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Flux Capacitor (Schematic) Physics 2102 Lecture 3phys.lsu.edu/~jdowling/PHYS21024SP07/lectures/  · PDF fileFlux Capacitor (Schematic) ... 4πR2E= −40π Nm2/C ... (180¡)=!EdA r

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Text of Flux Capacitor (Schematic) Physics 2102 Lecture 3phys.lsu.edu/~jdowling/PHYS21024SP07/lectures/...

  • Physics 2102Physics 2102Lecture 3Lecture 3

    GaussGauss Law ILaw IMichael Faraday

    1791-1867

    Version: 1/22/07

    Flux Capacitor (Schematic)

    Physics 2102

    Jonathan Dowling

  • What are we going to learn?What are we going to learn?A road mapA road map

    Electric charge Electric force on other electric charges Electric field, and electric potential

    Moving electric charges : current Electronic circuit components: batteries, resistors, capacitors Electric currents Magnetic field

    Magnetic force on moving charges Time-varying magnetic field Electric Field More circuit components: inductors. Electromagnetic waves light waves Geometrical Optics (light rays). Physical optics (light waves)

  • What? What? The Flux! The Flux!STRONGE-Field

    WeakE-Field

    Number of E-LinesThrough Differential

    Area dA is aMeasure of Strength

    dA

    AngleMatters Too

  • Electric Flux: Planar SurfaceElectric Flux: Planar Surface

    Given: planar surface, area A uniform field E E makes angle with NORMAL to

    plane

    Electric Flux: = EA = E A cos

    Units: Nm2/C Visualize: Flow of Wind

    Through Window

    E

    AREA = A=An

    normal

  • Electric Flux: General SurfaceElectric Flux: General Surface For any general surface: break up into

    infinitesimal planar patches

    Electric Flux = EdA Surface integral dA is a vector normal to each patch and

    has a magnitude = |dA|=dA CLOSED surfaces:

    define the vector dA as pointingOUTWARDS

    Inward E gives negative flux Outward E gives positive flux

    E

    dA

    dA

    EArea = dA

  • Electric Flux: ExampleElectric Flux: Example

    Closed cylinder of length L, radius R Uniform E parallel to cylinder axis What is the total electric flux through

    surface of cylinder?(a) (2RL)E(b) 2(R2)E(c) Zero

    Hint!Surface area of sides of cylinder: 2RLSurface area of top and bottom caps (each): R2

    L

    R

    E

    (R2)E(R2)E=0What goes in MUST come out!

    dA

    dA

  • Electric Flux: ExampleElectric Flux: Example

    Note that E is NORMALto both bottom and top cap

    E is PARALLEL tocurved surface everywhere

    So: = 1+ 2 + 3= R2E + 0 - R2E= 0!

    Physical interpretation:total inflow = totaloutflow! 3

    dA

    1dA

    2 dA

  • Electric Flux: ExampleElectric Flux: Example Spherical surface of radius R=1m; E is RADIALLY

    INWARDS and has EQUAL magnitude of 10 N/Ceverywhere on surface

    What is the flux through the spherical surface?

    (a) (4/3)R2 E = 13.33 Nm2/C

    (b) 2R2 E = 20 Nm2/C

    (c) 4R2 E= 40 Nm2/C

    What could produce such a field?

    What is the flux if the sphere is not centeredon the charge?

  • q

    r

    Electric Flux: ExampleElectric Flux: Example

    dr A = + dA( ) r

    r E d

    r A = EdAcos(180) = !EdA

    r E = !

    q

    r2

    r

    Since r is Constant on the Sphere RemoveE Outside the Integral!

    ! =r E "d

    r A # = $E dA = $ kq

    r2

    % & '

    ( ) * # 4+r2( )

    = $q

    4+,0

    4+( ) = $q /,0 Gauss Law:Special Case!

    (Outward!)

    Surface Area Sphere

    (Inward!)

  • GaussGauss Law: General Case Law: General Case

    Consider any ARBITRARYCLOSED surface S -- NOTE:this does NOT have to be areal physical object!

    The TOTAL ELECTRIC FLUXthrough S is proportional to theTOTAL CHARGEENCLOSED!

    The results of a complicatedintegral is a very simpleformula: it avoids longcalculations!

    !"r E # d

    r A =

    q

    $0Surface

    %

    S

    (One of Maxwells 4 equations!)

  • ExamplesExamples

    ! ="#$Surface 0

    %

    qAdErr

  • GaussGauss Law: Example Law: ExampleSpherical symmetrySpherical symmetry

    Consider a POINT charge q & pretend that youdont know Coulombs Law

    Use Gauss Law to compute the electric field at adistance r from the charge

    Use symmetry: draw a spherical surface of radius R centered

    around the charge q E has same magnitude anywhere on surface E normal to surface

    0!

    q="

    rq E

    24|||| rEAE !=="

    2204

    ||r

    kq

    r

    qE ==

    !"

  • GaussGauss Law: Example Law: ExampleCylindrical symmetryCylindrical symmetry

    Charge of 10 C is uniformly spreadover a line of length L = 1 m.

    Use Gauss Law to computemagnitude of E at a perpendiculardistance of 1 mm from the center ofthe line.

    Approximate as infinitely longline -- E radiates outwards.

    Choose cylindrical surface ofradius R, length L co-axial withline of charge.

    R = 1 mmE = ?

    1 m

  • GaussGauss Law: cylindrical Law: cylindricalsymmetry (cont)symmetry (cont)

    RLEAE !2|||| =="

    00!

    "

    !

    Lq==#

    Rk

    RRL

    LE

    !

    "#

    !

    "#

    !2

    22||

    00

    ===

    Approximate as infinitely longline -- E radiates outwards.

    Choose cylindrical surface ofradius R, length L co-axial withline of charge.

    R = 1 mmE = ?

    1 m

  • Compare with Example!Compare with Example!

    !" +

    =

    2/

    2/

    2/322 )(

    L

    L

    yxa

    dxakE #

    if the line is infinitely long (L >> a)

    224

    2

    Laa

    Lk

    +

    =!

    2/

    2/

    222

    L

    Laxa

    xak

    !"#

    $%&

    '

    +

    = (

    a

    k

    La

    LkEy

    !! 22

    2==

  • SummarySummary

    Electric flux: a surface integral (vector calculus!);useful visualization: electric flux lines caught by thenet on the surface.

    Gauss law provides a very direct way to computethe electric flux.

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