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Ley de Gauss. Física III. Flujo el é ctrico. El flujo eléctrico se representa por medio del número de líneas de campo eléctrico que penetran alguna superficie.

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Ley de Gauss

Física III

Flujo eléctrico

El 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 F.

F = EA

Área A


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.



F = EAcosq




El flujo DF a través de un pequeño elemento DAi es:

DF = EiDAi cos q = Ei• DAi

El flujo a través de toda la superficie es:

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:

Ley de Gauss

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

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 e0.





Aplicaciones de la ley de Gauss

Distribución esférica de carga





Esfera gaussiana

Esfera gaussiana

Reglas para la aplicación de la 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 electrostático

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 s/e0.

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

24 Gauss’ law

24-1 A New Look at Coulomb’s Law

Gauss’ law relates the electric fields

at points on a (closed) Gaussian

surface and the net charge enclosed

by that surface.

24-2 Flux

(a)The rate Φis equal to v·A



(d)A velocity field.Flux means the

product of an area and the field

across that area.

24-3 Flux of an Electric Field

A provisional definition

for the flux of the electric

field for the Gaussian

surface is

Electric flux through a Gaussian surface

The electric flux Φ through a Gaussian

surface is proportional to the net number

of electric field lines passing through that


Sample Problem 24-1

What is the flux Φ of

The electric field through

This closed surface?

Step one:

Step two:

Step three:

Sample Problem 24-2

What is the electric flux

through the right the face,

the left face,and the top


Right face:

Left face:

Top face:

24-4 Gauss’ Law

Gauss’ law and Coulomb’s law, although

expressed in different forms, are equivalent

ways of describing relation between charge

and electric field in static situations. Gauss’s

law is:


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 24-3

What is the net electric

flux through the surface

if Q1=q4=+3.1nC,


and q3=-3.1nC?

24-5 Gauss’ Law and Coulomb’s Law

Gauss’ law as:



Gauss’ law is equivalent to Coulomb’s law.

24-6 A Charged Isolated Conductor

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


An Isolated Conductor with a Cavity

There is no net charge on the cavity walls.

The Conductor Removed

The electric field is set up by the charges and

not by the conductor.The conductor simply

provides an initial pathway for the charges to

take up their position.

The External Electric Field

Conducting surface:

Sample Problem 24-4

Key idea

The electric flux through the Gaussian surface

must also be zero.The net charge enclosed by

the Gaussian surface must be zero.With a point

charge 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’ law:Cylindrical Symmetry

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

Where r is the perpendicular distance from

the line of charge to the point.

Sample Problem 24-5

If air molecules break down (ionize) in an

electric field exceeding 3×106N/C,what is the


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.

Can you think of another key idea?

24-8 Applying Gauss’ law:Planar Symmetry

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


Two Conducting Plates:

Sample Problem 24-6

Step one:

Step two:

24-9 Applying Gauss’ law: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.

Spherical shell,field at r ≥R

Spherical shell,field at r <R

Spherical distribution,field at r ≥R

Uniform charge,field at r ≤R


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