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Miss Millie Millie Gauss’s Law Electric Flux ~flux is the electric field passing through a given area~ = ! = = ⊥ = , = ! ≈ ! ! ! ! !!! ∙ Δ ! = ! ∙ ≠ , ≠ o Flux entering an enclosed volume is negative o Flux leaving the volume is positive o If all field lines that enter the volume also leave the volume then !!"# = 0 Gauss’s Law The precise relationship between the electric flux through a closed surface and the net charge, !"#$ ,enclosed within that surface is given by Gauss’s Law: ! ! ∙ = !"#$ ! Recall: ! = = 8.85 ∗ 10 !!" ! ! !! ! !"#$ = ℎ… it does not matter where or how the charge is distributed Gauss + Coulomb Coulomb’s Law from Gauss’s Law ! ! ∙ = ! ! = ! ! = (4 ! ) ∴ !"#$ = ∴ ! = ! ! ∙ = (4 ! ) As ! and are both perpendicular to the surface at all points, and cos = 1… ∴ = 4 ! ! (Electric Field form of Coulombs Law) Gauss’s Law from Coulomb Law ! = 1 4 ! ! (spherical surface) ∴ ! ! ∙ = ! 1 4 ! ! = 4 ! ! (4 ! ) = ! Applications Spherical Conductor A thin spherical shell… Determine the electric field ( ! ); Outside the shell ( r > r ! ) ∮ ! ∙ = (4 ! ) = ! !! = 1 4 ! ! Inside the shell (r < ! ) !"#$ = 0 ∴ = 0 Solid Sphere of Charge A solid sphere, with charge Q distributed uniformly… determine the electric field ( ! ); Outside the shell ( r > r ! ) ∮ ! ∙ = (4 ! ) = ! !! = ! !!!! ! ! ! Inside the shell (r < ! ) ∮ ! ∙ = (4 ! ) ! = !" !" = ℎ !"#$ = ! ! !! ! !! ! ! !! ! ! !! = ! ! !! ! ∴ (4 ! ) = !!"#$ !! = ! ! !! ! ! !! → = ! !!!! ! !! ! Non-uniform charged solid sphere ! = ! ; = Find in terms of ! = 4 ! = ∫ ! = ∫ ( ! )(4 ! ) = 4 ∫ ! = !!" ! ! ! !! ! !! ! ∴ = !! !!!! ! Find ! as a function of inside the sphere !"#$ = ∫ ! = ∫ ( ! )(4 ! ) = ∫ ! !! !!!! ! ! ! 4 ! = ! ! !! ! ! ! ! ! ! ! ∮ ! ∙ = !"#$ / ! ()(4 ! ) = ! ! !!!! ! = !! ! !!!!!! ! Long Uniform Line of Charge λ = charge per unit length Calculate E at points near but outside the wire ∮ ! ∙ = (2) = !!"#$ !! = !" !! = ! !!!! ! ! Infinite Plane of Charge Very large, but very thin flat plane of charge. σ = charge per unit area = dQ/dA Determine ! at points near the plane ∮ ! ∙ = 2 = !!"#$ !! = !" !! !"#$ = = ! !!! ! ! ! near any conducting surface ∮ ∙ = = !!"#$ !! = !" !! ∴ = ! !! → Conductor with a charge inside a cavity !"# = + (An empty cavity inside a charged conductor carries zero net charge) ! ! ! − − − − − − !

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Page 1: Gauss s Law - WordPress.com · Gauss’s Law !" The precise relationship between the electric flux through a closed surface and the net charge, !!"#$ ∴,enclosed within that

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!

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