When should I use $U=QV$ as opposed to $U=\frac{QV}2
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In either case, the correct formula is U=∫Q0Vdq. This is just a mathematical way of saying, "Add up the energy from building up the charge ... PhysicsStackExchangeisaquestionandanswersiteforactiveresearchers,academicsandstudentsofphysics.Itonlytakesaminutetosignup. Signuptojointhiscommunity Anybodycanaskaquestion Anybodycananswer Thebestanswersarevotedupandrisetothetop Home Public Questions Tags Users Unanswered Teams StackOverflowforTeams –Startcollaboratingandsharingorganizationalknowledge. CreateafreeTeam WhyTeams? Teams CreatefreeTeam Teams Q&Aforwork Connectandshareknowledgewithinasinglelocationthatisstructuredandeasytosearch. LearnmoreaboutTeams WhenshouldIuse$U=QV$asopposedto$U=\frac{QV}2$? AskQuestion Asked 4years,8monthsago Modified 4years,8monthsago Viewed 2ktimes 1 $\begingroup$ Inmyelectricitycourse,Iamhavingtroubleunderstandingthedifferenceinbetween$U=QV$and$U=\frac{QV}2$whentalkingaboutenergystoredinasystem. Myideawasthatwhenthepotentialiscreatedbythechargesarrivingtothesystem,wewoulduse$U=\frac{QV}2$,asthechargesthemselvesarebuildingthesystempotentialastheyarrive;ontheotherhand,whenapotentialisimposedfromtheoutside,wewoulduse$U=QV$. electrostaticschargepotentialpotential-energyvoltage Share Cite Improvethisquestion Follow editedDec27,2017at23:40 Qmechanic♦ 174k3434goldbadges440440silverbadges19951995bronzebadges askedDec27,2017at22:50 BeeBee 28911goldbadge55silverbadges1616bronzebadges $\endgroup$ 1 1 $\begingroup$ SayIhaveacell($EMF=\epsilon$),initiallyunchargedcapacitor($C$)andaswitch.Onclosingtheswitch,$W_{cell}=C\epsilon^{2}$whilechangeinenergyofcapacitoris$U=C\epsilon^{2}/2$.Followsfromyourquestion,justthoughtitwasinteresting. $\endgroup$ – ymuf Dec28,2017at5:22 Addacomment | 3Answers 3 Sortedby: Resettodefault Highestscore(default) Datemodified(newestfirst) Datecreated(oldestfirst) 5 $\begingroup$ Asyousaid,ifyouhaveforexampleaparticleofcharge$q$inanexternalelectricpotential$V$,thenitsenergyisgivenby $$E=qV$$ Ontheotherhand,takeacapacitorforexample.Thecharge$Q$thataccumulatesandthevoltageacrossit$V$satisfytherelation $$Q=CV$$ where$C$isthecapacitanceofthecapacitor-merelyaproportionalityconstant.Thenifyouchargethecapacitorfrom$Q=0$to$Q=q$,theenergyyougetisanintegralovertheinfinitesimalcontributions $$E=\int_{0}^{q}V{\rmd}Q=\frac{1}{C}\int_{0}^{q}Q{\rmd}Q=\frac{q^{2}}{2C}=\frac{qV}{2}$$ Ihopethoseconcreteexamplesmadeitclearer. Share Cite Improvethisanswer Follow editedDec27,2017at23:37 answeredDec27,2017at22:59 eranrecheseranreches 4,03411goldbadge1515silverbadges3030bronzebadges $\endgroup$ Addacomment | 3 $\begingroup$ Youuse$U=QV$whenVisbeingsuppliedbychargesotherthantheoneinyourformula(wecallthese"external").Whenyou'retalkingaboutthetheamountofenergystoredinchargesandthevoltageissuppliedbythesamechargesyouraskingabout,thenyouuse$U=\frac{QV}{2}$. Ineithercase,thecorrectformulais$U=\int_0^QV\operatorname{d}q$.Thisisjustamathematicalwayofsaying,"Adduptheenergyfrombuildingupthechargepiecebypiece(eachpieceis$\operatorname{d}q$)."When$V$isn'tchangedbythechargesyou'readding,thenit'saconstantthatcancomeoutsideoftheintegral,givingyou$U=QV$.Whenyou'readdingchargestoacapacitor,ontheotherhand,thenthevoltageandchargearerelatedby$V=q/C$,andwhenyouputthatintotheformulaanddotheintegral,youget$U=\frac{Q^2}{2C}=\frac{QV}{2}$.Foranotherexample,ifyoucansetupthesituationsomehowtomake$V=kq^2$thentheenergywouldbe$U=k\frac{Q^3}{3}$. Share Cite Improvethisanswer Follow editedDec28,2017at17:53 answeredDec27,2017at22:59 SeanE.LakeSeanE.Lake 20.2k22goldbadges3737silverbadges7676bronzebadges $\endgroup$ Addacomment | 0 $\begingroup$ Youhavealreadygiventhecorrectanswerinyourquestion. Share Cite Improvethisanswer Follow answeredDec28,2017at1:03 freecharlyfreecharly 15.4k22goldbadges1414silverbadges3737bronzebadges $\endgroup$ Addacomment | YourAnswer ThanksforcontributingananswertoPhysicsStackExchange!Pleasebesuretoanswerthequestion.Providedetailsandshareyourresearch!Butavoid…Askingforhelp,clarification,orrespondingtootheranswers.Makingstatementsbasedonopinion;backthemupwithreferencesorpersonalexperience.UseMathJaxtoformatequations.MathJaxreference.Tolearnmore,seeourtipsonwritinggreatanswers. Draftsaved Draftdiscarded Signuporlogin SignupusingGoogle SignupusingFacebook SignupusingEmailandPassword Submit Postasaguest Name Email Required,butnevershown PostYourAnswer Discard Byclicking“PostYourAnswer”,youagreetoourtermsofservice,privacypolicyandcookiepolicy Nottheansweryou'relookingfor?Browseotherquestionstaggedelectrostaticschargepotentialpotential-energyvoltageoraskyourownquestion. 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