Chapter 27. The Electric FieldThe Electric Field …electric field. Topics: • Electric Field...

Chapter 27. The Electric FieldThe Electric Field

Electric fields are

responsible for the electric

currents that flow through

your computer and the

nerves in your body. Electric

fields also line up polymer

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

fields also line up polymer

molecules to form the

images in a liquid crystal

display (LCD).

Chapter Goal: To learn how

to calculate and use the

electric field.

Topics:

• Electric Field Models

• The Electric Field of Multiple Point Charges

• The Electric Field of a Continuous Charge Distribution

• The Electric Fields of Rings, Disks, Planes, and

Chapter 27. The Electric FieldChapter 27. The Electric Field


• The Electric Fields of Rings, Disks, Planes, and Spheres

• The Parallel-Plate Capacitor

• Motion of a Charged Particle in an Electric Field

• Motion of a Dipole in an Electric Field

Chapter 27. Reading QuizzesChapter 27. Reading Quizzes


Chapter 27. Reading QuizzesChapter 27. Reading Quizzes

What device provides a practical

way to produce a uniform electric

field?


A. A long thin resistor

B. A Faraday cage

C. A parallel plate capacitor

D. A toroidal inductor

E. An electric field uniformizer

What device provides a practical

way to produce a uniform electric

field?


A. A long thin resistor

B. A Faraday cage

C. A parallel plate capacitor

D. A toroidal inductor

E. An electric field uniformizer

For charged particles, what is the

quantity q/m called?

A. Linear charge density

B. Charge-to-mass ratio



C. Charged mass density

D. Massive electric dipole

E. Quadrupole moment

For charged particles, what is the

quantity q/m called?

A. Linear charge density




C. Charged mass density

D. Massive electric dipole

E. Quadrupole moment

Which of these charge distributions did

not have its electric field determined in

Chapter 27?

A. A line of charge

B. A parallel-plate capacitor



C. A ring of charge

D. A plane of charge

E. They were all determined

Which of these charge distributions did

not have its electric field determined in

Chapter 27?

A. A line of charge




C. A ring of charge

D. A plane of charge

E. They were all determined

The worked examples of charged-

particle motion are relevant to

A. a transistor.


A. a transistor.

B. a cathode ray tube.

C. magnetic resonance imaging.

D. cosmic rays.

E. lasers.

The worked examples of charged-

particle motion are relevant to

A. a transistor.


A. a transistor.

B. a cathode ray tube.

C. magnetic resonance imaging.

D. cosmic rays.

E. lasers.

Chapter 27. Basic Content and ExamplesChapter 27. Basic Content and Examples


Chapter 27. Basic Content and ExamplesChapter 27. Basic Content and Examples

Electric Field ModelsElectric Field Models

The electric field of a point charge q at the origin, r = 0, is

where є0 = 8.85 × 10–12 C2/N m2 is the permittivity

constant.


constant.

The net electric field due to a group of point charges is

Electric Field ModelsElectric Field Models

where Ei is the field from point charge i.


ProblemProblem--Solving Strategy: The electric field of Solving Strategy: The electric field of

multiple point chargesmultiple point charges








Electric Dipole: Two Equal but Opposite Charges

Separated by a Small Distance


We can represent an electric dipole by two opposite charges

±q separated by the small distance s.

The dipole moment is defined as the vector

The Electric Field of a DipoleThe Electric Field of a Dipole

The dipole-moment magnitude p = qs determines the


The dipole-moment magnitude p = qs determines the

electric field strength. The SI units of the dipole moment

are C m.

The Electric Field of a Dipole

+

−+

−=+= −+ 2

2/2

2/0 )(

)(

)(4

1)()()(

sy

q

sy

qEEE yyydipole

πε

⋅+⋅−=

22

0 )5.0()5.0(

2

4)(

sysy

ysqE ydipole

πε

On the axis of an electric dipole


y>>s

3

0

2

4

1)(

y

qsE ydipole

πε≈

The electric field at a point on the axis of a dipole is

The Electric Field of a DipoleThe Electric Field of a Dipole

where r is the distance measured from the center of the

dipole.


dipole.

The electric field in the plane that bisects and is

perpendicular to the dipole is

This field is opposite to the dipole direction, and it is only

half the strength of the on-axis field at the same distance.


EXAMPLE 27.2 The electric field of a water EXAMPLE 27.2 The electric field of a water

moleculemolecule

QUESTION:



moleculemolecule



moleculemolecule


Tactics: Drawing and using electric field linesTactics: Drawing and using electric field lines


Tactics: Drawing and using electric field linesTactics: Drawing and using electric field lines


Continuous Charge Distribution: Charge Density

The total electric charge is Q.

Q

Q

Linear, length L

Volume V

Amount of charge in a

small volume dl:

Qdq dl dl

Lλλλλ= == == == =

Q

Lλλλλ ====

Linear charge density


small volume dV: Q


Q

Surface,

area A

Linear charge density


small volume dA:

Qdq dA dA

Aσσσσ= == == == =

Q

Aσσσσ ====

Surface charge density

small volume dV:

Qdq dV dV

Vρρρρ= == == == =

Q

Vρρρρ ====

Volume charge density

Electric Field: Continuous Charge Distribution

Procedure:

– Divide the charge distribution into small

elements, each of which contains ∆q

– Calculate the electric field due to one of

these elements at point P

– Evaluate the total field by summing the

contributions of all the charge elements


contributions of all the charge elements

Symmetry: take advantage of any symmetry to

simplify calculations

2ˆ

e

qk

r

∆∆ =E r

For the individual

charge elements

2 20ˆ ˆlim

i

ie i e

qi i

q dqk k

r r∆ →

∆= =∑ ∫E r r

Because the charge distribution

is continuous

Electric field due to three equal point charges?


Electric Field: Symmetry

1Errrr

2Errrr

Electric field

1 10q µC= 2 10q µC=

5m

5m

1 2E E E= += += += +r r rr r rr r rr r r2

ˆe

qE k r

r====

rrrr rrrr

Errrr 2

Errrr

ϕϕϕϕ

ϕϕϕϕ

12 cosE E φ=

1, 2,2 2z zE E E= =

1,zErrrr


++++

1 2

++++

6m

5m

++++++++

++++++++

1Errrr

1,zErrrr

Errrr

1, 24 4 coszE E E φ= =

1, 1 coszE E φ=

ϕϕϕϕ

ϕϕϕϕ

The symmetry gives us the direction

of resultant electric field

The Electric Field of a Continuous

Charge Distribution at a point in the plane that

bisects the rode



Charge Distribution


i

i

ixixir

QEE θ

πεθ cos

4

1cos)()(

2

0

∆==


Charge Distribution on a Thin Rod

i

i

ixixir

QEE θ

πεθ cos

4

1cos)()(

2

0

∆==

2222

04

1)(

ry

r

ry

QE

ii

xi

++

∆=

πε

1 Qr∆


2/322

0 )(4

1)(

ry

QrE

i

xi+

∆=

πε

∑+

∆=

i i

xry

yrLQE

2/322

0 )(4

/

πε

yLQyQ ∆=∆=∆ )/(λ

∫− +

=2/

2/

2/3220 )(4

/L

L

xry

rdyLQE

πε

How to evaluate this integral? ∫

− +=

2/

2/

2/3220 )(4

/L

L

xry

rdyLQE

πε-Trigonometric Substitution

http://en.wikipedia.org/wiki/Trigonometric_substitution


Electric Field due to a Thin Rod

∫− +

=2/

2/

2/3220 )(4

/L

L

xry

rdyLQE

πεhttp://integrals.wolfram.com/index.jsp

(to evaluate, x=a tanθ, dx=a sec2θ dθ

sinθ = x/(x2+a2)1/2

Use:


220 )2/(4

1

rLr

QEx

+=

πε

+−

−−

+=

+= −

22220

2/

2/22

0 )2/(

2/

)2/(

2/

4

/

4

/

rLr

L

rLr

LLQ

ryr

yLQE

L

Lxπεπε

sinθ = x/(x +a )

An infinite Line of Charge

220 )2/(4

1

rLr

QEx

+=

πε

220 )2/(4

1lim

rLr

QE

L

line

+=

∞→ πε


2/4

1

0 rL

QE line

πε=

rEline

λ

πε

2

4

1

0

=

negligible

An infinite Line of Charge

rEline

λ

πε

2

4

1

0

=

Unlike a point charge, for which the field decreases as


which the field decreases as 1/r2, the field of an infinitely long charged wire decreases more slowly – as only 1/r

Find electric field

Parallel-Plate Capacitor

0σ >

0σ− <

+ + + + ++ + + + +

+

− − − − −− − − − −

−


0σ >

E−

r

E+

r

02

σE

ε+ =

0σ− <

E+

r

E−

r

02

σE

ε− =

Find electric field

Er

Er


0σ >

0σ− <

+ + + + ++ + + + +

+

− − − − −− − − − −

−02

σE

ε+ =

02

σE

ε− =

E E E+ −= +r r r

0E E E= − =


0σ >E−

rE+

r

0σ− <E−

r

E+

r

E+

rE−

r

0E E E+ −= − =

0E E E+ −= − =

0

σE E E

ε+ −= + =


0σ >

0σ− <

+ + + + ++ + + + +

+

− − − − −− − − − −

−


0σ >

0σ− <

Er

0E =

0

σE

ε=

0E =

EXAMPLE 27.7 The electric field inside a EXAMPLE 27.7 The electric field inside a

capacitorcapacitor

QUESTIONS:



capacitorcapacitor



capacitorcapacitor



capacitorcapacitor


Electric Field: Continuous Charge Distribution on a ring

P

h

What is the electric field at point P ?


OR

o90

λλλλ - linear charge density

λλλλ - linear charge densityWhat is the electric field at point P ?

1. Symmetry determines the direction of the electric field.

Errrr


OR

o90

P

h


2. Divide the charge distribution into small elements, each

of which contains ∆q

Errrr

z

E∆∆∆∆rrrr

z

all elements all elements

E E E= ∆ = ∆= ∆ = ∆= ∆ = ∆= ∆ = ∆∑ ∑∑ ∑∑ ∑∑ ∑r r rr r rr r rr r r

2cos cos

z e

qE E k

rϕ ϕϕ ϕϕ ϕϕ ϕ

∆∆∆∆∆ = ∆ =∆ = ∆ =∆ = ∆ =∆ = ∆ =

z

all elements

E E= ∆= ∆= ∆= ∆∑∑∑∑

cosq

E k ϕϕϕϕ∆∆∆∆

==== ∑∑∑∑


OR

o90

P

h

q lλλλλ∆ = ∆∆ = ∆∆ = ∆∆ = ∆

l∆∆∆∆

ϕϕϕϕ

ϕϕϕϕ

E∆∆∆∆rrrr z

E∆∆∆∆rrrr

r

2cos

e

all elements

qE k

rϕϕϕϕ

∆∆∆∆==== ∑∑∑∑


3. Evaluate the total field by summing the contributions of

all the charge elements ∆q

Errrr

q l Rλ λ ψλ λ ψλ λ ψλ λ ψ∆ = ∆ = ∆∆ = ∆ = ∆∆ = ∆ = ∆∆ = ∆ = ∆

z

E∆∆∆∆rrrr

2cos

e

all elements

qE k

rϕϕϕϕ

∆∆∆∆==== ∑∑∑∑

l R ψψψψ∆ = ∆∆ = ∆∆ = ∆∆ = ∆

Rλ ψλ ψλ ψλ ψ∆∆∆∆∑∑∑∑


OR

o90

P

hl∆∆∆∆

ϕϕϕϕ

ϕϕϕϕ

E∆∆∆∆rrrr z

E∆∆∆∆rrrr

r

x

ψψψψψψψψ∆∆∆∆

2cos

e

all elements

RE k

r

λ ψλ ψλ ψλ ψϕϕϕϕ

∆∆∆∆==== ∑∑∑∑

Replace Sum by Integral

2

2

0

cose

RE d k

r

ππππ λλλλψ ϕψ ϕψ ϕψ ϕ==== ∫∫∫∫


4. Evaluate the integral

Errrr

z

E∆∆∆∆rrrr

2 2

2 2 2

0 0

cos cos 2 cose e e

R R RE d k k d k

r r r

π ππ ππ ππ πλ λ λλ λ λλ λ λλ λ λ

ψ ϕ ϕ ψ π ϕψ ϕ ϕ ψ π ϕψ ϕ ϕ ψ π ϕψ ϕ ϕ ψ π ϕ= = == = == = == = =∫ ∫∫ ∫∫ ∫∫ ∫

2 2cos

h h

r h Rϕϕϕϕ = == == == =

++++

2 2r h R= += += += +


OR

o90

P

hl∆∆∆∆

ϕϕϕϕ

ϕϕϕϕ

E∆∆∆∆rrrr z

E∆∆∆∆rrrr

r

x

ψψψψψψψψ∆∆∆∆

r h R++++

(((( ))))3/ 2

2 22

e

hRE k

h Rπλπλπλπλ====

++++

2/322

0 )(4

1

Rh

hQE

+=

πε

A Circular Disk of Charge

2/322

0 )(4

1

Rh

hQE

+=

πεElectric field due to a ring:

2R

Q

A

Q

πη ==

2/3220 )(4

1)(

i

izi

rz

QzE

+

∆=

πε

∑∑== +

∆==

N

i i

iN

i

zizdiskrz

QzEE

12/322

01 )(4)()(

πε

∑+

∆=

Ni

zdiskrz

rrzE

2/322 )(2)(

ε

η


∑= +

=i i

zdiskrz

E1

2/3220 )(2

)(ε

∫ +=

R

zdiskrz

rdrzE

0

2/322

0 )(2)(

ε

η

+−=

220

12

)(Rz

zE zdisk

ε

η

Quiz: If z>>R, then E = ?

Motion of a Charged Particle


Motion of Charged Particle

• When a charged particle is placed in an electric field, it experiences an

electrical force

• If this is the only force on the particle, it must be the net force

• The net force will cause the particle to accelerate according to Newton’s

second law

F qE====r rr rr rr r

- Coulomb’s law



F ma====rrrr rrrr

- Newton’s second law

- Coulomb’s law

qa E

m====

rrrrrrrr

Motion of Charged Particle

What is the final velocity?


F ma====rrrr rrrr

- Newton’s second law

- Coulomb’s law

| |y

qa E

m= −= −= −= −

fv


m

0

lt

v==== - travel time

Motion in x – with constant acceleration

0v

| |y

qa E

m= −= −= −= −

Motion in x – with constant velocity

After time t the velocity in y direction becomes

| |y y

qv a t Et

m= = −= = −= = −= = −

fv

0xv v====

yv

then

2

2

0f

qv v Et

m

= += += += +

Chapter 27. Summary SlidesChapter 27. Summary Slides


Chapter 27. Summary SlidesChapter 27. Summary Slides

General PrinciplesGeneral Principles


General PrinciplesGeneral Principles


ApplicationsApplications


ApplicationsApplications


Chapter 27. QuestionsChapter 27. Questions


Chapter 27. QuestionsChapter 27. Questions

At the

position of

the dot, the

electric field

points


A. Up.

B. Down.

C. Left.

D. Right.

E. The electric field is zero.

points

At the

position of

the dot, the

electric field

points


A. Up.

B. Down.

C. Left.

D. Right.

E. The electric field is zero.

points

A piece of plastic is uniformly charged with surface

charge density 1. The plastic is then broken into a large

piece with surface charge density 2 and a small piece

with surface charge density . Rank in order, from largest


with surface charge density 3. Rank in order, from largest

to smallest, the surface charge densities 1 to 3.

A. η2 = η3 > η1

B. η1 > η2 > η3

C. η1 > η2 = η3

D. η3 > η2 > η1

E. η1 = η2 = η3

A piece of plastic is uniformly charged with surface

charge density 1. The plastic is then broken into a large

piece with surface charge density 2 and a small piece

with surface charge density . Rank in order, from largest


with surface charge density 3. Rank in order, from largest

to smallest, the surface charge densities 1 to 3.

A. η2 = η3 > η1

B. η1 > η2 > η3

C. η1 > η2 = η3

D. η3 > η2 > η1

E. η1 = η2 = η3

Which of the following

actions will increase the

electric field strength at

the position of the dot?



A. Make the rod longer without changing the charge.

B. Make the rod fatter without changing the charge.

C. Make the rod shorter without changing the charge.

D. Remove charge from the rod.

E. Make the rod narrower without changing the charge.

Which of the following

actions will increase the

electric field strength at




A. Make the rod longer without changing the charge.

B. Make the rod fatter without changing the charge.

C. Make the rod shorter without changing the charge.

D. Remove charge from the rod.

E. Make the rod narrower without changing the charge.

Rank in order,

from largest to

smallest, the

electric field

strengths Ea to Ee

at these five points

near a plane of


near a plane of

charge.

A. Ea > Ec > Eb > Ee > Ed

B. Ea = Eb = Ec = Ed = Ee

C. Ea > Eb = Ec > Ed = Ee

D. Eb = Ec = Ed = Ee > Ea

E. Ee > Ed > Ec > Eb > Ea

Rank in order,

from largest to

smallest, the

electric field

strengths Ea to Ee

at these five points

near a plane of


A. Ea > Ec > Eb > Ee > Ed

B. Ea = Eb = Ec = Ed = Ee

C. Ea > Eb = Ec > Ed = Ee

D. Eb = Ec = Ed = Ee > Ea

E. Ee > Ed > Ec > Eb > Ea

near a plane of

charge.

Rank in order, from

largest to smallest, the

forces Fa to Fe a proton

would experience if

placed at points a – e in

this parallel-plate


this parallel-plate

capacitor.

A. Fa = Fb = Fd = Fe > Fc

B. Fa = Fb > Fc > Fd = Fe

C. Fa = Fb = Fc = Fd = Fe

D. Fe = Fd > Fc > Fa = Fb

E. Fe > Fd > Fc > Fb > Fa

Rank in order, from

largest to smallest, the

forces Fa to Fe a proton

would experience if

placed at points a – e in

this parallel-plate


A. Fa = Fb = Fd = Fe > Fc

B. Fa = Fb > Fc > Fd = Fe

C. Fa = Fb = Fc = Fd = Fe

D. Fe = Fd > Fc > Fa = Fb

E. Fe > Fd > Fc > Fb > Fa

this parallel-plate

capacitor.

Which electric field

is responsible for

the trajectory of the

proton?


Which electric field

is responsible for

the trajectory of the

proton?


Chapter 27. The Electric FieldThe Electric Field …electric field. Topics: • Electric Field...

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Transcript of Chapter 27. The Electric FieldThe Electric Field …electric field. Topics: • Electric Field...