn

Slope = n C x n-1

How is the derivative expressed?

After adding the constant back in and changing a to the variable x we get

f(x) =C xn

Slope = n C x n-1

How is the derivative expressed?

slope = ′f (x) =dydx


Summary of DerivativeSummary of Derivative

How is the derivative for a polynomial,

represented?

What about finding the area under a curve (or line)?

Find the integral.

For a polynomial:

How is the derivative for a polynomial,

represented?

What about finding the area under a curve (or line)?

Find the integral.

For a polynomial:

f (x) = Cxn∑ ,

slope = nCxn−1∑

area = f (x)∫ dx=1

n+1Cxn+1


So what is the relationship . . .

So what is the relationship . . .

between finding the slope and taking the first derivative?They are the same.

Other derivative representations

between finding the slope and taking the first derivative?They are the same.

Other derivative representations

€

′ f (x) = ′ f =dy

dx=

d f( (x))dx

=d

dxf( (x))


SummarySummary

After comparing the force constants for electrostatics and gravity, identify which Force is stronger.

HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 6) Derivative Practice

Future assignments: Electrostatics Lab #3: Lab Report (Due in 3 classes)

After comparing the force constants for electrostatics and gravity, identify which Force is stronger.

HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 6) Derivative Practice


How do we use Coulomb’s Law and the principle of superposition to determine the force that acts between point charges?

Friday (Day 12)Friday

(Day 12)Riemann SumsRiemann Sums

Warm-UpWarm-Up

Fri, Feb 6 Calculate the velocity of the electron moving around the hydrogen

nucleus (r = 0.53 x 10-10 m)

Place your homework on my desk:“Foundational Mathematics’ Skills of Physics” Packet (Page 6) Derivative Practice


Fri, Feb 6 Calculate the velocity of the electron moving around the hydrogen

nucleus (r = 0.53 x 10-10 m)

Place your homework on my desk:“Foundational Mathematics’ Skills of Physics” Packet (Page 6) Derivative Practice


Warm-Up ReviewWarm-Up Review

Calculate the velocity of the electron moving around the hydrogen nucleus (r = 0.53 x 10-10 m)

Calculate the velocity of the electron moving around the hydrogen nucleus (r = 0.53 x 10-10 m)



HOW DO WE DESCRIBE THE NATURE OF ELECTROSTATICS AND APPLY IT TO VARIOUS SITUATIONS?How do we describe and apply the concept of electric field?How do we describe and apply Coulomb’s Law and the Principle

of Superposition?



of Superposition?

VocabularyVocabulary Static Electricity Electric Charge Positive / Negative Attraction / Repulsion Charging / Discharging Friction Induction Conduction Law of Conservation of Electric

Charge Non-polar Molecules

Static Electricity Electric Charge Positive / Negative Attraction / Repulsion Charging / Discharging Friction Induction Conduction Law of Conservation of Electric


Polar Molecules Ion Ionic Compounds Force Derivative Integration (Integrals) Test Charge Electric Field Field Lines Electric Dipole Dipole Moment





11

26 3 11 16 16 21

213

147 4 12 17 17 8

322

238 5 13 18 18 9

424

†129 6 14 19 19 10

5 15 10 7 15 20 20 11


AgendaAgenda


Review Derivative Practice INTEGRAL PROOF USING RIEMANN SUMS Integral Practice

MONDAY:Discuss Electric Fields & Gravitational Field Apply Electric FieldsContinue with The Four Circles Graphic Organizer


Review Derivative Practice INTEGRAL PROOF USING RIEMANN SUMS Integral Practice

MONDAY:Discuss Electric Fields & Gravitational Field Apply Electric FieldsContinue with The Four Circles Graphic Organizer



y = 1/2 x




y = 1/4 x2


SummarySummary

Identify one section that in the Integral Proof using Riemann Sums that was confusing?

HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 7) Go through the Riemann sum derivation - determine what you do not

understand. Integral Practice


Identify one section that in the Integral Proof using Riemann Sums that was confusing?

HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 7) Go through the Riemann sum derivation - determine what you do not

understand. Integral Practice



Monday (Day 13)Monday (Day 13)

Riemann Sums (Day II)Section 21.6Section 21.8

Riemann Sums (Day II)Section 21.6Section 21.8

Warm-UpWarm-UpMon, Feb 9

1. If I measured the distance of each step I took and summed them all together, what would I have calculated?

2. If I was driving in a car on the turnpike at a constant speed and I multiplied my speed by the time I was traveling, what would I have calculated?

3. Now make it more complex, what if my speed was slowly changing and I1. Wrote down my velocity and the amount of time I was traveling at that velocity;2. Multiplied those two numbers together;3. Added those new numbers together; What would I have calculated?

Place your homework on my desk: “Foundational Mathematics’ Skills of Physics” Packet (Page 7) Integral Practice

1. For future assignments - check online at www.plutonium-239.com

Mon, Feb 91. If I measured the distance of each step I took and summed them all together, what

would I have calculated?2. If I was driving in a car on the turnpike at a constant speed and I multiplied my

speed by the time I was traveling, what would I have calculated?3. Now make it more complex, what if my speed was slowly changing and I

1. Wrote down my velocity and the amount of time I was traveling at that velocity;2. Multiplied those two numbers together;3. Added those new numbers together; What would I have calculated?

Place your homework on my desk: “Foundational Mathematics’ Skills of Physics” Packet (Page 7) Integral Practice

1. For future assignments - check online at www.plutonium-239.com

Warm-UpWarm-Up

Mon, Feb 91. If I measured the distance of each step I took and summed them all

together, what would I have calculated?2. If I was driving in a car on the turnpike at a constant speed and I

multiplied my speed by the time I was traveling, what would I have calculated?

3. Now make it more complex, what if my speed was slowly changing and I

1. Wrote down my velocity and the amount of time I was traveling at that velocity;

2. Multiplied those two numbers together;3. Added those new numbers together; What would I have calculated?

Mon, Feb 91. If I measured the distance of each step I took and summed them all

together, what would I have calculated?2. If I was driving in a car on the turnpike at a constant speed and I

multiplied my speed by the time I was traveling, what would I have calculated?

3. Now make it more complex, what if my speed was slowly changing and I

1. Wrote down my velocity and the amount of time I was traveling at that velocity;

2. Multiplied those two numbers together;3. Added those new numbers together; What would I have calculated?




of Superposition?



of Superposition?










11

26 3 11 16 16 21

213

147 4 12 17 17 8

322

238 5 13 18 18 9

424

†129 6 14 19 19 10

5 15 10 7 15 20 20 11


AgendaAgenda


Complete the Integral Proof using Riemann SumsReview Integral PracticeDiscuss

Electric FieldsGravitational FieldField Lines

Continue with The Four Circles Graphic OrganizerApply Electric Fields


Complete the Integral Proof using Riemann SumsReview Integral PracticeDiscuss

Electric FieldsGravitational FieldField Lines

Continue with The Four Circles Graphic OrganizerApply Electric Fields

Riemann Sums ProofRiemann Sums Proof

ADD RIEMANN SUMS PROOF HEREADD RIEMANN SUMS PROOF HERE

Riemann SumsRiemann Sums

Riemann Sums Related to RealityRiemann Sums Related to Reality

Big Picture Ideas and RelationshipsBig Picture Ideas and Relationships

If F(x) is written as x(t) (aka. Displacement as a function of time)

Then slope of the graph, F’(x), can be written as x’(t) (aka v(t), the velocity as a function of time)

And F(b) - F(a) is the really the just the xfinal - xinitial.

This is also equal to the summation of all velocity x time calculations [f(ci)(xi-xi-1)] or rewritten as [v(ci)(ti-ti-1)]

If F(x) is written as x(t) (aka. Displacement as a function of time)

Then slope of the graph, F’(x), can be written as x’(t) (aka v(t), the velocity as a function of time)

And F(b) - F(a) is the really the just the xfinal - xinitial.

This is also equal to the summation of all velocity x time calculations [f(ci)(xi-xi-1)] or rewritten as [v(ci)(ti-ti-1)]

The graph of y(x)Referred to as F(x) [or x(t)]

The graph of y(x)Referred to as F(x) [or x(t)]

x(a)

x(b)

x ti−1( )

x(ti )ti −ti−1

x(ti )−x ti−1( )ti −ti−1

′x (ci ) = v(ci )

Δt Δt

The graph of y’(x);Called F’ (x); [or x’ (t)]

The graph of y’(x);Called F’ (x); [or x’ (t)]

The graph of F’(x) is renamed f(x); [or x’ (t) is renamed v(t)]The graph of F’(x) is renamed f(x); [or x’ (t) is renamed v(t)]

Riemann Sum with only 1 approximation (Δt: large)Riemann Sum with only 1 approximation (Δt: large)

tt

x

x(a)

x(b)

x ti−1( )

x(ti )

ti −ti−1


′x (ci ) = v(ci )

x(ci ) only a reference point; not the "height"

Riemann Sum with only 2 approximations (Δt: still large)

Riemann Sum with only 2 approximations (Δt: still large)

t

x

x(a)

x(b)

x ti−1( )

x(ti )

ti −ti−1


′x (ci ) = v(ci )

Riemann Sum with 9 approximations (Δt: medium)

Riemann Sum with 9 approximations (Δt: medium)

t

x

x(a)

x(b)

x ti−1( )

x(ti )

ti −ti−1


′x (ci ) = v(ci )

Δt

Δt

Riemann Sum with 17 approximations (Δt: small)

Riemann Sum with 17 approximations (Δt: small)

t

x

x(a)

x(b)

x ti−1( )x(ti )

ti −ti−1


′x (ci ) = v(ci )

Δt

Riemann Sum with only 33 approximations (Δt: smaller)Riemann Sum with only 33 approximations (Δt: smaller)

t

x


Confusing Points:x(ci) is only a point of reference, not the “height” to which

the t is multiplied to get the area under the curve. In fact, it is the area under the v(t) graph that we are trying

to find in order to determine the total displacement.

Confusing Points:x(ci) is only a point of reference, not the “height” to which

the t is multiplied to get the area under the curve. In fact, it is the area under the v(t) graph that we are trying

to find in order to determine the total displacement.


As t decreases, your approximations become more accurate.

Note: Summing up all of the “slope of x vs t times t” (aka. “velocity x time”) calculations will equal the

total displacement (aka. The final position minus the starting position).

As t decreases, your approximations become more accurate.

Note: Summing up all of the “slope of x vs t times t” (aka. “velocity x time”) calculations will equal the

total displacement (aka. The final position minus the starting position).

Section 21.6Section 21.6

How do we describe and apply the concept of electric field?How do we define electric fields in terms of the

force on a test charge?

How do we describe and apply the concept of electric field?How do we define electric fields in terms of the

force on a test charge?


How do we describe and apply Coulomb’s Law and the Principle of Superposition?How do we use Coulomb’s Law to describe the

electric field of a single point charge?How do we use vector addition to determine

the electric field produced by two or more point charges?

How do we describe and apply Coulomb’s Law and the Principle of Superposition?How do we use Coulomb’s Law to describe the

electric field of a single point charge?How do we use vector addition to determine

the electric field produced by two or more point charges?

21.6 The Electric Field

The electric field is the force on a small charge, divided by the charge:


For a point charge:


Force on a point charge in an electric field:

Superposition principle for electric fields:


Problem solving in electrostatics: electric forces and electric fields

1. Draw a diagram; show all charges, with signs, and electric fields and forces with directions

2. Calculate forces using Coulomb’s law

3. Add forces vectorially to get result


How do we describe and apply Coulomb’s Law and the Principle of Superposition?How do we compare and contrast Coulomb’s

Law and the Universal Law of Gravitation?

How do we describe and apply Coulomb’s Law and the Principle of Superposition?How do we compare and contrast Coulomb’s

Law and the Universal Law of Gravitation?

21.8 Field Lines

The electric field can be represented by field lines. These lines start on a positive charge and end on a negative charge.

Electric Field created by a spherically charged objectElectric Field created by a spherically charged object

Electric Field created by a spherically charged objectElectric Field created by a spherically charged object

21.8 Field Lines

The number of field lines starting (ending) on a positive (negative) charge is proportional to the magnitude of the charge.

The electric field is stronger where the field lines are closer together.

21.8 Field Lines

Electric dipole: two equal charges, opposite in sign:

21.8 Field Lines

Summary of field lines:

1. Field lines indicate the direction of the field; the field is tangent to the line.

2. The magnitude of the field is proportional to the density of the lines.

3. Field lines start on positive charges and end on negative charges; the number is proportional to the magnitude of the charge.

21.8 Field Lines

Summary of field lines:

4. Field lines never cross because the electric field cannot have two values for the same point.

EM Field uses color to represent the field strength (ie. Red is stronger; blue is weaker). Each charge below is ±10q. EM Field uses color to represent the field strength (ie. Red is stronger; blue is weaker). Each charge below is ±10q.

SummarySummary

Using Newton’s Second Law, what the formula for force?

HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 16) Web Assign 21.5 - 21.7






Tuesday (Day 14)Tuesday (Day 14)



Warm-UpWarm-Up

Tues, Feb 10 Each charge on the next slide is ±q. What will happen to the lines if a 3rd

charge of +q is added to the (1) right side and (2) left side?

Place your homework on my desk: “Foundational Mathematics’ Skills of Physics” Packet (Page 16) Web Assign 21.5 - 21.7


Tues, Feb 10 Each charge on the next slide is ±q. What will happen to the lines if a 3rd

charge of +q is added to the (1) right side and (2) left side?



Field Example #1: Each charge below is ±q. What will happen to the lines if a 3rd charge of +q is added to the (1)

right side and (2) left side?

Field Example #1: Each charge below is ±q. What will happen to the lines if a 3rd charge of +q is added to the (1)


Essential Question(s)Essential Question(s) WHAT PRIOR FOUNDATIONAL MATHEMATICS’ SKILLS ARE

NECESSARY IN PHYSICS II? HOW DO WE DESCRIBE THE NATURE OF ELECTROSTATICS

AND APPLY IT TO VARIOUS SITUATIONS?How do we describe and apply the nature of electric fields in and

around conductors?How do we describe and apply the concept of induced charge and

electrostatic shielding? How do we describe and apply the concept of electric fields?


HOW DO WE DESCRIBE THE NATURE OF ELECTROSTATICS AND APPLY IT TO VARIOUS SITUATIONS?How do we describe and apply the nature of electric fields in and












11

26 3 11 16 16 21

213

147 4 12 17 17 8

322

238 5 13 18 18 9

424

†129 6 14 19 19 10

5 15 10 7 15 20 20 11


AgendaAgenda


Discuss Electric Fields and Conductors Motion of a Charged Particle in an Electric Field

Work on Web Assign


Discuss Electric Fields and Conductors Motion of a Charged Particle in an Electric Field

Work on Web Assign

Field Example #2: Each charge below is ±5q. What will happen to the lines if a 3rd charge of +q is added to the (1)









How do we describe and apply the nature of electric fields in and around conductors?How do we explain the mechanics responsible for the absence of

electric field inside of a conductor?Why must all of the excess charge reside on the surface of a

conductor? How do we prove that all excess charge on a conductor must

reside on its surface and the electric field outside of the conductor must be perpendicular to the surface?

How do we describe and apply the nature of electric fields in and around conductors?How do we explain the mechanics responsible for the absence of

electric field inside of a conductor?Why must all of the excess charge reside on the surface of a

conductor? How do we prove that all excess charge on a conductor must

reside on its surface and the electric field outside of the conductor must be perpendicular to the surface?


How do we describe and apply the concept of induced charge and electrostatic shielding?What is the significance of why there can be no electric

field in a charge-free region completely surrounded by a single conductor?

How do we describe and apply the concept of induced charge and electrostatic shielding?What is the significance of why there can be no electric

field in a charge-free region completely surrounded by a single conductor?

21.9 Electric Fields and Conductors

The static electric field inside a conductor is zero – if it were not, the charges would move.

The net charge on a conductor is on its surface.

Charge ball suspended in a hollow metal sphere

Charge ball suspended in a hollow metal sphere

ObservationsThe hollow sphere had a charge on

the outside.The charged ball still had a charge.

ConclusionsThe charged ball on the inside

induces an equal charge on the hollow sphere.


the outside.The charged ball still had a charge.

ConclusionsThe charged ball on the inside

induces an equal charge on the hollow sphere.

21.9 Electric Fields and Conductors

The electric field is perpendicular to the surface of a conductor – again, if it were not, charges would move.

Charge ball placed into a hollow metal sphere

Charge ball placed into a hollow metal sphere


the outside.The charged ball no longer had a

charge.Conclusions

The charge resides on the outside of a conductor.


the outside.The charged ball no longer had a

charge.Conclusions

The charge resides on the outside of a conductor.

Applications of E-fields and conductors: Faraday Cages

Applications of E-fields and conductors: Faraday Cages

Faraday cages protect you from lightning because there is no electrical field inside the metal cage (Notice (1) it completely surrounds him and (2) the size of the gaps in the fence (it is not a solid piece of metal).

Faraday cages protect you from lightning because there is no electrical field inside the metal cage (Notice (1) it completely surrounds him and (2) the size of the gaps in the fence (it is not a solid piece of metal).


How do we describe and apply the nature of electric fields in and around conductors?How do we determine the direction of the force

on a charged particle brought near an uncharged or grounded conductor?

How do we describe and apply the nature of electric fields in and around conductors?How do we determine the direction of the force



How do we describe and apply the concept of induced charge and electrostatic shielding?How do we determine the direction of the force


How do we describe and apply the concept of induced charge and electrostatic shielding?How do we determine the direction of the force



How do we describe and apply the concept of electric field?How do we calculate the magnitude and

direction of the force on a positive or negative charge in an electric field?

How do we analyze the motion of a particle of known mass and charge in a uniform electric field?

How do we describe and apply the concept of electric field?How do we calculate the magnitude and

direction of the force on a positive or negative charge in an electric field?

How do we analyze the motion of a particle of known mass and charge in a uniform electric field?

Electron accelerated by an electric fieldElectron accelerated by an electric field

An electron is accelerated in the uniform field E (E=2.0x104N/C) between two parallel charged plates. The separation of the plates is 1.5 cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate. (a) With what speed does it leave the hole? (b) Show that the gravitational force can be ignored. [NOTE: Assume the hole is so small that it does not affect the uniform field between the plates]

An electron is accelerated in the uniform field E (E=2.0x104N/C) between two parallel charged plates. The separation of the plates is 1.5 cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate. (a) With what speed does it leave the hole? (b) Show that the gravitational force can be ignored. [NOTE: Assume the hole is so small that it does not affect the uniform field between the plates]

Electron accelerated by an electric field (a) With what speed does it leave the hole?

Electron accelerated by an electric field (a) With what speed does it leave the hole?

F =qE F =ma⇒ a =F

m=

qE

m

v2 =v02

=0+ 2aΔx⇒ v = 2aΔx

v = 2qEm

⎛⎝⎜

⎞⎠⎟Δx

v = 21.60 x 10-19 C( ) 2.0 x 104 NC( )

9.1 x 10−31 kg( )0.015 m( )

v =1.0 x 107 ms

Electron accelerated by an electric field (b) Show that the gravitational force can be ignored. Electron accelerated by an electric field

(b) Show that the gravitational force can be ignored.

FE =qE

FE = 1.60 x 10-19 C( ) 2.0 x 104 NC( )

FE =3.5 x 10−15 N

FG =mg

FG = 9.1 x 10-31 kg( ) 9.8 ms2( )

FG =8.9 x 10-30 NNote that FE is 1014 times larger than the FG.Also note that the electric field due to the electron does not enter the problem since it cannot exert a force on itself.

Applications of an electron accelerated by an E-Field: Mass Spectrometer

Applications of an electron accelerated by an E-Field: Mass Spectrometer

Mass Spectrometers are used to separate isotopes of atoms. The charged isotopes (a.k.a. ions) are accelerated to a velocity by the parallel plates (located

from S to S1)

Mass Spectrometers are used to separate isotopes of atoms. The charged isotopes (a.k.a. ions) are accelerated to a velocity by the parallel plates (located

from S to S1)

Projectile Motion of a Charged Particle:Electron moving perpendicular to E

Projectile Motion of a Charged Particle:Electron moving perpendicular to E

Suppose an electron is traveling with a speed, v0 = 1.0x107m/s, enters a uniform field E at right angles to v0. Describe the motion by giving the equation of its path while in the electric field. Ignore gravity.

Suppose an electron is traveling with a speed, v0 = 1.0x107m/s, enters a uniform field E at right angles to v0. Describe the motion by giving the equation of its path while in the electric field. Ignore gravity.

F =qE=−eE F =ma⇒ a =F

m=−

eE

m

y =v0y=0

t+ 12 ayt

2=−eE

2mt 2

x =v0xt+ 12 ax

=0t2 ⇒ t =

x

v0 x

=x

v0

⇒ t 2 =x

v0

⎛

⎝⎜⎞

⎠⎟

2

=x2

v02

= −eE

qmv02 x2 This is the equation of a

parabola (i.e. projectile motion).

Electrons moving perpendicular to E: The discovery of the electron: J.J. Thomson’s

Experiment

Electrons moving perpendicular to E: The discovery of the electron: J.J. Thomson’s

Experiment J. J. Thomson’s famous experiment that allowed him to discover the electron. J. J. Thomson’s famous experiment that allowed him to discover the electron.

Applications of an electron moving perpendicular to E: Cathode Ray Tube (CRT)

Applications of an electron moving perpendicular to E: Cathode Ray Tube (CRT)

Television Sets & Computer Monitors (CRT)Television Sets & Computer Monitors (CRT)

Applications of an electron moving perpendicular to E: Mass Spectrometer

Applications of an electron moving perpendicular to E: Mass Spectrometer

Mass Spectrometers are used to separate isotopes of atoms. The charged isotopes (a.k.a. ions) are accelerated to a velocity by the parallel plates

(located at the - & + plates)

Mass Spectrometers are used to separate isotopes of atoms. The charged isotopes (a.k.a. ions) are accelerated to a velocity by the parallel plates

(located at the - & + plates)

Applications of an electron moving perpendicular to E: e/m ApparatusApplications of an electron moving perpendicular to E: e/m Apparatus

e/m Apparatus e/m Apparatus

Applications of an electron moving perpendicular to E: e/m ApparatusApplications of an electron moving perpendicular to E: e/m Apparatus

e/m Apparatus e/m Apparatus

SummarySummary

Using your kinematic equations, determine the equation that relates y to v0, g, , and x?


Future assignments: Electrostatics Lab #3: Lab Report (Due in 1 class)

Using your kinematic equations, determine the equation that relates y to v0, g, , and x?


Future assignments: Electrostatics Lab #3: Lab Report (Due in 1 class)


Wednesday (Day 15)

Wednesday (Day 15)

Work DayWork Day

Warm-UpWarm-Up

Wed, Feb 11 Write down the steps that you would use to explain how to open a door to a

blind person?



Wed, Feb 11 Write down the steps that you would use to explain how to open a door to a

blind person?





















11

26 3 11 16 16 21

213

147 4 12 17 17 8

322

238 5 13 18 18 9

424

†129 6 14 19 19 10

5 15 10 7 15 20 20 11


AgendaAgenda


Work Day Coulomb’s Law Web Assign

REVISE: Complete Graphic Organizer (up to Sections 21.10)


Work Day Coulomb’s Law Web Assign

REVISE: Complete Graphic Organizer (up to Sections 21.10)

SummarySummary



Future assignments:



Future assignments:




Warm-UpWarm-Up

Thurs, Feb 12

Complete Graphic Organizers for Sections 21-8 & 21-10.



Thurs, Feb 12

Complete Graphic Organizers for Sections 21-8 & 21-10.





AND APPLY IT TO VARIOUS SITUATIONS?How do we compare and contrast the basic properties of an

insulator and a conductor?How do we describe and apply the concept of electric field?


HOW DO WE DESCRIBE THE NATURE OF ELECTROSTATICS AND APPLY IT TO VARIOUS SITUATIONS?How do we compare and contrast the basic properties of an

insulator and a conductor?How do we describe and apply the concept of electric field?










11

26 3 11 16 16 21

213

147 4 12 17 17 8

322

238 5 13 18 18 9

424

†129 6 14 19 19 10

5 15 10 7 15 20 20 11


AgendaAgenda


Discuss Torque, factors that affect torque, r X F Electric Dipoles Electric Dipoles in an Electric Field The electric field produced by a dipole Calculations: Dipoles in an electric field


Discuss Torque, factors that affect torque, r X F Electric Dipoles Electric Dipoles in an Electric Field The electric field produced by a dipole Calculations: Dipoles in an electric field


How do we compare and contrast the basic properties of an insulator and a conductor?What are characteristics and classification(s) of

electrically . . .conductive atoms?insulative atoms?semi-conductive atoms?conductive compounds?insulative compounds?semi-conductive compounds?

How do we compare and contrast the basic properties of an insulator and a conductor?What are characteristics and classification(s) of

electrically . . .conductive atoms?insulative atoms?semi-conductive atoms?conductive compounds?insulative compounds?semi-conductive compounds?


How do we describe and apply the concept of electric field?How do we calculate the net force and torque

on a collection of charges in an electric field?

How do we describe and apply the concept of electric field?How do we calculate the net force and torque

on a collection of charges in an electric field?

How do we calculate the net force and torque on a collection of charges in an electric field?

TorqueTo make an object start rotating, a force is needed; the position and direction of the force matter as well.

The perpendicular distance from the axis of rotation to the line along which the force acts is called the lever arm.


Torque

A longer lever arm is very helpful in rotating objects.


Torque

Here, the lever arm for FA is the distance from the knob to the hinge; the lever arm for FD is zero; and the lever arm for FC is as shown.


Torque

The torque is defined as:

= rF

or

= r F

Torque, Torque, Torque is perpendicular to the direction of the rotation. Right-hand rule - The direction of the positive torque is

in the direction of increasing angle) In general, if we define torque as

= r x F = r F sin Also, torque can be defined about any point using

net = (ri x Fi) where ri is the position vector of the ith particle and Fi is

the net force on the ith particle.

Torque is perpendicular to the direction of the rotation. Right-hand rule - The direction of the positive torque is

in the direction of increasing angle) In general, if we define torque as

= r x F = r F sin Also, torque can be defined about any point using

net = (ri x Fi) where ri is the position vector of the ith particle and Fi is

the net force on the ith particle.


Electric DipolesElectric Dipoles

•The combination of two equal charges of opposite sign, +Q and -Q , separated by a distance l, is referred to as an electric dipole. The quantity Ql is called the dipole moment, p. The dipole moment points from the negative to the positive charge. Many molecules have a dipole moment and are referred to as polar molecules. •It is interesting to note that the value of the separated charges may be less than that of a single electron or proton but cannot be isolated.


Electric DipolesElectric Dipoles

• Electric Dipoles: The combination of two equal charges of opposite sign, +Q and -Q, separated by a distance l.

• The dipole moment, p: The quantity Ql. • The dipole moment points from the negative to the positive charge. • Many molecules have a dipole moment and are referred to as polar molecules. • It is interesting to note that the value of the separated charges may be less than that

of a single electron or proton, but they cannot be isolated.

• Electric Dipoles: The combination of two equal charges of opposite sign, +Q and -Q, separated by a distance l.

• The dipole moment, p: The quantity Ql. • The dipole moment points from the negative to the positive charge. • Many molecules have a dipole moment and are referred to as polar molecules. • It is interesting to note that the value of the separated charges may be less than that

of a single electron or proton, but they cannot be isolated.

Dipole in an External FieldDipole in an External Field


F±

-q

+q F+

F±

F−

A dipole, p = Ql, is placed in an electric field E. First, let us analyze the angle , for torque and about its bisector at point O.

Point O sinPoint O F± sinF±

0° 180°

45° 135°

90° 90°

135° 45°

180° 0°

0 0

22

22

1 1

22

22

0 0

Note that the choice of the angle does not change our value for sin point O will be used for all reference angles instead of the rxF angle to relate the direction of the dipole moment to the E-Field.


A dipole, p = Ql, is placed in an electric field E. Next, let us analyze the direction of the torque force to the change angle , Note: By definition, positive torque always increases the value of (I.e. move the dipole in the counterclockwise direction).


It is also important to note that the applied torque force will cause the angle decrease (in the clockwise direction) instead of increase (counter-clockwise) about point O. about point O is negative.


A dipole p = Ql is placed in an electric field E.


=r × F =rF sinθ

net = r × F∑ = rF sinθ∑=rF+ sinθ + rF− sinθ

net =l

2QE( )sinθ +

l

2QE( )sinθ

net = QlE sinθ =pE sinθ =p × E

net = τ + + τ −


The effect of the torque is to try to turn the dipole so p is parallel to E. The work done on the dipole by the electric field to change the angle from 1 to 2 , is

W = d1

2

∫Because the direction of the torque is opposite to the direction of increasing , we write the torque as

Then the dipole so p is parallel to E. The work done on the dipole by the electric field to change the angle from 1 to 2 , is

=−p × E or τ = − pE sinθ

W =−pE sind1

2

∫ =pE cosθθ1

θ2 =pE cosθ2 − cosθ1( )



Positive work done by the field decreases the potential energy, U, of the dipole in the field. If we choose U = 0 when p is perpendicular to E (that is choosing 1 = 90º so cos 1 = 0), and setting 2 = then

W =pE cos2

={ −cos1

⎛

⎝⎜

⎞

⎠⎟ =pE cosθ

U =−W =−pE cosθ =−p ⋅E


Torque with respect to the Dipole’s Orientation

Torque with respect to the Dipole’s Orientation


Electric Field Produced by a Dipole


To determine the electric field produced by a dipole in the absence of an external field along the midpoint or perpendicular bisector of the dipole.

hh hr

–

+

h = r2 + L2 4 E =E+ +E−


E+ = E− =E =1

4πε 0

Q

h2

Enet =E+ cos +E−cos =2E cosθ =2EL

2r2 + L2 4

Enet =EL

r2 + L2 4=

1

4πε 0

Q

r2 + L2 4( )

L

r2 + L2 4

Enet =1

4πε0

p

r2 + L2 4( )32

Enet =1

4πε0

pr3 at r >> L



It is interesting to note that at r >> l, the electric field decreases more rapidly for a dipole (1/r3) than for a single point charge (1/r2). This is due to the fact that at large distances the two opposite charges neutralize each other due to their close proximity At distances where r >> l, this 1/r3 dependence also applies for points that are not on the perpendicular bisector of the dipole.

E =1

4πε0

pr3 For a dipole at r >> l

hh hr

–

+

E =1

4πε0

qr2 For a single point charge


∝1

r3

∝1

r2

Table of Dipole Moment Values

Table of Dipole Moment Values


Dipoles in an Electric FieldDipoles in an Electric Field

The dipole moment of a water molecule is 6.1 x 10-30 C•m. A water molecule is placed in a uniform electric field with magnitude 2.0 x 105 N/C.

• What is the magnitude of the maximum torque that electric field can exert on the molecule?

• What is the potential energy when the torque is at its maximum?

• What is the dipole moment, p1 and p2, for a single O-H bond (where 2 = 104.5°)? Note: Let


max = p × E =pE sin θ

=90°}

=1{ =pE

= 6.1 x 10−30 Cgm( ) 2.0 x 105 N

C( ) =1.2 x 10−24 Ngm

U =−p⋅E =−pE cosθ =−pE cos 90o( ) =0

p1 =p2 =pOH .

p1 cos

p2 cos

p =pnet =p1 cosθ + p2 cosθ =2 pOH cos θ 2( )

⇒ pOH =p

2cos θ 2( ) =

6.1 x 10−30 Cgm( )

2 cos 104.5° 2( ) = 4.98 x 10−30 Cgm


The dipole moment of a water molecule is 6.1 x 10-30 C•m. A water molecule is placed in a uniform electric field with magnitude 2.0 x 105 N/C.

• In what position will the potential energy take on its greatest value?

• Why is this different than the position where the torque is maximized?

The potential energy will be maximized when cos = –1, so = 180°, which means p and E are antiparallel. The potential energy is maximized when the dipole moment is oriented so that it has to rotate through the largest angle, 180°, to reach equilibrium at = 0°.

The torque is maximized when the electric forces are perpendicular to p.



The carbonyl group (C=O) dipole. The distance between the carbon (+) and oxygen ( –) atoms in the carbonyl group which occurs in many organic molecules is about 1.2 x 10-10 m and the dipole moment of this group is about 8.0 x 10-30 C•m. A formaldehyde molecule, CH2O, is placed in a uniform electric field with magnitude 2.0 x 105 N/C.

• What the direction of the dipole moment, p?

• What is the magnitude of the maximum torque that electric field can exert on the molecule?

• What is the potential energy when the torque is at its maximum?


p

max = p × E =pE sin θ

=90°}

=1{ =pE

= 8.0 x 10−30 Cgm( ) 2.0 x 105 N

C( ) =1.6 x 10−24 Ngm

U =−p⋅E =−pE cosθ =−pE cos 90o( ) =0



• What is the partial charge (±) of the carbon (+) and oxygen ( –) atoms in the carbonyl group?

• (a) How much of the quantized charge of an electron/proton is the partial charge of the carbonyl group to? (b) What is this value in percent?

p =Ql ⇒ Q =p

l =

8.0 x 10−30 Cgm( )

1.2 x 10−10 m( )= 6.7 x 10−20 C

(a)n =

Qδ±

Q±

=0.42


(b) Percentage of an electron's charge =n x 100%=.42 x 100% = 42%



• In what position will the potential energy take on its greatest value?

• Why is this different than the position where the torque is maximized?

The potential energy will be maximized when cos = –1, so = 180°, which means p and E are antiparallel. The potential energy is maximized when the dipole moment is oriented so that it has to rotate through the largest angle, 180°, to reach equilibrium at = 0°.

The torque is maximized when the electric forces are perpendicular to p.


SummarySummary

How does positive torque relate to the change in the angle?

HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 19) Web Assign 21.15

Future assignments:

How does positive torque relate to the change in the angle?

HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 19) Web Assign 21.15

Future assignments:


Supplementary NotesSupplementary Notes

Vector Cross ProductVector Cross Product

Known as the vector product or cross productThe cross product of two vectors A and B is

defined as another vector C = A x B whose magnitude is C = |A x B| = AB sin where < 180º between A and B and whose direction is perpendicular to both A and B.

Right hand rules for cross products

Known as the vector product or cross productThe cross product of two vectors A and B is

defined as another vector C = A x B whose magnitude is C = |A x B| = AB sin where < 180º between A and B and whose direction is perpendicular to both A and B.

Right hand rules for cross products

Vector Cross ProductVector Cross Product

The cross product of two vectors A = Axi + Ayj + AzkB = Bxi + Byj + Bzk

Can be written as

A x B = (AyBz-AzBy)i + (AzBx-AxBz)j + (AxBy-AyBx)k

The cross product of two vectors A = Axi + Ayj + AzkB = Bxi + Byj + Bzk

Can be written as

A x B = (AyBz-AzBy)i + (AzBx-AxBz)j + (AxBy-AyBx)k

€

A × B =

i j k

Ax Ay Az

Bx By Bz

Properties of Vector Cross Products

Properties of Vector Cross Products

A x A = 0A x B = -B x A A x (B + C) = (A x B) + (A x C) .

A x A = 0A x B = -B x A A x (B + C) = (A x B) + (A x C) .

€

d

dtA × B( ) =

dA

dt× B + A ×

dB

dt

NOT FINISHEDSTART HERE

NOT FINISHEDSTART HERE

Review Riemann Sums Proof (see notes)

Review Riemann Sums Proof (see notes)

Friday (Day 17)Friday

(Day 17)

Warm-UpWarm-Up

Fri, Feb 13

Complete Graphic Organizer for Section 21-11.

Place your homework on my desk: “Foundational Mathematics’ Skills of Physics” Packet (Page 19) Web Assign 21.15


Fri, Feb 13

Complete Graphic Organizer for Section 21-11.

Place your homework on my desk: “Foundational Mathematics’ Skills of Physics” Packet (Page 19) Web Assign 21.15




AND APPLY IT TO VARIOUS SITUATIONS?How do we apply integration and the Principle of Superposition to

uniformly charged objects?How do we identify and apply the fields of highly symmetric

charge distributions?How do we describe and apply the electric field created by

uniformly charged objects?


HOW DO WE DESCRIBE THE NATURE OF ELECTROSTATICS AND APPLY IT TO VARIOUS SITUATIONS?How do we apply integration and the Principle of Superposition to













11

26 3 11 16 16 21

213

147 4 12 17 17 8

322

238 5 13 18 18 9

424

†129 6 14 19 19 10

5 15 10 7 15 20 20 11


AgendaAgenda


Steps to Determine the E-Field Created by a Uniform Charge Distributions

Calculate the electric field for continuous charge distributions for the following:Uniformly Charged Ring ( 0 2π) Uniformly Charged Vertical Wire (–∞ +∞) Uniformly Charged Vertical Wire (–L/2 +L/2)

Distribute E-Field Derivation Rubrics



Calculate the electric field for continuous charge distributions for the following:Uniformly Charged Ring ( 0 2π) Uniformly Charged Vertical Wire (–∞ +∞) Uniformly Charged Vertical Wire (–L/2 +L/2)

Distribute E-Field Derivation Rubrics

Section 21.7Section 21.7 How do we apply integration and the Principle of Superposition to

uniformly charged objects? How do we use integration and the principle of superposition to calculate

the electric field of a straight, uniform charge wire? How do we use integration and the principle of superposition to calculate

the electric field of a thin ring of charge on the axis of the ring? How do we use integration and the principle of superposition to

calculate the electric field of a semicircle of charge at its center? How do we use integration and the principle of superposition to calculate

the electric field of a uniformly charged disk on the axis of the disk?

How do we apply integration and the Principle of Superposition to uniformly charged objects? How do we use integration and the principle of superposition to calculate

the electric field of a straight, uniform charge wire? How do we use integration and the principle of superposition to calculate

the electric field of a thin ring of charge on the axis of the ring? How do we use integration and the principle of superposition to

calculate the electric field of a semicircle of charge at its center? How do we use integration and the principle of superposition to calculate

the electric field of a uniformly charged disk on the axis of the disk?


How do we identify and apply the fields of highly symmetric charge distributions?How do we identify situations in which the direction of the electric

field produced by highly symmetric charge distributions can be deduced from symmetry considerations?




How do we describe and apply the electric field created by uniformly charged objects?How do we describe the electric field of parallel

charged plates? How do we describe the electric field of a long,

uniformly charged wire? How do we use superposition to determine the electric

fields of parallel charged plates?





21-7 The Field of a Continuous Distribution

To find the field of a continuous distribution of charge, treat it as a collection of near-point charges:

Summing over the infinitesimal fields:

How do we apply integration and the Principle of Superposition to uniformly charged objects? How do we identify and apply the fields of highly symmetric charge distributions?

How do we describe and apply the electric field created by uniformly charged objects?


Finally, making the charges infinitesimally small and integrating rather than summing:




Constant linear charge density :

Some types of charge distribution are relatively simple.




Constant surface charge density :




Constant volume charge density :





Graphical: Draw a picture of the object and 3-D plane. Label the partial length, area, or volume that is creating the partial E-field. Determine the distance from the charged object to the location of the desired E-Field and label all

components and lengths. Mathematical:

Write the formulas for dE and the component of the E-field that contributes to the net E-field (i.e. does not cancel due to symmetry).

Write the total charge density and solve it for Q. Write the charge density in relation to the partial charge and solve it for the partial charge (dq). Set up the integral by determining what key component(s) change. †Solve the integral and write the answer in a concise manner.

†See the instructor, AP Calculus BC students, or Schaum’s Mathematical Handbook.






Uniformly Charged Ring ( 0 2π)

Uniformly Charged Ring ( 0 2π)

dl

rφ

z

h = r2 + z2

dE

dEz=dE cosθ

dE =1

4πε0

dqh2 ; dEz =dEcos =dE

z

h=

1

4πε 0

z

h3 dq

λ =Q

l=

Q

2π r=

Q

rφ⇒ Q = 2π rλ = λ rφ

λ =dq

dl=

dq

rdφ⇒ dq = λ rdφ

Ez = dEz0

Etot

∫ =

1

4πε 0

z

h3 dq=λ rdφ{0

Qtot

∫

Ez =1

4πε0

zh

z2+r2{

3 λrdφ0

2π

∫ =1

4πε 0

zλ r

z2 + r2( )

32

dφ0

2π

∫ =1

4πε 0

zλ r

z2 + r2( )

32

φ[ ]0

2π

Ez =1

4πε0

zλr

z2 + r2( )322π −0( )

=1

4πε 0

z

z2 + r2( )

32

λ 2π r=Q

{ =1

4πε 0

zQ

z2 + r2( )

32

Uniformly ChargedVertical Wire (–∞+∞)


-∞

∞dy

y

x

h = x2 + y2

dE

dEx=dE cosθ

dE =1

4πε0

dqh2 ; dEx =dEcos =dE

x

h=

1

4πε 0

x

h3 dq

λ =Q

l=

Q

y⇒ Q = λ y

λ =dq

dl=

dq

dy⇒ dq = λ dy

Ex = dEx0

Etot

∫ =

1

4πε 0

x

h3 dq=λ dy{0

Qtot

∫

=1

4πε 0

x

hx2 +y2

{3 λ dy

−∞

∞

∫ =1

4πε 0

λ x1

x2 + y2( )

32

dy−∞

∞

∫

Ex =1

4πε0

λxy

x2 x2 + y2

⎡

⎣⎢⎢

⎤

⎦⎥⎥−∞

∞

=1

4πε 0

λ

x

y

x2 + y2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−∞

∞

=1

2πε 0

λ

x

Ex =1

2πε0

λx

yy

=1

2πε 0

Q

xy

Uniformly ChargedVertical Wire (–L/2+L/2)


−L

2

L

2dy

y

x

h = x2 + y2

dE

dEx=dE cosθ

dE =1

4πε0


x

h=

1

4πε 0

x

h3 dq

λ =Q

l=

Q

y⇒ Q = λ y

λ =dq

dl=

dq

dy⇒ dq = λ dy

Ex = dEx0

Etot

∫ =

1

4πε 0

x

h3 dq=λ dy{0

Qtot

∫

=1

4πε 0

x

hx2 +y2

{3 λ dy

−L 2

L 2

∫ =1

4πε 0

λ x1

x2 + y2( )

32

dy−L 2

L 2

∫

Ex =1

4πε0

λxy

x2 x2 + y2

⎡

⎣⎢⎢

⎤

⎦⎥⎥−L 2

L 2

=1

4πε 0

λ

x

y

x2 + y2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−L 2

L 2

Ex =1

4πε0

λx

L 2

x2 + L 2( )2−

−L 2

x2 + −L 2( )2

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥ =

1

2πε 0

Lλ

x

1

4x2 + L2

SummarySummary How did symmetry help to reduce our calculations?

HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 20)For each of the following, complete 1 derivation with reasons for

each mathematical step and 3 additional derivations: (*Refer to rubric)Uniformly Charged Ring ( 0 2π) Uniformly Charged Vertical Wire (–∞ +∞) *Uniformly Charged Vertical Wire (–L/2 +L/2)

Web Assign 21.8 - 21.11

How did symmetry help to reduce our calculations?

HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 20)For each of the following, complete 1 derivation with reasons for

each mathematical step and 3 additional derivations: (*Refer to rubric)Uniformly Charged Ring ( 0 2π) Uniformly Charged Vertical Wire (–∞ +∞) *Uniformly Charged Vertical Wire (–L/2 +L/2)

Web Assign 21.8 - 21.11



Tuesday (Day 18)Tuesday (Day 18)

Warm-UpWarm-Up

Tues, Feb 17 Identify the parts of the derivation that were confusing.

Place your homework on my desk: “Foundational Mathematics’ Skills of Physics” Packet (Page 20) For each of the following, 1 derivation with reasons for each

mathematical step and 3 additional derivations: (*Refer to rubric)Uniformly Charged Ring ( 0 2π) Uniformly Charged Vertical Wire (–∞ +∞) *Uniformly Charged Vertical Wire (–L/2 +L/2)

Web Assign 21.8 - 21.11


Tues, Feb 17 Identify the parts of the derivation that were confusing.


mathematical step and 3 additional derivations: (*Refer to rubric)Uniformly Charged Ring ( 0 2π) Uniformly Charged Vertical Wire (–∞ +∞) *Uniformly Charged Vertical Wire (–L/2 +L/2)

Web Assign 21.8 - 21.11






















11

26 3 11 16 16 21

213

147 4 12 17 17 8

322

238 5 13 18 18 9

424

†129 6 14 19 19 10

5 15 10 7 15 20 20 11


AgendaAgenda


Calculate the electric field for continuous charge distributions for the following: †Uniformly Charged Horizontal Wire (dd+l) Uniformly Charged Disk (0 R)

Complete 4 Derivations + Reasoning for the above problems

Complete Web Assign Problems 21.8 - 21.11


Calculate the electric field for continuous charge distributions for the following: †Uniformly Charged Horizontal Wire (dd+l) Uniformly Charged Disk (0 R)




How do we apply integration and the Principle of Superposition to uniformly charged objects?How do we use integration and the principle of superposition to

calculate the electric field of a straight, uniform charge wire? How do we use integration and the principle of superposition to

calculate the electric field of a thin ring of charge on the axis of the ring?

How do we use integration and the principle of superposition to calculate the electric field of a semicircle of charge at its center?

How do we use integration and the principle of superposition to calculate the electric field of a uniformly charged disk on the axis of the disk?






















Graphical: Draw a picture of the object and 3-D plane. Label the partial length, area, or volume that is creating the partial E-field. Determine the distance from the charged object to the location of the desired E-Field

and label all components and lengths. Mathematical:


Write the total charge density and solve it for Q. Write the charge density in relation to the partial charge and solve it for the partial

charge (dq). Set up the integral by determining what key component(s) change. †Solve the integral and write the answer in a concise manner.


Graphical: Draw a picture of the object and 3-D plane. Label the partial length, area, or volume that is creating the partial E-field. Determine the distance from the charged object to the location of the desired E-Field

and label all components and lengths. Mathematical:


Write the total charge density and solve it for Q. Write the charge density in relation to the partial charge and solve it for the partial

charge (dq). Set up the integral by determining what key component(s) change. †Solve the integral and write the answer in a concise manner.


†Uniformly ChargedHorizontal Wire (dd+l)

†Uniformly ChargedHorizontal Wire (dd+l)

l

x

d

dx dEx

dE =1

4πε0

dqh2 ; dE =dEx =

1

4πε 0

1

x2 dq

λ =Q

l⇒ Q = λ l

λ =dq

dl=

dq

dx⇒ dq = λ dx

Ex = dEx0

Etot

∫ =

1

4πε 0

1

x2 dq=λ dx{0

Qtot

∫ =1

4πε 0

λ

x2 dxd

d+l

∫ =1

4πε 0

λ1

x2 dxd

d+l

∫

=1

4πε 0

λ1

x⎡⎣⎢⎤⎦⎥d+l

d

=1

4πε 0

λd + l − d

d d + l( )

⎡

⎣⎢

⎤

⎦⎥

Ex =1

4πε0

λ ld d+ l( )

⎡

⎣⎢

⎤

⎦⎥ =

1

4πε 0

Q

d d + l( )

⎡

⎣⎢

⎤

⎦⎥

Uniformly Charged Disk (0R)


R

dr

r

z

h = r2 + z2

dE

dEz=dE cosθ

dE =1

4πε0


z

h=

1

4πε 0

z

h3 dq

σ =Q

A=

Q

π r2 ⇒ Q = π r2σ

dA

dr=2πr ⇒ dA = 2π rdr

σ =dq

dA=

dq

2π rdr⇒ dq = 2π rσ dr

Ez = dEz0

Etot

∫ =

1

4πε 0

z

h3 dq=2π rσ dr

{0

Qtot

∫

=1

4πε 0

z

hz2 +r2

{3 2π rσ dr

0

R

∫

Ez =1

4πε0

2πσzr

z2 + r2( )32dr

0

R

∫ =1

2ε 0

σ z −1

z2 + r2

⎡

⎣⎢

⎤

⎦⎥

0

R

=1

2ε 0

σ z1

z2 + r2

⎡

⎣⎢

⎤

⎦⎥

R

0

Ez =12ε0

σz1z−

1

z2 +R2

⎡

⎣⎢

⎤

⎦⎥ =

σ2ε 0

z

z−

z

z2 + R2

⎡

⎣⎢

⎤

⎦⎥ =

σ2ε 0

1−z

z2 + R2

⎡

⎣⎢

⎤

⎦⎥

SummarySummary


HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 21) For each of the following, complete 1 derivation with reasons for each

mathematical step and 3 additional derivations: (*Refer to rubric) †Uniformly Charged Horizontal Wire (dd+l) Uniformly Charged Disk (0 R)

Web Assign 21.8 - 21.11


HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 21) For each of the following, complete 1 derivation with reasons for each

mathematical step and 3 additional derivations: (*Refer to rubric) †Uniformly Charged Horizontal Wire (dd+l) Uniformly Charged Disk (0 R)

Web Assign 21.8 - 21.11



Wednesday (Day 19)

Wednesday (Day 19)

Warm-UpWarm-Up

Wed, Feb 18 Identify the parts of the derivation that were confusing.

Place your homework on my desk:“Foundational Mathematics’ Skills of Physics” Packet (Page 21) For each of the following, 1 derivation with reasons for each

mathematical step and 3 additional derivations: (*Refer to rubric)†Uniformly Charged Horizontal Wire (dd+l) Uniformly Charged Disk (0 R)

Web Assign 21.8 - 21.11


Wed, Feb 18 Identify the parts of the derivation that were confusing.

Place your homework on my desk:“Foundational Mathematics’ Skills of Physics” Packet (Page 21) For each of the following, 1 derivation with reasons for each

mathematical step and 3 additional derivations: (*Refer to rubric)†Uniformly Charged Horizontal Wire (dd+l) Uniformly Charged Disk (0 R)

Web Assign 21.8 - 21.11






















11

26 3 11 16 16 21

213

147 4 12 17 17 8

322

238 5 13 18 18 9

424

†129 6 14 19 19 10

5 15 10 7 15 20 20 11


AgendaAgenda


Calculate the electric field for continuous charge distributions for the following:Uniformly Charged Disk (0 ∞) Uniformly Charged Hoop (R1 R2) Uniformly Charged Infinite Plate




Calculate the electric field for continuous charge distributions for the following:Uniformly Charged Disk (0 ∞) Uniformly Charged Hoop (R1 R2) Uniformly Charged Infinite Plate








































Uniformly Charged Disk (0∞)

Uniformly Charged Disk (0∞)

R

dr

r

z

h = r2 + z2

dE

dEz=dE cosθ

dE =1

4πε0


z

h=

1

4πε 0

z

h3 dq

σ =Q

A=

Q

π r2 ⇒ Q = π r2σ

dA


σ =dq

dA=

dq


Ez = dEz0

Etot

∫ =

1

4πε 0

z

h3 dq=2π rσ dr

{0

Qtot

∫

=1

4πε 0

z

hz2 +r2

{3 2π rσ dr

0

∞

∫

Ez =1

4πε0

2πσzh3

r

z2 + r2( )32dr

0

∞

∫ =1

2ε 0

σ z −1

z2 + r2

⎡

⎣⎢

⎤

⎦⎥

0

∞

=1

2ε 0

σ z1

z2 + r2

⎡

⎣⎢

⎤

⎦⎥

∞

0

Ez =12ε0

σz1z−

1

z2 +∞2

⎡

⎣⎢

⎤

⎦⎥ =

σ2ε 0

z

z⎡⎣⎢⎤⎦⎥=

σ2ε 0

Uniformly Charged Hoop (R1R2)

Uniformly Charged Hoop (R1R2)

R1

R2

dr

r

z

h = r2 + z2

dE

dEz=dE cosθ

dE =1

4πε0


z

h=

1

4πε 0

z

h3 dq

σ =Q

A=

Q

π r2 ⇒ Q = π r2σ

dA


σ =dq

dA=

dq


Ez = dEz0

Etot

∫ =

1

4πε 0

z

h3 dq=2π rσ dr

{0

Qtot

∫

=1

4πε 0

z

hz2 +r2

{3 2π rσ dr

R1

R2

∫

Ez =1

4πε0

2πσzh3

r

z2 + r2( )32dr

R1

R2

∫ =1

2ε 0

σ z −1

z2 + r2

⎡

⎣⎢

⎤

⎦⎥

R1

R2

=1

2ε 0

σ z1

z2 + r2

⎡

⎣⎢

⎤

⎦⎥

R2

R1

Ez =12ε0

σz1

z2 +R12−

1

z2 +R22

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

σ2ε 0

z

z2 + R12

−z

z2 + R22

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Uniformly ChargedInfinite Plate

Uniformly ChargedInfinite Plate

-∞ ∞-∞

∞

dxdyx

yr = x2 + y2

z h = r2 + z2

= x2 + y2 + z2

dE

dEz=dE cosθdE =

14πε0


z

h=

1

4πε 0

z

h3 dq

σ =Q

A=

Q

xy⇒ Q = σ xy

dA =dxdy

σ =dq

dA=

dq

dx dy⇒ dq = σ dx dy

Ez = dEz0

Etot

∫ =

1

4πε 0

z

h3 dq=σ dx dy{0

Qtot

∫

Ez =1

4πε0

zh

x2+y2+z2{

3 σ dx−∞

∞

∫−∞

∞

∫ dy=1

4πε 0

σ z1

x2 + y2 + z2( )

32

dx−∞

∞

∫ dy−∞

∞

∫

Ez =1

4πε0

σz2

y2 + z2dy

−∞

∞

∫ =1

4πε 0

σ z2π

z⎡⎣⎢

⎤⎦⎥=

σ2ε 0

No radius?!? What does that mean?!?

21.8 Field Lines

The electric field between two closely spaced, oppositely charged parallel plates is constant.


From the electric field due to a uniform sheet of charge, we can calculate what would happen if we put two oppositely-charged sheets next to each other:

The individual fields:The superposition:

The result:



SummarySummary How does the result of the Uniformly Charged Disk (0∞) & the Uniformly Charged

Infinite Plate compare? Why is this the case when one is circular and the other is a rectangle?

HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 8)

For each of the following, complete 1 derivation with reasons for each mathematical step and 3 additional derivations: (*Refer to rubric)*Uniformly Charged Disk (0∞) *Uniformly Charged Hoop (R1R2) Uniformly Charged Infinite Plate

Web Assign 21.8 - 21.11

How does the result of the Uniformly Charged Disk (0∞) & the Uniformly Charged Infinite Plate compare? Why is this the case when one is circular and the other is a rectangle?

HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 8)

For each of the following, complete 1 derivation with reasons for each mathematical step and 3 additional derivations: (*Refer to rubric)*Uniformly Charged Disk (0∞) *Uniformly Charged Hoop (R1R2) Uniformly Charged Infinite Plate

Web Assign 21.8 - 21.11




Warm-UpWarm-UpThurs, Feb 19

Identify the parts of the derivation that were confusing.


mathematical step and 3 additional derivations: (*Refer to rubric)*Uniformly Charged Disk (0 ∞) *Uniformly Charged Hoop (R1 R2) Uniformly Charged Infinite Plate

Web Assign 21.8 - 21.11


Thurs, Feb 19 Identify the parts of the derivation that were confusing.


mathematical step and 3 additional derivations: (*Refer to rubric)*Uniformly Charged Disk (0 ∞) *Uniformly Charged Hoop (R1 R2) Uniformly Charged Infinite Plate

Web Assign 21.8 - 21.11




















11

26 3 11 16 16 21

213

147 4 12 17 17 8

322

238 5 13 18 18 9

424

†129 6 14 19 19 10

5 15 10 7 15 20 20 11


AgendaAgenda


Work Day Web Assign Problems and Final Copy


Work Day Web Assign Problems and Final Copy

SummarySummary On the 3-2-1 Sheets, write down:

3 things you already knew about static electricity. 2 things that you learned about static electricity. 1 thing you would like to know about static electricity.

HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 9) Web Assign 21.8 - 21.11 Web Assign Final Copies

Future assignments: Web Assign Final Copies are due in 2 classes ALL “5 Derivations” are due in 2 classes

TOMORROW: CHAPTER 22: ELECTRIC FLUX & GAUSS’S LAW

On the 3-2-1 Sheets, write down: 3 things you already knew about static electricity. 2 things that you learned about static electricity. 1 thing you would like to know about static electricity.

HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 9) Web Assign 21.8 - 21.11 Web Assign Final Copies

Future assignments: Web Assign Final Copies are due in 2 classes ALL “5 Derivations” are due in 2 classes

TOMORROW: CHAPTER 22: ELECTRIC FLUX & GAUSS’S LAW



Summary of Chapter 21

• Charge is quantized in units of e

• Objects can be charged by conduction or induction

• Coulomb’s law:

• Electric field is force per unit charge:

Summary of Chapter 21

• Electric field of a point charge:

• Electric field can be represented by electric field lines

• Static electric field inside conductor is zero; surface field is perpendicular to surface

CHAPTER 21CHAPTER 21

- el fin -- el fin -

TEMPLATES FOR Derivations

TEMPLATES FOR Derivations



dE =1

4πε0


z

h=

1

4πε 0

z

h3 dq

σ =Q

A=

Q

π r2 ⇒ Q = π r2σ

dA


σ =dq

dA=

dq


Ez = dEz0

Etot

∫ =

1

4πε 0

z

h3 dq=2π rσ dr

{0

Qtot

∫

=1

4πε 0

z

hz2 +r2

{3 2π rσ dr

0

R

∫

Ez =1

4πε0

2πσzh3

r

z2 + r2( )32dr

0

R

∫ =1

2ε 0

σ z −1

z2 + r2

⎡

⎣⎢

⎤

⎦⎥

0

R

=1

2ε 0

σ z1

z2 + r2

⎡

⎣⎢

⎤

⎦⎥

R

0

Ez =12ε0

σz1z−

1

z2 +R2

⎡

⎣⎢

⎤

⎦⎥ =

σ2ε 0

z

z−

z

z2 + R2

⎡

⎣⎢

⎤

⎦⎥ =

σ2ε 0

1−z

z2 + R2

⎡

⎣⎢

⎤

⎦⎥

Uniformly Charged Ring (02π)

Uniformly Charged Ring (02π)

dE =1

4πε0

dqh2 ; dEz =dEcosφ=dE

z

h=

1

4πε 0

z

h3 dq

λ =Q

l=

Q

2π r=

Q

rθ⇒ Q = 2π rλ = λ rθ

λ =dq

dl=

dq

rdθ⇒ dq = λ rdθ

Ez = dEz0

Etot

∫ =

1

4πε 0

z

h3 dq=λ rdθ{0

Qtot

∫ =1

4πε 0

z

h3 λ r dθ0

2π

∫ =1

4πε 0

zλ r

h3 dθ0

2π

∫

=1

4πε 0

zλ r

h3 2π − 0( ) =

1

4πε 0

z

h3 λ 2π r=Q

{ =1

4πε 0

zQ

h3



dE =1

4πε0


x

h=

1

4πε 0

x

h3 dq

λ =Q

l=

Q

y⇒ Q = λ y

λ =dq

dl=

dq

dy⇒ dq = λ dy

Ex = dEx0

Etot

∫ =

1

4πε 0

x

h3 dq=λ dy{0

Qtot

∫

=1

4πε 0

x

hx2 +y2

{3 λ dy

−∞

∞

∫ =1

4πε 0

λ x1

x2 + y2( )

32

dy−∞

∞

∫

Ex =1

4πε0

λxy

x2 x2 + y2

⎡

⎣⎢⎢

⎤

⎦⎥⎥−∞

∞

=1

4πε 0

λ

x

y

x2 + y2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−∞

∞

=1

4πε 0

2λ

x

Ex =1

4πε0

2λx

yy

=1

4πε 0

2Q

xy



dE =1

4πε0


x

h=

1

4πε 0

x

h3 dq

λ =Q

l=

Q

y⇒ Q = λ y

λ =dq

dl=

dq

dy⇒ dq = λ dy

Ex = dEx0

Etot

∫ =

1

4πε 0

x

h3 dq=λ dy{0

Qtot

∫

=1

4πε 0

x

hx2 +y2

{3 λ dy

−∞

∞

∫ =1

4πε 0

λ x1

x2 + y2( )

32

dy−∞

∞

∫

Ex =1

4πε0

λxy

x2 x2 + y2

⎡

⎣⎢⎢

⎤

⎦⎥⎥−∞

∞

=1

4πε 0

λ

x

y

x2 + y2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−∞

∞

=1

4πε 0

2λ

x

Ex =1

4πε0

2λx

yy

=1

4πε 0

2Q

xy



dE =1

4πε0


x

h=

1

4πε 0

x

h3 dq

λ =Q

l=

Q

y⇒ Q = λ y

λ =dq

dl=

dq

dy⇒ dq = λ dy

Ex = dEx0

Etot

∫ =

1

4πε 0

x

h3 dq=λ dy{0

Qtot

∫

=1

4πε 0

x

hx2 +y2

{3 λ dy

−L 2

L 2

∫ =1

4πε 0

λ x1

x2 + y2( )

32

dy−L 2

L 2

∫

Ex =1

4πε0

λxy

x2 x2 + y2

⎡

⎣⎢⎢

⎤

⎦⎥⎥−L 2

L 2

=1

4πε 0

λ

x

y

x2 + y2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−L 2

L 2

Ex =1

4πε0

λx

L 2

x2 + L 2( )2−

−L 2

x2 + −L 2( )2

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥ =

1

4πε 0

2Lλ

x

1

4x2 + L2

Uniformly Charged Vertical Wire

Uniformly Charged Vertical Wire

Uniformly Charged Horizontal Wire

Uniformly Charged Horizontal Wire

Documents

Physics II: Electricity & Magnetism Binomial Expansions, Riemann Sums, Sections 21.6 to 21.11