electric potential - wordpress.com · 2017. 3. 28. · electric potential due to point charges! the...
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Miss Millie Millie
Applications ElectricFieldobtainedfromvoltage FindE 𝐸 = !!"
!= !"!
!.!"!!= 1000 !
!
ChargedConductingSphereDetermine𝑉 atadistancerfromthecentreofa chargedconductingsphereofradius𝑟!; 𝑟 > 𝑟!
𝑉! − 𝑉! = ! 𝐸!⃗!!
!!
∙ 𝑑𝑙 = −𝑄4𝜋𝜀!
!𝑑𝑟𝑟!=
𝑄4𝜋𝜀!
!1𝑟!−1𝑟!!
!!
!!
𝑠𝑒𝑡 𝑑𝑙 = 𝑑𝑟 ,𝑉! = 0 𝑎𝑡 𝑟! = ∞ 𝑉 = !
!!!!
!
!
𝑟 = 𝑟! 𝑉 = !
!!!!
!
!!
𝑟 ≤ 𝑟! 𝑉 = !
!!!!
!
!!
Workrequiredtobringtwopositivechargesclosetogether𝑊 = 𝑞(𝑉! − 𝑉!) = 𝑞 !!"
!!− !"
!!!
PotentialabovetwochargesCalculatetheelectricpotentialatpointAWeaddthepotentialsatpointAduetoeachcharge𝑄! and𝑄!
𝑉! = 𝑉!! + 𝑉!! = 𝑘𝑄!𝑟!+ 𝑘
𝑄!𝑟!
PotentialduetoaringofchargeDetermineVatapointPontheaxisofaringadistancexfromitscentre
𝑉 =1
4𝜋𝜀!!𝑑𝑞𝑟=
14𝜋𝜀!
1(𝑥! + 𝑅!)!/!
! 𝑑𝑞 =1
4𝜋𝜀!𝑄
(𝑥! + 𝑅!)!!
PotentialduetoachargeddiscAthinflatdiskofradius𝑅! hasauniformlydistributed𝑄.Determine𝑉atpoint𝑃ontheaxisofthedisk,adistance𝑥fromitscentre.
𝑑𝑞𝑄=2𝜋𝑅 𝑑𝑅𝜋𝑅!!
→ ∴ 𝑑𝑞 = 𝑄(2𝜋𝑅)(𝑑𝑅)
𝜋𝑅!!=2𝑄𝑅 𝑑𝑅𝑅!!
𝑟 = (𝑥! + 𝑅!)!!
𝑉 =1
4𝜋𝜀!!
𝑑𝑞
(𝑥! + 𝑅!)!!=
2𝑄4𝜋𝜀!𝑅!!
!𝑅 𝑑𝑅
(𝑥! + 𝑅!)!!=
!!
!|
𝑄2𝜋𝜀!𝑅!!
(𝑥! + 𝑅!)!!|!!!!!!!
=𝑄
2𝜋𝜀!𝑅!!!(𝑥! + 𝑅!!)
!! − 𝑥!
Note:For𝑥 ≫ 𝑅! thisformulareducesto;
𝑉 ≈𝑄
2𝜋𝜀!𝑅!!!𝑥 !1 +
12
𝑅!!
𝑥! ! –𝑥! =
𝑄4𝜋𝜀!𝑥
Electric Potential �Electric Potential Energy�
~Thechangeinelectricpotentialenergyequalsthenegativeoftheworkdonebytheelectrostaticforce~
𝑊 = 𝐹𝑑 = 𝑞𝐸𝐷𝑈! − 𝑈! = −𝑊 = −𝑞𝐸𝐷
�Electric Potential vs. Potential Energy� ElectricPotential:potentialenergyperunitcharge
𝑉! =𝑈!𝑞
AtsomepointawithchargeqPotentialDifference:
𝑉!" = ∆𝑉 = 𝑉! − 𝑉! =𝑈! − 𝑈!
𝑞= −
𝑊!"
𝑞
Changeinpotentialenergy:
∆𝑈 = 𝑈! − 𝑈! = 𝑞(𝑉! − 𝑉!) = 𝑞𝑉!"
�Electric Potential and Electric Field�
𝑉!" = 𝑉! − 𝑉! = −! 𝐸!⃗!
! ⋅ 𝑑𝑙
(GeneralRelationship)
𝑉!" = −𝐸𝑑(OnlyifEisuniform)
EdeterminedfromV
𝐸!⃗ (𝑥, 𝑦, 𝑧) = 𝐸!⃗! + 𝐸!⃗! + 𝐸!⃗ ! = −∇𝑉(𝑥, 𝑦, 𝑧)
𝑉(𝑥, 𝑦, 𝑧)
⎩⎪⎪⎨
⎪⎪⎧𝐸! =
𝜕𝑉𝜕𝑥
𝐸! =𝜕𝑉𝜕𝑦
𝐸! =𝜕𝑉𝜕𝑧
�Electric Potential due to Point Charges� Theelectricpotentialatadistance𝑟fromasinglepointcharge𝑄canbederiveddirectlyfrom𝑉! − 𝑉! = − ∫𝐸!⃗ ∙ 𝑑𝑙𝐸 = !
!!!!
!!!= 𝑘 !
!! (ElectricFieldduetoasinglepointcharge)
𝑉! − 𝑉! = −! 𝐸!⃗!!
!!∙ 𝑑𝑙 = −
𝑄4𝜋𝜀!
!1𝑟!
𝑑𝑟 =1
4𝜋𝜀!!𝑄𝑟!−𝑄𝑟!!
!!
!!
𝑉 =1
4𝜋𝜀!𝑄𝑟
(Singlepointcharge:𝑉 = 0 @ 𝑟 = ∞)
�Electric Potential due to Any Charge Distribution
Ifwehavenindividualpointcharges,thepotentialatsomepoint(relativeto 𝑉 = 0 @ 𝑟 = ∞)is
𝑉! = !𝑉!
!
!!!
=1
4𝜋𝜀!!
𝑄!𝑟!"
!
!!!
Where𝑟!"isthedistancefromthe𝑖𝑡ℎcharge(𝑄! )tothepointa.
𝑉 =1
4𝜋𝜀!!𝑑𝑞𝑟
Where𝑟isthedistancefrom𝑑𝑞tothepointwhere𝑉isbeingdetermined
← 𝑉!" →= 50𝑉
← 𝑑 →= 5.0𝑐𝑚
𝐸 = ?
𝑄! 𝑄!
𝐴
𝑟!𝑟!
𝑅
(𝑥 !+ 𝑅 !) !!𝑃
𝑥
𝑑𝑞
𝑅
(𝑥 !+ 𝑅 !) !!𝑃
𝑥
𝑑𝑅
𝑅!