Review & Summary. Field Due to a Point Charge The magnitude of the electric set up by a point charge q at a distance r from the charge is

Size: px
Start display at page:

Download "Review & Summary. Field Due to a Point Charge The magnitude of the electric set up by a point charge q at a distance r from the charge is"

Transcription

1 CHATE 22 ELECTIC FIELDS viw & Summr Elctric Fil To lin th lctrosttic forc btwn two chrgs, w ssum tht ch chrg sts u n lctric fil in th sc roun it. Th forc cting on ch chrg is thn u to th lctric fil st u t its loction b th othr chrg. Dfinition of Elctric Fil Th lctric fil E : t n oint is fin in trms of th lctrosttic forc F : tht woul b rt on ositiv tst chrg q 0 lc thr: E : F: q 0. (22-1) Elctric Fil Lins Elctric fil lins rovi mns for visuliing th irction n mgnitu of lctric fils. Th lctric fil vctor t n oint is tngnt to fil lin through tht oint. Th nsit of fil lins in n rgion is roortionl to th mgnitu of th lctric fil in tht rgion. Fil lins origint on ositiv chrgs n trmint on ngtiv chrgs. Fil Du to oint Chrg Th mgnitu of th lctric fil E : st u b oint chrg q t istnc r from th chrg is E q r 2. (22-3) Th irction of is w from th oint chrg if th chrg is ositiv n towr it if th chrg is ngtiv. E : Fil Du to n Elctric Diol An lctric iol consists of two rticls with chrgs of qul mgnitu q but oosit sign, srt b smll istnc. Thir lctric iol momnt : hs mgnitu q n oints from th ngtiv chrg to th ositiv chrg. Th mgnitu of th lctric fil st u b th iol t istnt oint on th iol is (which runs through both chrgs) is E 1 (22-9) 2 0, 3 whr is th istnc btwn th oint n th cntr of th iol. Fil Du to Continuous Chrg Distribution Th lctric fil u to continuous chrg istribution is foun b trting chrg lmnts s oint chrgs n thn summing, vi intgrtion, th lctric fil vctors rouc b ll th chrg lmnts to fin th nt vctor.

2 QUESTIONS 651 Fil Du to Chrg Disk Th lctric fil mgnitu t oint on th cntrl is through uniforml chrg isk is givn b E s , 2 (22-26) whr is th istnc long th is from th cntr of th isk, is th rius of th isk, n s is th surfc chrg nsit. Forc on oint Chrg in n Elctric Fil Whn oint chrg q is lc in n trnl lctric fil E :, th lctrosttic forc F : tht cts on th oint chrg is F : qe :. (22-28) F : Forc hs th sm irction s if q is ositiv n th oosit irction if q is ngtiv. Diol in n Elctric Fil Whn n lctric iol of iol momnt : is lc in n lctric fil E :, th fil rts torqu t : on th iol: t : : E :. (22-34) Th iol hs otntil nrg U ssocit with its orinttion in th fil: U : E :. (22-38) This otntil nrg is fin to b ro whn : is rniculr to E : ; it is lst ( U E) whn : is lign with E : n grtst ( U E) whn : is irct oosit E :. E : Qustions 1 Figur shows thr rrngmnts of lctric fil lins. In ch rrngmnt, roton is rls from rst t oint A n is thn cclrt through oint B b th lctric fil. oints A n B hv qul srtions in th thr rrngmnts. nk th rrngmnts ccoring to th linr momntum of th roton t oint B, grtst first. A B A B A B (c) Figur Qustion 1. 2 Figur shows two +6q 2q +3q squr rrs of chrg rticls. Th squrs, which r cntr on oint, r mislign. 3q +2q Th rticls r s- rt b ithr or /2 2q q long th rimtrs of th squrs. Wht r th mgnitu q q n irction of th nt lctric fil t? +2q 3q 3 In Fig , two rticls of chrg q r rrng smmtricll bout th is; ch roucs n lctric +3q 2q Figur Qustion 2. +6q fil t oint on tht is. Ar th mgnitus of th fils t qul? Is ch lctric fil irct towr or w from th chrg roucing it? (c) Is th mgnitu of th nt lctric fil t qul to th sum of th mgnitus E of th two fil vctors (is it qul to 2E)? () Do th comonnts of thos two fil vctors or cncl? () Do thir comonnts or cncl? (f) Is q q th irction of th nt fil t tht of th cncling comonnts or th ing comonnts? (g) Wht is th irction of th nt fil? Figur Qustion 3. 4 Figur shows four situtions in which four chrg rticls r vnl sc to th lft n right of cntrl oint. Th chrg vlus r inict. nk th situtions ccor - ing to th mgnitu of th nt lctric fil t th cntrl oint, grtst first. (1) (2) (3) (4) : 5 Figur shows two chrg rticls fi in lc on n is. Whr Figur Qustion 4. on th is (othr thn t n infinit istnc) is thr oint t which thir nt lctric fil is ro: btwn +q 3q th chrgs, to thir lft, or to thir right? Is thr oint of Figur Qustion 5. ro nt lctric fil nwhr off th is (othr thn t n infinit istnc)? 6 In Fig , two inticl circulr nonconucting rings r cntr on th sm lin with thir lns rniculr to th lin. Ech ring 1 ing A 2 3 ing B hs chrg tht is uniforml istribut Figur Qustion 6. long its circumfrnc. Th rings ch rouc lctric fils t oints long th lin. For thr situtions, th chrgs on rings A n B r, rsctivl, (1) q 0 n q 0, (2) q 0 n q 0, n (3) q 0 n q 0. nk th situtions ccoring to th mgnitu of th nt lctric fil t oint 1 miw btwn th rings, oint 2 t th cntr of ring B, n (c) oint 3 to th right of ring B, grtst first. 7 Th otntil nrgis ssocit with four orinttions of n lctric iol in n lctric fil r (1) 5U 0, (2) 7U 0, (3) 3U 0, n (4) 5U 0, whr U 0 is ositiv. nk th orinttions ccoring to th ngl btwn th lctric iol momnt n th lctric fil E : n th mgnitu of th torqu on th lctric iol, grtst first. 8 In Chckoint 4, if th iol rotts from orinttion 1 to orinttion 2, is th work on on th iol b th fil ositiv, ngtiv, or ro? If, inst, th iol rotts from orinttion 1 to orinttion 4, is th work on b th fil mor thn, lss thn, or th sm s in?

3 652 CHATE 22 ELECTIC FIELDS 9 Figur shows two isks n flt ring, ch with th sm uniform chrg Q. nk th objcts ccoring to th mgnitu of th lctric fil th crt t oints (which r t th sm vrticl hights), grtst first. 10 In Fig , n lctron trvls through smll hol in lt A n thn towr lt B. A uniform lctric fil in th rgion btwn th lts thn slows th lctron +q 2 without flcting it. Wht is th +q 1 q irction of th fil? Four othr 3 n rticls similrl trvl through A B smll hols in ithr lt A or lt B n thn into th rgion btwn Figur Qustion 10. th lts. Thr hv chrgs q 1, q 2, n q 3. Th fourth (lbl n) is nutron, which is lctricll nutrl. Dos th s of ch of thos four othr rticls incrs, crs, or rmin th sm in th rgion btwn th lts? 11 In Fig , circulr lstic ro with uniform chrg Q roucs n lctric fil of mgnitu E t th cntr of 2 (c) Figur Qustion 9. 2 curvtur (t th origin). In Figs b, c, n, mor circulr ros, ch with inticl uniform chrgs Q, r until th circl is comlt. A fifth rrngmnt (which woul b lbl ) is lik tht in ct th ro in th fourth qurnt hs chrg Q. nk th fiv rrngmnts ccoring to th mgnitu of th lctric fil t th cntr of curvtur, grtst first. 12 Whn thr lctric iols r nr ch othr, th ch rinc th lctric fil of th othr two, n th thr-iol sstm hs crtin otntil nrg. Figur shows two rrngmnts in which thr lctric iols r si b si. Ech iol hs th sm mgnitu of lctric iol momnt, n th scings btwn jcnt iols r inticl. In which rrngmnt is th otntil nrg of th thr-iol sstm grtr? Figur Qustion Figur shows thr ros, ch with th sm chrg Q sr uniforml long its lngth. os (of lngth L) n b (of lngth L/2) r stright, n oints r lign with thir mioints. o c (of lngth L/2) forms comlt circl bout oint. nk th ros ccoring to th mgnitu of th lctric fil th crt t oints, grtst first. L L/2 L/2 (c) Figur Qustion Figur shows fiv rotons tht r lunch in uniform lctric fil E : ; th mgnitu n irction of th lunch vlocitis r inict. nk th rotons ccoring to th mgnitu of thir cclrtions u to th fil, grtst first. E (c) () 3 m/s b c 10 m/s 16 m/s 5 m/s 7 m/s Figur Qustion 11. Figur Qustion 14. roblms SSM Tutoring roblm vilbl (t instructor s iscrtion) in WilLUS n WbAssign Work-out solution vilbl in Stunt Solutions Mnul WWW Work-out solution is t Numbr of ots inicts lvl of roblm ifficult ILW Intrctiv solution is t Aitionl informtion vilbl in Th Fling Circus of hsics n t flingcircusofhsics.com htt:// Moul 22-1 Th Elctric Fil 1 Sktch qulittivl th lctric fil lins both btwn n outsi two concntric conucting shricl shlls whn uniform ositiv chrg q 1 is on th innr shll n uniform ngtiv chrg q 2 is on th outr. Consir th css q 1 q 2, q 1 q 2, n q 1 q 2.

4 OBLEMS In Fig th lctric fil lins on th lft hv twic th srtion of thos on th right. If th mgnitu of th fil t A is 40 N/C, wht is th mgnitu of th forc on roton t A? Wht is th mgnitu of th fil t B? B A Figur roblm Figur shows two chrg rticls fi in lc on n is with srtion L. Th rtio q 1 /q 2 of thir chrg mgnitus is Figur 22-38b shows th comonnt E nt, of thir nt lctric fil long th is just to th right of rticl 2.Th is scl is st b s 30.0 cm. At wht vlu of 0 is E nt, mimum? If rticl 2 hs chrg q 2 3, wht is th vlu of tht mimum? Moul 22-2 Th Elctric Fil Du to Chrg rticl 3 SSM Th nuclus of lutonium-239 tom contins 94 rotons. Assum tht th nuclus is shr with rius 6.64 fm n with th chrg of th rotons uniforml sr through th shr. At th surfc of th nuclus, wht r th mgnitu n irction (rill inwr or outwr) of th lctric fil rouc b th rotons? 4 Two chrg rticls r ttch to n is: rticl 1 of chrg C is t osition 6.00 cm n rticl 2 of chrg C is t osition 21.0 cm. Miw btwn th rticls, wht is thir nt lctric fil in unit-vctor nottion? 5 SSM A chrg rticl roucs n lctric fil with mgnitu of 2.0 N/C t oint tht is 50 cm w from th rticl. Wht is th mgnitu of th rticl s chrg? 6 Wht is th mgnitu of oint chrg tht woul crt n lctric fil of 1.00 N/C t oints 1.00 m w? 7 SSM ILW WWW In Fig , th q 1 q 2 four rticls form squr of g lngth 5.00 cm n hv chrgs q nc, q nc, q nc, n q nc. In unitvctor nottion, wht nt lctric fil q 4 q 3 o th rticls rouc t th squr s cntr? 8 In Fig , th four rticls Figur roblm 7. r fi in lc n hv chrgs q 1 q 2 5, q 3 3, n q Distnc 5.0 mm. Wht is th q 4 mgnitu of th nt lctric fil t q 1 oint u to th rticls? q 3 9 Figur shows two chrg rticls on n is: q q C t 3.00 m n 2 q C t 3.00 m. Wht r th mgnitu n Figur roblm 8. irction (rltiv to th ositiv irction of th is) of th nt lctric fil rouc t oint t 4.00 m? q q Figur roblm 9. +q 1 q 2 L 11 SSM Two chrg rticls r fi to n is: rticl 1 of chrg q C is t osition 20 cm n rticl 2 of chrg q q 1 is t osition 70 cm.at wht coorint on th is (othr thn t infinit) is th nt lctric fil rouc b th two rticls qul to ro? E nt, (10 8 N/C) Figur roblm Figur shows n unvn rrngmnt of lctrons () n rotons () on circulr rc of rius r 2.00 cm, with ngls u , u , u , n u Wht r th mgnitu n irction (rltiv to th ositiv irction of th is) of th nt lctric fil rouc t th cntr of th rc? 13 Figur shows roton () on th cntrl is through isk with uniform chrg nsit u to css lctrons. Th isk is sn from n g-on viw. Thr of thos lctrons r shown: lctron c t th isk cntr n lctrons s t oosit sis of th isk, t rius from th cntr. Th roton is initill t s (cm) 2 1 s Figur roblm 12. c Figur roblm 13. istnc 2.00 cm from th isk. At tht loction, wht r th mgnitus of th lctric fil E : u to lctron c n th nt lctric fil E : c s,nt u to lctrons s? Th roton is thn mov to /10.0. Wht thn r th mgnitus of : : (c) E c n () E s,nt t th roton s loction? () From n (c) w s tht s th roton gts nrr to th isk, th mgnitu of E : c incrss, s ct. Wh os th mgnitu of E : s,nt from th two si lctrons crs, s w s from n ()? s q 1 q 2 L 14 In Fig , rticl 1 of chrg q q n rticl 2 of chrg q q r fi to n is. As Figur roblm 14. multil of istnc L, t wht coorint on th is is th nt lctric fil of th rticls ro? Sktch th nt lctric fil lins btwn n roun th rticls.

5 654 CHATE 22 ELECTIC FIELDS 15 In Fig , th thr rticls r fi in lc n hv chrgs q 1 q 2 n q 3 2. Distnc 6.00 mm. Wht r th mgnitu n irction of th nt lctric fil t oint u to th rticls? 16 Figur shows lstic ring of rius 50.0 cm. Two smll chrg bs r on th ring: B 1 of chrg 2.00 mc is fi in lc t th lft si; b 2 of chrg 6.00 mc cn b mov long th ring. Th two bs rouc nt lctric fil of ing mgnitu E t th cntr of th ring. At wht ositiv n ngtiv vlu of ngl u shoul B 1 b 2 b osition such tht E N/C? 17 Two chrg bs r on th lstic ring in Fig B Figur roblm 16. 2, which is not shown, is fi in lc on th ring, which hs rius 60.0 cm. B 1, which is not fi in lc, is initill on th is t ngl u 0. It is thn mov to th oosit si, t ngl u 180, through th first n scon qurnts of th coorint sstm. Figur 22-44b givs th comonnt of th nt lctric fil rouc t th origin b th two bs s function of u, n Fig c givs th comonnt of tht nt lctric fil. Th vrticl is scls r st b E s N/C n E s N/C. At wht ngl u is b 2 loct? Wht r th chrgs of b 1 n (c) b 2? E (10 4 N/C) E s E s ing B 1 (c) E (10 4 N/C) Figur roblm 17. E s Figur roblm 15. B 2 2 Moul 22-3 Th Elctric Fil Du to Diol 18 Th lctric fil of n lctric iol long th iol is is roimt b Eqs n If binomil nsion is m of Eq. 22-7, wht is th nt trm in th rssion for th iol s lctric fil long th iol is? Tht is, wht is E nt in th rssion E 1 q E nt? 19 Figur shows n lctric iol. Wht r th mgnitu n irction (rltiv to th ositiv irction of th is) of th iol s lctric fil t oint, loct t istnc r? +q /2 /2 q + Figur roblm 19. r 20 Equtions 22-8 n 22-9 r roimtions of th mgnitu of th lctric fil of n lctric iol, t oints long th iol is. Consir oint on tht is t istnc 5.00 from th iol cntr ( is th srtion istnc btwn th rticls of th iol). Lt E r b th mgnitu of th fil t oint s roimt b Eqs n Lt E ct b th ctul mgnitu.wht is th rtio E r /E ct? 21 SSM Elctric quruol. Figur shows gnric lctric quruol. It consists of two iols with iol momnts tht r qul in + + +q q q +q mgnitu but oosit in irction. Show tht th vlu of E on th is of th quruol for oint istnc from its cntr (ssum + Figur roblm 21. ) is givn b E 3Q 4 0 4, in which Q ( 2q 2 ) is known s th quruol momnt of th chrg istribution. Moul 22-4 Th Elctric Fil Du to Lin of Chrg 22 Dnsit, nsit, nsit. A chrg 300 is uniforml istribut long circulr rc of rius 4.00 cm, which subtns n ngl of 40. Wht is th linr chrg nsit long th rc? A chrg 300 is uniforml istribut ovr on fc of circulr isk of rius 2.00 cm. Wht is th surfc chrg nsit ovr tht fc? (c) A chrg 300 is uniforml istribut ovr th surfc of shr of rius 2.00 cm. Wht is th surfc chrg nsit ovr tht surfc? () A chrg 300 is uniforml sr through th volum of shr of rius 2.00 cm. Wht is th volum chrg nsit in tht shr? 23 Figur shows two rlll nonconucting rings with thir 1 q 2 q ing 1 ing 2 cntrl s long common lin. ing 1 hs uniform chrg q 1 n rius ; ring 2 hs uniform chrg q 2 n th sm rius. Th rings r srt b istnc 3.00.Th nt lctric fil t oint on th Figur roblm 23. common lin, t istnc from ring 1, is ro.wht is th rtio q 1 /q 2? 24 A thin nonconucting ro with uniform istribution of ositiv chrg Q is bnt into comlt circl of rius

6 OBLEMS 655 (Fig ). Th cntrl rniculr is through th ring is is, with th origin t th cntr of th ring. Wht is th mgnitu of th lctric fil u to th ro t 0 n? (c) In trms of,t wht ositiv vlu of is tht mgnitu mimum? () If 2.00 cm n Q 4.00 mc, wht is th mimum mgnitu? Figur roblm Figur shows thr circulr rcs cntr on th origin of co- 4Q +9Q 3 2 orint sstm. On ch rc, th uniforml istribut chrg is givn in +Q trms of Q 2.00 mc. Th rii r givn in trms of 10.0 cm. Wht r th mgnitu n irction (rltiv to th ositiv irction) of th nt lctric fil t th ori- Figur roblm 25. gin u to th rcs? 26 ILW In Fig , thin glss ro +q forms smicircl of rius r 5.00 cm. Chrg is uniforml istribut long th ro, with q 4.50 C in th ur hlf n q 4.50 C in th lowr hlf.wht r th q mgnitu n irction (rltiv to th ositiv irction of th is) of th lctric Figur fil E : t, th cntr of th smicircl? roblm In Fig , two curv lstic ros, on of chrg q n th othr of chrg q, form circl of rius +q 8.50 cm in n ln. Th is sss through both of th conncting oints, n th chrg is istribut uniforml on both ros. If q 15.0 C, wht r th q mgnitu n irction (rltiv to th ositiv irction of th is) of th lctric fil E : rouc t, th Figur cntr of th circl? roblm Chrg is uniforml istribut roun ring of rius 2.40 cm, n th rsulting lctric fil mgnitu E is msur long th ring s cntrl is (rniculr to th ln of th ring). At wht istnc from th ring s cntr is E mimum? 29 Figur shows nonconucting ro with uniforml istribut chrg Q. Th ro forms hlf-circl with rius n roucs n lctric fil of mgnitu E rc t its cntr of curvtur. If th rc is colls to oint t istnc from (Fig b), b wht fctor is th mgnitu of th lctric fil t multili? +Q Figur roblm 29. +Q r 30 Figur shows two concntric rings, of rii n 3.00, tht li on th sm ln. oint lis on th cntrl D is, t istnc D 2.00 from th cntr of th rings. Th smllr ring hs uniforml istribut chrg Q. In trms of Q, wht is th uniforml ' istribut chrg on th lrgr ring if th nt lctric fil t is ro? 31 SSM ILW WWW In Fig , Figur roblm 30. nonconucting ro of lngth L 8.15 cm hs chrg q 4.23 fc q uniforml istribut long its lngth. Wht is th linr chrg nsit of th ro? Wht r th mgnitu L n (c) irction (rltiv to th ositiv irction of th is) Figur roblm 31. of th lctric fil rouc t oint, t istnc 12.0 cm from th ro? Wht is th lctric fil mgnitu rouc t istnc 50 m b () th ro n () rticl of chrg q 4.23 fc tht w us to rlc th ro? (At tht istnc, th ro looks lik rticl.) 32 In Fig , ositiv chrg q 7.81 C is sr uniforml long thin nonconucting ro of lngth L 14.5 cm. Wht r th mgnitu n irction (rltiv to th ositiv irction of th is) of th lctric fil rouc t oint, t istnc 6.00 cm from th ro long its rniculr bisctor? 33 In Fig , smiinfinit nonconucting ro (tht is, infinit in on irction onl) hs uniform linr chrg nsit l. Show tht th lctric fil E : t oint mks n ngl of 45 with th ro n tht this rsult is innnt of th istnc. (Hint: Srtl fin Figur roblm 33. th comonnt of E : rlll to th ro n th comonnt rniculr to th ro.) Moul 22-5 Th Elctric Fil Du to Chrg Disk 34 A isk of rius 2.5 cm hs surfc chrg nsit of 5.3 mc/m 2 on its ur fc. Wht is th mgnitu of th lctric fil rouc b th isk t oint on its cntrl is t istnc 12 cm from th isk? 35 SSM WWW At wht istnc long th cntrl rniculr is of uniforml chrg lstic isk of rius m is th mgnitu of th lctric fil qul to on-hlf th mgnitu of th fil t th cntr of th surfc of th isk? 36 A circulr lstic isk with rius 2.00 cm hs uniforml istribut chrg Q ( ) on on fc. A circulr ring of with 30 mm is cntr on tht fc, with th cntr of tht with t rius r 0.50 cm. In coulombs, wht chrg is contin within th with of th ring? L Figur roblm 32.

7 656 CHATE 22 ELECTIC FIELDS 37 Suos ou sign n rtus in which uniforml chrg isk of rius is to rouc n lctric fil. Th fil mgnitu is most imortnt long th cntrl rniculr is of th isk, t oint t istnc 2.00 from th isk (Fig ). Cost nlsis suggsts tht ou switch to ring of th sm outr rius but with innr rius /2.00 (Fig b). Assum Figur roblm 37. tht th ring will hv th sm surfc chrg nsit s th originl isk. If ou switch to th ring, b wht rcntg will ou crs th lctric fil mgnitu t? 38 Figur shows circulr isk tht is uniforml chrg. Th cntrl is is rniculr to th isk fc, with th origin t th isk. Figur 22-58b givs th mgnitu of th lctric fil long tht is in trms of th mimum mgnitu E m t th isk surfc. Th is scl is st b s 8.0 cm.wht is th rius of th isk? E m 0.5E m 0 Figur roblm 38. (cm) Moul 22-6 A oint Chrg in n Elctric Fil 39 In Millikn s rimnt, n oil ro of rius 1.64 mm n nsit g/cm 3 is susn in chmbr C (Fig ) whn ownwr lctric fil of N/C is li. Fin th chrg on th ro, in trms of. 40 An lctron with s of cm/s ntrs n lctric fil of mgnitu N/C, trvling long fil lin in th irction tht rtrs its motion. How fr will th lctron trvl in th fil bfor stoing momntril, n how much tim will hv ls? (c) If th rgion contining th lctric fil is 8.00 mm long (too short for th lctron to sto within it), wht frction of th lctron s initil kintic nrg will b lost in tht rgion? 41 SSM A chrg clou sstm roucs n lctric fil in th ir nr Erth s surfc. A rticl of chrg C is ct on b ownwr lctrosttic forc of N whn lc in this fil. Wht is th mgnitu of th lctric fil? Wht r th mgnitu n (c) irction of th lctrosttic forc F : l on th roton lc in this fil? () Wht is th mgnitu of th grvittionl forc F : g on th roton? () Wht is th rtio F l /F g in this cs? 42 Humi ir brks own (its molculs bcom ioni) in n lctric fil of N/C. In tht fil, wht is th mgnitu of th lctrosttic forc on n lctron n n ion with singl lctron missing? 43 SSM An lctron is rls from rst in uniform lctric fil of mgnitu N/C. Clcult th cclrtion of th lctron. (Ignor grvittion.) s 44 An lh rticl (th nuclus of hlium tom) hs mss of kg n chrg of 2.Wht r th mgnitu n irction of th lctric fil tht will blnc th grvittionl forc on th rticl? 45 ILW An lctron on th is of n lctric iol is 25 nm from th cntr of th iol. Wht is th mgnitu of th lctrosttic forc on th lctron if th iol momnt is C m? Assum tht 25 nm is much lrgr thn th srtion of th chrg rticls tht form th iol. 46 An lctron is cclrt stwr t m/s 2 b n lctric fil. Dtrmin th fil mgnitu n irction. 47 SSM Bms of high-s rotons cn b rouc in guns using lctric fils to cclrt th rotons. Wht cclrtion woul roton rinc if th gun s lctric fil wr N/C? Wht s woul th roton ttin if th fil cclrt th roton through istnc of 1.00 cm? 48 In Fig , n lctron () is to b rls from rst on th cntrl is of uniforml chrg isk of rius. Th surfc chrg nsit on th isk is 4.00 mc/m 2. Wht is th mgnitu of th lctron s initil cclrtion if it is rls t istnc, /100, Figur roblm 48. n (c) /1000 from th cntr of th isk? () Wh os th cclrtion mgnitu incrs onl slightl s th rls oint is mov closr to th isk? 49 A 10.0 g block with chrg of C is lc in n lctric fil E : (3000î 600ĵ) N/C. Wht r th mgnitu n irction (rltiv to th ositiv irction of th is) of th lctrosttic forc on th block? If th block is rls from rst t th origin t tim t 0, wht r its (c) n () coorints t t 3.00 s? 50 At som instnt th vlocit comonnts of n lctron moving btwn two chrg rlll lts r v m/s n v m/s. Suos th lctric fil btwn th lts is uniform n givn b E : (120 N/C)ĵ. In unit-vctor nottion, wht r th lctron s cclrtion in tht fil n th lctron s vlocit whn its coorint hs chng b 2.0 cm? 51 Assum tht honb is shr of imtr cm with chrg of 45.0 C uniforml sr ovr its surfc. Assum lso tht shricl olln grin of imtr 40.0 mm is lctricll hl on th surfc of th b bcus th b s chrg inucs chrg of 1.00 C on th nr si of th grin n chrg of 1.00 C on th fr si. Wht is th mgnitu of th nt lctrosttic forc on th grin u to th b? Nt, ssum tht th b brings th grin to istnc of mm from th ti of flowr s stigm n tht th ti is rticl of chrg 45.0 C. Wht is th mgnitu of th nt lctrosttic forc on th grin u to th stigm? (c) Dos th grin rmin on th b or os it mov to th stigm? 52 An lctron ntrs rgion of uniform lctric fil with n initil vlocit of 40 km/s in th sm irction s th lctric fil, which hs mgnitu E 50 N/C. Wht is th s of th lctron 1.5 ns ftr ntring this rgion? How fr os th lctron trvl uring th 1.5 ns intrvl?

8 OBLEMS Two lrg rlll cor lts r 5.0 cm rt n hv uniform lctric fil btwn thm s ict in Fig An lctron is rls from th ngtiv lt t th sm tim tht roton is rls from th ositiv lt. Nglct th forc of th rticls on ch othr n fin thir istnc from th ositiv lt whn th ositiv lt Ngtiv lt Figur roblm 53. ss ch othr. (Dos it surris ou tht ou n not know th lctric fil to solv this roblm?) 54 In Fig , n lctron is E Dtcting shot t n initil s of v 0 scrn v m/s, t ngl u from n is. It movs through uniform lctric fil E : (5.00 N/C)ĵ. A scrn for tcting lctrons is osition r- Figur roblm 54. lll to th is, t istnc 3.00 m. In unit-vctor nottion, wht is th vlocit of th lctron whn it hits th scrn? 55 ILW A uniform lctric fil ists in rgion btwn two oositl chrg lts. An lctron is rls from rst t th surfc of th ngtivl chrg lt n striks th surfc of th oosit lt, 2.0 cm w, in tim s. Wht is th s of th lctron s it striks th scon lt? Wht is th mgnitu of th lctric fil? Moul 22-7 A Diol in n Elctric Fil 56 An lctric iol consists of chrgs 2 n 2 srt b 0.78 nm. It is in n lctric fil of strngth N/C. Clcult th mgnitu of th torqu on th iol whn th iol momnt is rlll to, rniculr to, n (c) ntirlll to th lctric fil. 57 SSM An lctric iol consisting of chrgs of mgnitu 1.50 nc srt b 6.20 mm is in n lctric fil of strngth 1100 N/C. Wht r th mgnitu of th lctric iol momnt n th iffrnc btwn th otntil nrgis for iol orinttions rlll n ntirlll to E :? 58 A crtin lctric iol is U s lc in uniform lctric fil E : of mgnitu 20 N/C. Figur givs th otntil nrg U of th iol 0 vrsus th ngl u btwn E : n th iol momnt :. Th vrticl is scl is st b U s J.Wht U s is th mgnitu of :? Figur roblm How much work is rquir to turn n lctric iol 180 in uniform lctric fil of mgnitu E 46.0 N/C if th iol momnt hs mgnitu of C m n th initil ngl is 64? 60 A crtin lctric iol is lc in uniform lctric fil E : of mgnitu 40 N/C. Figur givs th mgnitu t of th torqu on th iol vrsus th ngl u btwn fil E : n th iol momnt :.Th vrticl is scl is st b t s N m. Wht is th mgnitu of :? E : U (10 28 J) (10 28 N m) s E 0 Figur roblm Fin n rssion for th oscilltion frqunc of n lctric iol of iol momnt : n rottionl inrti I for smll mlitus of oscilltion bout its quilibrium osition in uniform lctric fil of mgnitu E. Aitionl roblms 62 Wht is th mgnitu of n lctron s cclrtion in uniform lctric fil of mgnitu N/C? How long woul th lctron tk, strting from rst, to ttin on-tnth th s of light? (c) How fr woul it trvl in tht tim? 63 A shricl wtr ro 1.20 mm in imtr is susn in clm ir u to ownwr-irct tmoshric lctric fil of mgnitu E 462 N/C. Wht is th mgnitu of th grvittionl forc on th ro? How mn css lctrons os it hv? 64 Thr rticls, ch with ositiv chrg Q, form n quiltrl tringl, with ch si of lngth.wht is th mgnitu of th lctric fil rouc b th rticls t th mioint of n si? 65 In Fig , rticl of chrg Q roucs n lctric fil of mgnitu E rt t oint, t istnc from th rticl. In Fig b, tht sm mount of chrg is sr uniforml long circulr rc tht hs rius n subtns n ngl u. Th chrg on th rc roucs n lctric fil of mgnitu E rc t its cntr of curvtur. For wht vlu of u os E rc 0.500E rt? (Hint: You will robbl rsort to grhicl solution.) +Q +Q 66 A roton n n lctron form two cornrs of n quiltrl tringl of si lngth m. Wht is th mgnitu of th nt lctric fil ths two rticls rouc t th thir cornr? 67 A chrg (uniform linr nsit 9.0 nc/m) lis on string tht is strtch long n is from 0 to 3.0 m. Dtrmin th mgnitu of th lctric fil t 4.0 m on th is. 68 In Fig , ight rticls form squr in which istnc 2.0 cm. Th chrgs r q 1 3, q 1 q 2 q 3 q 2, q 3 5, q 4 2, q 5 3, q 6, q 7 5,n q 8. In unitvctor nottion, wht is th nt lctric fil t th squr s cntr? q 8 q 4 69 Two rticls, ch with chrg of mgnitu 12 nc, r t two of th vrtics of n quiltrl tringl with g lngth 2.0 m. Wht is th mgnitu of th lctric fil t th thir 7 q 6 q q 5 vrt if both chrgs r ositiv Figur n on chrg is ositiv n th roblm 68. othr is ngtiv? 70 Th following tbl givs th chrg sn b Millikn t iffrnt tims on singl ro in his rimnt. From th t, clcult th lmntr chrg C C C C C C C C C /2 /2 Figur roblm 65.

9 658 CHATE 22 ELECTIC FIELDS 71 A chrg of 20 nc is uniforml istribut long stright ro of lngth 4.0 m tht is bnt into circulr rc with rius of 2.0 m. Wht is th mgnitu of th lctric fil t th cntr of curvtur of th rc? 72 An lctron is constrin to th cntrl is of th ring of chrg of rius in Fig , with. Show tht th lctrosttic forc on th lctron cn cus it to oscillt through th ring cntr with n ngulr frqunc v A q 4 0 m 3 whr q is th ring s chrg n m is th lctron s mss. 73 SSM Th lctric fil in n ln rouc b ositivl chrg rticl is 7.2(4.0î 3.0ĵ) N/C t th oint (3.0, 3.0) cm n 100î N/C t th oint (2.0, 0) cm.wht r th n coorints of th rticl? (c) Wht is th chrg of th rticl? 74 Wht totl (css) chrg q must th isk in Fig hv for th lctric fil on th surfc of th isk t its cntr to hv mgnitu N/C, th E vlu t which ir brks own lctricll, roucing srks? Tk th isk rius s 2.5 cm. Suos ch surfc tom hs n ffctiv cross-sctionl r of nm 2. How mn toms r n to mk u th isk surfc? (c) Th chrg clcult in rsults from som of th surfc toms hving on css lctron. Wht frction of ths toms must b so chrg? 75 In Fig , rticl 1 (of 3 chrg 1.00 mc), rticl 2 (of chrg 1.00 mc), n rticl 3 (of chrg Q) form n quiltrl tringl of g lngth. For wht 1 2 vlu of Q (both sign n mgnitu) os th nt lctric fil ro- Figur roblms 75 uc b th rticls t th cntr n 86. of th tringl vnish? 76 In Fig , n lctric iol swings from n initil orinttion i (u i 20.0 ) to E finl orinttion f (u f 20.0 ) in uniform trnl lctric fil E :. Th lctric iol f momnt is C m; th fil mgnitu f is N/C. Wht is th chng in i th iol s otntil nrg? 77 A rticl of chrg q 1 is t th origin of n is. At wht loction on th is shoul rticl of chrg 4q 1 b lc so tht th nt lctric fil is ro t 2.0 mm on th is? If, inst, rticl of chrg 4q 1 is lc t tht loction, wht is th irction (rltiv to th ositiv irction of th is) of th nt lctric fil t 2.0 mm? 78 Two rticls, ch of ositiv chrg q, r fi in lc on is, on t n th othr t. Writ n rssion tht givs th mgnitu E of th nt lctric fil t oints on th is givn b. Grh E vrsus for th rng 0 4. From th grh, trmin th vlus of tht giv (c) th mimum vlu of E n () hlf th mimum vlu of E. 79 A clock fc hs ngtiv oint chrgs q, 2q, 3q,..., 12q fi t th ositions of th corrsoning numrls. Th clock hns o not rturb th nt fil u to th oint chrgs.at, i Figur roblm 76. wht tim os th hour hn oint in th sm irction s th lctric fil vctor t th cntr of th il? (Hint: Us smmtr.) 80 Clcult th lctric iol momnt of n lctron n roton 4.30 nm rt. 81 An lctric fil E : with n vrg mgnitu of bout 150 N/C oints ownwr in th tmoshr nr Erth s surfc. W wish to flot sulfur shr wighing 4.4 N in this fil b chrging th shr. Wht chrg (both sign n mgnitu) must b us? Wh is th rimnt imrcticl? 82 A circulr ro hs rius of curvtur 9.00 cm n uniforml istribut ositiv chrg Q 6.25 C n subtns n ngl u 2.40 r. Wht is th mgnitu of th lctric fil tht Q roucs t th cntr of curvtur? 83 SSM An lctric iol with iol momnt : (3.00î 4.00ĵ)( C m) is in n lctric fil E : (4000 N/C)î. Wht is th otntil nrg of th lctric iol? Wht is th torqu cting on it? (c) If n trnl gnt turns th iol until its lctric iol momnt is : ( 4.00î 3.00ĵ)( C m), how much work is on b th gnt? 84 In Fig , uniform, uwr lctric fil E : of mgnitu 2.00 E 10 3 v N/C hs bn st u btwn two 0 horiontl lts b chrging th lowr lt ositivl n th ur L lt ngtivl. Th lts hv Figur roblm 84. lngth L 10.0 cm n srtion 2.00 cm. An lctron is thn shot btwn th lts from th lft g of th lowr lt. Th initil : vlocit v 0 of th lctron mks n ngl u 45.0 with th lowr lt n hs mgnitu of m/s. Will th lctron strik on of th lts? If so, which lt n how fr horiontll from th lft g will th lctron strik? 85 For th t of roblm 70, ssum tht th chrg q on th ro is givn b q n, whr n is n intgr n is th lmntr chrg. Fin n for ch givn vlu of q. Do linr rgrssion fit of th vlus of q vrsus th vlus of n n thn us tht fit to fin. 86 In Fig , rticl 1 (of chrg 2.00 C), rticl 2 (of chrg 2.00 C), n rticl 3 (of chrg 5.00 C) form n quiltrl tringl of g lngth 9.50 cm. ltiv to th ositiv irction of th is, trmin th irction of th forc F : 3 on rticl 3 u to th othr rticls b sktching lctric fil lins of th othr rticls. Clcult th mgnitu of F : In Fig , rticl 1 of chrg q C n rticl 2 of chrg q C r fi t istnc 5.00 cm rt. In unit-vctor nottion, wht is th nt lctric fil t oints A, B, n (c) C? () Sktch th lctric fil lins A 1 B 2 C Figur roblm 87.

Page 1. Question 19.1b Electric Charge II Question 19.2a Conductors I. ConcepTest Clicker Questions Chapter 19. Physics, 4 th Edition James S.

Page 1. Question 19.1b Electric Charge II Question 19.2a Conductors I. ConcepTest Clicker Questions Chapter 19. Physics, 4 th Edition James S. ConTst Clikr ustions Chtr 19 Physis, 4 th Eition Jms S. Wlkr ustion 19.1 Two hrg blls r rlling h othr s thy hng from th iling. Wht n you sy bout thir hrgs? Eltri Chrg I on is ositiv, th othr is ngtiv both

More information

SOLVED EXAMPLES. be the foci of an ellipse with eccentricity e. For any point P on the ellipse, prove that. tan

SOLVED EXAMPLES. be the foci of an ellipse with eccentricity e. For any point P on the ellipse, prove that. tan LOCUS 58 SOLVED EXAMPLES Empl Lt F n F th foci of n llips with ccntricit. For n point P on th llips, prov tht tn PF F tn PF F Assum th llips to, n lt P th point (, sin ). P(, sin ) F F F = (-, 0) F = (,

More information

Floating Point Number System -(1.3)

Floating Point Number System -(1.3) Floting Point Numbr Sstm -(.3). Floting Point Numbr Sstm: Comutrs rrsnt rl numbrs in loting oint numbr sstm: F,k,m,M 0. 3... k ;0, 0 i, i,...,k, m M. Nottions: th bs 0, k th numbr o igits in th bs xnsion

More information

Floating Point Number System -(1.3)

Floating Point Number System -(1.3) Floting Point Numbr Sstm -(.3). Floting Point Numbr Sstm: Comutrs rrsnt rl numbrs in loting oint numbr sstm: F,k,m,M 0. 3... k ;0, 0 i, i,...,k, m M. Nottions: th bs 0, k th numbr o igts in th bs xnsion

More information

Chem 104A, Fall 2016, Midterm 1 Key

Chem 104A, Fall 2016, Midterm 1 Key hm 104A, ll 2016, Mitrm 1 Ky 1) onstruct microstt tl for p 4 configurtion. Pls numrt th ms n ml for ch lctron in ch microstt in th tl. (Us th formt ml m s. Tht is spin -½ lctron in n s oritl woul writtn

More information

Case Study VI Answers PHA 5127 Fall 2006

Case Study VI Answers PHA 5127 Fall 2006 Qustion. A ptint is givn 250 mg immit-rls thophyllin tblt (Tblt A). A wk ltr, th sm ptint is givn 250 mg sustin-rls thophyllin tblt (Tblt B). Th tblts follow on-comprtmntl mol n hv first-orr bsorption

More information

ME 522 PRINCIPLES OF ROBOTICS. FIRST MIDTERM EXAMINATION April 19, M. Kemal Özgören

ME 522 PRINCIPLES OF ROBOTICS. FIRST MIDTERM EXAMINATION April 19, M. Kemal Özgören ME 522 PINCIPLES OF OBOTICS FIST MIDTEM EXAMINATION April 9, 202 Nm Lst Nm M. Kml Özgörn 2 4 60 40 40 0 80 250 USEFUL FOMULAS cos( ) cos cos sin sin sin( ) sin cos cos sin sin y/ r, cos x/ r, r 0 tn 2(

More information

, between the vertical lines x a and x b. Given a demand curve, having price as a function of quantity, p f (x) at height k is the curve f ( x,

, between the vertical lines x a and x b. Given a demand curve, having price as a function of quantity, p f (x) at height k is the curve f ( x, Clculus for Businss nd Socil Scincs - Prof D Yun Finl Em Rviw vrsion 5/9/7 Chck wbsit for ny postd typos nd updts Pls rport ny typos This rviw sht contins summris of nw topics only (This rviw sht dos hv

More information

Elliptical motion, gravity, etc

Elliptical motion, gravity, etc FW Physics 130 G:\130 lctur\ch 13 Elliticl motion.docx g 1 of 7 11/3/010; 6:40 PM; Lst rintd 11/3/010 6:40:00 PM Fig. 1 Elliticl motion, grvity, tc minor xis mjor xis F 1 =A F =B C - D, mjor nd minor xs

More information

PHYS ,Fall 05, Term Exam #1, Oct., 12, 2005

PHYS ,Fall 05, Term Exam #1, Oct., 12, 2005 PHYS1444-,Fall 5, Trm Exam #1, Oct., 1, 5 Nam: Kys 1. circular ring of charg of raius an a total charg Q lis in th x-y plan with its cntr at th origin. small positiv tst charg q is plac at th origin. What

More information

Section 3: Antiderivatives of Formulas

Section 3: Antiderivatives of Formulas Chptr Th Intgrl Appli Clculus 96 Sction : Antirivtivs of Formuls Now w cn put th is of rs n ntirivtivs togthr to gt wy of vluting finit intgrls tht is ct n oftn sy. To vlut finit intgrl f(t) t, w cn fin

More information

Chapter 16. 1) is a particular point on the graph of the function. 1. y, where x y 1

Chapter 16. 1) is a particular point on the graph of the function. 1. y, where x y 1 Prctic qustions W now tht th prmtr p is dirctl rltd to th mplitud; thrfor, w cn find tht p. cos d [ sin ] sin sin Not: Evn though ou might not now how to find th prmtr in prt, it is lws dvisl to procd

More information

CSE 373: AVL trees. Warmup: Warmup. Interlude: Exploring the balance invariant. AVL Trees: Invariants. AVL tree invariants review

CSE 373: AVL trees. Warmup: Warmup. Interlude: Exploring the balance invariant. AVL Trees: Invariants. AVL tree invariants review rmup CSE 7: AVL trs rmup: ht is n invrint? Mihl L Friy, Jn 9, 0 ht r th AVL tr invrints, xtly? Disuss with your nighor. AVL Trs: Invrints Intrlu: Exploring th ln invrint Cor i: xtr invrint to BSTs tht

More information

The Angular Momenta Dipole Moments and Gyromagnetic Ratios of the Electron and the Proton

The Angular Momenta Dipole Moments and Gyromagnetic Ratios of the Electron and the Proton Journl of Modrn hysics, 014, 5, 154-157 ublishd Onlin August 014 in SciRs. htt://www.scir.org/journl/jm htt://dx.doi.org/.436/jm.014.51415 Th Angulr Momnt Diol Momnts nd Gyromgntic Rtios of th Elctron

More information

This Week. Computer Graphics. Introduction. Introduction. Graphics Maths by Example. Graphics Maths by Example

This Week. Computer Graphics. Introduction. Introduction. Graphics Maths by Example. Graphics Maths by Example This Wk Computr Grphics Vctors nd Oprtions Vctor Arithmtic Gomtric Concpts Points, Lins nd Plns Eploiting Dot Products CSC 470 Computr Grphics 1 CSC 470 Computr Grphics 2 Introduction Introduction Wh do

More information

Potential Due to Point Charges The electric potential due to a single point charge at a distance r from that point charge is

Potential Due to Point Charges The electric potential due to a single point charge at a distance r from that point charge is 646 CHATE 24 ELECTIC OTENTIAL Electric otentil Energ The chnge U in the electric potentil energ U of point chrge s the chrge moves from n initil point i to finl point f in n electric fiel is U U f U i

More information

Polygons POLYGONS.

Polygons POLYGONS. Polgons PLYGNS www.mthltis.o.uk ow os it work? Solutions Polgons Pg qustions Polgons Polgon Not polgon Polgon Not polgon Polgon Not polgon Polgon Not polgon f g h Polgon Not polgon Polgon Not polgon Polgon

More information

Review & Summary. Electric Potential The electric potential V at a point P in the electric field of a charged object is

Review & Summary. Electric Potential The electric potential V at a point P in the electric field of a charged object is 77 eview & Summr Electric otentil The electric potentil V t point in the electric fiel of chrge object is W V W q U q, (24-2) where is the work tht woul be one b the electric force on positive test chrge

More information

Integration Continued. Integration by Parts Solving Definite Integrals: Area Under a Curve Improper Integrals

Integration Continued. Integration by Parts Solving Definite Integrals: Area Under a Curve Improper Integrals Intgrtion Continud Intgrtion y Prts Solving Dinit Intgrls: Ar Undr Curv Impropr Intgrls Intgrtion y Prts Prticulrly usul whn you r trying to tk th intgrl o som unction tht is th product o n lgric prssion

More information

Instructions for Section 1

Instructions for Section 1 Instructions for Sction 1 Choos th rspons tht is corrct for th qustion. A corrct nswr scors 1, n incorrct nswr scors 0. Mrks will not b dductd for incorrct nswrs. You should ttmpt vry qustion. No mrks

More information

UNIT # 08 (PART - I)

UNIT # 08 (PART - I) . r. d[h d[h.5 7.5 mol L S d[o d[so UNIT # 8 (PRT - I CHEMICL INETICS EXERCISE # 6. d[ x [ x [ x. r [X[C ' [X [[B r '[ [B [C. r [NO [Cl. d[so d[h.5 5 mol L S d[nh d[nh. 5. 6. r [ [B r [x [y r' [x [y r'

More information

TOPIC 5: INTEGRATION

TOPIC 5: INTEGRATION TOPIC 5: INTEGRATION. Th indfinit intgrl In mny rspcts, th oprtion of intgrtion tht w r studying hr is th invrs oprtion of drivtion. Dfinition.. Th function F is n ntidrivtiv (or primitiv) of th function

More information

# 1 ' 10 ' 100. Decimal point = 4 hundred. = 6 tens (or sixty) = 5 ones (or five) = 2 tenths. = 7 hundredths.

# 1 ' 10 ' 100. Decimal point = 4 hundred. = 6 tens (or sixty) = 5 ones (or five) = 2 tenths. = 7 hundredths. How os it work? Pl vlu o imls rprsnt prts o whol numr or ojt # 0 000 Tns o thousns # 000 # 00 Thousns Hunrs Tns Ons # 0 Diml point st iml pl: ' 0 # 0 on tnth n iml pl: ' 0 # 00 on hunrth r iml pl: ' 0

More information

Ch 1.2: Solutions of Some Differential Equations

Ch 1.2: Solutions of Some Differential Equations Ch 1.2: Solutions of Som Diffrntil Equtions Rcll th fr fll nd owl/mic diffrntil qutions: v 9.8.2v, p.5 p 45 Ths qutions hv th gnrl form y' = y - b W cn us mthods of clculus to solv diffrntil qutions of

More information

( ) Geometric Operations and Morphing. Geometric Transformation. Forward v.s. Inverse Mapping. I (x,y ) Image Processing - Lesson 4 IDC-CG 1

( ) Geometric Operations and Morphing. Geometric Transformation. Forward v.s. Inverse Mapping. I (x,y ) Image Processing - Lesson 4 IDC-CG 1 Img Procssing - Lsson 4 Gomtric Oprtions nd Morphing Gomtric Trnsformtion Oprtions dpnd on Pil s Coordints. Contt fr. Indpndnt of pil vlus. f f (, ) (, ) ( f (, ), f ( ) ) I(, ) I', (,) (, ) I(,) I (,

More information

Walk Like a Mathematician Learning Task:

Walk Like a Mathematician Learning Task: Gori Dprtmnt of Euction Wlk Lik Mthmticin Lrnin Tsk: Mtrics llow us to prform mny usful mthmticl tsks which orinrily rquir lr numbr of computtions. Som typs of problms which cn b on fficintly with mtrics

More information

Lecture contents. Bloch theorem k-vector Brillouin zone Almost free-electron model Bands Effective mass Holes. NNSE 508 EM Lecture #9

Lecture contents. Bloch theorem k-vector Brillouin zone Almost free-electron model Bands Effective mass Holes. NNSE 508 EM Lecture #9 Lctur contnts Bloch thorm -vctor Brillouin zon Almost fr-lctron modl Bnds ffctiv mss Hols Trnsltionl symmtry: Bloch thorm On-lctron Schrödingr qution ch stt cn ccommo up to lctrons: If Vr is priodic function:

More information

b. How many ternary words of length 23 with eight 0 s, nine 1 s and six 2 s?

b. How many ternary words of length 23 with eight 0 s, nine 1 s and six 2 s? MATH 3012 Finl Exm, My 4, 2006, WTT Stunt Nm n ID Numr 1. All our prts o this prolm r onrn with trnry strings o lngth n, i.., wors o lngth n with lttrs rom th lpht {0, 1, 2}.. How mny trnry wors o lngth

More information

CIVL 8/ D Boundary Value Problems - Rectangular Elements 1/7

CIVL 8/ D Boundary Value Problems - Rectangular Elements 1/7 CIVL / -D Boundr Vlu Prolms - Rctngulr Elmnts / RECANGULAR ELEMENS - In som pplictions, it m mor dsirl to us n lmntl rprsnttion of th domin tht hs four sids, ithr rctngulr or qudriltrl in shp. Considr

More information

PARTICLE MOTION IN UNIFORM GRAVITATIONAL and ELECTRIC FIELDS

PARTICLE MOTION IN UNIFORM GRAVITATIONAL and ELECTRIC FIELDS VISUAL PHYSICS ONLINE MODULE 6 ELECTROMAGNETISM PARTICLE MOTION IN UNIFORM GRAVITATIONAL and ELECTRIC FIELDS A fram of rfrnc Obsrvr Origin O(,, ) Cartsian coordinat as (X, Y, Z) Unit vctors iˆˆj k ˆ Scif

More information

INTEGRALS. Chapter 7. d dx. 7.1 Overview Let d dx F (x) = f (x). Then, we write f ( x)

INTEGRALS. Chapter 7. d dx. 7.1 Overview Let d dx F (x) = f (x). Then, we write f ( x) Chptr 7 INTEGRALS 7. Ovrviw 7.. Lt d d F () f (). Thn, w writ f ( ) d F () + C. Ths intgrls r clld indfinit intgrls or gnrl intgrls, C is clld constnt of intgrtion. All ths intgrls diffr y constnt. 7..

More information

HIGHER ORDER DIFFERENTIAL EQUATIONS

HIGHER ORDER DIFFERENTIAL EQUATIONS Prof Enriqu Mtus Nivs PhD in Mthmtis Edution IGER ORDER DIFFERENTIAL EQUATIONS omognous linr qutions with onstnt offiints of ordr two highr Appl rdution mthod to dtrmin solution of th nonhomognous qution

More information

(2) If we multiplied a row of B by λ, then the value is also multiplied by λ(here lambda could be 0). namely

(2) If we multiplied a row of B by λ, then the value is also multiplied by λ(here lambda could be 0). namely . DETERMINANT.. Dtrminnt. Introution:I you think row vtor o mtrix s oorint o vtors in sp, thn th gomtri mning o th rnk o th mtrix is th imnsion o th prlllppi spnn y thm. But w r not only r out th imnsion,

More information

Paths. Connectivity. Euler and Hamilton Paths. Planar graphs.

Paths. Connectivity. Euler and Hamilton Paths. Planar graphs. Pths.. Eulr n Hmilton Pths.. Pth D. A pth rom s to t is squn o gs {x 0, x 1 }, {x 1, x 2 },... {x n 1, x n }, whr x 0 = s, n x n = t. D. Th lngth o pth is th numr o gs in it. {, } {, } {, } {, } {, } {,

More information

Last time: introduced our first computational model the DFA.

Last time: introduced our first computational model the DFA. Lctur 7 Homwork #7: 2.2.1, 2.2.2, 2.2.3 (hnd in c nd d), Misc: Givn: M, NFA Prov: (q,xy) * (p,y) iff (q,x) * (p,) (follow proof don in clss tody) Lst tim: introducd our first computtionl modl th DFA. Tody

More information

Review & Summary. Conservation of Charge The net electric charge of any isolated system is always conserved.

Review & Summary. Conservation of Charge The net electric charge of any isolated system is always conserved. HPTER OULOM S LW Rviw & Smmar Elctric harg Th strngth o a articl s lctrical intraction with objcts aron it ns on its lctric charg (sall rrsnt as q), which can b ithr ositiv or ngativ. Particls with th

More information

Nefertiti. Echoes of. Regal components evoke visions of the past MULTIPLE STITCHES. designed by Helena Tang-Lim

Nefertiti. Echoes of. Regal components evoke visions of the past MULTIPLE STITCHES. designed by Helena Tang-Lim MULTIPLE STITCHES Nrtiti Ehos o Rgl omponnts vok visions o th pst sign y Hln Tng-Lim Us vrity o stiths to rt this rgl yt wrl sign. Prt sping llows squr s to mk roun omponnts tht rp utiully. FCT-SC-030617-07

More information

Module graph.py. 1 Introduction. 2 Graph basics. 3 Module graph.py. 3.1 Objects. CS 231 Naomi Nishimura

Module graph.py. 1 Introduction. 2 Graph basics. 3 Module graph.py. 3.1 Objects. CS 231 Naomi Nishimura Moul grph.py CS 231 Nomi Nishimur 1 Introution Just lik th Python list n th Python itionry provi wys of storing, ssing, n moifying t, grph n viw s wy of storing, ssing, n moifying t. Bus Python os not

More information

Mathematics 1110H Calculus I: Limits, derivatives, and Integrals Trent University, Summer 2018 Solutions to the Actual Final Examination

Mathematics 1110H Calculus I: Limits, derivatives, and Integrals Trent University, Summer 2018 Solutions to the Actual Final Examination Mathmatics H Calculus I: Limits, rivativs, an Intgrals Trnt Univrsity, Summr 8 Solutions to th Actual Final Eamination Tim-spac: 9:-: in FPHL 7. Brought to you by Stfan B lan k. Instructions: Do parts

More information

Winter 2016 COMP-250: Introduction to Computer Science. Lecture 23, April 5, 2016

Winter 2016 COMP-250: Introduction to Computer Science. Lecture 23, April 5, 2016 Wintr 2016 COMP-250: Introduction to Computr Scinc Lctur 23, April 5, 2016 Commnt out input siz 2) Writ ny lgorithm tht runs in tim Θ(n 2 log 2 n) in wors cs. Explin why this is its running tim. I don

More information

CSE 373: More on graphs; DFS and BFS. Michael Lee Wednesday, Feb 14, 2018

CSE 373: More on graphs; DFS and BFS. Michael Lee Wednesday, Feb 14, 2018 CSE 373: Mor on grphs; DFS n BFS Mihl L Wnsy, F 14, 2018 1 Wrmup Wrmup: Disuss with your nighor: Rmin your nighor: wht is simpl grph? Suppos w hv simpl, irt grph with x nos. Wht is th mximum numr of gs

More information

12. Traffic engineering

12. Traffic engineering lt2.ppt S-38. Introution to Tltrffi Thory Spring 200 2 Topology Pths A tlommunition ntwork onsists of nos n links Lt N not th st of nos in with n Lt J not th st of nos in with j N = {,,,,} J = {,2,3,,2}

More information

CSI35 Chapter 11 Review

CSI35 Chapter 11 Review 1. Which of th grphs r trs? c f c g f c x y f z p q r 1 1. Which of th grphs r trs? c f c g f c x y f z p q r . Answr th qustions out th following tr 1) Which vrtx is th root of c th tr? ) wht is th hight

More information

Construction 11: Book I, Proposition 42

Construction 11: Book I, Proposition 42 Th Visul Construtions of Euli Constrution #11 73 Constrution 11: Book I, Proposition 42 To onstrut, in givn rtilinl ngl, prlllogrm qul to givn tringl. Not: Equl hr mns qul in r. 74 Constrution # 11 Th

More information

MASTER CLASS PROGRAM UNIT 4 SPECIALIST MATHEMATICS SEMESTER TWO 2014 WEEK 11 WRITTEN EXAMINATION 1 SOLUTIONS

MASTER CLASS PROGRAM UNIT 4 SPECIALIST MATHEMATICS SEMESTER TWO 2014 WEEK 11 WRITTEN EXAMINATION 1 SOLUTIONS MASTER CLASS PROGRAM UNIT SPECIALIST MATHEMATICS SEMESTER TWO WEEK WRITTEN EXAMINATION SOLUTIONS FOR ERRORS AND UPDATES, PLEASE VISIT WWW.TSFX.COM.AU/MC-UPDATES QUESTION () Lt p ( z) z z z If z i z ( is

More information

Oppgavesett kap. 6 (1 av..)

Oppgavesett kap. 6 (1 av..) Oppgvstt kp. 6 (1 v..) hns.brnn@go.uio.no Problm 1 () Wht is homognous nucltion? Why dos Figur 6.2 in th book show tht w won't gt homognous nucltion in th tmosphr? ˆ Homognous nucltion crts cloud droplts

More information

Lecture 4. Conic section

Lecture 4. Conic section Lctur 4 Conic sction Conic sctions r locus of points whr distncs from fixd point nd fixd lin r in constnt rtio. Conic sctions in D r curvs which r locus of points whor position vctor r stisfis r r. whr

More information

Using the Printable Sticker Function. Using the Edit Screen. Computer. Tablet. ScanNCutCanvas

Using the Printable Sticker Function. Using the Edit Screen. Computer. Tablet. ScanNCutCanvas SnNCutCnvs Using th Printl Stikr Funtion On-o--kin stikrs n sily rt y using your inkjt printr n th Dirt Cut untion o th SnNCut mhin. For inormtion on si oprtions o th SnNCutCnvs, rr to th Hlp. To viw th

More information

MASSACHUSETTS INSTITUTE OF TECHNOLOGY HAYSTACK OBSERVATORY WESTFORD, MASSACHUSETTS

MASSACHUSETTS INSTITUTE OF TECHNOLOGY HAYSTACK OBSERVATORY WESTFORD, MASSACHUSETTS VSRT MEMO #05 MASSACHUSETTS INSTITUTE OF TECHNOLOGY HAYSTACK OBSERVATORY WESTFORD, MASSACHUSETTS 01886 Fbrury 3, 009 Tlphon: 781-981-507 Fx: 781-981-0590 To: VSRT Group From: Aln E.E. Rogrs Subjct: Simplifid

More information

CONIC SECTIONS. MODULE-IV Co-ordinate Geometry OBJECTIVES. Conic Sections

CONIC SECTIONS. MODULE-IV Co-ordinate Geometry OBJECTIVES. Conic Sections Conic Sctions 16 MODULE-IV Co-ordint CONIC SECTIONS Whil cutting crrot ou might hv noticd diffrnt shps shown th dgs of th cut. Anlticll ou m cut it in thr diffrnt ws, nml (i) (ii) (iii) Cut is prlll to

More information

SPH4U Electric Charges and Electric Fields Mr. LoRusso

SPH4U Electric Charges and Electric Fields Mr. LoRusso SPH4U lctric Chargs an lctric Fils Mr. LoRusso lctricity is th flow of lctric charg. Th Grks first obsrv lctrical forcs whn arly scintists rubb ambr with fur. Th notic thy coul attract small bits of straw

More information

COHORT MBA. Exponential function. MATH review (part2) by Lucian Mitroiu. The LOG and EXP functions. Properties: e e. lim.

COHORT MBA. Exponential function. MATH review (part2) by Lucian Mitroiu. The LOG and EXP functions. Properties: e e. lim. MTH rviw part b Lucian Mitroiu Th LOG and EXP functions Th ponntial function p : R, dfind as Proprtis: lim > lim p Eponntial function Y 8 6 - -8-6 - - X Th natural logarithm function ln in US- log: function

More information

13. Binary tree, height 4, eight terminal vertices 14. Full binary tree, seven vertices v 7 v13. v 19

13. Binary tree, height 4, eight terminal vertices 14. Full binary tree, seven vertices v 7 v13. v 19 0. Spnning Trs n Shortst Pths 0. Consir th tr shown blow with root v 0.. Wht is th lvl of v 8? b. Wht is th lvl of v 0? c. Wht is th hight of this root tr?. Wht r th chilrn of v 0?. Wht is th prnt of v

More information

2008 AP Calculus BC Multiple Choice Exam

2008 AP Calculus BC Multiple Choice Exam 008 AP Multipl Choic Eam Nam 008 AP Calculus BC Multipl Choic Eam Sction No Calculator Activ AP Calculus 008 BC Multipl Choic. At tim t 0, a particl moving in th -plan is th acclration vctor of th particl

More information

Lecture 11 Waves in Periodic Potentials Today: Questions you should be able to address after today s lecture:

Lecture 11 Waves in Periodic Potentials Today: Questions you should be able to address after today s lecture: Lctur 11 Wvs in Priodic Potntils Tody: 1. Invrs lttic dfinition in 1D.. rphicl rprsnttion of priodic nd -priodic functions using th -xis nd invrs lttic vctors. 3. Sris solutions to th priodic potntil Hmiltonin

More information

Massachusetts Institute of Technology Department of Mechanical Engineering

Massachusetts Institute of Technology Department of Mechanical Engineering Massachustts Institut of Tchnolog Dpartmnt of Mchanical Enginring. Introduction to Robotics Mid-Trm Eamination Novmbr, 005 :0 pm 4:0 pm Clos-Book. Two shts of nots ar allowd. Show how ou arrivd at our

More information

Chapter 8: Electron Configurations and Periodicity

Chapter 8: Electron Configurations and Periodicity Elctron Spin & th Pauli Exclusion Principl Chaptr 8: Elctron Configurations and Priodicity 3 quantum numbrs (n, l, ml) dfin th nrgy, siz, shap, and spatial orintation of ach atomic orbital. To xplain how

More information

Case Study Vancomycin Answers Provided by Jeffrey Stark, Graduate Student

Case Study Vancomycin Answers Provided by Jeffrey Stark, Graduate Student Cas Stuy Vancomycin Answrs Provi by Jffry Stark, Grauat Stunt h antibiotic Vancomycin is liminat almost ntirly by glomrular filtration. For a patint with normal rnal function, th half-lif is about 6 hours.

More information

a b c cat CAT A B C Aa Bb Cc cat cat Lesson 1 (Part 1) Verbal lesson: Capital Letters Make The Same Sound Lesson 1 (Part 1) continued...

a b c cat CAT A B C Aa Bb Cc cat cat Lesson 1 (Part 1) Verbal lesson: Capital Letters Make The Same Sound Lesson 1 (Part 1) continued... Progrssiv Printing T.M. CPITLS g 4½+ Th sy, fun (n FR!) wy to tch cpitl lttrs. ook : C o - For Kinrgrtn or First Gr (not for pr-school). - Tchs tht cpitl lttrs mk th sm souns s th littl lttrs. - Tchs th

More information

Errata for Second Edition, First Printing

Errata for Second Edition, First Printing Errt for Scond Edition, First Printing pg 68, lin 1: z=.67 should b z=.44 pg 71: Eqution (.3) should rd B( R) = θ R 1 x= [1 G( x)] pg 1: Eqution (.63) should rd B( R) = x= R = θ ( x R) p( x) R 1 x= [1

More information

Errata for Second Edition, First Printing

Errata for Second Edition, First Printing Errt for Scond Edition, First Printing pg 68, lin 1: z=.67 should b z=.44 pg 1: Eqution (.63) should rd B( R) = x= R = θ ( x R) p( x) R 1 x= [1 G( x)] = θp( R) + ( θ R)[1 G( R)] pg 15, problm 6: dmnd of

More information

Fisika Departement Physics Department

Fisika Departement Physics Department Surnm/Vn: Mmo Initils/Voorlttrs: Stunt No/ Stunt Nr: PHY 4 / 63 Totl/Totl : / 3 Th unrsign hrby confirms tht sh/h is fmilir with n ccpts th xmintion rgultions of th Univrsity of Prtori. / Onrgtkn bvstig

More information

The Transfer Function. The Transfer Function. The Transfer Function. The Transfer Function. The Transfer Function. The Transfer Function

The Transfer Function. The Transfer Function. The Transfer Function. The Transfer Function. The Transfer Function. The Transfer Function A gnraliation of th frquncy rsons function Th convolution sum scrition of an LTI iscrt-tim systm with an imuls rsons h[n] is givn by h y [ n] [ ] x[ n ] Taing th -transforms of both sis w gt n n h n n

More information

I. The Connection between Spectroscopy and Quantum Mechanics

I. The Connection between Spectroscopy and Quantum Mechanics I. Th Connction twn Spctroscopy nd Quntum Mchnics On of th postults of quntum mchnics: Th stt of systm is fully dscrid y its wvfunction, Ψ( r1, r,..., t) whr r 1, r, tc. r th coordints of th constitunt

More information

22.615, MHD Theory of Fusion Systems Prof. Freidberg Lecture 8: Effect of a Vertical Field on Tokamak Equilibrium

22.615, MHD Theory of Fusion Systems Prof. Freidberg Lecture 8: Effect of a Vertical Field on Tokamak Equilibrium .65, MHD Thory of usion Systms Prof. ridrg Lctur 8: Effct of Vrticl ild on Tokmk Equilirium Toroidl orc lnc y Mns of Vrticl ild. Lt us riw why th rticl fild is imortnt. 3. or ry short tims, th cuum chmr

More information

Math 61 : Discrete Structures Final Exam Instructor: Ciprian Manolescu. You have 180 minutes.

Math 61 : Discrete Structures Final Exam Instructor: Ciprian Manolescu. You have 180 minutes. Nm: UCA ID Numr: Stion lttr: th 61 : Disrt Struturs Finl Exm Instrutor: Ciprin nolsu You hv 180 minuts. No ooks, nots or lultors r llow. Do not us your own srth ppr. 1. (2 points h) Tru/Fls: Cirl th right

More information

Examples and applications on SSSP and MST

Examples and applications on SSSP and MST Exampls an applications on SSSP an MST Dan (Doris) H & Junhao Gan ITEE Univrsity of Qunslan COMP3506/7505, Uni of Qunslan Exampls an applications on SSSP an MST Dijkstra s Algorithm Th algorithm solvs

More information

Garnir Polynomial and their Properties

Garnir Polynomial and their Properties Univrsity of Cliforni, Dvis Dprtmnt of Mthmtis Grnir Polynomil n thir Proprtis Author: Yu Wng Suprvisor: Prof. Gorsky Eugny My 8, 07 Grnir Polynomil n thir Proprtis Yu Wng mil: uywng@uvis.u. In this ppr,

More information

Lectures 2 & 3 - Population ecology mathematics refresher

Lectures 2 & 3 - Population ecology mathematics refresher Lcturs & - Poultio cology mthmtics rrshr To s th mov ito vloig oultio mols, th olloig mthmtics crisht is suli I i out r mthmtics ttook! Eots logrithms i i q q q q q q ( tims) / c c c c ) ( ) ( Clculus

More information

DUET WITH DIAMONDS COLOR SHIFTING BRACELET By Leslie Rogalski

DUET WITH DIAMONDS COLOR SHIFTING BRACELET By Leslie Rogalski Dut with Dimons Brlt DUET WITH DIAMONDS COLOR SHIFTING BRACELET By Lsli Roglski Photo y Anrw Wirth Supruo DUETS TM from BSmith rt olor shifting fft tht mks your work tk on lif of its own s you mov! This

More information

(A) the function is an eigenfunction with eigenvalue Physical Chemistry (I) First Quiz

(A) the function is an eigenfunction with eigenvalue Physical Chemistry (I) First Quiz 96- Physcl Chmstry (I) Frst Quz lctron rst mss m 9.9 - klogrm, Plnck constnt h 6.66-4 oul scon Sp of lght c. 8 m/s, lctron volt V.6-9 oul. Th functon F() C[cos()+sn()] s n gnfuncton of /. Th gnvlu s (A)

More information

Minimum Spanning Trees

Minimum Spanning Trees Minimum Spnning Trs Minimum Spnning Trs Problm A town hs st of houss nd st of rods A rod conncts nd only houss A rod conncting houss u nd v hs rpir cost w(u, v) Gol: Rpir nough (nd no mor) rods such tht:

More information

The van der Waals interaction 1 D. E. Soper 2 University of Oregon 20 April 2012

The van der Waals interaction 1 D. E. Soper 2 University of Oregon 20 April 2012 Th van dr Waals intraction D. E. Sopr 2 Univrsity of Orgon 20 pril 202 Th van dr Waals intraction is discussd in Chaptr 5 of J. J. Sakurai, Modrn Quantum Mchanics. Hr I tak a look at it in a littl mor

More information

Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 16 CHAPTER 16

Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 16 CHAPTER 16 CHAPTER 16 1. The number of electrons is N = Q/e = ( 30.0 10 6 C)/( 1.60 10 19 C/electrons) = 1.88 10 14 electrons.. The mgnitude of the Coulomb force is Q /r. If we divide the epressions for the two forces,

More information

MATHEMATICS FOR MANAGEMENT BBMP1103

MATHEMATICS FOR MANAGEMENT BBMP1103 Objctivs: TOPIC : EXPONENTIAL AND LOGARITHM FUNCTIONS. Idntif pnntils nd lgrithmic functins. Idntif th grph f n pnntil nd lgrithmic functins. Clcult qutins using prprtis f pnntils. Clcult qutins using

More information

How much air is required by the people in this lecture theatre during this lecture?

How much air is required by the people in this lecture theatre during this lecture? 3 NTEGRATON tgrtio is us to swr qustios rltig to Ar Volum Totl qutity such s: Wht is th wig r of Boig 747? How much will this yr projct cost? How much wtr os this rsrvoir hol? How much ir is rquir y th

More information

Additional Math (4047) Paper 2 (100 marks) y x. 2 d. d d

Additional Math (4047) Paper 2 (100 marks) y x. 2 d. d d Aitional Math (07) Prpar b Mr Ang, Nov 07 Fin th valu of th constant k for which is a solution of th quation k. [7] Givn that, Givn that k, Thrfor, k Topic : Papr (00 marks) Tim : hours 0 mins Nam : Aitional

More information

Review & Summary. e e : g g (annihilation). (21-14) e e

Review & Summary. e e : g g (annihilation). (21-14) e e 6 HPTER OULOM S LW orts Lawrnc rkl Laborator Figr -0 hotograh o trails o bbbls lt in a bbbl chambr b an lctron an a ositron.th air o articls was roc b a gamma ra that ntr th chambr irctl rom th bottom.

More information

1 Introduction to Modulo 7 Arithmetic

1 Introduction to Modulo 7 Arithmetic 1 Introution to Moulo 7 Arithmti Bor w try our hn t solvin som hr Moulr KnKns, lt s tk los look t on moulr rithmti, mo 7 rithmti. You ll s in this sminr tht rithmti moulo prim is quit irnt rom th ons w

More information

Physics 2135 Exam 1 September 23, 2014

Physics 2135 Exam 1 September 23, 2014 Exm Totl Physics 2135 Exm 1 September 23, 2014 Key Printe Nme: 200 / 200 N/A Rec. Sec. Letter: Five multiple choice questions, 8 points ech. Choose the best or most nerly correct nswer. B 1. Object A hs

More information

Week 7: Ch. 11 Semiconductor diodes

Week 7: Ch. 11 Semiconductor diodes Wk 7: Ch. 11 Smiconductor diods Principls o Scintilltion Countrs Smiconductor Diods bsics o smiconductors pur lmnts & dopnts 53 Mtrils ion collction, lkg currnt diod structur, pn, np junctions dpltion

More information

CSC Design and Analysis of Algorithms. Example: Change-Making Problem

CSC Design and Analysis of Algorithms. Example: Change-Making Problem CSC 801- Dsign n Anlysis of Algorithms Ltur 11 Gry Thniqu Exmpl: Chng-Mking Prolm Givn unlimit mounts of oins of nomintions 1 > > m, giv hng for mount n with th lst numr of oins Exmpl: 1 = 25, 2 =10, =

More information

Payroll Direct Deposit

Payroll Direct Deposit Payroll Dirct Dposit Dirct Dposit for mploy paychcks allows cntrs to avoi printing an physically istributing papr chcks to mploys. Dirct posits ar ma through a systm known as Automat Claring Hous (ACH),

More information

CONTINUITY AND DIFFERENTIABILITY

CONTINUITY AND DIFFERENTIABILITY MCD CONTINUITY AND DIFFERENTIABILITY NCERT Solvd mpls upto th sction 5 (Introduction) nd 5 (Continuity) : Empl : Chck th continuity of th function f givn by f() = + t = Empl : Emin whthr th function f

More information

PH427/PH527: Periodic systems Spring Overview of the PH427 website (syllabus, assignments etc.) 2. Coupled oscillations.

PH427/PH527: Periodic systems Spring Overview of the PH427 website (syllabus, assignments etc.) 2. Coupled oscillations. Dy : Mondy 5 inuts. Ovrviw of th PH47 wsit (syllus, ssignnts tc.). Coupld oscilltions W gin with sss coupld y Hook's Lw springs nd find th possil longitudinl) otion of such syst. W ll xtnd this to finit

More information

The Derivative of the Natural Logarithmic Function. Derivative of the Natural Exponential Function. Let u be a differentiable function of x.

The Derivative of the Natural Logarithmic Function. Derivative of the Natural Exponential Function. Let u be a differentiable function of x. Th Ntrl Logrithmic n Eponntil Fnctions: : Diffrntition n Intgrtion Objctiv: Fin rivtivs of fnctions involving th ntrl logrithmic fnction. Th Drivtiv of th Ntrl Logrithmic Fnction Lt b iffrntibl fnction

More information

Binomials and Pascal s Triangle

Binomials and Pascal s Triangle Binomils n Psl s Tringl Binomils n Psl s Tringl Curriulum R AC: 0, 0, 08 ACS: 00 www.mthltis.om Binomils n Psl s Tringl Bsis 0. Intif th prts of th polnomil: 8. (i) Th gr. Th gr is. (Sin is th highst

More information

Phys 201 Midterm 1 S. Nergiz, E.Oğuz, C. Saçlıoğlu T. Turgut Fall '01 No :. Name :. Total Grade :. Grade :. y=a. x=-a +q. x=a -q +q. +Q r.

Phys 201 Midterm 1 S. Nergiz, E.Oğuz, C. Saçlıoğlu T. Turgut Fall '01 No :. Name :. Total Grade :. Grade :. y=a. x=-a +q. x=a -q +q. +Q r. Phs 0 Miterm S. Nergiz, E.ğuz, C. Sçlıoğlu T. Turgut Fll '0 No :. Nme :. Totl Gre :. Gre :. Question : Three point chrges re plce s shown in figure. ) Clculte the coulom force on ech chrge. ) Fin the electrosttic

More information

Constants and Conversions:

Constants and Conversions: EXAM INFORMATION Radial Distribution Function: P 2 ( r) RDF( r) Br R( r ) 2, B is th normalization constant. Ordr of Orbital Enrgis: Homonuclar Diatomic Molculs * * * * g1s u1s g 2s u 2s u 2 p g 2 p g

More information

Exam 1 Solution. CS 542 Advanced Data Structures and Algorithms 2/14/2013

Exam 1 Solution. CS 542 Advanced Data Structures and Algorithms 2/14/2013 CS Avn Dt Struturs n Algorithms Exm Solution Jon Turnr //. ( points) Suppos you r givn grph G=(V,E) with g wights w() n minimum spnning tr T o G. Now, suppos nw g {u,v} is to G. Dsri (in wors) mtho or

More information

Multiple Short Term Infusion Homework # 5 PHA 5127

Multiple Short Term Infusion Homework # 5 PHA 5127 Multipl Short rm Infusion Homwork # 5 PHA 527 A rug is aministr as a short trm infusion. h avrag pharmacokintic paramtrs for this rug ar: k 0.40 hr - V 28 L his rug follows a on-compartmnt boy mol. A 300

More information

MSLC Math 151 WI09 Exam 2 Review Solutions

MSLC Math 151 WI09 Exam 2 Review Solutions Eam Rviw Solutions. Comput th following rivativs using th iffrntiation ruls: a.) cot cot cot csc cot cos 5 cos 5 cos 5 cos 5 sin 5 5 b.) c.) sin( ) sin( ) y sin( ) ln( y) ln( ) ln( y) sin( ) ln( ) y y

More information

Quantum Mechanics & Spectroscopy Prof. Jason Goodpaster. Problem Set #2 ANSWER KEY (5 questions, 10 points)

Quantum Mechanics & Spectroscopy Prof. Jason Goodpaster. Problem Set #2 ANSWER KEY (5 questions, 10 points) Chm 5 Problm St # ANSWER KEY 5 qustios, poits Qutum Mchics & Spctroscopy Prof. Jso Goodpstr Du ridy, b. 6 S th lst pgs for possibly usful costts, qutios d itgrls. Ths will lso b icludd o our futur ms..

More information

cycle that does not cross any edges (including its own), then it has at least

cycle that does not cross any edges (including its own), then it has at least W prov th following thorm: Thorm If a K n is drawn in th plan in such a way that it has a hamiltonian cycl that dos not cross any dgs (including its own, thn it has at last n ( 4 48 π + O(n crossings Th

More information

Calculus II (MAC )

Calculus II (MAC ) Calculus II (MAC232-2) Tst 2 (25/6/25) Nam (PRINT): Plas show your work. An answr with no work rcivs no crdit. You may us th back of a pag if you nd mor spac for a problm. You may not us any calculators.

More information

ECE COMBINATIONAL BUILDING BLOCKS - INVEST 13 DECODERS AND ENCODERS

ECE COMBINATIONAL BUILDING BLOCKS - INVEST 13 DECODERS AND ENCODERS C 24 - COMBINATIONAL BUILDING BLOCKS - INVST 3 DCODS AND NCODS FALL 23 AP FLZ To o "wll" on this invstition you must not only t th riht nswrs ut must lso o nt, omplt n onis writups tht mk ovious wht h

More information

MA1506 Tutorial 2 Solutions. Question 1. (1a) 1 ) y x. e x. 1 exp (in general, Integrating factor is. ye dx. So ) (1b) e e. e c.

MA1506 Tutorial 2 Solutions. Question 1. (1a) 1 ) y x. e x. 1 exp (in general, Integrating factor is. ye dx. So ) (1b) e e. e c. MA56 utorial Solutions Qustion a Intgrating fator is ln p p in gnral, multipl b p So b ln p p sin his kin is all a Brnoulli quation -- st Sin w fin Y, Y Y, Y Y p Qustion Dfin v / hn our quation is v μ

More information

ME311 Machine Design

ME311 Machine Design ME311 Machin Dsign Lctur 4: Strss Concntrations; Static Failur W Dornfld 8Sp017 Fairfild Univrsit School of Enginring Strss Concntration W saw that in a curvd bam, th strss was distortd from th uniform

More information

September 23, Honors Chem Atomic structure.notebook. Atomic Structure

September 23, Honors Chem Atomic structure.notebook. Atomic Structure Atomic Structur Topics covrd Atomic structur Subatomic particls Atomic numbr Mass numbr Charg Cations Anions Isotops Avrag atomic mass Practic qustions atomic structur Sp 27 8:16 PM 1 Powr Standards/ Larning

More information

PHYS-333: Problem set #2 Solutions

PHYS-333: Problem set #2 Solutions PHYS-333: Problm st #2 Solutions Vrsion of March 5, 2016. 1. Visual binary 15 points): Ovr a priod of 10 yars, two stars sparatd by an angl of 1 arcsc ar obsrvd to mov through a full circl about a point

More information