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Gauss s Law - WordPress.com · Gauss’s Law !" The precise relationship between the electric flux...
Transcript of Gauss s Law - WordPress.com · Gauss’s Law !" The precise relationship between the electric flux...
Miss Millie Millie
Gauss’s Law �Electric Flux�
~fluxistheelectricfieldpassingthroughagivenarea~
𝐹𝑙𝑢𝑥 = 𝜙! = 𝐸𝐴 𝑐𝑜𝑠 𝜃 = 𝐸 ⊥ 𝐴𝐸 = 𝑢𝑛𝑖𝑓𝑜𝑟𝑚, 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 = 𝑓𝑙𝑎𝑡
𝜙! ≈! 𝐸!⃗ !
!
!!!
∙ Δ𝐴! = ! 𝐸 ∙ 𝑑𝐴
𝐸 ≠ 𝑢𝑛𝑖𝑓𝑜𝑟𝑚, 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 ≠ 𝑓𝑙𝑎𝑡
o Fluxenteringanenclosedvolumeisnegativeo Fluxleavingthevolumeispositive
o Ifallfieldlinesthatenterthevolumealsoleavethe
volumethen𝜙!!"# = 0
�Gauss’s Law� Thepreciserelationshipbetweentheelectricfluxthroughaclosedsurfaceandthenetcharge,𝑄!"#$ ,enclosedwithinthatsurfaceisgivenbyGauss’sLaw:
! 𝐸!⃗ ∙ 𝑑𝐴 =𝑄!"#$𝜖!
Recall:𝜖! = 𝑝𝑒𝑟𝑚𝑖𝑡𝑖𝑣𝑖𝑡𝑦 𝑜𝑓 𝑓𝑟𝑒𝑒 𝑠𝑝𝑎𝑐𝑒= 8.85 ∗ 10!!" !!
!!!𝑄!"#$ = 𝑛𝑒𝑡 𝑐ℎ𝑎𝑟𝑔𝑒…itdoesnotmatterwhereorhowthechargeisdistributed
�Gauss + Coulomb Coulomb’sLawfromGauss’sLaw
! 𝐸!⃗ ∙ 𝑑𝐴 = ! 𝐸!⃗ 𝐴 = 𝐸!⃗ !𝑑𝐴 = 𝐸(4𝜋𝑟!)
∴ 𝑄!"#$ = 𝑄
∴𝑄𝜀!= ! 𝐸!⃗ ∙ 𝑑𝐴 = 𝐸(4𝜋𝑟!)
As𝐸!⃗ and𝑑𝐴arebothperpendiculartothesurfaceatallpoints,andcos 𝜃 = 1…
∴ 𝐸 =𝑄
4𝜋𝜀!𝑟!
(ElectricFieldformofCoulombsLaw)Gauss’sLawfromCoulombLaw
𝐸!⃗ =1
4𝜋𝜀!𝑄𝑟!
(sphericalsurface)
∴ !𝐸!⃗ ∙ 𝑑𝐴 = !1
4𝜋𝜀!𝑄𝑟! 𝑑𝐴
=𝑄
4𝜋𝜀!𝑟! (4𝜋𝑟!)
=𝑄𝜀!
�Applications SphericalConductorAthinsphericalshell…Determinetheelectricfield(𝐸!⃗ ); Outsidetheshell(r > r!) ∮𝐸!⃗ ∙ 𝑑𝐴 = 𝐸(4𝜋𝑟!) = !
!!
𝐸 =1
4𝜋𝜀!𝑄𝑟!
Insidetheshell(r < 𝑟!) 𝐸!"#$ = 0 ∴ 𝐸 = 0SolidSphereofChargeAsolidsphere,withchargeQdistributeduniformly…determinetheelectricfield(𝐸!⃗ );
Outsidetheshell(r > r!) ∮𝐸!⃗ ∙ 𝑑𝐴 = 𝐸(4𝜋𝑟!) = !
!!
𝐸 = !
!!!!
!
!!
Insidetheshell(r < 𝑟!) ∮𝐸!⃗ ∙ 𝑑𝐴 = 𝐸(4𝜋𝑟!)𝜌! =
!"
!"= 𝑐ℎ𝑎𝑟𝑔𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
𝑄!"#$ =!!!!
!!!!!!!!
!!!𝑄 = !!
!!! 𝑄
∴ 𝐸(4𝜋𝑟!) = !!"#$!!
= !!
!!!!
!!→ 𝐸 = !
!!!!
!
!!! 𝑟
Non-uniformchargedsolidsphere𝜌! = 𝛼𝑟!; 𝛼 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Find𝛼intermsof𝑄 𝑎𝑛𝑑 𝑟! 𝑑𝑉 = 4𝜋𝑟!𝑑𝑟 𝑄 = ∫ 𝜌!𝑑𝑉 = ∫ (𝛼𝑟!)(4𝜋𝑟!𝑑𝑟) = 4𝜋𝛼 ∫ 𝑟!𝑑𝑟 = !!"
! 𝑟!!
!!!
!!!
∴ 𝛼 = !!
!!!!!
Find𝐸!⃗ asafunctionof𝑟insidethesphere 𝑄!"#$ = ∫ 𝜌!𝑑𝑉 = ∫ (𝛼𝑟!)(4𝜋𝑟!𝑑𝑟) = ∫ ! !!
!!!!! 𝑟!! 4𝜋𝑟!𝑑𝑟 = 𝑄 !!
!!!
!!
!!
!!
∮𝐸!⃗ ∙ 𝑑𝐴 = 𝑄!"#$/𝜀! (𝐸)(4𝜋𝑟!) = 𝑄 !!
!!!!!
𝐸 = !!!
!!!!!!!
LongUniformLineofChargeλ = charge per unit length CalculateEatpointsnearbutoutsidethewire ∮𝐸!⃗ ∙ 𝑑𝐴 = 𝐸(2𝜋𝑅𝑙) = !!"#$
!!= !"
!!
𝐸 = !
!!!!
!
!
InfinitePlaneofChargeVerylarge,butverythinflatplaneofcharge.σ = charge per unit area = dQ/dA Determine𝐸!⃗ atpointsneartheplane ∮𝐸!⃗ ∙ 𝑑𝐴 = 2𝐸𝐴 = !!"#$
!!= !"
!!
𝑄!"#$ = 𝜎𝐴 𝐸 = !
!!!
𝐸 !!!⃗ nearanyconductingsurface ∮𝐸 ∙ 𝑑𝐴 = 𝐸𝐴 = !!"#$
!!= !"
!! ∴ 𝐸 = !
!!→ 𝑎𝑡 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑜𝑓 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟
Conductorwithachargeinsideacavity
𝑄!"# = 𝑄 + 𝑞
(Anemptycavityinsideachargedconductorcarrieszeronetcharge)
𝑟!
𝑟!
𝑅
𝐸
𝑙
𝑞!
−
−
−−
−
−
𝑄!