physics ii: electricity & magnetism chapter 22. friday (day 1) add 21.9 figure: e-field in a...

Physics II:Electricity & Magnetism

Physics II:Electricity & Magnetism

Chapter 22Chapter 22

Friday (Day 1)Friday (Day 1)

Add 21.9 Figure: E-field in a conductor.

Fix derivations for uniform charge density (dr is wrong, etc)

Add EM Field Activities

Add 21.9 Figure: E-field in a conductor.

Fix derivations for uniform charge density (dr is wrong, etc)

Add EM Field Activities

Warm-UpWarm-Up

Fri, Feb 20

Write down the steps you give a visually-disabled individual to explain to how to fill a coffee cup with water.

Place your homework on my desk: “Foundational Mathematics’ Skills of Physics” Packet (Page 9)

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

Fri, Feb 20

Write down the steps you give a visually-disabled individual to explain to how to fill a coffee cup with water.

Place your homework on my desk: “Foundational Mathematics’ Skills of Physics” Packet (Page 9)


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 concept of electric field?How do we describe and apply the electric field created by uniformly

charged objects?How do we describe and apply the relationship between the electric

field and electric flux?How do we describe and apply Gauss’s Law?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?

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 concept of electric field?How do we describe and apply the electric field created by uniformly





VocabularyVocabulary

FluxElectric FluxFlux LinesSurface AreaPerpendicularOrthogonalNormalDot Product


Gauss’s LawCharge Density


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 9) with answer guide.

Review electric fields, electrons, and conductorsDiscuss the following:

Electric Flux

Work Day:Chapter 21 Web Assign ProblemsUniform Charge Distribution DerivationsChapter 22 Web Assign Problems 22.1 - 22.4


Review electric fields, electrons, and conductorsDiscuss the following:

Electric Flux

Work Day:Chapter 21 Web Assign ProblemsUniform Charge Distribution DerivationsChapter 22 Web Assign Problems 22.1 - 22.4

A Review of Electric Fields and Conductors

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

Chapter 22

Gauss’s Law

Main Points of Chapter 22

• Electric flux

• Gauss’ law

• Using Gauss’ law to determine electric fields

• Conductors and electric fields

• Testing Gauss’ and Coulomb’s laws

Section 22.1Section 22.1

Given a diagram where the electric field is represented by flux lines, how do wedetermine the direction of the field at a given point? identify the locations where the field is strong and where it is

weak? identify where the positive and negative charges must be

present?

How do we use the relationship between the electric field and electric flux to calculate the flux of a uniform electric field E through an arbitrary surface?

Given a diagram where the electric field is represented by flux lines, how do wedetermine the direction of the field at a given point? identify the locations where the field is strong and where it is

weak? identify where the positive and negative charges must be

present?

How do we use the relationship between the electric field and electric flux to calculate the flux of a uniform electric field E through an arbitrary surface?


How do we use the relationship between the electric field and electric flux to calculate the flux of a uniform electric field E through a curved surface when E is uniform in magnitude and perpendicular to the surface?

How do we use the relationship between the electric field and electric flux to calculate the flux of a uniform electric field E through a rectangle when E is perpendicular to the rectangle and a function of one coordinate only?

How do we state and apply the relationship of between electric flux and lines of force?

How do we use the relationship between the electric field and electric flux to calculate the flux of a uniform electric field E through a curved surface when E is uniform in magnitude and perpendicular to the surface?

How do we use the relationship between the electric field and electric flux to calculate the flux of a uniform electric field E through a rectangle when E is perpendicular to the rectangle and a function of one coordinate only?

How do we state and apply the relationship of between electric flux and lines of force?

22.1 Electric Flux

Electric flux:

Electric flux through an area is proportional to the total number of field lines crossing the area.

A

ΦE = E Acosθ

ΦE = E⊥A =E A⊥

A

22.1 Electric Flux

Electric flux:

ΦE = EgA

ΦE = E Acosθ

When = 0°, the flux (field lines) passing through the area is maximized and when = 90°, the flux (field lines) is zero. Therefore, for mathematical simplicity, it is important to note that the area A of a surface will represented by a vector A whose magnitude is A but whose direction is perpendicular to it surface.

A

22.1 Electric Flux

Flux through a closed surface:

EM Field 6EM Field 6

Using EM Field, determine the flux through a closed surface (represented by a 2-D cross-section) by counting the number of field lines entering the area and subtracting it from the number of field lines exiting the area.This is because the area vector A for a

closed surface is always directed outward. i.e. cos (0°) = 1 and cos (180°) = -1

Using EM Field, determine the flux through a closed surface (represented by a 2-D cross-section) by counting the number of field lines entering the area and subtracting it from the number of field lines exiting the area.This is because the area vector A for a

closed surface is always directed outward. i.e. cos (0°) = 1 and cos (180°) = -1


How can we draw a surface that has more field lines entering or exiting the closed surface?

How can we draw a surface that has more field lines entering or exiting the closed surface?


The following are examples of the Flux calculations using EM Field 6.

Besides its relation to the number of field lines entering and exiting the closed surface, how does the flux calculation relate to charge?

The following are examples of the Flux calculations using EM Field 6.

Besides its relation to the number of field lines entering and exiting the closed surface, how does the flux calculation relate to charge?

SummarySummary

How does flux relate to the charge enclosed by a closed surface?

HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 10) The final copies for all Uniformly Charged Objects: 1 derivation with

reasons for each mathematical step and 3 additional derivations: (*Refer to rubric)

Web Assign Final Copies: Chapter 21 Web Assign Problems: Chapter 22.1 & 22.2

How does flux relate to the charge enclosed by a closed surface?

HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 10) The final copies for all Uniformly Charged Objects: 1 derivation with

reasons for each mathematical step and 3 additional derivations: (*Refer to rubric)

Web Assign Final Copies: Chapter 21 Web Assign Problems: Chapter 22.1 & 22.2

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?

Monday (Day 2)Monday (Day 2)

Warm-UpWarm-UpMon, Feb 23

For an electric field raining straight down into an imaginary box . . . What is the direction of E? What is the direction of A for the (1) bottom, (2) top, and (3) side surfaces? What is the angle, , between E and A for the (1) bottom, (2) top, and (3) side

surfaces? What is value of cos for the (1) bottom, (2) top, and (3) side surfaces?

Place your homework on my desk: “Foundational Mathematics’ Skills of Physics” Packet (Page 10) - POSTPONED The final copies for all Uniformly Charged Objects: 1 derivation with reasons for

each mathematical step and 3 additional derivations: (*Refer to rubric) Web Assign Final Copies: Chapter 21


Mon, Feb 23 For an electric field raining straight down into an imaginary box . . .

What is the direction of E? What is the direction of A for the (1) bottom, (2) top, and (3) side surfaces? What is the angle, , between E and A for the (1) bottom, (2) top, and (3) side

surfaces? What is value of cos for the (1) bottom, (2) top, and (3) side surfaces?

Place your homework on my desk: “Foundational Mathematics’ Skills of Physics” Packet (Page 10) - POSTPONED The final copies for all Uniformly Charged Objects: 1 derivation with reasons for

each mathematical step and 3 additional derivations: (*Refer to rubric) Web Assign Final Copies: Chapter 21


Application:Electric Flux


Warm-up: What is the direction of E? What is the direction of A

for the (1) bottom, (2) top, and (3) side surfaces?

What is the angle, , between E and A for the (1) bottom, (2) top, and (3) side surfaces?

What is value of cos for the (1) bottom, (2) top, and (3) side surfaces?

Warm-up: What is the direction of E? What is the direction of A

for the (1) bottom, (2) top, and (3) side surfaces?

What is the angle, , between E and A for the (1) bottom, (2) top, and (3) side surfaces?

What is value of cos for the (1) bottom, (2) top, and (3) side surfaces?

E

A

E



Eout

Abottom

ΦEbottom= Eout Abottom cosθ

Abottom

The bottom square flux is

ΦE =EAcos 0°( )= EA

Ein

Atop

ΦEtop= Ein Atop cosθ

Atop

= −EA

ΦEtop= Ein Aside cosθ

Aside = 0

The net flux, Φnet, isΦnet = Φ top +Φbottom + Φsides

Φnet = −EA + EA + 0 = 0




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 10) with answer guide. - POSTPONED

Review electric fluxDiscuss the following:

Gauss’s Law Applications of Gauss’s Law

Spherical Conductor, Point Charge, & Line Charges

Work Time: Web Assign Problems 22.1 - 22.5

Review “Foundational Mathematics’ Skills of Physics” Packet (Page 10) with answer guide. - POSTPONED

Review electric fluxDiscuss the following:

Gauss’s Law Applications of Gauss’s Law

Spherical Conductor, Point Charge, & Line Charges



How do we state Gauss’s Law in integral form and apply it qualitatively to relate electric flux and electric charge for a specified surface?

How do we state Gauss’s Law in integral form and apply it qualitatively to relate electric flux and electric charge for a specified surface?

22-2 What Does Gauss’ Law Do?

Imagine field lines emanating from a positive charge.

Now imagine a sphere of tissue paper around the charge. How many field lines penetrate the tissue? It doesn’t really matter how many we draw in the first place, as long as we are consistent; they all go through.

Now imagine the charge being off-center; all the lines still go through:


Suppose the tissue is some shape other than spherical, but still surrounds the charge.

All the field lines still go through:

Now, imagine the paper is crinkled and overlaps itself; how shall we deal with the lines that pierce the tissue three times?

Notice that they go out twice and in once – if we subtract the “ins” from the “outs” we are left with one line going out, which is consistent with the other situations.


Now, look at an open (flat) sheet. If it is perpendicular to the field, the maximum number of lines penetrates:

If it is at an angle, fewer lines penetrate:


The number of field lines piercing the surface is proportional to the surface area, the orientation, and the field strength. If we stop counting lines and just use the field strength itself, we can define the electric flux through an infinitesimal area:

Integrating gives the total flux:


For a closed surface, we can uniquely define the direction of the normal to the surface as pointing outwards and define:

Then:

Note the circle on the integral sign, which means that the integration is over a closed surface.


Note that the surface does not have to be made of real matter – it is a surface that we can imagine, but that does not have to exist in reality.

This kind of imaginary surface is called a Gaussian surface. We can imagine it to be any shape we want; it is very useful to choose one that makes the problem you are trying to solve as easy as possible.

22-2 Gauss’ Law

The electric flux through a closed surface that encloses a net charge is equal to the net

charge divided by the permittivity of free space.

Without further ado, we can state Gauss’ law:

22-2 Gauss’ Law

The electric flux through a closed surface that encloses no net charge is zero.

22.2 Gauss’s Law

Restating Gauss’ Law, the net number of field lines through the surface is proportional to the charge enclosed, and also to the flux:

Gauss’ Law can easily be used to find the electric field in situations with a high degree of symmetry.

ΦE = E⋅dA—∫ =

Qencl

ε0

E-Field for a Sphere or Point Charge

E-Field for a Sphere or Point Charge

Spherical ConductorSpherical Conductor

+

+

+

++

+

+

+

E

A1

r1

At radius r1 r > r0( ),

E⋅dA—∫

dAsphere

=EA1 cosθ =EA1=E 4π r2( ) =

Qencl

ε 0

Therefore, Eoutside =1

4πε0

Qr12

To calculate the electric field inside of a

conductor, we look at radius r2 r < r0( ),

r0

A2

r2

E⋅dA—∫ =EA2 cosθ =EA2=E 4π r2

( ) =Qencl

ε 0Because the enclosed charge inside

of a conductor is zero, Einside =0

22-2 Gauss’ Law for a Point Charge

The electric field through a Gaussian sphere with a single point charge at the center is easily calculated using:

ΦE = E ⋅dA—∫ =

Qencl

ε 0

ΦE = E 4π r2( ) =

Qencl

ε 0

Recall that F =qE.Therefore,

F =q21

4πε0

q1

r2

⎛

⎝⎜⎞

⎠⎟⇒ F =

1

4πε 0

q1q2

r2

This is Coulomb's Law!

22-2 Gauss’ LawBut the result would be the same if the surface was not spherical, or if the charge was anywhere inside it!

Therefore, we can quickly generalize this to any surface and any charge distribution; all can be considered as a collection of point or infinitesimal charges:

Here, Q is the total net charge enclosed by the surface.

Applications of Gauss’ LawApplications of Gauss’ Law

22-3 Using Gauss’ Law to Determine Electric Fields

Problem-solving techniques:

1. Make a sketch.

2. Identify any symmetries.

3. Choose a Gaussian surface that matches the symmetry – that is, the electric field is either parallel to the surface or constant and perpendicular to it.

4. The correct choice in 3 should allow you to get the field outside the integral. Then solve.


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

How do we describe the electric field of a thin spherical shell?

How do we use superposition to determine the electric fields of coaxial cylinders?

How do we use superposition to determine the electric fields of concentric spheres?

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

How do we describe the electric field of a thin spherical shell?

How do we use superposition to determine the electric fields of coaxial cylinders?

How do we use superposition to determine the electric fields of concentric spheres?


How do we apply Gauss’s Law, along with symmetry arguments, to determine the electric field inside a uniformly charged long cylinder? outside a uniformly charged long cylinder? inside a uniformly charged cylindrical shell? outside a uniformly charged cylindrical shell? inside a uniformly charged sphere? outside a uniformly charged sphere? inside a uniformly charged spherical shell?

outside a uniformly charged spherical shell?

How do we apply Gauss’s Law, along with symmetry arguments, to determine the electric field inside a uniformly charged long cylinder? outside a uniformly charged long cylinder? inside a uniformly charged cylindrical shell? outside a uniformly charged cylindrical shell? inside a uniformly charged sphere? outside a uniformly charged sphere? inside a uniformly charged spherical shell?

outside a uniformly charged spherical shell?

Section 22.3Section 22.3 How do we apply Gauss’s Law to determine the charge density on a surface in

terms of the electric field near the surface? How do we apply Gauss’s Law to determine the total charge on a surface in

terms of the electric field near the surface? How do we prove and apply the relationship between the surface charge

density on a conductor and the electric field strength near its surface? How do we qualitatively explain why there can be no electric field in a

charge-free region completely surrounded by a single conductor? How do we qualitatively explain why the electric field outside of a closed

conducting surface cannot depend on the precise location of charge in the space enclosed by the conductor?

What is the significance of why the electric field outside of a closed conducting surface cannot depend on the precise location of charge in the space enclosed by the conductor?

How do we apply Gauss’s Law to determine the charge density on a surface in terms of the electric field near the surface?

How do we apply Gauss’s Law to determine the total charge on a surface in terms of the electric field near the surface?

How do we prove and apply the relationship between the surface charge density on a conductor and the electric field strength near its surface?

How do we qualitatively explain why there can be no electric field in a charge-free region completely surrounded by a single conductor?

How do we qualitatively explain why the electric field outside of a closed conducting surface cannot depend on the precise location of charge in the space enclosed by the conductor?


E-Field for a Line of Charge

E-Field for a Line of Charge

Long Uniform Line of Charge

Long Uniform Line of Charge

E

E

E⋅dA—∫ =EAtop

dAtop

+EAtube

dAtube

+EAbottom

dAbottom

Note: There is no electric field lines

going through the top or bottom of

the cylinder.

=E Atube =E 2π rL( )

=Qencl

ε 0

E 2πrL( ) =Qencl

ε0

⇒ Ewire =1

2πε 0

Q

rL

Uniformly ChargedVertical Wire (–∞+∞)

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=

1

2πε 0

Q

rLNote: "r" replaced "x" and "L" replaced "y"( )

Do they agree?

SummarySummary

What does the closed integral sign mean in Gauss’ Law?

HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 10)†“Foundational Mathematics’ Skills of Physics” Packet (Page 11)Web Assign 22.1 - 22.5

What does the closed integral sign mean in Gauss’ Law?

HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 10)†“Foundational Mathematics’ Skills of Physics” Packet (Page 11)Web Assign 22.1 - 22.5



Tuesday (Day 3)

Tuesday (Day 3)

Warm-UpWarm-Up

Tues, Feb 24

Calculate the electric flux for a charged object on the following slide.

Place your homework on my desk: “Foundational Mathematics’ Skills of Physics” Packet (Page 10) †“Foundational Mathematics’ Skills of Physics” Packet (Page 11) -

POSTPONED Web Assign Problems 22.1 - 22.5


Tues, Feb 24

Calculate the electric flux for a charged object on the following slide.

Place your homework on my desk: “Foundational Mathematics’ Skills of Physics” Packet (Page 10) †“Foundational Mathematics’ Skills of Physics” Packet (Page 11) -

POSTPONED Web Assign Problems 22.1 - 22.5


Application: Electric FluxApplication: Electric Flux

Two objects, O1 and O2, have charges +1 C and -2.0 C, respectively, and a third object, O3, is electrically neutral. (a) What is the electric flux through the surface A1 that encloses all three objects?(b) What is the electric flux through the surface A2 that encloses the third object only?

ΦA1

= E ⋅d—∫ A =Qencl

ε 0

=1.0μC + −2.0μC( )( )

8.85 x 10−12 C2 N ⋅m2 =1.13 x 105 N ⋅m2 C

ΦA2

= E ⋅d—∫ A=Qencl

ε 0

=0 μC( )

8.85 x 10−12 C2 N ⋅m2 =0 N ⋅m2 C




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 the following:Applications of Gauss’ Law

Charged Plate(s), Charged Surfaces, & Charged Spherical Insulators

Conductors in Electric FieldsExperimental Basis of Gauss’ and Coulomb’s Law



Discuss the following:Applications of Gauss’ Law

Charged Plate(s), Charged Surfaces, & Charged Spherical Insulators

Conductors in Electric FieldsExperimental Basis of Gauss’ and Coulomb’s Law



How do we apply Gauss’s Law, along with symmetry arguments, to determine the electric field? near a large, uniformly charged plane?

How do we apply Gauss’s Law, along with symmetry arguments, to determine the electric field? near a large, uniformly charged plane?

Section 22.3Section 22.3 How do we apply Gauss’s Law to determine the charge density on a surface in

terms of the electric field near the surface? How do we apply Gauss’s Law to determine the total charge on a surface in

terms of the electric field near the surface? How do we prove and apply the relationship between the surface charge

density on a conductor and the electric field strength near its surface? How do we qualitatively explain why there can be no electric field in a

charge-free region completely surrounded by a single conductor? How do we qualitatively explain why the electric field outside of a closed

conducting surface cannot depend on the precise location of charge in the space enclosed by the conductor?


How do we apply Gauss’s Law to determine the charge density on a surface in terms of the electric field near the surface?

How do we apply Gauss’s Law to determine the total charge on a surface in terms of the electric field near the surface?

How do we prove and apply the relationship between the surface charge density on a conductor and the electric field strength near its surface?

How do we qualitatively explain why there can be no electric field in a charge-free region completely surrounded by a single conductor?

How do we qualitatively explain why the electric field outside of a closed conducting surface cannot depend on the precise location of charge in the space enclosed by the conductor?


Infinite Plane of Charge(nonconducting uniform )


EA

E

E⋅dA—∫

dA

dA

=Qencl

ε 0

=A

ε 0

Also Note: The electric field, E,

is in the direction of dA and exists

on both sides of the charged plate.

Therefore,

Anet = Atop + Abottom = 2Acircle

E 2A( ) =Aε0

⇒ E =σ

2ε 0

for each plate

Enet = En∑ =E1 + E2=

2ε 0

+σ

2ε 0

=ε0

dAtube



EA

E

E⋅dA—∫

dA

dA

=Qencl

ε 0

=A

ε 0

Also Note: The electric field, E,

is in the direction of dA and exists

on both sides of the charged plate.

Therefore, dAtube

ΦE = E ⋅dA

=EAtop

{top∫ + E ⋅dA

=0{

tube∫ + E ⋅dA

=EAbottom

{bottom∫

ΦE = 2EAcircle

2EA =Aε0

⇒ E =σ

2ε 0

for each side

For two oppositely charged plates:

Enet = En∑ =E1 + E2 =

2ε 0

+σ

2ε 0

=ε0

Uniformly Charged Disk (0∞)

Uniformly Charged Disk (0∞)

R

dr

r

z

h = r2 + z2

dE

dEz=dE cosθ

dE =1

4πε0

dqh2 ; dEz =dEcos =dE

z

h=

1

4πε 0

z

h3 dq

=Q

A=

Q

π r2 ⇒ Q = π r2σ

dA

dr=2πr ⇒ dA = 2π rdr

=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

∞

∫

Ez =1

4πε0

2πzr

z2 + r2( )32dr

0

∞

∫ =1

2ε 0

σ z −1

z2 + r2

⎡

⎣⎢

⎤

⎦⎥

0

∞

=1

2ε 0

σ z1

z2 + r2

⎡

⎣⎢

⎤

⎦⎥

∞

0

Ez =12ε0

z1z−

1

z2 +∞2

⎡

⎣⎢

⎤

⎦⎥ =

2ε 0

z

z⎡⎣⎢

⎤⎦⎥=

2ε 0

Uniformly ChargedInfinite Plate

Uniformly ChargedInfinite Plate

-∞ ∞-∞

∞

dxdyx

yr = x2 + y2

z h = r2 + z2

= x2 + y2 + z2

dE

dEz=dE cosθdE =

14πε0

dqh2 ; dEz =dEcos =dE

z

h=

1

4πε 0

z

h3 dq

=Q

A=

Q

xy⇒ Q = σ xy

dA =dxdy

=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 dx−∞

∞

∫−∞

∞

∫ dy=1

4πε 0

σ z1

x2 + y2 + z2( )

32

dx−∞

∞

∫ dy−∞

∞

∫

Ez =1

4πε0

z2

y2 + z2dy

−∞

∞

∫ =1

4πε 0

σ z2π

z⎡⎣⎢

⎤⎦⎥=

2ε 0

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

21-7 The Field of a Continuous Distribution

From the electric field due to a nonconducting 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:



21.8 Field Lines

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

E-Field at the Surface of a Conductor

E-Field at the Surface of a Conductor

Surface of a ConductorSurface of a Conductor

+

+ +

+ + +

+ + + +

+ + +

+ +

+

E

A

E⋅dA—∫

dA

=Qencl

ε 0

=A

ε 0

Note: Since we are Gaussian surface

is just below the surface of the

conductor, the electric field, E,

is only in the direction of dA and

does not exist within the conductor.

= EAcircle

EA =Aε0

⇒ E =σ

ε 0

for the surface

dAtube

So why are the E-Fields different for the plate and

the surface?

So why are the E-Fields different for the plate and

the surface?Nonconducting: Charge remains

localized (ie. +2Q remains fixed)Conducting: Like charges repel (ie.

+2Q total -> +Q move to each side and conduct=1/2 insulator )

Therefore, both E-fields will have the same magnitude and direction.

Nonconducting: Charge remains localized (ie. +2Q remains fixed)

Conducting: Like charges repel (ie. +2Q total -> +Q move to each side and conduct=1/2 insulator )

Therefore, both E-fields will have the same magnitude and direction.


Find the field lines for:2 Parallel Oppositely Charged Plates2 Parallel Plates of the Same Charge

Find the field lines for:2 Parallel Oppositely Charged Plates2 Parallel Plates of the Same Charge

Spherical Insulator (Uniformly distributed charge)

Spherical Insulator (Uniformly distributed charge)

++

+

+

++

++

E

A1

r1

At radius r1 r > r0( ),

E⋅dA—∫

dA1= dAsphere


Qencl

ε 0

Therefore, Eoutside =1

4πε0

Qr12

At radius r2 r < r0( ),r0

A2

r2

E⋅dA—∫

+

+

+

++ +

+

+

+dA2


Qencl

ε 0Note: Since the enclosed charge density is

evenly distributed, ρE =Qtot

Vtot

=dQ

dV=constant

Then, ρE =Qtot

Vtot

=Qencl

Vencl

⇒ Qencl =Vencl

Vtot

Qtot =r3

r03 Qtot

E⋅dA—∫ =E 4π r2

( )=Qencl

ε 0

=r3

r03

Qtot

ε 0

Therefore, Einside =1

4πε0

Qtot

r03 r

22-3 Conductors and Electric Fields

In conductors, the charges are free to move if there is an external electric field exerting a force on them.

• Therefore, in equilibrium, there is no static field inside a conductor. This also means that the external electric field is perpendicular to the conductor at its surface.


What if the conductor is charged – where does the excess charge go?

• By making a Gaussian surface very close to, but just under, the surface of the conductor, we see that any excess charge must lie on the outside of the conductor.


What if there is a cavity inside the conductor, and that cavity has charges in it?

The field inside the conductor must still be zero, so charges will be induced on the inner surface of the cavity and the outer surface of the conductor:

22-3 Conductors and Electric FieldsElectrostatic Fields Near Conductors

Looking at the electrostatic field very near a conductor, we find:

and therefore:

The electric field is perpendicular to the surface, and where the charge density is higher, the field is larger.


1. The electrostatic field inside a conductor is zero.

2. The electrostatic field immediately outside a conductor is perpendicular to the surface and has the value σ/ε0 where σ is the local surface charge density.

3. A conductor in electrostatic equilibrium—even one that contains nonconducting cavities—can have charge only on its outer surface, as long as the cavities contain no net charge. If there is a net charge within the cavity, then an equal and opposite charge will be distributed on the surface of the conductor that surrounds the cavity.

To summarize:

22-4 Are Gauss’ and Coulomb’s Laws Correct?

An experiment to validate Gauss’ law (that there is no charge within a conductor) can be done as follows:

Need a hollow conducing sphere, a small conducting ball on an insulating rod, and an electroscope attached to the surface of the conductor.


Charge the small sphere and hold it inside the shell without touching. Induced charge will be on outside of shell and on electroscope.

Now touch the inside of the shell with the small sphere. Charge will flow onto it until it is neutral, leaving the shell with a net positive charge.


Finally, remove the rod. The electroscope leaves do not move, indicating that the excess charge resides on the outside of the shell.


This table shows the results of such experiments looking for a deviation from an inverse-square law:


One problem with the above experiments is that they have all been done at short range, 1 meter or so.

Other experiments, more sensitive to cosmic-scale distances, have been done, testing whether Coulomb’s law has the form:

No evidence for a nonzero μ has been found.

Summary of Chapter 22

• Electric flux due to field intersecting a surface S:

• Gauss’ law relates flux through a closed surface to charge enclosed:

• Can use Gauss’ law to find electric field in situations with a high degree of symmetry

ΦS = E ⋅dAS∫

Summary of Chapter 22

• Electric flux:

• Gauss’s law:

E⋅dA—∫ =

Qencl

ε0

Summary of Chapter 22 (con’t)

• Properties of conductors:

1. Electric field is zero inside

2. Field just outside conductor is perpendicular to surface

3. Excess charge resides on the outside of a conductor, unless there is a nonconducting cavity in it; in that case, there is an induced charge on both surfaces

• Gauss’ law has been verified to a very high degree of accuracy

SummarySummary

In a conductor, what happens to an electron in the presence of an electric field?

What happens to the electric field in the presence of an conductor?

HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 10)†“Foundational Mathematics’ Skills of Physics” Packet (Page 11)

LAST ONE!!!!!!!!

Web Assign 22.6 - 22.15

In a conductor, what happens to an electron in the presence of an electric field?

What happens to the electric field in the presence of an conductor?

HW (Place in your agenda): “Foundational Mathematics’ Skills of Physics” Packet (Page 10)†“Foundational Mathematics’ Skills of Physics” Packet (Page 11)

LAST ONE!!!!!!!!

Web Assign 22.6 - 22.15



Wednesday (Day 4)

Wednesday (Day 4)

Warm-UpWarm-Up

Wed, Feb 25

Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain a uniform surface charge densities 1 and 2, respectively. Determine the electric field for (a) r < r1, (b) r1 < r < r2, and (c) r > r2. (d) Under what conditions will E = 0 for r > r2? (e) Under what conditions will E = 0 for r1 < r < r2?

Place your homework on my desk: †“Foundational Mathematics’ Skills of Physics” Packet (Page 11) Web Assign Problems 22.6 - 22.15


Wed, Feb 25

Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain a uniform surface charge densities 1 and 2, respectively. Determine the electric field for (a) r < r1, (b) r1 < r < r2, and (c) r > r2. (d) Under what conditions will E = 0 for r > r2? (e) Under what conditions will E = 0 for r1 < r < r2?

Place your homework on my desk: †“Foundational Mathematics’ Skills of Physics” Packet (Page 11) Web Assign Problems 22.6 - 22.15


Essential Question(s)Essential Question(s)


How do we describe and apply the concept of electric field?How do we describe and apply the electric field created by

uniformly charged objects?How do we describe and apply the relationship between the

electric field and electric flux?How do we describe and apply Gauss’s Law?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?







AgendaAgenda


Complete the Gauss’ Law lab using EM Field 6Complete Web Assign Problems 22.6 - 22.15



SummarySummary

What did you learn about drawing Gaussian surfaces when using EM Field?

HW (Place in your agenda): Electrostatics Lab #4 Report (Due in 5 Classes) Web Assign 22.6 - 22.15

What did you learn about drawing Gaussian surfaces when using EM Field?




Thursday (Day 5)

Thursday (Day 5)

Warm-UpWarm-Up

Thurs, Feb 26

Complete Chapter 22 Graphic Organizers

Place your homework on your desk: Web Assign Problems 22.6 - 22.15


Thurs, Feb 26

Complete Chapter 22 Graphic Organizers

Place your homework on your desk: Web Assign Problems 22.6 - 22.15


Essential Question(s)Essential Question(s)













AgendaAgenda



SummarySummary

Would Gauss’ Law be helpful in determining the electric field due to an electric dipole?


Would Gauss’ Law be helpful in determining the electric field due to an electric dipole?




physics ii: electricity & magnetism chapter 22. friday (day 1) add 21.9 figure: e-field in a...

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