Physics for Scientists and Engineers II, Summer Semester 2009 1 Lecture 29: August 5 th 2009 Physics for Scientists and Engineers II

Physics for Scientists and Engineers II , Summer Semester 2009

1

Lecture 29: August 5th 2009

Physics for Scientists and Engineers II


2

Gauss’s Law

Very long straight cable.

1) Inner cylinder isa conductor.Radius R1=1.00cmCharge per unit length = - 350C/m

2) Outer cylinder is insulatorOuter radius R2= 2.00cmCharge density = 750x10-2 C/m3

Determine E(r) (magn. and direction)Determine surface charge density onInner cylinder.


3

Gauss’s Law

Symmetry dictates that the electric field must be pointing radial in or out:Ez(r) =0 z

Gaussian closed surface

Inside the conductor:

00 rEq

AdEsurface o

in

All the charge is on the surface of the conductor!

22111

011.0105.02

350

222 m

C

mm

C

RLR

L

LR

q

Inside the conductor (r < R1)


4

Gauss’s Law

Inside the insulator (R1<r < R2)

2

212

622

2

212

32

21

21

2

21

2

21

221

2

10854.82

10350105.0

10854.82

10750

222

)()(

)(2)(

)()(

NmCr

mC

r

mr

NmCm

C

rr

Rr

Lr

LRrLrE

LRrLLrrE

LRrLLLRLrqVqAdE

ooo

o

ooo

conductor

surface o

in


5

Gauss’s Law

Note: The field points inwards up to a certain radius. Then the field is 0 whenr is such that the charge on the conductor and the enclosed charge in the insulatorIs zero.

cmmrm

Cm

CmRrRr

Rrrr

Rr

oo

63.00063.0

10750

10350105.0

0022

32

6222

12

12

21

22

1

outwards points E :0.0063mr

inwards points:0063.0

Emr


6

Gauss’s Law

Outside the insulator (R2<r)

ooo

o

oo

o

conductorinsulator

o

conductorinsulator

surface o

in

rRR

rLr

LRRLrE

LRRLLrrE

LRRLLLRLR

qVqqqAdE

222

)()(

)(2)(

)()(

21

22

21

22

21

22

21

22

21

22


7

Ray Tracing Convex Mirror

C

R=0.80m0.60m


8

Ray Tracing

C

R=0.80m0.60m

F


9

Calculation of Image

p = + 0.60m (positive because object is real, in front of the mirror)f= - R/2 = - 0.40m (negative because it is a diverging mirror)

negative) is q (becausemirror thebehind virtual,is Image

24.00.1

24.0

40.060.0

40.060.0

111111

2

mm

m

mm

mm

fp

pfq

pf

fp

pfqfqp

)1M(object an smaller thbut positive) is M (becauseupright is Image

4.060.0

24.0'

m

m

p

q

h

hM


10

Two Lenses – Characterize the Image

mmf 1501 mmf 3502

mm250 mm50


11

Image due to Lens 1

mmf 1501 mmf 3502

mm250 mm50

)(5.1250

375M

1) lens ofright theTo(

375100

150250

150250

150250q

object) real(250

1

11

11

111

1

invertedmm

mm

p

q

mmmm

mmmm

mmmm

mmmm

fp

fp

mmp

O2

I1=O2

mm375

mm325


12

Image due to Lens 2mmf 1501 mmf 3502

mm250 mm50

real. is image final The

image). (inverted 0.78- is lenses two the todueion magnificat overall The

78.052.05.1

52.0325

169M

image) real 2, lens ofright theTo(

169675

350325

350325

350325q

lens) thebehind -object virtual(325

21

2

22

22

222

2

MMM

mm

mm

p

q

mmmm

mmmm

mmmm

mmmm

fp

fp

mmp

total

O1

O2

mm325

I2


13

• We did not get to the following problems in the lecture, but I’ll post them anyhow in case you would like to have some extra practice.


14

Forces on Charges

nCq 0.14

nCq 0.62

nCq 0.21 nCq 0.23

cm0.5cm0.5

4545

Find the force on 1.0nC charge.


15

Forces on Charges

nCq 0.14

nCq 0.62

nCq 0.21 nCq 0.23

cm0.5cm0.5

4545

3424144 FFFF

14F34F

24F

y

x

45


16

Forces on Charges

ji

m

nCnCkji

d

qqkjiFF ee

ˆ2

1ˆ2

1

100.5

0.10.2ˆ2

1ˆ2

1ˆ45cosˆ45sin 22214

411414

jm

nCnCkj

d

qqkjFF ee

ˆ100.5

0.10.6ˆˆ222

24

422424

ji

m

nCnCkji

d

qqkjiFF ee

ˆ2

1ˆ2

1

100.5

0.10.2ˆ2

1ˆ2

1ˆ45cosˆ45sin 22234

433434

Nm

nCnCnC

C

Nm

m

nCnCnCkj

m

nCnCj

m

nCnCkF ee

5222

29

2222224

101.1100.5

0.10.60.221099.8

100.5

0.10.60.22ˆ100.5

0.10.6ˆ2

2

100.5

0.10.2


17

Circuits and Induction

0.500m 0.500m

0.500m00.6 00.3

00.5

mr 20.02

resistor?

eachin current theis What 100T/s. of rate aat increases B

5.0V


18


0.500m 0.500m

0.500m00.6 00.3

00.5

mr 20.02

it. caused that one theopposes that changeflux ain results that loop

right in thecurrent ckwisecounterclo a cause wouldaloneit Such that

law. sLenz' from deduced becan direction emf Induced

1I 2I

1I 2I

3I 5.0Vind


19


000.300.6 :1 Loop 31 II

000.300.50.507.7 :2 Loop 32 IIVV

Vs

Tm

dt

dBr

dt

d Bind 07.7100150.0 22

22

321321 0 :ruleJunction IIIIII

32

32332

5.1

000.900.6000.300.6

II

IIIII

AV

IIIV 197.05.10

07.2000.35.100.507.2 333

AAAIII

AAII

099.0197.0296.0

296.0197.05.15.1

321

32


20

Ampere’s Law – calculate B

I2

I1 .conductorsthin density wicurrent uniform Assume

r:Outside

rr:)(Iconductor Outer

rr :)(IConductor First

rr :holeInner

3

322

211

1

r

r

r

)r2

)(;0)B( (e.g.,

r22

rrr :ConductorFirst

0020

rr :holeInner

1021

21

22

21

21

021

22

21

2

100

21

0

1

IrBr

rr

rrIB

rr

rrIrBIsdB

BrBIsdB

inside

inside


21

Ampere’s Law – calculate B

I2

I1 .conductorsthin density wicurrent uniform Assume

r:Outside

rr:)(Iconductor Outer

rr :)(IConductor First

rr :holeInner

3

322

211

1

r

r

r

ccw means - and clockwise means :on here From rule). hand(right Clockwise :Direction

)r2

)(;0)B( (e.g.,

r22

rrr :ConductorFirst

0020

rr :holeInner

1021

21

22

21

21

021

22

21

2

100

21

0

1

IrBr

rr

rrIB

rr

rrIrBIsdB

BrBIsdB

inside

inside


22

Ampere’s Law

21

210

2100

3

22

23

22

2

210

22

23

22

2

2100

32

I and I of sizeon dependsDirection

r2

2

rr :Outside

r

2

r2

rrr :conductorOuter

IIB

IIrBIsdB

rr

rII

rB

rr

rIIrBIsdB

inside

inside

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Physics for Scientists and Engineers II, Summer Semester 2009 1 Lecture 29: August 5 th 2009 Physics for Scientists and Engineers II