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Chapter25:ElectricPotentialEnergy Equations: π = βπ !!πΈ β ππ ! π π = !π π ππ βπ = π! ! π ! βπ = β ! πΈ β ππ ! ! !! βπ = βπΈπ π = ! ππΈ β ππ !! π=π π= π! π! π π! π Background: Energyconceptsallowustoconsiderthebehaviorofchargesinanelectricfield. ElectricPotentialEnergy: PotentialEnergyforaSystem: Similartotheπ! .Inthiscase,the chargedparticlegainsπ! whenitis closertoitsownchargeinthefield. Fora+particle,π! ishighnear postivepartofEfield. Inasystemofchargedparticles,the particlesgiveoffforcesonthemsleves andcreatepotentialenergy. π! π! π=π π ElectricPotential ElectricPotential(V)ishowmuchenergyapariclewillaquiretravellingthrougha givenfield.ThefollowingequationshowstheraltionshipofVtoachangingEfield: βπ = β ! πΈ ! β ππ = π! !" ! VinaConstantEField: TheElectricPotentialofapositive chargeexperiencesadecreasewhen itmovesfromapostivetoanegative charge.Theequationforaconstant fieldisdifferentthenwhenits changing: βπ = βπΈπ AbsoluteEPotential: Whenathereisapositioninspace wheretheamountWorkneededto bringapositivechargefrominfinity toapointinspace: π! π! π=π π EquipotentialLines: Showtheareasofhighandlowelectricpotential Question1: AchargeofβuCislocatedonthey-axis,1mfromtheorigin,while secondchargeof+1uCislocatedonthex-axis1mfromtheorigin. Determinethe(a)magnitudeoftheEfiled,(b)theelectricpotential (assumingthepotentialis0ataninfinitedistance),and(c)theenergy neededtobringthe+1uCchargetothispositionfrominfinitelyfar away. Question2: Arodoflengthlislocatedalongthex-axisandhasatotalchargeofQ andalinearchargedensityπ.FindtheEpotentialatapointPlocatedon they-axiswithadistanceafromtheorigin. Question3: Anelectronacceleratesfromrestthroughapotentialdifferenceof2500 V.Whatisthe(a)changeisUoftheelectronand(b)thefinalspeedof theelectron? Question1: (a) ππ πΈ = ! π (!!!)(!!!!) πΈ= (!)! πΈ = 9πΈ3 π/πΆ πΈ = 2 β 9000 π¬ = π. πππ¬π π΅/πͺ (b) ππ π= π (9πΈ9)(1πΈ β 6) π= 1 π = 9000V π = 9000 β 9000 = π π½ (c) π! π= π π! = ππ π! = (1πΈ β 6)(0) π! = π π± Thechargesareidenticalinmagnitude andequallyfarfromtheorigin.One computationisdoneforbothcharges. Efieldlinescomefrompositive chargesandtonegativecharges.This resultsinanEfieldthatpointsleftand up.Theseformthetwolegsofa45-4590triangle.Theratiois1:1: β2. Question2: ! ππ π = π! ! π ππ = πππ₯ ππ πππ₯ ππ = π! = π! π π! + π₯ ! ! πππ₯ π= π! π! + π₯ ! ! ! ππ₯ π = π! π π! + π₯ ! ! π + π π + ππ π = ππ π(π₯π§ ) π Question3: (a) π! π π! = ππ π! = (1.602πΈ β 19)(2500) πΌπ =4E-16V (b) βπ = βπΎ 1 ππβ = ππ£ ! 2 1 4πΈ β 16 = (9.11πΈ β 31)π£ ! 2 π = π. πππ¬π π/π π=