Ley de Gauss Física III.

Presentación del tema: "Ley de Gauss Física III."— Transcripción de la presentación:

1 Ley de Gauss Física III

2 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 E

3 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 q F = EAcos q

4 El flujo DF a través de un pequeño elemento DAi es:
DF = Ei DAi cos q = Ei • DAi DAi Ei q 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:

5 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. dA E r q

6 Aplicaciones de la ley de Gauss
Distribución esférica de carga r a r a Esfera gaussiana Esfera gaussiana

7 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.

8 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.

9 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.

10 24-2 Flux (a)The rate Φis equal to v·A (b) (c)
(d)A velocity field.Flux means the product of an area and the field across that area.

11 24-3 Flux of an Electric Field
A provisional definition for the flux of the electric field for the Gaussian surface is

12 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 surface.

13 Sample Problem 24-1 What is the flux Φ of The electric field through
This closed surface? Step one: Step two:

14 Step three:

15 Sample Problem 24-2 What is the electric flux
through the right the face, the left face,and the top face? Right face:

16 Left face: Top face:

17 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:

18 or 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.

19 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.

20 Sample Problem 24-3 What is the net electric flux through the surface
if Q1=q4=+3.1nC, q2=q5=-5.9nC, and q3=-3.1nC?

21 24-5 Gauss’ Law and Coulomb’s Law
Gauss’ law as: Coulomb’s law Gauss’ law is equivalent to Coulomb’s law.

22 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 conductor.

23 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.

24 The External Electric Field
Conducting surface:

25 Sample Problem 24-4

26 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?

27 24-7 Applying Gauss’ law:Cylindrical Symmetry

28 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.

29 Sample Problem 24-5 If air molecules break down (ionize) in an
electric field exceeding 3×106N/C,what is the column? 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.

30 Can you think of another key idea?

31 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 magnitude

32 Two Conducting Plates:

33 Sample Problem 24-6

34 Step one: Step two:

35 24-9 Applying Gauss’ law:Spherical Symmetry

36 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.

37 Spherical shell,field at r ≥R

38 Spherical distribution,field at r ≥R
Uniform charge,field at r ≤R 主菜单