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Divide shell into rings of charge, each delimited by the angle and the angle + Use polar coordinates (r, ,). Distance from center: d=(r-Rcos) Surface area of ring: Integration R R Rcos Rsin r d E = 1 4 0 ( Q ) d ( d 2 + ( R sin ) 2 ) 3/2 E = 1 4 0 Q 2 ( r - R cos ) ( r - R cos ) 2 + ( R sin ) 2 3/2 0 sin d 2 ( R sin ) R Q =Q 2 R sin R 4 R 2

Integration

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Integration. Divide shell into rings of charge, each delimited by the angle  and the angle + Use polar coordinates ( r , ,) . Distance from center: d =( r - R cos ) Surface area of ring:. R . . R. R sin . . R cos . d. r. Chapter 22. Patterns of Fields in Space. - PowerPoint PPT Presentation

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E 1

40

(Q)d

(d 2 (Rsin )2 )3/2

E 1

40

Q

2

(r Rcos)

(r Rcos )2 (Rsin)2 3/2

0

sind

Divide shell into rings of charge, each delimited by the angle and the angle +Use polar coordinates (r, ,).Distance from center: d=(r-Rcos)Surface area of ring:

2 (Rsin)R

Q Q2RsinR

4R2

Integration

R

R

Rcos

Rsin

r

d

Chapter 22

Patterns of Fields in Space

• Electric flux• Gauss’s law

What is in the box?

no charges? vertical charged plate?

Patterns of Fields in Space

Box versus open surface

Seem to be able to tellif there are charges inside

…no clue…

Gauss’s law: If we know the field distribution on closed surface we can tell what is inside.

Patterns of Fields in Space

Need a way to quantify pattern of electric field on surface: electric flux

1. Direction

flux>0 : electric field comes outflux<0 : electric field goes in

+1 -10Relate flux to the angle between outward-going normal and E:

flux ~ cos()

Electric Flux: Direction of E

2. Magnitude

flux ~ E

flux ~ Ecos()

Electric Flux: Magnitude of E

𝑓𝑙𝑢𝑥 𝐸 ∙ ��

3. Surface area

flux through small area:

AnEflux ˆ~

Definition of electric flux on a surface:

surface

AnE ˆ

Electric Flux: Surface Area

Perpendicular field

cosˆ AEAnE

AEAnE ˆ

Perpendicular area

coscosˆ yxEAEAnE

x y

AEAnE ˆ

Electric Flux: Perpendicular Field or Area

q

surface

AnE ˆ

dAnE ˆ

Ad

AdE

AdE

surface closed a on flux electric

Adding up the Flux

0

ˆ

inside

surface

qAnE

0

ˆ

insideqdAnE

Features:1. Proportionality constant2. Size and shape independence3. Independence on number of charges inside4. Charges outside contribute zero

Gauss’s Law

0

ˆ

inside

surface

qAnE

204

1

r

QE

surface

Anrr

Qˆˆ

4

12

0

surface

Ar

Q2

04

1

0

22

0

44

1

Q

rr

Q

What if charge is negative?

Works at least for one charge and spherical surface

1. Gauss’s Law: Proportionality Constant

0

ˆ

inside

surface

qAnE

204

1

r

QE

2

1~

rE

2~ rA

2

1~

rE universe would be

much different ifexponent was not exactly 2!

2. Gauss’s Law: The Size of the Surface

0

ˆ

inside

surface

qAnE

E nA

surface EA

surface

All elements of the outer surface can be projected onto corresponding areas on the inner sphere with the same flux

3. Gauss’s Law: The Shape of the Surface

A2 / A1 r22 / r1

2

E2A2 / E1A1 1

A2 R2 (r2 tan)2 r22

∆ 𝐴1⊥=𝜋 𝑟12

0

ˆ

inside

surface

qAnE

surfacesurface

AEAnE ˆ

2~ rA

2

1~

rE 2211 EAEA –

Outside charges contribute 0 to total flux

4. Gauss’s Law: Outside Charges

0

11 ˆ

Q

AnEsurface

0

22 ˆ

Q

AnEsurface

0ˆ3 surface

AnE

0

ˆ

inside

surface

qAnE

5. Gauss’s Law: Superposition

0

ˆ

inside

surface

qAnE

0

ˆ

insideqdAnE

Features:1. Proportionality constant2. Size and shape independence3. Independence on number of charges inside4. Charges outside contribute zero

Gauss’s Law and Coulomb’s Law?

204

1

r

QE

Can derive one from another

Gauss’s law is more universal:works at relativistic speeds

Gauss’s Law

0

ˆ

inside

surface

qAnE

1. Knowing E can conclude what is inside2. Knowing charges inside can conclude what is E

Applications of Gauss’s Law

Symmetry: Field must be perpendicular to surfaceEleft=Eright

0

ˆ

inside

surface

qAnE

2EAbox Q / A Abox

0

E Q / A 20

The Electric Field of a Large Plate

Symmetry: 1. Field should be radial2. The same at every location

on spherical surface

0

ˆ

inside

surface

qAnE

A. Outer Dashed Sphere:

0

24

QrE 2

04

1

r

QE

B. Inner Dashed Sphere:

0

2 04

rE 0E

The Electric Field of a Uniform Spherical Shell of Charge

0

ˆ

inside

surface

qAnE

Is Gauss’s law still valid?

Can we find E using Gauss’s law?

The Electric Field of a Uniform Cube

Without symmetry, Gauss’s law loses much of its power.

Yes, it’s always valid.

Gauss’s Law for Electric Dipole

No symmetry

Direction and Magnitude of E varies

NumericalSolution

Clicker Question

What is the net electric flux through the box?

A) 0 VmB) 0.36 VmC) 0.84 VmD) 8.04 VmE) 8.52 Vm

Can we have excess charge inside a metal that is in static equilibrium?

Proof by contradiction:

0

ˆ

inside

surface

qAnE

=0

00

insideq

Gauss’s Law: Properties of Metal

0

ˆ

inside

surface

qAnE

=0

00

insideq

Gauss’s Law: Hole in a Metal

+5nC

0

ˆ

inside

surface

qAnE

=0

00

insideq

0 insidesurface qq

nC 5 surfaceq

Gauss’s Law: Charges Inside a Hole

Review for Midterm