Ley de Gauss Física III Flujo el é ctrico El flujo eléctrico se representa por medio del número...

Ley de GaussLey de Gauss

Física III

Flujo eléctricoFlujo eléctricoEl flujo eléctrico se representa por medio del número de líneas de campo eléctrico que penetran alguna superficie.

El número de líneas que penetra una superficie es proporcional a EA. Al producto de la intensidad del campo E por el área de la superficie perpendicular A se le llama flujo eléctrico .

= EA

Área A

E

Si la superficie no es perpendicular al campo, el flujo es igual al producto de la magnitud del campo por el área por el coseno del ángulo entre el campo y la normal a la superficie.

Normal

= EAcos

Ai

Ei

El flujo a través de un pequeño elemento Ai es:

= Ei Ai cos = Ei • Ai

El flujo a través de toda la superficie es:

Superficie

ii di

AEAEA 0lim

Si sin más las líneas que salen, el flujo neto es positivo. Si son más las líneas que entran, el flujo neto es negativo.

Si la superficie es cerrada el flujo es: AE d

Ley de GaussLey de Gauss

r

EdA

q

Considere una carga puntual q. El flujo en una esfera de radio r será:

0

22 44

q

qkrr

qkdAEd e

e AE

La ley de Gauss establece que el flujo eléctrico neto a través de una superficie cerrada es igual a la carga neta dentro de la superficie dividida por 0.

Aplicaciones de la ley de Aplicaciones de la ley de GaussGauss

ra

Esfera gaussiana

r a

Esfera gaussiana

Distribución esférica de carga

Reglas para la aplicación de Reglas para la aplicación de la ley de Gaussla ley de Gauss

1. El valor del campo eléctrico puede considerarse, por simetría, como constante sobre toda la superficie.

2. El producto punto E dA puede escribirse como EdA.

3. El producto punto E dA es cero porque E y dA son perpendiculares.

4. Puede decirse que el campo sobre la superficie es cero.

Conductores en equilibrio Conductores en equilibrio electrostáticoelectrostático

Los conductores tienen las siguientes propiedades:

El campo eléctrico es cero en cualquier punto del interior del conductor.

Cualquier carga reside en su superficie.

El campo eléctrico en la superficie es perpendicular a la superficie y tiene una magnitud de /0.

La carga tiende a acumularse en las partes con radio de curvatura más grande.

24 Gauss’ law24 Gauss’ law

Gauss’ law relates the electric fields at points on a (closed) Gaussian surface and the net charge enclosed by that surface.

24-1 A New Look at Coulomb’s Law

24-2 Flux24-2 Flux

(a)The rate Φis equal to v·A

Av cos(b)

AvvA

cos(c)

(d)A velocity field.Flux means the product of an area and the field across that area.

24-3 Flux of an Electric Field24-3 Flux of an Electric Field

A provisional definitionfor the flux of the electricfield for the Gaussiansurface is

AE

AE

Electric flux through a Gaussian surface

AdE

AdE

The electric flux Φ through a Gaussian surface is proportional to the net number of electric field lines passing through that surface.

Sample Problem 24-1Sample Problem 24-1What is the flux Φ of The electric field throughThis closed surface?

cba

AdEAdEAdEAE

EAdAEdAEAdEa

0180cos

EAdAEAdEc

0cos

Step one:Step one:

Step two:Step two:

00 EAEA 00 EAEA

090cos 0 dAEAdEb

090cos 0 dAEAdE

b

Step three:Step three:

Sample Problem 24-2Sample Problem 24-2

What is the electric fluxthrough the right the face,the left face,and the top face?Right face:

idAAd ˆ

idAAd ˆ

Left face: CmNl /12 2 CmNl /12 2

Top face:

CmN

jdAjixt

/16

ˆˆ0.4ˆ0.3

2

CmN

jdAjixt

/16

ˆˆ0.4ˆ0.3

2

CmN

dAdAxdA

idAjixAdEr

/36

0.90.30.30.3

ˆˆ0.4ˆ0.3

2

CmN

dAdAxdA

idAjixAdEr

/36

0.90.30.30.3

ˆˆ0.4ˆ0.3

2

24-4 Gauss’ Law24-4 Gauss’ Law

Gauss’ law and Coulomb’s law, although expressed in different forms, are equivalentways of describing relation between charge and electric field in static situations. Gauss’s law is:

encq0encq0

or encqAdE

0 encqAdE

0

Surface S1

The electric field is outward for all point on this surface.

Surface S2

The electric field is inward for all point on this surface.

Surface S3

This surface encloses no charge,and thus qenc=0

Surface S4

This surface encloses no net charge,because the enclosed positive and negative charges have equal magnitudes.

Sample Problem 24-3Sample Problem 24-3

What is the net electricflux through the surface if Q1=q4=+3.1nC,q2=q5=-5.9nC,and q3=-3.1nC?

CmNqqqqenc /670 2

0

321

0

CmN

qqqqenc /670 2

0

321

0

24-5 Gauss’ Law and Coulomb’s 24-5 Gauss’ Law and Coulomb’s LawLawGauss’ law as:

encqEdAAdE 00

encqEdAAdE 00

qdAE 0 qdAE 0

qrE 20 4 qrE 2

0 4 204

1

r

qE

204

1

r

qE

Coulomb’s

law

Gauss’ law is equivalent to Coulomb’s law.

24-6 A Charged Isolated 24-6 A Charged Isolated ConductorConductor

If an excess charge is placed on an isolated conductor,that amount of charge will move entirely to the surface of the conductor .None of the excess charge will be found within the body of the conductor.

An Isolated Conductor with a Cavity

There is no net charge on the cavity walls.

The Conductor RemovedThe electric field is set up by the charges andnot by the conductor.The conductor simply provides an initial pathway for the charges totake up their position.

The External Electric Field

Conducting surface:

0

E0

E

Sample Problem 24-4Sample Problem 24-4

Key idea

The electric flux through the Gaussian surface must also be zero.The net charge enclosed bythe Gaussian surface must be zero.With a pointcharge of -5.0μC within the shell,a charge of +5.0 μC must lie on the inner wall of the shell.

Can you think of another key idea?

24-7 Applying Gauss’ 24-7 Applying Gauss’ law:Cylindrical Symmetrylaw:Cylindrical Symmetry

)2(0cos)2(

cos

rhErhE

EA

)2(0cos)2(

cos

rhErhE

EA

encq0encq0

hrhE 20 hrhE 20

rE

02

r

E02

The electric field at any point due to an infinite line of charge with uniform linear charge density λis perpendicular to the line of charge and has magnitude

rE

02

r

E02

Where r is the perpendicular distance from the line of charge to the point.

Sample Problem 24-5Sample Problem 24-5

If air molecules break down (ionize) in an electric field exceeding 3×106N/C,what is thecolumn?

Key idea

The surface of the column of charge must be at The radius r where the magnitude of is 3 ×106N/C,because air molecules within that Radius ionize while those farther out do not.

E

mE

r 62 0

mE

r 62 0

Can you think of another key idea?

24-8 Applying Gauss’ law:Planar 24-8 Applying Gauss’ law:Planar SymmetrySymmetry

nonconducting sheet The electric field due to an infinite nonconducting sheet with uniform surface charge density σis perpendicular to the plane of the sheet and has magnitude

02

E02

E

00

12

E00

12

E

Two Conducting Plates:

Sample Problem 24-6Sample Problem 24-6

CNEEEl /104.1 5 CNEEEl /104.1 5

CNEEEb /103.6 5 CNEEEb /103.6 5

CNE /1084.3

25

0

CNE /1084.32

5

0

CNE /1043.2

25

0

CNE /1043.22

5

0

Step one:Step one:

Step two:Step two:

24-9 Applying Gauss’ 24-9 Applying Gauss’ law:Spherical Symmetrylaw:Spherical Symmetry

A shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell’s charge were concentrated at the center of the shell.

A shell of uniform charge exerts no electrostatic force on a charged particle that is located inside the shell.

204

1

r

qE

204

1

r

qE

Spherical shell,field at r ≥R

Spherical shell,field at r ＜ R

0E 0E

2

'

04

1

r

qE

2

'

04

1

r

qE

Spherical distribution,field at r ≥R

Uniform charge,field at r ≤R

rR

qE )

4(

30

rR

qE )

4(

30