Charge in a Cavity of Conductor

Transcription

1 Tdy s Pln Electic Ptentil Enegy (mesued in Jules Electic Ptentil Ptentil Enegy pe unit Chge (mesued in Vlts). Recll tht the electic field E is fce F pe unit chge. Cpcitnce

2 BB Chge in Cvity f Cnduct A pticle with chge +Q is plced in the cente f n unchged cnducting hllw sphee. Hw much chge will be induced n the inne nd ute sufces f the sphee? A) inne = Q, ute = +Q B) inne = Q/2, ute = +Q/2 C) inne = 0, ute = 0 D) inne = +Q/2, ute = -Q/2 E) inne = +Q, ute = -Q Q Guss Lw: E da Since E=0 in cnduct Q = enc Q enc = 0 ε 0 Wht Gussin sufce shuld we use? 11

3 BB Infinite Cylindes A lng thin wie hs unifm psitive chge density f 2.5 C/m. Cncentic with the wie is lng thick cnducting cylinde, with inne dius 3 cm, nd ute dius 5 cm. The cnducting cylinde hs net line chge density f -4 C/m. Wht is the line chge density f the induced chge n the inne sufce f the cnducting cylinde? A) -6.5 C/m B) -4 C/m C) -2.5 C/m D) -1.5 C/m E) 0 Tw pts Wht is the line chge density f the induced chge n the ute sufce f the cnducting cylinde? A) -4 C/m B) -2.5 C/m C) -1.5 C/m D) 0 C/m E) +2.5 C/m 15 Wht Gussin sufce(s) shuld we use?

4 Electic Ptentil Enegy nd Electic Ptentil Cnsevtive Fces nd Enegy Cnsevtin Ttl enegy is cnstnt nd is the sum f kinetic(k) nd ptentil enegies(u); E= K+U A vey pweful tl in mechnics (Physics 170) Electic Ptentil Enegy Electic Ptentil

5 Cnsevtin f Enegy (f pticle) fm Phys 170 Kinetic Enegy (K) nn-eltivistic Ptentil Enegy (U) detemined by fce lw K = 1 mv 2 f Cnsevtive Fces: K+U is cnstnt ttl enegy is lwys cnstnt exmples f cnsevtive fces gvity; gvittinl ptentil enegy spings; ciled sping enegy (Hke s Lw): U(x)=½kx 2 electic; electic ptentil enegy (tdy s tpic!) exmples f nn-cnsevtive fces (het) fictin viscus dmping (teminl velcity) 2 U ( x, y, z )

6 Exmple: Gvity t sufce f eth. Gvittinl fce is cnsevtive. F = mg = cnst. U = mgy gvittinl ptentil enegy U + K = cnstnt dp bll fm : U lge; K smll t b: K lge; U smll eveywhee: U + K = cnst. U + K = U b + K b Enegy cnseved. y b F = mg Cn slve this pblem tw wys: using fces nd Newtn s Lws using cnsevtin f E. Abslute ptentil enegy is bity; nly diffeences in ptentil enegy hve physicl mening me ptentil enegy

7 Exmple: Chge in unifm E field. Electicl fce ls cnsevtive. F unifm electic field: F = E = cnst. U = Ey electicl ptentil enegy elese chge fm est t : E nd F dwnwd s it cceletes dwn, K inceses; U deceses. E y F = E me ptentil enegy U + K = cnstnt b Unifm gvity U= mgy Unifm E field U=Ey Tw pblems vey simil.

8 Wk nd ptentil enegy. W b = wk dne by fce in ging fm t b lng pth. W b b = F dl = U = (U b U ) = - W b = + W b b Fdl csθ F θ dl b F u exmple: W b = b F dl = b E dl = b Edl cs0 0 E = b Edl = E dl = E( y U U b = + W b = E(y y b ) b y b ) F = E let y b = 0 nd chse U = 0 t y = 0 U = Ey Abslute ptentil enegy is bity; nly diffeences in ptentil enegy hve physicl mening y b dl

9 Electic ptentil enegy Imgine tw psitive chges, ne with chge, the the with chge 0 : 0 Initilly the chges e vey f pt, s we sy tht the initil ptentil enegy U i is ze (we e fee t define the enegy ze smewhee) b Cnside fixed nd mves dilly fm t b b b b W = F dl = E d = E d cs0 b = Cn identify: b 1 4πε U 2 d = = 4πε 1 4πε 1 ( b 1 ) Like: = U U U b Gmm = And independent f pth!

10 Electic ptentil enegy Wht if is psitive nd is negtive? Pticles tend t mve t smlle U. like chges epel. unlike chges ttct. U = 4πε 1 Cn ls think f the U s the wk dne by n utside gent t ssemble the chge distibutin stting with the pticles t infinity. F O = - F E W O = -W E = U F O F E

11 Electic Ptentil Enegy Exmple: Wht is the ptentil enegy f this cllectin f chges? Step 1: Bing in +2 fm infinity. This csts nthing. +2 d - d 2d - Step 2: Bing in ne - chge. The fce is ttctive! The wk euied is negtive: U = (2)( ) 4πε d 0 Step 3: Bing in 2nd - chge. It is ttcted t the +2, but epelled fm the the - chge. The ttl wk (ll 3 chges) is 2 (2 )( ) (2 )( ) ( )( ) 1 U = + + = 4 4πε 0d 4πε 0d 4πε 4 0 2d πε0d 2 A negtive munt f wk ws euied t bing these chges fm infinity t whee they e nw (i.e., the ttctive fces between the chges e lge thn the epulsive nes).

12 UIL5A1 Cnside the 3 cllectins f pint chges shwn belw. Which cllectin hs the smllest ptentil enegy? -Q d -Q d d -Q d +Q d d -Q d +Q d -Q +Q () (b) (c) +Q

13 UIL5A1 Cnside the 3 cllectins f pint chges shwn belw. Which cllectin hs the smllest ptentil enegy? -Q d -Q d d -Q d +Q d d -Q d +Q d -Q +Q () (b) (c) +Q We hve t d psitive wk t ssemble the chges in () since they ll hve the sme chge nd will ntully epel ech the. In (b) nd (c), it s nt cle whethe we hve t d psitive negtive wk since thee e 2 ttctive pis nd ne epulsive pi. () U = πε 0 Q d 2 (b) U = 1 4πε 0 Q d 2 (c) U = 1 4πε 0 Q 2 2 d (b) (c) 0 () U

14 UI5PF 5: A Tw chges which e eul in mgnitude, but ppsite in sign e plced t eul distnces fm pint A. 2) If thid chge is dded t the system nd plced t pint A, hw des the electic ptentil enegy f the chge cllectin chnge? ) inceses b) deceses c) desn t chnge

15 Electic ptentil Cnside tht we hve thee chges fixed in spce. The ptentil enegy f n dded test chge 0 t pint P is just Q Q Q Uf 0 t P = 0 k + k + k 1 p 2 p 3 p Q 1 Q 2 Q 3 2p 1p 3p Nte tht this fcts: 0 x (the effects f ll the chges) 0 Just s we peviusly defined the electic field s the fce/chge, we nw define the electic ptentil s the ptentil enegy/chge: V(x,y,z) = U(x,y,z)/ 0 (U = V) U depends n, but V is independent f (cn be + -) Units f electic ptentil e vlts: 1 V = 1 J/C V(x,y,z) is scl field, defined eveywhee in spce.

16 ELECTRIC POTENTIAL f 1 pint chge V ( ) U = ( ) 1 = 4πε ) Wht is electic ptentil distnce 1m fm +1C chge? 1m Q=1C ELECTRIC POTENTIAL f 2 pint chges: supepsitin! ) Wht is electic ptentil distnce L fm fm tw +1C chges? L L L b) Wht if the tw chges e +1C nd 1C??

17 ELECTRIC POTENTIAL b We cn ls clculte the wk t mve the chge fm t b s, W b = U = ( U U ) b = ( Vb V ) = V Vb SO, the wk pe unit chge t mve fm t b euls, b b Als W b = F dl = E dl W b b = V Vb = E dl The ptentil diffeence, V V b, is line integl V V b

18 Exmple; Clculte vltge diffeence between tw psitins eltive t pint chge Use the line integl definitin + V V b = b E = K E dl 2 ˆ b ANSWER = K [ 1/ 1/ b ]

19 A E B C Tw pts 4) Pints A, B, nd C lie in unifm electic field. Wht is the ptentil diffeence between pints A nd B? V AB = V B - V A ) VAB > 0 b) V AB = 0 c) V AB < 0 5) Pint C is t highe ptentil thn pint A. Tue Flse

20 9) A psitive chge is elesed fm est in egin f electic field. The chge mves: ) twds egin f smlle electic ptentil b) lng pth f cnstnt electic ptentil c) twds egin f gete electic ptentil

21 13) If yu wnt t mve in egin f electic field withut chnging yu electic ptentil enegy. Hw wuld yu chse the diectin? Yu wuld hve t mve pependicul t the field if yu wish t mve withut chnging electic ptentil.

22 Clicke execise A single chge ( Q = -1µC) is fixed t the igin. Define pint A t x = + 5m nd pint B t x = +2m. Wht is the sign f the ptentil diffeence -1µC B Á x between A nd B? ( V AB V B - V A ) () V AB < 0 (b) V AB = 0 (c) V AB > 0

23 Clicke execise A single chge ( Q = -1µC) is fixed t the igin. Define pint A t x = + 5m nd pint B t x = +2m. Wht is the sign f the ptentil diffeence -1µC B Á x between A nd B? ( V AB V B - V A ) () V AB < 0 (b) V AB = 0 (c) V AB > 0 The simplest wy t get the sign f the ptentil diffeence is t imgine plcing psitive chge t pint A nd detemining which wy it wuld mve. Remembe tht psitive chge will lwys fll t lwe ptentil. A psitive chge t A wuld be ttcted t the -1µC chge; theefe NEGATIVE wk wuld be dne t mve the chge fm A t B. Yu culd ls detemine the sign diectly fm the definitin: B V V V = E dl AB B A A E dl > 0 Since, V AB <0!!

24 V AB is Independent f Pth = B WAB VAB VB VA E dl 0 A 0 A E B The integl is the sum f the tngentil (t the pth) cmpnent f the electic field lng pth fm A t B. This integl des nt depend upn the exct pth chsen t mve fm A t B. (tue f ny cnsevtive fce) V AB is the sme f ny pth chsen t mve fm A t B (becuse electic fces e cnsevtive).

25 Pth Indepence: Des it elly wk? Cnside cse f cnstnt field: Diect: A - B B V B V A = E dl = A Eh B h A dl θ C E Lng wy und: A - C B V B V A = C A E dl B C E dl = C A ( E( dl ) sinθ ) 0 V B V A = E (sin θ ) = Eh S hee we hve t lest ne exmple f cse in which the integl is the sme f BOTH pths. In fct, it wks f ll pths.

26 E A B C 7) Cmpe the ptentil diffeences between pints A t C nd pints B t C. ) V AC > V BC b) V AC = V BC c) V AC < V BC

27 Clicke Execise 3 A psitive chge Q is mved fm A t B lng the pth shwn by the w. Wht is the sign f the wk dne t mve the chge fm A t B? A B () W AB < 0 (b) W AB = 0 (c) W AB > 0

28 Clicke execise 3 A psitive chge Q is mved fm A t B lng the pth shwn by the w. Wht is the sign f the wk dne t mve the chge fm A t B? A B () W AB < 0 (b) W AB = 0 (c) W AB > 0 A diect clcultin f the wk dne t mve psitive chge fm pint A t pint B is nt esy. Neithe the mgnitude n the diectin f the field is cnstnt lng the stight line fm A t B. But, yu DO NOT hve t d the diect clcultin. Remembe: ptentil diffeence is INDEPENDENT OF THE PATH!! Theefe we cn tke ny pth we wish. Chse pth lng the c f E cicle dl centeed t the chge. Alng this pth = 0 t evey pint!!

29 Electic Ptentil: whee is it ze? S f we hve nly cnsideed ptentil diffeences. Define the electic ptentil f pint in spce s the ptentil diffeence between tht pint nd efeence pint. gd efeence pint is infinity... we ften set V = 0 the electic ptentil is then defined s: V ( ) V V f pint chge t igin, integte in fm infinity lng sme xis, e.g., the x-xis hee is distnce t igin V( ) V V 1 = 4πε Ptentil fm pint chge 0 V( ) V( ) = dl E line integl

30 Ptentil fm N chges The ptentil fm cllectin f N chges is just the lgebic sum f the ptentil due t ech chge septely (this is much esie t clculte thn the net electic field, which wuld be vect sum). = = N V( ) = E dl = E dl = = n= 1 n Q 1 Q 2 Q 3 1p V( ) = N V ( ) = n n= πε 0 N n= 1 n n 2p 3p V t P = 1 4πε Q 1 1p + 1 4πε Q 2 2 p + 1 4πε Q 3 3p P

31 +5 µc -3 µc A 11) Tw chges 1 = + 5 µc, 2 = -3µC e plced t eul distnces fm the pint A. Wht is the electic ptentil t pint A? ) V A < 0 b) V A = 0 c) V A > 0

32 Clicke execise Which f the fllwing chge distibutins pduces V(x) = 0 f ll pints n the x-xis? (we define V(x) 0 t x = ) +2µC +1µC +2µC +1µC +2µC -2µC x x x -2µC () -1µC -1µC (b) -2µC -1µC (c) +1µC

33 Clicke execise Which f the fllwing chge distibutins pduces V(x) = 0 f ll pints n the x-xis? (we define V(x) 0 t x = ) +2µC +1µC +2µC +1µC +2µC -2µC x x x -2µC () -1µC -1µC (b) -2µC -1µC (c) +1µC The key hee is t elize tht t clculte the ttl ptentil t pint, we must nly mke n ALGEBRAIC sum f the individul cntibutins. Theefe, t mke V(x)=0 f ll x, we must hve the +Q nd -Q cntibutins cncel, which mens tht ny pint n the x-xis must be euidistnt fm +2µC nd -2µC nd ls fm +1µC nd -1µC. This cnditin is met nly in cse ()!

34 Shuld lightening ds hve smll lge dius f cuvtue? V is pptinl t 1/R. If yu wnt high vltge t pss thugh the d then use smll dius f cuvtue. Ai is nmlly n insult, hweve f lge E fields (E>3 x 10 6 V/m) it stts cnducting. Empie Stte Building, NYC

35 Summy Ptentil enegy sted in sttic chge distibutin wk we d t ssemble the chges Electic ptentil enegy f chge in the pesence f set f suce chges ptentil enegy f the test chge euls the ptentil fm the suces times the test chge: U = V If we knw the electic field E, B V V = E dl B llws us t clculte the ptentil functin V eveywhee (define V A = 0 bve) A A Ptentil due t n chges: V( ) = N V ( ) = n n= πε 0 N n= 1 n n