From last times. MTE1 results. Quiz 1. GAUSS LAW for any closed surface. What is the Electric Flux? How to calculate Electric Flux?

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Allen Hutchinson
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om last times MTE1 esults Mean 75% = 90/120 Electic chages and foce Electic

on Gauss law and conductos in electostatic equilibium Wok, potential eneg

an o a D, talk to us and ou Ts fo suggestions on how to impove What is the

to suface) x (suface aea) Electic flux " E though a suface: (component of

diected espect to each piece of suface though which ou have to calculate it

closed suface depends ONLY on the chage enclosed b suface ut ou need to

that cosses the suface because of the scala poduct in the suface integal 2

1 om last times MTE1 esults Mean 75% = 90/120 Electic chages and foce Electic ield and was to calculate it Motion of chages in E-field Gauss Law Toda: Moe on Gauss law and conductos in electostatic equilibium Wok, potential eneg Electic potential Scoes available at Lean@UW, ou Ts have exams If ou scoe is an o a D, talk to us and ou Ts fo suggestions on how to impove What is the Electic lux? (Component of E-field How to calculate Electic lux? to suface) x (suface aea) Electic flux " E though a suface: (component of E-field to suface) x (suface aea) So ou need to undestand how the E-field is diected espect to each piece of suface though which ou have to calculate it Ecos is component of E-field pependicula to suface Wh do we need to calculate it? to use Gauss law to detemine E sˆ uiz 1 GUSS LW fo an closed suface lux thu closed suface depends ONLY on the chage enclosed b suface ut ou need to calculate the lux to detemine E The lux depends on the nomal component of E that cosses the suface because of the scala poduct in the suface integal 2 3) 2) 1) Rank fluxes though the blue dashed spheical sufaces fo the 3 cases: 3 < < ) 1)<2)=3) a 3"0 "0 "0 ) 3)<1)<2) lux thu sphees: C) 1) = 3) < 2) 4 " 2 E() D) 1) < 2) < 3)

Conductos in equilibium electostatic equilibium Popet 2: Chage on onl conducto suface chose a gaussian suface inside the conducto (as close to the suface as desied) In a conducto

eside on the suface (cannot be inside) Conducto slab in an extenal field E: if E-field not null inside the conducto, fee electons would be acceleated Ein These electons would not

When the extenal field is applied, the Etot =0 Etot = EEin= 0 E=0 electons edistibute until the magnitude of the intenal field equals the magnitude of the extenal field The total

acceleate along the suface NO equilibium ppling Gauss s law 1st Newton s law: n object at est will emain at est and a moving object will continue to move at constant velocit on a

2nd Newton s law(supeposition pinciple applies): net = ma moving paticle has kinetic eneg Wok: dw = " ds = dscos# ield lines ds 1 K= mv 2 2 Infinitesimal displacement >0: oce is

to lift the block done b extenal agent (): W = mgh = mg(f-i) > 0 (the agent has to make an effot to move the block) Wok done b gavitational field: Wg = -mgh Consevative oces: wok

2 Conductos in equilibium electostatic equilibium Popet 2: Chage on onl conducto suface chose a gaussian suface inside the conducto (as close to the suface as desied) In a conducto in electostatic equilibium thee is no net motion of chage Thee is no net flux though the gaussian suface (since E=0) Popet 1: E=0 evewhee inside the conducto n net chage must eside on the suface (cannot be inside) Conducto slab in an extenal field E: if E-field not null inside the conducto, fee electons would be acceleated Ein These electons would not be in equilibium. When the extenal field is applied, the Etot =0 Etot = EEin= 0 E=0 electons edistibute until the magnitude of the intenal field equals the magnitude of the extenal field The total field inside the conducto is zeo Remembe fom P207 Magnitude and diection of E-ield E-field pependicula to the suface: paallel component to E would put foce on chages chages would acceleate along the suface NO equilibium ppling Gauss s law 1st Newton s law: n object at est will emain at est and a moving object will continue to move at constant velocit on a staight line if =0. 2nd Newton s law(supeposition pinciple applies): net = ma moving paticle has kinetic eneg Wok: dw = " ds = dscos# ield lines ds 1 K= mv 2 2 Infinitesimal displacement >0: oce is in diection of motion <0: oce is opposite to diection of motion =0: oce is pependicula to diection of motion Total wok and wok-eneg theoem: W = # " ds = K $ K Potential eneg Wok to lift the block done b extenal agent (): W = mgh = mg(f-i) > 0 (the agent has to make an effot to move the block) Wok done b gavitational field: Wg = -mgh Consevative oces: wok done b the foce is independent on the path and depends onl on the stating and ending locations (if the initial and final point coincide => W = 0) f Conside binging two positive chages togethe The epel each othe oce is consevative Pushing them togethe equies wok Stop afte some distance It is possible to define the potential eneg U Wconsevative = - U = Uinitial - Ufinal = K h i Electic foce wok g = mg "Ug = Ug - U0 = -mgh Ug = U0 mg fo gavitational foce and U0 =U(=0) = 0 How much wok was done?

3 Calculating the wok E.g. Keep fixed, push at constant velocit Net foce on? Zeo oce fom hand on? Total wok done b hand oce in diection of motion 1 4"# o R 2 " dx # = k 1 e ( R $ x) dx = # k 1 2 e R $ x xinitial Coulomb R initial final = R $ $ k e R $ = final $ k e initial > 0 final < initial " 1 final > 1 initial x fo like chages Potential eneg of 2 point chages If we initiall put 1 ve fa awa (initial = ) and move 1 at distance final =12 fom 2: W ext final " k e initial 12 The extenal agent makes positive wok (because chages epel) and changes the potential eneg of the sstem Wext = U = Ufinal - Uinitial >0 => U inceases. The eneg is stoed in the electic field as electic potential eneg. We can set: Uinitial = U() = 0 (at infinite distance foce becomes null). Hence, electic potential eneg of 2 chages at distance : U elec 1 2 Moe about U of 2 chages Like chages U > 0 and wok must be done to bing the chages togethe since the epel (W>0) Unlike chages U < 0 and wok is done to keep the chages apat since the attact one the othe (W<0) Two balls of equal mass and equal chage ae held fixed a distance 12 apat, then suddenl eleased. The fl awa fom each othe, each ending up moving at some constant speed. If the initial distance between them is educed b a facto of fou, thei final speeds ae. Two times bigge. ou times bigge C. Two times smalle D. ou times smalle E. None of the above W el = "#U = U initial "U final =0 initiall fixed =0 ve fa awa W el = "K = K final # K initial = K final = 1 2 mv 2 12 if 12 = 1/4 12 => K final = 4 Kfinal and v final = 2 vfinal 12 U with multiple chages If thee ae moe than two chages, then find U fo each pai of chages and add them o thee chages: How much wok would it take YOU to assemble 3 negative chages?. W = 19.8 mj. W = 0 mj C. W = mj 3µC 1µC 2µC

Equipotential lines Electic Potential Wok done b E-field poduced b some souce chage on a test chage q to move it along a path: W el = "#U = Define: Coulomb $ ds = % qe $ds = q % E $ ds "U /q # V =

souce chage sstem V connected to E due to souce chage (moe in this week laboato) Lines of constant potential In 3D, sufaces of constant potential (ou Lab) Electic potential at point chage In the

4 Equipotential lines Electic Potential Wok done b E-field poduced b some souce chage on a test chage q to move it along a path: W el = "#U = Define: Coulomb $ ds = % qe $ds = q % E $ ds "U /q # V = Electic potential V scala quantit and units volts 1 V = 1 J/C V ceated b some souce chage o chage distibution and independent on test chage V can be used to detemine potential eneg of chage q and souce chage sstem V connected to E due to souce chage (moe in this week laboato) Lines of constant potential In 3D, sufaces of constant potential (ou Lab) Electic potential at point chage In the igue, q1 is a negative souce chage and q2 is a test chage. If q2 is initiall positive and then is changed to a negative chage of the same magnitude, the potential at the position of q2 due to q1 (.) inceases (.) deceases (C.) emains the same Conside one chage as ceating electic potential, the othe chage as expeiencing it q - Uq ( ) Vq ( ) = q Uq ( ) = ke q nswe: (c). The potential is established onl b the souce chage and is independent of the test chage. Electic Potential of point chage Eve point in space has a numeical value fo the electic potential V= k Electic potential eneg=qov % Distance fom souce chage Potential eneg: U = q0v (q0 test chage) V > V Means that wok must be done to move the test chage q0 fom to to ovecome the Coulomb epulsive foce. Wok done = qov-qov = " Coulomb d l #( ) Diffeential fom: qo dv = "Coulomb d l Two points in space and have electic potential V=20 volts and V=100 volts. How much wok does it take to move a 100"C chage fom to?. 2 mj. -20 mj qo > 0 x C. 8 mj D. 100 mj E mj

Supeposition: electic potential of dipole x=-a Supeposition of potential fom

chages V = ke " i Van de Gaaff geneato The high-voltage electode is a hollow

emit light Conside a small chage element dq.

metal dome mounted on an insulated column Chage is deliveed continuousl to

epeated tips of the belt Potons can be acceleated though such lage potentials

5 Supeposition: electic potential of dipole x=-a Supeposition of potential fom potential fom - x=a - Eg:V(=0) = k/a-k/a=0 In geneal: fo a goup of point chages V = ke " i Van de Gaaff geneato The high-voltage electode is a hollow "U = q"v = #"K = K f Spaks: electons fom od to dome excite ai molecules that emit light Conside a small chage element dq. The potential at some point due to this chage element is Total potential: = metal dome mounted on an insulated column Chage is deliveed continuousl to dome b a moving belt of insulating mateial Lage potentials can be developed b epeated tips of the belt Potons can be acceleated though such lage potentials Electic Potential of a continuous chage distibution V qi i This value of V used the efeence that V = 0 at infinite distance

MTE1 results. Mean 75% = 90/120

MTE1 results. Mean 75% = 90/120 MTE1 results Mean 75% = 90/120 Scores available at Learn@UW, your TAs have exams If your score is an F or a D, talk to us and your TAs for suggestions on how to improve From last times Electric charges