5 Curl-free fields and electrostatic potential

Transcription

1 5 Curl-fr filds and lctrstatic tntial Mathmaticall, w can gnrat a curl-fr vctr fild E(,, ) as E = ( V, V, V ), b taking th gradint f an scalar functin V (r) =V (,, ). Th gradint f V (,, ) is dfind t b th vctr V ( V, V, V ), inting in th dirctin f incrasing V ; in abbrviatd ntatin, curlfr filds E can b indicatd as E = V. Vrificatin: Curl f vctr V is ˆ ŷ ẑ ( V )= V V V =ˆ0 ŷ0 ẑ0 =0. If E = V rrsnts an lctrstatic fild, thn V is calld th lctrstatic tntial. Siml dimnsinal analsis indicats that units f lctrstatic tntial must b vlts (V). 1

2 Th rscritin E = V, including th minus sign (tinal, but takn b cnvntin in lctrstatics), nsurs that lctrstatic fild E ints frm rgins f high tntial t lw tntial as illustratd in th nt aml. Elctrstatic filds E int frm rgins f high V t lw V Eaml 1: Givn an lctrstatic tntial V (,, ) = 2 6 V in a crtain rgin f sac, dtrmin th crrsnding lctrstatic fild E = V in th sam rgin. Slutin: Th lctrstatic fild is E = ( 2 6) = (,, )(2 6) =( 2, 6, 0) = ˆ 2 +ŷ6 V/m. Nt that this fild is dirctd frm rgins f high tntial t lw tntial. Als nt that lctric fild vctrs ar rndicular vrwhr t quitntial cnturs Givn an lctrstatic tntial V (,, ), finding th crrsnding lctrstatic fild E(,, ) is a straightfrward rcdur (taking th ngativ gradint) as alrad illustratd in Eaml 1. Th rvrs ratin f finding V (,, ) frm a givn E(,, ) can b accmlishd b rfrming a vctr lin intgral E dl Light clrs indicat high V dark clrs lw V

3 in 3D sac, sinc, as shwn blw, such intgrals ar ath indndnt fr curl-fr filds E = V. Th vctr lin intgral E dl vr an intgratin ath C tnding frm a int =(,, ) in 3D sac t sm thr int =(,, ) is dfind t b th limiting valu f th sum f dt rducts E j l j cmutd vr all sub-lmnts f ath C having incrmntal lngths l j and unit vctrs l j / l j dirctd frm twards th limiting valu is btaind as all l j arach r (i.., with incrasingl finr subdivisin f C int l j lmnts). Cmutatin f th intgral (s aml blw) invlvs th us f infinitsimal dislacmnt vctrs and vctr dt rduct dl =ˆd +ŷd +ẑd =(d, d, d) E dl =(E,E,E ) (d, d, d) =E d + E d + E d. E j C C l j =(,, ) =(,, ) Th intgral E dl = (E d + E d + E d) 3

4 will in gnral b ath dndnt ct fr whn E is curl-fr. Curl-fr: ath-indndnt Eaml 2: Th fild E =ˆ ± ŷ is curl-fr with th + sign, but nt with as vrifid blw b cmuting E. Calculat th lin intgral f E (fr bth signs, ±) frm int = (0, 0, 0) t int = (1, 1, 0) fr tw diffrnt aths C ging thrugh ints u = (0, 1, 0) and l = (1, 0, 0), rsctivl (s margin). Slutin: First w nt that (ˆ ± ŷ) = ˆ ŷ ẑ ± 0 =ẑ(±1 1) which cnfirms that E =ˆ ± ŷ is curl-fr with with + sign, but nt with. In ithr cas, th intgral t b rfrmd is E dl = (E d + E d + E d) = Fr th first ath C u ging thrugh u = (0, 1, 0), w hav ( d ± d) = 1 =0 (±) d =0 + 1 =0 Fr th scnd ath C l ging thrugh l = (1, 0, 0), w hav ( d ± d) = 1 =0 d =0 ± 1 =0 ( d ± d). d =1 = = 1. d =1 =0± 1=±1. Clarl, th rsult shws that th lin intgral E dl is ath indndnt fr E =ˆ +ŷ which is curl-fr, and ath dndnt fr E =ˆ ŷ in which cas E 0. lin intgrals C u Curl : ath-dndnt lin intgrals C u C l C l 4

5 Th mathmatical rasn wh curl-fr filds hav ath-indndnt lin intgrals is bcaus in ths ccasins th intgrals can b writtn in trms f act diffrntials: fr curl-fr E =ˆ +ŷ w hav E dl as an act diffrntial d + d = d() f th functin, in which cas E dl = = ( ) = 1 vr all aths. fr E = ˆ ŷ with E = 2ẑ 0, n th thr hand, E dl = d d ds nt frm an act diffrntial dv, and thus thr is n ath-indndnt intgral V, nr an undrling tntial functin V. E dl is guarantd t b an act diffrntial if E = V =( V, V, V ), sinc in that cas th diffrntial f V (,, ), naml dv V d+ V d+ V d, is rcisl E d E d E d = E dl. E(r) dl V =0 V = E dl In that cas E dl = E dl = dv = V = V V is indndnt f intgratin ath; thus, if w w call th grund, and st V =0, thn V = E dl dnts th tntial dr frm (an) int t grund. 5

6 Th hsical rasn wh this intgral frmula fr tntial V wrks with an intgratin ath is th rincil f nrg cnsrvatin: intgral E dl rrsnts th wrk dn b fild E r unit charg mvd frm t, s if th lin intgral wr athdndnt thr wuld b was f crating nt nrg b making a charg q fllw scial aths within th lctrstatic fild E, in vilatin f th gnral rincil f nrg cnsrvatin (that rmits nrg cnvrsin but nt cratin r dstructin). E(r) dl V =0 V = E dl As lng as E is curl-fr, lin intgral is ath-indndnt and rducs th vltag dr frm int t "grund". Eaml 3: Givn that V = V (0, 0, 0) = 0 and E =2ˆ +3ŷ + 3( + 1)ẑ V m, dtrmin th lctrstatic tntial V = V (X, Y, Z) at int =(X, Y, Z) in vlts. Slutin: Assuming that th fild is curl-fr (it is), s that an intgratin ath can b usd, w find that X Z Y V = E dl V = = X E dl = 0 2 d,=0 E dl = Y = X 2 0 3(Y + 1)Z. 0 3 d =X,=0 (2 d +3 d + 3( + 1) d) Z 0 3( + 1) d =X,=Y This imlis V (,, ) = 2 3( + 1) V. 6

7 Nt that ( 2 3( + 1)) = ( 2 + 3( + 1)) = ˆ2 +ŷ3 +ẑ3( + 1) ilds th riginal fild E, which is an indicatin that E is indd curl-fr. Eaml 5: Accrding t Culmb s law lctrstatic fild f a rtn with charg Q = (whr is lctrnic charg) lcatd at th rigin is givn as E = 4πɛ r ˆr, 2 whr r = (,, ) and ˆr =. r Dtrmin th lctrstatic tntial fild V stablishd b charg Q = with th rvisin that V 0 as r (i.., grund at infinit). Slutin: Fild E and its tntial V will hibit shrical smmtr in this rblm. Thrfr, with n lss f gnralit, w can calculat th lin intgral frm a int at a distanc r frm th rigin t a int at (th scifid grund) alng, sa, th -ais. Araching th rblm that wa, th tntial dr frm r t is V (r) = =r 4πɛ 2ẑ ẑd = 4πɛ r = 4πɛ r. = r 7

8 T cnvrt lctrstatic tntial V (in vlts) at an int t tntial nrg f a charg q brught t th sam int, it is sufficint t multil V with q (r just th sign f q, dnding n which nrg units w want t us s th nt aml). Eaml 6: In viw f Eaml 5, what ar th tntial nrgis f a rtn and an lctrn lacd at distanc r = a awa frm th rtn at th rigin, whr distanc a 4πɛ 2 = m 2 m stands fr Bhr radius it is th man distanc f th grund stat lctrn in a hdrgn atm frm th cntr f th atm. Rcall that = C and ɛ 10 9 /36π F/m. Slutin: Lt s first valuat th tntial V (r) at r = a: V (a) = 4πɛ a ( )36π 10 9 = π = 27.2 V. Fr th rtn, tntial nrg in Juls is calculatd b multiling V (a) = 27.2 V with q = = C. Hwvr, b rfrring t J f nrg as 1 V (lctrn-vlt), it is mr cnvnint t rfr t tntial nrg V (a) f th rtn at r = a as V (a) = 27.2 V. Likwis, fr a articl with charg q =, i.., an lctrn, tntial nrg at th sam lcatin is V (a) = 27.2 V. ± = a 8