Download Chapter 25: Electric Potential Energy

Chapter25:ElectricPotentialEnergy
Equations:
𝑈 = −𝑞 !!𝐸 ∙ 𝑑𝑠
!
𝑞
𝑉 = !𝑘 𝑟
𝑑𝑞
∆𝑉 = 𝑘! ! 𝑟
!
∆𝑉 = − ! 𝐸 ∙ 𝑑𝑠
!
!
!!
∆𝑉 = −𝐸𝑑
𝑊 = ! 𝑞𝐸 ∙ 𝑑𝑠
!!
𝑈=𝑘
𝑉=
𝑞! 𝑞!
𝑟
𝑈!
𝑞
Background:
Energyconceptsallowustoconsiderthebehaviorofchargesinanelectricfield.
ElectricPotentialEnergy:
PotentialEnergyforaSystem:
Similartothe𝑈! .Inthiscase,the
chargedparticlegains𝑈! whenitis
closertoitsownchargeinthefield.
Fora+particle,𝑈! ishighnear
postivepartofEfield.
Inasystemofchargedparticles,the
particlesgiveoffforcesonthemsleves
andcreatepotentialenergy.
𝑞! 𝑞!
𝑈=𝑘
𝑟
ElectricPotential
ElectricPotential(V)ishowmuchenergyapariclewillaquiretravellingthrougha
givenfield.ThefollowingequationshowstheraltionshipofVtoachangingEfield:
∆𝑉 = −
!
𝐸
!
∙ 𝑑𝑠 = 𝑘!
!"
!
VinaConstantEField:
TheElectricPotentialofapositive
chargeexperiencesadecreasewhen
itmovesfromapostivetoanegative
charge.Theequationforaconstant
fieldisdifferentthenwhenits
changing:
∆𝑉 = −𝐸𝑑
AbsoluteEPotential:
Whenathereisapositioninspace
wheretheamountWorkneededto
bringapositivechargefrominfinity
toapointinspace:
𝑞! 𝑞!
𝑈=𝑘
𝑟
EquipotentialLines:
Showtheareasofhighandlowelectricpotential
Question1:
Achargeof–uCislocatedonthey-axis,1mfromtheorigin,while
secondchargeof+1uCislocatedonthex-axis1mfromtheorigin.
Determinethe(a)magnitudeoftheEfiled,(b)theelectricpotential
(assumingthepotentialis0ataninfinitedistance),and(c)theenergy
neededtobringthe+1uCchargetothispositionfrominfinitelyfar
away.
Question2:
Arodoflengthlislocatedalongthex-axisandhasatotalchargeofQ
andalinearchargedensity𝜆.FindtheEpotentialatapointPlocatedon
they-axiswithadistanceafromtheorigin.
Question3:
Anelectronacceleratesfromrestthroughapotentialdifferenceof2500
V.Whatisthe(a)changeisUoftheelectronand(b)thefinalspeedof
theelectron?
Question1:
(a)
𝑘𝑞
𝐸 = !
𝑟
(!!!)(!!!!)
𝐸=
(!)!
𝐸 = 9𝐸3 𝑁/𝐶
𝐸 = 2 ∗ 9000
𝑬 = 𝟏. 𝟐𝟕𝑬𝟑 𝑵/𝑪
(b)
𝑘𝑞
𝑉= 𝑟
(9𝐸9)(1𝐸 − 6)
𝑉=
1
𝑉 = 9000V
𝑉 = 9000 − 9000 = 𝟎 𝑽
(c)
𝑈!
𝑉= 𝑞
𝑈! = 𝑞𝑉
𝑈! = (1𝐸 − 6)(0)
𝑈! = 𝟎 𝑱
Thechargesareidenticalinmagnitude
andequallyfarfromtheorigin.One
computationisdoneforbothcharges.
Efieldlinescomefrompositive
chargesandtonegativecharges.This
resultsinanEfieldthatpointsleftand
up.Theseformthetwolegsofa45-4590triangle.Theratiois1:1: √2.
Question2:
!
𝑑𝑞
𝑉 = 𝑘!
! 𝑟
𝑑𝑞 = 𝜆𝑑𝑥
𝑑𝑞
𝜆𝑑𝑥
𝑑𝑉 = 𝑘!
= 𝑘!
𝑟
𝑎! + 𝑥 !
!
𝜆𝑑𝑥
𝑉=
𝑘!
𝑎! + 𝑥 !
!
!
𝑑𝑥
𝑉 = 𝑘! 𝜆
𝑎! + 𝑥 !
!
𝒍 + 𝒂 𝟐 + 𝒍𝟐
𝑉 = 𝒌𝒆 𝝀(𝐥𝐧
)
𝒂
Question3:
(a)
𝑈!
𝑞
𝑈! = 𝑞𝑉
𝑈! = (1.602𝐸 − 19)(2500)
𝑼𝒆 =4E-16V
(b)
∆𝑈 = ∆𝐾
1
𝑚𝑔ℎ = 𝑚𝑣 ! 2
1
4𝐸 − 16 = (9.11𝐸 − 31)𝑣 ! 2
𝒗 = 𝟒. 𝟎𝟓𝑬𝟕 𝒎/𝒔
𝑉=