Ly de Gauss 畢業(yè)的設(shè)計(jì)

1、ley de gaussfsica iiiflujo elctricoel flujo elctrico se representa por medio del nmero de lneas de campo elctrico que penetran alguna superficie.el nmero de lneas que penetra una superficie es proporcional a ea. al producto de la intensidad del campo e por el rea de la superficie perpendicular a se

2、le llama flujo elctrico f.f = earea aesi la superficie no es perpendicular al campo, el flujo es igual al producto de la magnitud del campo por el rea por el coseno del ngulo entre el campo y la normal a la superficie.qnormalf = eacos qdaieiqel flujo df a travs de un pequeo elemento dai es: df = ei

3、dai cos q = ei daiel flujo a travs de toda la superficie es:dfdsuperficieiidiaeaea0limsi sin ms las lneas que salen, el flujo neto es positivo. si son ms las lneas que entran, el flujo neto es negativo.si la superficie es cerrada el flujo es:fae dley de gaussredaqconsidere una carga puntual q. el fl

4、ujo en una esfera de radio r ser:02244qqkrrqkdaedeefaela ley de gauss establece que el flujo elctrico neto a travs de una superficie cerrada es igual a la carga neta dentro de la superficie dividida por 0.aplicaciones de la ley de gaussraesfera gaussianaraesfera gaussianadistribucin esfrica de carga

5、reglas para la aplicacin de la ley de gauss1. el valor del campo elctrico puede considerarse, por simetra, 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 sob

6、re la superficie es cero.conductores en equilibrio electrostticolos conductores tienen las siguientes propiedades:el campo elctrico es cero en cualquier punto del interior del conductor.cualquier carga reside en su superficie.el campo elctrico en la superficie es perpendicular a la superficie y tien

7、e una magnitud de s/0.la carga tiende a acumularse en las partes con radio de curvatura ms grande.24 gauss lawgauss 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 coulombs law24-2 flux(a)the rate is equal to vaa

8、vqcosf(b)avvafqcos(c)(d)a velocity field.flux means the product of an area and the field across that area.24-3 flux of an electric fielda provisional definitionfor the flux of the electricfield for the gaussiansurface is electric flux through a gaussian surfacethe electric flux through a gaussian su

9、rface is proportional to the net number of electric field lines passing through that surface.sample problem 24-1what is the flux of the electric field throughthis closed surface?fcbaadeadeadeaeeadaedaeadea0180coseadaeadec0cosstep one:step two:step three:sample problem 24-2what is the electric fluxth

10、rough the right the face,the left face,and the top face?right face:left face:top face:24-4 gauss lawgauss law and coulombs law, although expressed in different forms, are equivalentways of describing relation between charge and electric field in static situations. gausss law is: orsurface s1the elec

11、tric field is outward for all point on this surface.surface s2the electric field is inward for all point on this surface.surface s3this surface encloses no charge,and thus qenc=0surface s4this surface encloses no net charge,because the enclosed positive and negative charges have equal magnitudes.sam

12、ple problem 24-3what is the net electricflux through the surface if q1=q4=+3.1nc,q2=q5=-5.9nc,and q3=-3.1nc?24-5 gauss law and coulombs lawgauss law as:coulombslawgauss law is equivalent to coulombs law.24-6 a charged isolated conductorif an excess charge is placed on an isolated conductor,that amou

13、nt 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 cavitythere is no net charge on the cavity walls.the conductor removedthe electric field is set up by the charges andnot by the condu

14、ctor.the conductor simply provides an initial pathway for the charges totake up their position. the external electric fieldconducting surface:sample problem 24-4key ideathe electric flux through the gaussian surface must also be zero.the net charge enclosed bythe gaussian surface must be zero.with a

15、 pointcharge of -5.0c within the shell,a charge of +5.0 c must lie on the inner wall of the shell.can you think of another key idea?24-7 applying gauss law:cylindrical symmetrythe electric field at any point due to an infinite line of charge with uniform linear charge density is perpendicular to the

16、 line of charge and has magnitude where r is the perpendicular distance from the line of charge to the point. sample problem 24-5if air molecules break down (ionize) in an electric field exceeding 3106n/c,what is thecolumn?key ideathe surface of the column of charge must be at the radius r where the

17、 magnitude of is 3 106n/c,because air molecules within that radius ionize while those farther out do not.ecan you think of another key idea?24-8 applying gauss law:planar symmetrynonconducting sheet the electric field due to an infinite nonconducting sheet with uniform surface charge density is perpendicular to the plane of the sheet and has magnitude two conducting plates:sample problem 24-6step one:step two:24-9 applying gauss law:spherical symmetrya shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shells