ap lecture ch 16

8/3/2019 AP Lecture Ch 16

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January 19, 2012

Notes on Electric Charge (R2)

Question of the Day (L1):

Explain why the balloons repel each other?


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Chapter 16

Electric Charge andElectric Field


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16.1 Static Electricity;

Electric Charge and Its

Conservation

Charge comes in two

types, positive andnegative; like charges

repel and opposite

charges attract


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16.1 Static Electricity; Electric Charge

and Its Conservation

Electric charge is conserved the

arithmetic sum of the total charge cannotchange in any interaction.


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16.2 Electric Charge in the Atom

Atom:

Nucleus (small,

massive, positive

charge)Electron cloud (large,

very low density,

negative charge)


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Atom is electrically neutral.

Rubbing charges objects by moving electrons

from one to the other.


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Question

Why do the balloons eventually lose charge

and no longer repel?


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Polar molecule: neutral overall, but charge not

evenly distributed


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16.3 Insulators and Conductors

Conductor:

Charge flows freely

Metals

Insulator:

Almost no charge flows

Most other materials

Some materials are semiconductors.


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16.4 Induced Charge

Metal objects can be charged by conduction:


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16.4 Induced Charge

They can also be charged by induction:


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16.4 Induced Charge

Nonconductors wont become charged byconduction or induction, but will experience

charge separation:


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16.4 Electroscope

The electroscopecan be used for

detecting charge:


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16.4 Electroscope

The electroscope can be charged either by

conduction or by induction.


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16.4 Electroscope

The charged electroscope can then be used todetermine the sign of an unknown charge.


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January 20, 2012

ConcepTest Review

Coulombs Law (R3)

Practice Problems

Question of the Day (L2):

Whats behind the phenomena of these

demonstrations? Explain.


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1) have opposite charges

2) have the same charge

3) all have the same charge

4) one ball must be neutral (no charge)

From theFrom the

picture, whatpicture, whatcan youcan you

conclude aboutconclude about

the charges?the charges?


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1) have opposite charges

2) have the same charge

3) all have the same charge

4) one ball must be neutral (no charge)

From theFrom the

picture, whatpicture, whatcan youcan you

conclude aboutconclude about

the charges?the charges?

The TAN and PURPLE balls must

have the same charge, since they

repel each other. TheYELLOW

ball also repels the TAN, so it must

also have the same charge as the

TAN (and the PURPLE).


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Ametal ball hangs from the

ceiling by an insulating thread.

The ball is attracted to a positive-

charged rod held near the ball.

The charge of the ball must be:

1) positive

2) negative

3) neutral

4) positive or neutral

5) negative or neutral


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Clearly, the ball will be attracted if its

charge is negativenegative. However, even if

the ball is neutralneutral, the charges in theball can be separated by inductioninduction

(polarization), leading to a net

attraction.

1) positive

2) negative

3) neutral

4) positive or neutral

5) negative or neutral

Ametal ball hangs from the

ceiling by an insulating thread.

The ball is attracted to a positive-

charged rod held near the ball.

The charge of the ball must be:

rememberthe ball is aconductor!

FollowFollow--up:up: What happens if theWhat happens if the metal ballmetal ball is replaced by ais replaced by a plastic ballplastic ball??


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Two neutral conductors are connected

by a wire and a charged rod is brought

near, but does not touch. The wire is

taken away, and then the charged rod is

removed. What are the charges on the

conductors?

1) 0 0

2) +

3) +

4) + +

5)

0 0

? ?


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While the conductors are connected,positivepositive

charge will flow from the blue to the greencharge will flow from the blue to the green

ball due to polarizationball due to polarization. Once disconnected,

the charges will remain on the separatecharges will remain on the separate

conductorsconductors even when the rod is removed.

Two neutral conductors are connected

by a wire and a charged rod is broughtnear, but does not touch. The wire is

taken away, and then the charged rod is

removed. What are the charges on the

conductors?

1) 0 0

2) +

3) +

4) + +

5)

0 0

? ?

FollowFollow--up:up: What will happen when theWhat will happen when the

conductors are reconnected with a wire?conductors are reconnected with a wire?


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16.5 Coulombs Law

Experiment shows that the electric forcebetween two charges is proportional to the

product of the charges and inversely

proportional to the distance between them.


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16.5 Coulombs Law

Coulombs law:

(16-1)

This equation gives the magnitude of

the force.


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16.5 Coulombs Law

The force is along the line connecting the

charges, and is attractive if the charges areopposite, and repulsive if they are the same.


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16.5 Coulombs Law

Unit of charge: coulomb, C

The proportionality constant in Coulombs

law is then:

Charges produced by rubbing are

typically around a microcoulomb:


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16.5 Coulombs Law

Charge on the electron:

Electric charge is quantized in units

of the electron charge.


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16.5 Coulombs Law Practice

Calculate the magnitude of the forcebetween two point charges 9.3 cm

apart.

C-60.3 Q


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Calculate the magnitude of the forcebetween two point charges 9.3 cm

apart.

C-60.3 Q

26

9 2 21 2

222

3.60 10 C

8.988 10 N m C 13.47 N 13 N9.3 10 m

Q Q

F k r

v

! ! v ! }v


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Opening Question: Practice

How many electrons make up a charge of

C?0.30 Q


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How many electrons make up a charge of

C?0.30 Q

6 14

19

1 electron30.0 10 C 1.87 10 electrons

1.602 10 C

v ! v v


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16.5 Coulombs Law

The proportionality constant kcan also bewritten in terms of , the permittivity of free

space:

(16-2)


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16.5 Coulombs Law

Coulombs law strictly applies only to point charges.

Superposition: for multiple point charges, the forceson each charge from every other charge can be

calculated and then added as vectors.


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16.6 Solving Problems Involving

Coulombs Law and Vectors

The net force on a charge is the vector

sum of all the forces acting on it.


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16.6 Solving Problems Involving

Coulombs Law and Vectors

Vector addition review:


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Three charged particles are placed at thecorners of an equilateral triangle of side 1.20

m (Fig. 1653). The charges are

and Calculate the magnitude and

direction of the net force on each due to theother two.

C,0.8C,0.4 QQ

C.0.6 Q


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1Q

2Q

3Q

12

F

&

d

13F

&

d

d

32F

&

31F

&

21F

&

23F

&

First draw a picture.


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1Q

2Q

3Q

12

F

&

d

13F

&

d

d

32F

&

31F

&

21F

&

23F

&

6 6

9 2 21 2

12 22

21

6 6

9 2 21 3

13 22

31

4.0 10 C 8.0 10 C8.988 10 N m C

1.20 m

0.1997 N

4.0 10 C 6.0 10 C8.988 10 N m C

1.20 m

0.1498N

Q QF k

d

F

Q QF k

d

F

v v! ! v

! !

v v! ! v

! !

Then calculate the magnitude ofeach individual force.


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1Q

2Q

3Q

12

F

&

d

13F

&

d

d

32F

&

31F

&

21F

&

23F

&

6 6

9 2 21 2

12 22

21

6 6

9 2 21 3

13 22

31

4.0 10 C 8.0 10 C8.988 10 N m C

1.20 m

0.1997 N

4.0 10 C 6.0 10 C8.988 10 N m C

1.20 m

0.1498N

Q QF k

d

F

Q QF k

d

F

v v! ! v

! !

v v! ! v

! !

First calculate the magnitude of eachindividual force.


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1Q

2Q

3Q

12

F

&

d

13F

&

d

d

32F

&

31F

&

21F

&

23F

&

6 6

9 2 21 2

12 22

21

6 6

9 2 21 3

13 22

31

4.0 10 C 8.0 10 C8.988 10 N m C

1.20 m

0.1997 N

4.0 10 C 6.0 10 C8.988 10 N m C

1.20 m

0.1498N

Q QF k

d

F

Q QF k

d

F

v v! ! v

! !

v v! ! v

! !

6 6

9 2 22 3

23 3222

8.0 10 C 6.0 10 C8.988 10 N m C 0.2996 N

1.20 m

Q Q F k F

d

v v! ! v ! !

First calculate the magnitude of eachindividual force.


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Now calculate the net force on each charge

o o 2

1 12 13

o o 1

1 12 13

1

12 2 1 1 o

1 1 1 1 2

1

0.1997 N cos 60 0.1498N cos 60 2.495 10 N

0.1997 N sin 60 0.1498N sin 60 3.027 10 N

3.027 10 N

0.30 N tan tan 2652.495 10 N

x x x

y y y

y

x y

x

F F F

F F F

F

F F F FU

! ! ! v

! ! ! v

v

! ! ! ! ! v

1Q

2Q

3Q

12F

&

13F

&


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Now calculate direction of that net force, using

components.

o o 2

13

o o 1

13

1

12 1 1 o

1 1 2

1

0.1997 N cos 60 0.1498 N cos 60 2.495 10 N

0.1997 N sin 60 0.1498N sin 60 3.027 10 N

3.027 10 N0.30 N tan tan 2652.495 10 N

x

y

y

y

x

F

F

FFF

U

! ! v

! ! v

v ! ! ! ! v

1Q

2Q

3Q

12F

&

13F

&


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16.7 The Electric Field

The electric field is theforce on a small charge,

divided by the charge:

(16-3)


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For a point charge:

(16-4a)

(16-4b)


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Force on a point charge in an electric field:

(16-5)

Superposition principle for electric fields:


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Problem solving in electrostatics: electricforces and electric fields

1. Draw a diagram; show all charges, with

signs, and electric fields and forces with

directions

2. Calculate forces using Coulombs law

3. Add forces vectorially to get result


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16.8 Field Lines

The electric field can be represented by field

lines. These lines start on a positive charge

and end on a negative charge.


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16.8 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.


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16.8 Field Lines

Electric dipole: two equal charges, opposite insign:


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16.8 Field Lines

The electric field between

two closely spaced,

oppositely charged parallel

plates is constant.


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16.8 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 thecharge.


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16.9 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.


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16.9 Electric Fields and Conductors

The electric field isperpendicular to the

surface of a conductor

again, if it were not,

charges would move.

16 12 Ph t M hi d


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16.12 Photocopy Machines and

Computer Printers Use Electrostatics

Photocopy machine:

drum is charged positively

image is focused on drum

only black areas stay charged and

therefore attract toner particles

image is transferred to paper and sealed by

heat

16 12 Ph t M hi d


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16 12 Ph t M hi d


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Laser printer is similar, except a computercontrols the laser intensity to form the image

on the drum

Summary of Chapter 16


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Two kinds of electric charge positive andnegative

Charge is conserved

Charge on electron:

Conductors: electrons free to move

Insulators: nonconductors




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Charge is quantized in units ofe

Objects can be charged by conduction or

induction

Coulombs law:

Electric field is force per unit charge:



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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

Electric flux:

Gausss law:

ap lecture ch 16

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