( ) + is the distance from the point of interest to the location of the charge q i
|
|
- Lesley Thornton
- 6 years ago
- Views:
Transcription
1
2
3 Elctcal Engy and apactanc 57. Bcaus lctc ocs a consvatv, th kntc ngy gand s qual to th dcas n lctcal potntal ngy, o KE PE q( ).. so th coct choc s (a).. Fom consvaton o ngy, KE + PE KE + PE, o mv mv + q q o v q( ) v + m ( 6. m s) + (.. ) kg Thus, th coct answ s choc m s 4. In a unom lctc ld, th chang n lctc potntal s E ( x), gvng Ex x x x ( ) 5 and t s sn that th coct choc s (d). ( 9 ) (. m. m) 5 m 5 N 5. Wth th gvn spccatons, th capactanc o ths paalll plat capacto wll b ( ) N m. cm d. m κ F F pf and th coct choc s (a). 4 x m cm 6. Th total potntal at a pont du to a st o pont chags q s kq, wh s th dstanc om th pont o ntst to th locaton o th chag q. Not that n ths cas, th pont at th cnt o th ccl s qudstant om th 4 pont chags locatd on th m o th ccl. Not also that q + q + q4 ( ) µ, so w hav kq kq kq kq4 k cnt q + q + q ( q ) ( q + ) + k 4 k q o th total potntal at th cnt o th ccl s just that du to th st chag alon, and th coct answ s choc. 7. In a ss combnaton o capactos, th quvalnt capactanc s always lss than any ndvdual capactanc n th combnaton, manng that choc (a) s als. lso, o a ss combnaton o capactos, th magntud o th chag s th sam on all plats o capactos n th combnaton, makng both chocs (d) and () als. Th potntal dnc acoss th capactanc s Q, wh Q s th common chag on ach capacto n th combnaton. Thus, th lagst potntal dnc (voltag) appas acoss th capacto wth th last capactanc, makng choc th coct answ. 8. Kpng th capacto connctd to th batty mans that th potntal dnc btwn th plats s kpt at a constant valu qual to th voltag o th batty. Snc th capactanc o a paalll plat capacto s κ d, doublng th plat spaaton d, whl holdng oth chaactstcs o th capacto constant, mans th capactanc wll b dcasd by a acto o.
4 58 hapt 6 Th ngy stod n a capacto may b xpssd as U, so whn th potntal dnc s hld constant whl th capactanc s dcasd by a acto o, th stod ngy dcass by a acto o, makng (c) th coct choc o ths quston. 9. Whn th batty s dsconnctd, th s no long a path o chags to us n movng onto o o o th plats o th capacto. Ths mans that th chag Q s constant. Th capactanc o a paalll plat capacto s κ d and th dlctc constant s κ whn th capacto s a lld. Whn a dlctc wth dlctc constant κ s nstd btwn th plats, th capactanc s doubld ( ). Thus, wth Q constant, th potntal dnc btwn th plats, Q, s dcasd by a acto o, manng that choc (a) s a tu statmnt. Th lctc ld btwn th plats o a paalll plat capacto s E d and dcass whn dcass, makng choc () als and lavng (a) as th only coct choc o ths quston.. Onc th capacto s dsconnctd om th batty, th s no path o chags to mov onto o o o th plats, so th chags on th plats a constant, and choc () can b lmnatd. Th capactanc o a paalll plat capacto s κ d, so th capactanc dcass whn th plat spaaton d s ncasd. Wth Q constant and dcasng, th ngy stod n th capacto, U Q ncass, makng choc (a) als and choc tu. Th potntal dnc btwn th plats, Q Q d κ, ncass and th lctc ld btwn th plats, E d Q κ, s constant. Ths mans that both chocs (c) and (d) a als and lavs choc as th only coct spons.. apactancs connctd n paalll all hav th sam potntal dnc acoss thm and th quvalnt capactanc, + q + + L, s lag than th capactanc o any on o th capactos n th combnaton. Thus, choc (c) s a tu statmnt. Th chag on a capacto s Q ( ), so wth constant, but th capactancs dnt, th capactos all sto dnt chags that a popotonal to th capactancs, makng chocs (a),, (d), and () all als. Tho, (c) s th only coct answ.. Fo a ss combnaton o capactos, th magntud o th chag s th sam on all plats o capactos n th combnaton. lso, th quvalnt capactanc s always lss than any ndvdual capactanc n th combnaton. Tho, choc (a) s tu whl chocs and (c) a both als. Th potntal dnc acoss a capacto s Q, so wth Q constant, capactos havng dnt capactancs wll hav dnt potntal dncs acoss thm, wth th lagst potntal dnc bng acoss th capacto wth th smallst capactanc. Ths mans that choc (d) s als and choc () s tu. Thus, both chocs (a) and () a tu statmnts. NSWERS TO EEN NUMBERED ONEPTUL QUESTIONS. hangng th aa wll chang th capactanc and maxmum chag but not th maxmum voltag. Th quston dos not allow you to ncas th plat spaaton. You can ncas th maxmum opatng voltag by nstng a matal wth hgh dlctc stngth btwn th plats. 4. Elctc potntal s a masu o th potntal ngy p unt chag. Elctcal potntal ngy, PE Q, gvs th ngy o th total chag Q. 6. shap pont on a chagd conducto would poduc a lag lctc ld n th gon na th pont. n lctc dschag could most asly tak plac at th pont.
5 Elctcal Engy and apactanc Th a ght dnt combnatons that us all th capactos n th ccut. Ths combnatons and th quvalnt capactancs a: ll th capactos n ss - q + + ll th capactos n paalll - + q + On capacto n ss wth a paalll combnaton o th oth two: q + +, q + +, q + + On capacto n paalll wth a ss combnaton o th oth two: q + +, q + +, q + +. Nothng happns to th chag th ws a dsconnctd. I th ws a connctd to ach oth, th chag apdly combns, lavng th capacto unchagd.. ll connctons o capactos a not smpl combnatons o ss and paalll ccuts. s an xampl o such a complx ccut, consd th ntwok o v capactos,,, 4, and 5 shown blow. Ths combnaton cannot b ducd to a smpl quvalnt by th tchnqus o combnng ss and paalll capactos. 4. Th matal o th dlctc may b abl to wthstand a lag lctc ld than a can wthstand bo bakng down to pass a spak btwn th capacto plats. PROBLEM SOLUTIONS 6. (a) Bcaus th lcton has a ngatv chag, t xpncs a oc n th dcton oppost to th ld and, whn lasd om st, wll mov n th ngatv x dcton. Th wok don on th lcton by th ld s 9 8 W F ( x) ( qe ) x N. m. 9 J x x ( ) Th chang n th lctc potntal ngy s th ngatv o th wok don on th patcl by th ld. Thus, PE W. 9 8 J contnud on nxt pag
6 6 hapt 6 (c) Snc th oulomb oc s a consvatv oc, consvaton o ngy gvs KE + PE, o KE m v PE PE, and v 8 PE. 9 J m 9. kg. 5 6 m s n th x dcton 6. (a) Th chang n th lctc potntal ngy s th ngatv o th wok don on th patcl by th ld. Thus, PE W qex x qey y + q( ) x ( + ) 7 N. m J Th chang n th lctcal potntal s th chang n lctc potntal ngy p unt chag, o PE J q Th wok don by th agnt movng th chag out o th cll s W W PE q nput ld J 9. 4 J PE 6.4 PE q( ) q( ), so q 7. 9 J. +6. J 9 5 J 6.5 E. 7 d. 5 m 6 N 6.6 Snc potntal dnc s wok p unt chag W, th wok don s q W q( ). 6 J 4. J 6 J 6.7 (a) E. d 5. m 5 N ( ) F q E. 6. N. 8 N (c) W F s cosθ. 8 4 N m cos J
7 Elctcal Engy and apactanc (a) Usng consvaton o ngy, KE + PE, wth KE snc th patcl s stoppd, w hav PE KE m + ( 9 v. kg) ( m s ) +. 7 J Th qud stoppng potntal s thn PE +. 7 q J.. k Bng mo massv than lctons, potons tavlng at th sam ntal spd wll hav mo ntal kntc ngy and qu a gat magntud stoppng potntal. (c) PE Snc KE m q stoppng q v, th ato o th stoppng potntal o a poton to q that o an lcton havng th sam ntal spd s p mp v ( + ) m m v ( ) p m 6.9 (a) Us consvaton o ngy ( KE + PE + PE ) ( KE + PE + PE ) s s + + o KE PE PE Thus, s ( KE ) snc th block s at st at both bgnnng and nd. ( PEs ) kxmax, wh x max s th maxmum sttch o th spng. max PE W QE x x kxmax QE xmax, gvng + max ( ) 6 4 QE m k 78. N m t qulbum, ΣF Fs + F, o kxq + QE Tho, x QE x k q max. 8 cm 4. 6 m 4. 6 cm Th ampltud s th dstanc om th qulbum poston to ach o th tunng ponts at x and x 4. 6 cm, so 8 cm x.. max contnud on nxt pag
8 6 hapt 6 (c) Fom consvaton o ngy, KE PEs PE kxmax Q x, ths gvs max kx k max Q Q 6. Usng y v yt + ayt o th ull lght gvs v y t + a y t, o a Thn, usng v y v y t o k Q v + a ( y) o th upwad pat o th lght gvs y y y y y y t ( y) v v v. m s 4. s max a v t 4 4 y ( y ). 6 m ΣFy mg qe qe Fom Nwton s scond law, ay g + m m m. Equatng ths to th al qe y sult gvs ay g + m v, so th lctc ld stngth s t. Snc m y E g v q t. kg m s 4. s 9. 8 m s. 95 N 6. (a) ( ) Thus, y E max max. 6 m. 95 N k B kq kq B 9 9 ( N m ) m 9 9 ( N m ) m B + (c) Th ognal lcton wll b plld by th ngatvly chagd patcl whch suddnly appas at pont. Unlss th lcton s xd n plac, t wll mov n th oppost dcton, away om ponts and B, thby lowng th potntal dnc btwn ths ponts. 6. (a) t th ogn, th total potntal s ogn contnud on nxt pag kq kq + 9 ( N m ) m.8 m. 6
9 Elctcal Engy and apactanc 6 t pont B locatd at (. 5 cm, ), th ndd dstancs a x x y y B B cm. 5 cm. 95 cm and x x y y B B + + gvng B kq kq N m 6. (a) allng th. µ chag q,. 5 cm. 8 cm. 4 cm m. 4 m. 6 k q q q k + + q N m m. m m Rplacng. by. 6 6 n pat (a) ylds. 6 ( ) 6.4 W q( ) q, and snc th 8. µ s nnt dstanc om oth chags. q q k N m m (. ) + (.6 ) m. 5 6 ( ) 6 6 Thus, W J
10 64 hapt (a) k q N m m.75 m PE k q q N m 9 9 ( 5. )..5 m J Th ngatv sgn mans that postv wok must b don to spaat th chags (that s, bng thm up to a stat o zo potntal ngy). 6.6 Th potntal at dstanc. m om a chag Q s kq 9 9 ( N m ) 9.. m +7 Thus, th wok qud to cay a chag q. 9 om nnty to ths locaton s W q J 6.7 Th Pythagoan thom gvs th dstanc om th mdpont o th bas to th chag at th apx o th tangl as ( 4. ) (. ) 5 5 cm cm cm m Thn, th potntal at th mdpont o th bas s k q, o N m 7.. m 4.. k ( 7. ) + + ( 7. ). m 5 m
11 Elctcal Engy and apactanc Outsd th sphcal chag dstbuton, th potntal s th sam as o a pont chag at th cnt o th sph, Thus, kq, wh Q. 9 PE q kq and om consvaton o ngy, ( KE) ( PE ), o k Q mv kq. Ths gvs v m, o v 9 N m (. ) kg. m. m v m s 6.9 (a) Whn th chag conguaton conssts o only th two potons ( q and q n th sktch), th potntal ngy o th conguaton s PE a kqq 9 9 ( N m ) m o PE a J Whn th alpha patcl ( q n th sktch) s addd to th conguaton, th a th dstnct pas o patcls, ach o whch posssss potntal ngy. Th total potntal ngy o th conguaton s now PE b k q q k q q k q q k + + PEa + ( ) wh us has bn mad o th acts that q q q q and (. m ) + (. m ) 4. 4 m m. lso, not that th st tm n ths computaton s just th potntal ngy computd n pat (a). Thus, PE b 4k PEa + ( ) N m J m. 55 J contnud on nxt pag
12 66 hapt 6 (c) I w stat wth th th-patcl systm o pat and allow th alpha patcl to scap to nnty [thby tunng us to th two-patcl systm o pat (a)], th chang n lctc potntal ngy wll b 4 PE PE PE. 84 J. 55 J. 7 J a b (d) onsvaton o ngy, KE + PE, gvs th spd o th alpha patcl at nnty n th stuaton o pat (c) as m v PE, o α α v α ( PE ). α ( J) 7 7 m kg m s () Whn, statng wth th th-patcl systm, th two potons a both allowd to scap to nnty, th wll b no manng pas o patcls and hnc no manng potntal ngy. Thus, PE PEb PEb, and consvaton o ngy gvs th chang n kntc ngy as KE PE + PE b. Snc th potons a dntcal patcls, ths ncas n kntc ngy s splt qually btwn thm gvng KE m v PE poton p p b PEb o v p m p. 55 J 7. 4 m s 7.67 kg 6. (a) I a poton and an alpha patcl, ntally at st 4. m apat, a lasd and allowd to cd to nnty, th nal spds o th two patcls wll d bcaus o th dnc n th masss o th patcls. Thus, attmptng to solv o th nal spds by us o consvaton o ngy alon lads to a stuaton o havng on quaton wth two unknowns, and dos not pmt a soluton. In th stuaton dscbd n pat (a) abov, on can obtan a scond quaton wth th two unknown nal spds by usng consvaton o lna momntum. Thn, on would hav two quatons whch could b solvd smultanously both unknowns. contnud on nxt pag
13 Elctcal Engy and apactanc 67 (c) Fom consvaton o ngy: mα v + mpv α p kqα qp kqα qp N m.. 6 o mα vα + mpvp 5 4. m ( ) yldng m α v α m p v + p. J [] Fom consvaton o lna momntum, m α v α + m p v p o vα m v p m α p [] Substtutng Equaton [] nto Equaton [] gvs m α m m p α p mp p v + v. J o m m p α + mp p v. J and. J. J α kg v p mp m + mp. 5 ( ) 7 m s Thn, Equaton [] gvs th nal spd o th alpha patcl as v α m v p. 67 p m α 7 7 kg 7 6 kg (. 5 m s ). 64 m s kq 6. so kq ( N m )( 8. 9 ) 7. 9 m Fo, 5., and 5.,. 79 m,.44 m, and.88 m Th ad a nvsly popotonal to th potntal. 6. By dnton, th wok qud to mov a chag om on pont to any oth pont on an qupotntal suac s zo. Fom th dnton o wok, W ( F cosθ ) s, th wok s zo only s o F cosθ. Th dsplacmnt s cannot b assumd to b zo n all cass. Thus, on must qu that F cosθ. Th oc F s gvn by F qe and nth th chag q no th ld stngth E can b assumd to b zo n all cass. Tho, th only way th wok can b zo n all cass s cosθ. But cosθ, thn θ 9 o th oc (and hnc th lctc ld) must b ppndcula to th dsplacmnt s (whch s tangnt to th suac). That s, th ld must b ppndcula to th qupotntal suac at all ponts on that suac.
14 68 hapt 6 ( + ), whch gvs 6. Fom consvaton o ngy, KE + PE KE PE kqq + mα + v o kqq k 79 m v m v α α N m ( 58 ) kg. m s 9 9 ( ) ( ) m 6.4 (a) Th dstanc om any on o th cons o th squa to th pont at th cnt s on hal th lngth o th dagonal o th squa, o dagonal a + a a a Snc th chags hav qual magntuds and a all th sam dstanc om th cnt o th squa, thy mak qual contbutons to th total potntal. Thus, total kq kq 4sngl 4 4 chag a 4 k Q a Th wok qud to cay chag q om nnty to th pont at th cnt o th squa s qual to th ncas n th lctc potntal ngy o th chag, o W PE PE q q k Q k cnt total 4 a 4 qq a 6.5 (a) d N m 6 (. m ). 8 8 m F Q E d max max max 8 6 (. F) (. N )( 8 m ) 7 Q 7. µ 6.6 (a) 9.. µ F Q ( ) (. µ F )(. ) 6. µ 6.7 (a) Th capactanc o ths a lld dlctc constant, κ. paalll plat capacto s k d 4 ( N m )( m ) m. 6 F. 6 pf contnud on nxt pag
15 Elctcal Engy and apactanc 69 Q ( ). 6 F p (c). E 8. m 8. N d.5 m (a) Q ( ) 4. F µ 6 6 Q ( ) 4. F µ. 6.9 (a) E. 4 m. k m dctd towad th ngatv plat d.8 m d 4 ( N m ) 7. 6 m.8 m. 74 F. 74 pf (c) Q ( ). 74 F p on on plat and p on th oth plat. 6., so d d ( N m ). m 5 6. F 9 Å d. m. Å m. 9 m 6. (a) ssumng th capacto s a-lld ( κ ), th capactanc s d ( N m )(. m ) m F 9 Q ( ) 5. 9 F (c) 6. E. m. N d. m Q. 54 (d) σ. m () m Incasng th dstanc spaatng th plats dcass th capactanc, th chag stod, and th lctc ld stngth btwn th plats. Ths mans that all o th pvous answs wll b dcasd.
16 7 hapt 6 6. ΣF mg y T cos 5. mg o T cos 5. ΣF x qe T sn 5. mg tan 5. o E mg tan 5. q mgd tan 5. Ed q ( 5 6 kg )( 9. 8 m s )(. 4 m) tan 5... k (a) apactos n a ss combnaton sto th sam chag, Q q ( ), wh q s th quvalnt capactanc and s th potntal dnc mantand acoss th ss combnaton. Th quvalnt capactanc o th gvn ss combnaton s +, o q +, gvng q. 5 µ F 6. 5 µ F q. 79 µf. 5 µ F µ F so th chag stod on ach capacto n th ss combnaton s Q q ( ) (. 79 µ F )( 6. ). 7 µ Whn connctd n paalll, ach capacto has th sam potntal dnc, 6., mantand acoss t. Th chag stod on ach capacto s thn Fo. 5 µ F: Q ( ) (. 5 µ F )( 6. ) 5. µ Fo 6. 5 µ F: Q ( ) ( 6. 5 µ F )( 6. ) 7. 5 µ 6.4 (a) Whn connctd n ss, th quvalnt capactanc s +, o q q + 4. µ F 8. 5 µ F 4. µ F µ F. 8 µ F Whn connctd n paalll, th quvalnt capactanc s q + 4. µ F µ F. 7 µ F
17 Elctcal Engy and apactanc (a) Fst, w plac th paalll combnaton btwn ponts b and c by ts quvalnt capactanc, bc. µ F + 6. µ F 8. µ F. Thn, w hav th capactos n ss btwn ponts a and d. Th quvalnt capactanc o ths ccut s tho µ q ab bc cd F 8. µ F gvng q. 67 µ F Th chag stod on ach capacto n th ss combnaton s Qab Qbc Qcd q ( ad ) (. 67 µ F )( 9. ) 4. µ Thn, not that ccut s: bc Qbc 4. µ.. Th chag on ach capacto n th ognal 8. µ F bc On th 8. µ F btwn a and b: Q8 Q 4. µ ab On th 8. µ F btwn c and d: Q8 Q 4. µ cd On th. µ F btwn b and c: Q ( bc ) (. µ F )(. ) 6. µ On th 6. µ F btwn b and c: Q6 6 ( bc ) ( 6. µ F )(. ) 8. µ (c) Not that ab Qab ab 4. µ 8. µ F., and that cd Qcd cd 4. µ 8. µ F.. W al ound that bc., so w conclud that th potntal dnc acoss ach capacto n th ccut s paalll + 9. pf 9. pf [] + ss + ss Thus, usng Equaton [], ss +. pf ( 9. pf ) 9. pf +. pf, whch ducs to 9. pf 8. pf, o 6. pf. pf ( ) Tho, th 6. pf and, om Equaton [],. pf o. pf and 6. pf. W conclud that th two capactancs a. pf and 6. pf.
18 7 hapt (a) Th quvalnt capactanc o th ss combnaton n th upp banch s upp F + 6 F +. µ. µ 6. µ F. mf 6. mf o upp. µ F. mf 4. mf Lkws, th quvalnt capactanc o th ss combnaton n th low banch s + + low. µ F 4. µ F 4. µ F o low. µ F 9. Ths two quvalnt capactancs a connctd n paalll wth ach oth, so th quvalnt capactanc o th nt ccut s +. µ F +. µ F. µ F q upp low Not that th sam potntal dnc, qual to th potntal dnc o th batty, xsts acoss both th upp and low banchs. Th chag stod on ach capacto n th ss combnaton n th upp banch s Q Q6 Qupp upp ( ) (. µ F )( 9. ) 8 µ and th chag stod on ach capacto n th ss combnaton n th low banch s Q Q6 Qlow low ( ) (. µ F )( 9. ) µ (c) Th potntal dnc acoss ach o th capactos n th ccut s: Q µ. µ F Q 8 µ. µ F Q4 µ 4. µ F 4 Q6 8 µ 6. µ F 6..
19 Elctcal Engy and apactanc (a) Th quvalnt capactanc o th ss combnaton n th ghtmost banch o th ccut s ght 4 F + 8 F +. µ. µ 4. µ F o ght 6. µ F Fgu P6.8 Th quvalnt capactanc o th th capactos now connctd n paalll wth ach oth and wth th batty s q 4. µ F +. µ F + 6. µ F. µ F Dagam (c) Th total chag stod n ths ccut s Q. µf 6. total q (d) o Q total 4 µ Th chags on th th capactos shown n Dagam a: Q4 4 ( ) ( 4. µ F )( 6. ) 44 µ Dagam Q ( ) (. µ F )( 6. ) 7 µ Qght ght ( ) ( 6. µ F )( 6. ) 6 µ Ys. Q + Q + Q Q 4 ght total as t should. () Th chag on ach capacto n th ss combnaton n th ghtmost banch o th ognal ccut (Fgu P6.8) s Q Q Q 4 8 ght 6 µ () (g) 4 Q4 6 µ µ F 8 4 Q8 6 µ 8 7. Not that as t should. 8. µ F
20 74 hapt Fgu Fgu Fgu Th ccut may b ducd n stps as shown abov. 4. µ F µ Usng Fgu, Q ac Thn, n Fgu, Qac 96. µ ( ) 6. ab 6. µ F ab and bc ac ab 8. Fnally, usng Fgu, Q ( ) ab (. µ F )( 6. ) 6. µ Q 5 ( 5. µ F )( ) ab 8. µ, Q 8 ( 8. µ F )( ) bc 64. µ and Q 4 ( 4. µ F )( ) bc. µ 6.4 Fom Q, th ntal chag o ach capacto s Q (. µ F )(. ) µ and Q x x t th capactos a connctd n paalll, th potntal dnc acoss ach s., and th total chag o Q Q + Q x µ s dvdd btwn th two capactos as Q. µ F.. µ and Qx Q Q µ. µ 9. µ Thus, x Qx 9. µ.. µ F
21 Elctcal Engy and apactanc (a) Fom Q ( ), Q 5 ( 5. µ F )( 5. ). 5 µ. 5 m and Q 4 ( 4. µ F )( 5. ). µ. m Whn th two capactos a connctd n paalll, th quvalnt capactanc s q + 5. µ F + 4. µ F 65. µ F. Snc th ngatv plat o on was connctd to th postv plat o th oth, th total chag stod n th paalll combnaton s Q Q Q. µ. 5 µ 75 µ 4 5 Th potntal dnc acoss ach capacto o th paalll combnaton s Q 75 µ 65. µ F q. 5 and th nal chag stod n ach capacto s Q5 ( ) ( 5. µ F )(. 5 ) 88 µ and Q4 Q Q5 75 µ 88 µ 46 µ 6.4 (a) Th ognal ccut ducs to a sngl quvalnt capacto n th stps shown blow. s µ F. µ F. µ F p s + + s (. µ F ) +. µ F 8.66 µ F p + (. µ F). µ F q +.66 F + 8 µ. µ F p p 6. 4 µ F contnud on nxt pag
22 76 hapt 6 Th total chag stod btwn ponts a and b s Q total q ( ) ab ( 6. 4 µ F )( 6. ) 6 µ Thn, lookng at th thd gu, obsv that th chags o th ss capactos o that gu a Qp Qp Qtotal 6 µ. Thus, th potntal dnc acoss th upp paalll combnaton shown n th scond gu s Qp 6 µ ( ) 4. 8 p 8.66 µ F p Fnally, th chag on s Q ( ) p (. µ F )( 4. 8 ) 8. 6 µ 6.4 Fom Q, th ntal chag o ach capacto s Q (. µ F )(. ). µ and Q (. µ F)( ) t th capactos a connctd n paalll, th potntal dnc acoss on s th sam as that acoss th oth. Ths gvs Q Q. µ F. µ F o Q Q [] Fom consvaton o chag, Q + Q Q + Q. µ. Thn, substtutng om Equaton [], ths bcoms Q + Q. µ, gvng Q µ Fnally, om Equaton [], Q µ 6.44 Rcognz that th 7. µ F and th 5. µ F o th cnt banch a connctd n ss. Th total capactanc o that banch s s +. 9 µ F Thn cognz that ths capacto, th 4. µ F capacto, and th 6. µ F capacto a all connctd n paalll btwn ponts a and b. Thus, th quvalnt capactanc btwn ponts a and b s q 4. µ F +. 9 µ F + 6. µ F. 9 µ F
23 Elctcal Engy and apactanc 77 Q Engy stod ( ) ( 4. 5 F) (. ). 4 4 J 6.46 (a) Th quvalnt capactanc o a ss combnaton o and s q 8 F + 6 F +. µ. µ 6. µ F o q. µ F Whn ths ss combnaton s connctd to a.- batty, th total stod ngy s Total ngy stod 4. F 8 64 J 6 q ( ) ( )(. ). Th chag stod on ach o th two capactos n th ss combnaton s Q Q Qtotal q ( ) (. µ F )(. ) 44 µ and th ngy stod n ach o th ndvdual capactos s Engy stod n 4 Q (. 44 ) 6 8. F J and Engy stod n 4 Q (. 44 ) 6 6. F J Engy stod n + Engy stod n J J J, whch s th sam as th total stod ngy ound n pat (a). Ths must b tu th computd quvalnt capactanc s tuly quvalnt to th ognal combnaton. (c) I and had bn connctd n paalll ath than n ss, th quvalnt capactanc would hav bn q + 8. µ F + 6. µ F 54. µ F. I th total ngy stod q ( ) n ths paalll combnaton s to b th sam as was stod n th ognal ss combnaton, t s ncssay that. Total ngy stod J q F Snc th two capactos n paalll hav th sam potntal dnc acoss thm, th ngy stod n th ndvdual capactos ( ) s dctly popotonal to th capactancs. Th lag capacto,, stos th most ngy n ths cas.
24 78 hapt (a) Th ngy ntally stod n th capacto s Q ( Engy stod ) ( ) (. µf )( 6. ) 54. µ J Whn th capacto s dsconnctd om th batty, th stod chag bcoms solatd wth no way o th plats. Thus, th chag mans constant at th valu Q as long as th capacto mans dsconnctd. Snc th capactanc o a paalll plat capacto s κ d, whn th dstanc d spaatng th plats s doubld, th capactanc s dcasd by a acto o (..,. 5 µ F). Th stod ngy (wth Q unchangd) bcoms ( Engy stod) Q Q Q ( ) Engy stod 8 µ J (c) Whn th capacto s connctd to th batty, th potntal dnc btwn th plats s stablshd at th ognal valu o ( ) 6., whl th capactanc mans at. 5 µ F. Th ngy stod und ths condtons s ( Engy stod ) ( F). 5 µ ( 6. ) 7. µ J 6.48 Th ngy tansd to th wat s.. W Q 5 8 Thus, m s th mass o wat bold away, W m c( T ) + Lv bcoms. 5 7 J J. 5 7 J m 486. ( ) J kg kg 7. 5 J gvng m.55 J kg kg 6.49 (a) Not that th chag on th plats mans constant at th ognal valu, Q, as th dlctc s nstd. Thus, th chang n th potntal dnc, Q, s du to a chang n capactanc alon. Th ato o th nal and ntal capactancs s κ d κ d and Q Q ( ) ( ) ( ) Thus, th dlctc constant o th nstd matal s κ. 4, and th matal s pobably nylon (s Tabl 6.). I th dlctc only patally lld th spac btwn th plats, lavng th manng spac a-lld, th quvalnt dlctc constant would b somwh btwn κ. (a) and κ. 4. Th sultng potntal dnc would thn l somwh btwn ( ) 85. and ( ) 5..
25 Elctcal Engy and apactanc (a) Th capactanc o th capacto whl a-lld s d 4 ( N m )( 5. m ). 5 Th ognal chag stod on th plats s m. 48 F. 48 pf Q ( ). 48 F. 5 7 ( ) 7 p Snc dstlld wat s an nsulato, ntoducng t btwn th solatd capacto plats dos not allow th chag to chang. Thus, th nal chag s Q 7 p. t mmson dstlld wat ( κ 8 s Tabl 6. ), th nw capactanc s κ ( 8)(. 48 pf ) 8 pf and th nw potntal dnc s Q 7 p 8 pf. 4. (c) Th ngy stod n a capacto s: Engy stod Q. Thus, th chang n th stod ngy du to mmson n th dstlld wat s E Q Q Q J J nj 8 F. 48 F 6.5 (a) Th dlctc constant o Tlon s κ., so th capactanc s d ( ) N m 75 m κ F 8. nf.4 m Fo Tlon, th dlctc stngth s E max 6. 6 m, so th maxmum voltag s 6 E d 6. m. 4 m max max ( ) max k 6.5 Bo th capacto s olld, th capactanc o ths paalll plat capacto s w L κ κ d d wh s th suac aa o on sd o a ol stp. Thus, th qud lngth s L ( ) ( ) 8 d w 9. 5 F. 5 m.7) N m 7. m κ (. 4 m
26 8 hapt 6 m. kg 6.5 (a) 9. 9 ρ kg m 6 m Snc 4 π, th adus s 4π 4π 4π 4π, and th suac aa s 9. 9 m 4π 4π m κ d ( N m )( m ) m. F ( ) (c) Q ( ). F. and th numb o lctonc chags s n Q Snc th capactos a n paalll, th quvalnt capactanc s + + q d d d d o q wh + + d 6.55 Snc th capactos a n ss, th quvalnt capactanc s gvn by d d d d + d + d q o q wh d d + d + d d 6.56 (a) Plas to th soluton o Poblm 6.7, wh th ollowng sults w obtand:. mf 6. mf. µ q F Q Q6 8 µ Q Q4 µ Th total ngy stod n th ull ccut s thn 6 ( Engy stod) total q ( ). F J. 5 J. 5 mj. mf mf contnud on nxt pag
27 Elctcal Engy and apactanc 8 (c) Th ngy stod n ach ndvdual capacto s 6 Q ( ) Fo. µ F: ( Engy stod) 6. F Fo. µ F: Engy stod Fo 4. µ F: Engy stod Fo 6. µ F: Engy stod J. 6 mj 6 Q ( 8 ) 6. F 5. 4 J 5. 4 mj 6 Q ( ) F 4. 8 J. 8 mj 6 Q ( 8 ) F Th total ngy stod n th ndvdual capactos s 6. 7 J. 7 mj Engy stod mj. 5 mj Engy stod total Thus, th sums o th ngs stod n th ndvdual capactos quals th total ngy stod by th systm In th absnc o a dlctc, th capactanc o th paalll plat capacto s d Wth th dlctc nstd, t lls on-thd o th gap btwn th plats as shown n sktch (a) at th ght. W modl ths stuaton as consstng o a pa o capactos, and, connctd n ss as shown d d d k d k c n sktch at th ght. In alty, th low plat o and th upp plat o a on and th sam, consstng o th low suac o th dlctc shown n sktch (a). Th capactancs n th modl o sktch a gvn by: (a) d c κ κ d d and d d and th quvalnt capactanc o th ss combnaton s q d κ d d + + κ κ + d κ + d κ + κ κ κ and κ q + κ
28 8 hapt Fo th paalll combnaton: p + whch gvs p [] Fo th ss combnaton: + o s s s s Thus, w hav p s s s s and quatng ths to Equaton [] abov gvs o + W wt ths sult as : + p p s s s p p s and us th quadatc omula to obtan ± 4 Thn, Equaton [] gvs 4 p p p s p p p s 6.59 Th chag stod on th capacto by th batty s Q ( ) ( ) Ths s also th total chag stod n th paalll combnaton whn ths chagd capacto s connctd n paalll wth an unchagd. -µ F capacto. Thus, acoss th paalll combnaton, Q p ( ) gvs ( + ). µ F. o 7... µ F and. (. µ F ) 4. 9 µ F 7. s th sultng voltag 6.6 (a) Th. -µ s locatd.5 m om pont P, so ts contbuton to th potntal at P s k q N m ( ) 6..5 m. 8 4 Th potntal at P du to th. -µ chag locatd.5 m away s k q N m ( )..5 m (c) Th total potntal at pont P s P (d) Th wok qud to mov a chag q. µ to pont P om nnty s 6 4 W q q J P ( )
29 Elctcal Engy and apactanc Th stags o th ducton o ths ccut a shown blow. Thus, q 6. 5 µ F 6.6 (a) Du to sphcal symmty, th chag on ach o th concntc sphcal shlls wll b unomly dstbutd ov that shll. Insd a sphcal suac havng a unom chag dstbuton, th lctc ld du to th chag on that suac s zo. Thus, n ths gon, th potntal du to th chag on that suac s constant and qual to th potntal at th suac. Outsd a sphcal suac havng a unom chag dstbuton, th potntal du kq to th chag on that suac s gvn by, wh s th dstanc om th cnt o that suac and q s th chag on that suac. In th gon btwn a pa o concntc sphcal shlls, wth th nn shll havng chag + Q and th out shll havng adus b and chag Q, th total lctc potntal s gvn by + du to nn shll du to out shll k Q k + Q b kq b Th potntal dnc btwn th two shlls s tho k Q k Q a b a b b b k Q b a ab Th capactanc o ths dvc s gvn by Q ab k b a Whn b >> a, thn b a b. Thus, n th lmt as b, th capactanc ound abov bcoms ab k b a k π Th ngy stod n a chagd capacto s W a. Hnc, W J k 6. F
30 84 hapt Fom Q, th capactanc o th capacto wth a btwn th plats s Q 5 µ t th dlctc s nstd, th potntal dnc s hld to th ognal valu, but th chag changs to Q Q + µ 5 µ. Thus, th capactanc wth th dlctc slab n plac s Q 5 µ Th dlctc constant o th dlctc slab s tho µ κ 5 5 µ Th chags ntally stod on th capactos a Q ( ) ( 6. µ F )( 5 ). 5 µ and Q ( ) (. µ F )( 5 ) 5. µ Whn th capactos a connctd n paalll, wth th ngatv plat o on connctd to th postv plat o th oth, th nt stod chag s Q Q Q. 5 µ 5. µ. µ Th quvalnt capactanc o th paalll combnaton s q + 8. µ F. Thus, th nal potntal dnc acoss ach o th capactos s Q ( ). µ 5 8. µ F q and th nal chag on ach capacto s Q ( 6. µ F )( 5 ) 75 µ. 75 m and Q ( ) ( µ ) µ. F m 6.66 Th ngy qud to mlt th lad sampl s W m c ( T ) + L Pb ( kg 8 J kg ) J kg. 8 J Th ngy stod n a capacto s W, so th qud potntal dnc s W. 8 J 6 5. F
31 Elctcal Engy and apactanc Whn xcss chag sds on a sphcal suac that s a movd om any oth chag, ths xcss chag s unomly dstbutd ov th sphcal suac, and th lctc potntal at th suac s th sam as all th xcss chag w concntatd at th cnt o th sphcal suac. In th gvn stuaton, w hav two chagd sphs, ntally solatd om ach oth, wth chags and potntals o: Q + 6. µ, kq R wh R. cm, Q 4. µ, and k Q R wth R 8. cm. Whn ths sphs a thn connctd by a long conductng thad, th chags a dstbutd ( yldng chags o Q and Q, spctvly) untl th two suacs com to a common potntal ( kq R kq R ). Whn qulbum s stablshd, w hav: Fom consvaton o chag: Q + Q Q + Q Q + Q +. µ [] Fom qual potntals: kq kq R R R Q R Q o Q. 5Q [] Substtutng Equaton [] nto [] gvs: +. µ Q. 8 µ. 5 Thn, Equaton [] gvs: Q µ. µ 6.68 Th lctc ld btwn th plats s dctd downwad wth magntud E y d m N m Snc th gavtatonal oc xpncd by th lcton s nglgbl n compason to th lctcal oc actng on t, th vtcal acclaton s a y Fy qey N m m m 9. kg ( ) m s (a) t th closst appoach to th bottom plat, v y. Thus, th vtcal dsplacmnt om pont O s ound om v v a y as y a y + y y y 5 ( m s ) 6 v sn θ 5. 6 m s sn mm Th mnmum dstanc abov th bottom plat s thn D d + y. mm. 89 mm. mm Th tm o th lcton to go om pont O to th upp plat s ound om y v yt + ayt as 6 +. m 5. 6 m sn 45 t + s m t s Solvng o t gvs a postv soluton o t. 9 s. Th hozontal dsplacmnt om pont O at ths tm s 6 9 x v xt 5. 6 m s cos 45. s 4. 4 mm
Homework: Due
hw-.nb: //::9:5: omwok: Du -- Ths st (#7) s du on Wdnsday, //. Th soluton fom Poblm fom th xam s found n th mdtm solutons. ü Sakua Chap : 7,,,, 5. Mbach.. BJ 6. ü Mbach. Th bass stats of angula momntum
More informationDiffraction. Diffraction: general Fresnel vs. Fraunhofer diffraction Several coherent oscillators Single-slit diffraction. Phys 322 Lecture 28
Chapt 10 Phys 3 Lctu 8 Dffacton Dffacton: gnal Fsnl vs. Faunhof dffacton Sval cohnt oscllatos Sngl-slt dffacton Dffacton Gmald, 1600s: dffacto, dvaton of lght fom lna popagaton Dffacton s a consqunc of
More informationGRAVITATION. (d) If a spring balance having frequency f is taken on moon (having g = g / 6) it will have a frequency of (a) 6f (b) f / 6
GVITTION 1. Two satllits and o ound a plant P in cicula obits havin adii 4 and spctivly. If th spd of th satllit is V, th spd of th satllit will b 1 V 6 V 4V V. Th scap vlocity on th sufac of th ath is
More information5- Scattering Stationary States
Lctu 19 Pyscs Dpatmnt Yamou Unvsty 1163 Ibd Jodan Pys. 441: Nucla Pyscs 1 Pobablty Cunts D. Ndal Esadat ttp://ctaps.yu.du.jo/pyscs/couss/pys641/lc5-3 5- Scattng Statonay Stats Rfnc: Paagaps B and C Quantum
More informationDiffraction. Diffraction: general Fresnel vs. Fraunhofer diffraction Several coherent oscillators Single-slit diffraction. Phys 322 Lecture 28
Chapt 10 Phys 3 Lctu 8 Dffacton Dffacton: gnal Fsnl vs. Faunhof dffacton Sval cohnt oscllatos Sngl-slt dffacton Dffacton Gmald, 1600s: dffacto, dvaton of lght fom lna popagaton Dffacton s a consqunc of
More informationLoad Equations. So let s look at a single machine connected to an infinite bus, as illustrated in Fig. 1 below.
oa Euatons Thoughout all of chapt 4, ou focus s on th machn tslf, thfo w wll only pfom a y smpl tatmnt of th ntwok n o to s a complt mol. W o that h, but alz that w wll tun to ths ssu n Chapt 9. So lt
More informationThe Random Phase Approximation:
Th Random Phas Appoxmaton: Elctolyts, Polym Solutons and Polylctolyts I. Why chagd systms a so mpotant: thy a wat solubl. A. bology B. nvonmntally-fndly polym pocssng II. Elctolyt solutons standad dvaton
More informationCHARACTERISTICS OF MAGNETICALLY ENHANCED CAPACITIVELY COUPLED DISCHARGES*
CHARACTERISTICS OF MAGNETICALLY ENHANCED CAPACITIVELY COUPLED DISCHARGES* Alx V. Vasnkov and Mak J. Kushn Unvsty of Illnos 1406 W. Gn St. Ubana, IL 61801 vasnkov@uuc.du k@uuc.du http://uglz.c.uuc.du Octob
More informationMassachusetts Institute of Technology Introduction to Plasma Physics
Massachustts Insttut of Tchnology Intoducton to Plasma Physcs NAME 6.65J,8.63J,.6J R. Pak Dcmb 5 Fnal Eam :3-4:3 PM NOTES: Th a 8 pags to th am, plus on fomula sht. Mak su that you copy s complt. Each
More informationGRAVITATION 4) R. max. 2 ..(1) ...(2)
GAVITATION PVIOUS AMCT QUSTIONS NGINING. A body is pojctd vtically upwads fom th sufac of th ath with a vlocity qual to half th scap vlocity. If is th adius of th ath, maximum hight attaind by th body
More informationPeriod vs. Length of a Pendulum
Gaphcal Mtho n Phc Gaph Intptaton an Lnazaton Pat 1: Gaphng Tchnqu In Phc w u a vat of tool nclung wo, quaton, an gaph to mak mol of th moton of objct an th ntacton btwn objct n a tm. Gaph a on of th bt
More information( V ) 0 in the above equation, but retained to keep the complete vector identity for V in equation.
Cuvlna Coodnats Outln:. Otogonal cuvlna coodnat systms. Dffntal opatos n otogonal cuvlna coodnat systms. Dvatvs of t unt vctos n otogonal cuvlna coodnat systms 4. Incompssbl N-S quatons n otogonal cuvlna
More informationCHAPTER 5 CIRCULAR MOTION
CHAPTER 5 CIRCULAR MOTION and GRAVITATION 5.1 CENTRIPETAL FORCE It is known that if a paticl mos with constant spd in a cicula path of adius, it acquis a cntiptal acclation du to th chang in th diction
More informationPhysics 111. Lecture 38 (Walker: ) Phase Change Latent Heat. May 6, The Three Basic Phases of Matter. Solid Liquid Gas
Physics 111 Lctu 38 (Walk: 17.4-5) Phas Chang May 6, 2009 Lctu 38 1/26 Th Th Basic Phass of Matt Solid Liquid Gas Squnc of incasing molcul motion (and ngy) Lctu 38 2/26 If a liquid is put into a sald contain
More informationSTATISTICAL MECHANICS OF DIATOMIC GASES
Pof. D. I. ass Phys54 7 -Ma-8 Diatomic_Gas (Ashly H. Cat chapt 5) SAISICAL MECHAICS OF DIAOMIC GASES - Fo monatomic gas whos molculs hav th dgs of fdom of tanslatoy motion th intnal u 3 ngy and th spcific
More informationApplications of Lagrange Equations
Applcaton of agang Euaton Ca Stuy : Elctc Ccut ng th agang uaton of oton, vlop th athatcal ol fo th ccut hown n Fgu.Sulat th ult by SIMI. Th ccuty paat a: 0.0 H, 0.00 H, 0.00 H, C 0.0 F, C 0. F, 0 Ω, Ω
More informationElectrical Energy and Capacitance
haptr 6 Elctrical Enrgy and apacitanc Quick Quizzs. (b). Th fild xrts a forc on th lctron, causing it to acclrat in th dirction opposit to that of th fild. In this procss, lctrical potntial nrgy is convrtd
More informationPhysics 202, Lecture 5. Today s Topics. Announcements: Homework #3 on WebAssign by tonight Due (with Homework #2) on 9/24, 10 PM
Physics 0, Lctu 5 Today s Topics nnouncmnts: Homwok #3 on Wbssign by tonight Du (with Homwok #) on 9/4, 10 PM Rviw: (Ch. 5Pat I) Elctic Potntial Engy, Elctic Potntial Elctic Potntial (Ch. 5Pat II) Elctic
More informationPhysics Exam II Chapters 25-29
Physcs 114 1 Exam II Chaptes 5-9 Answe 8 of the followng 9 questons o poblems. Each one s weghted equally. Clealy mak on you blue book whch numbe you do not want gaded. If you ae not sue whch one you do
More informationCIVL 7/ D Boundary Value Problems - Axisymmetric Elements 1/8
CIVL 7/8 -D Bounday Valu Poblms - xsymmtc Elmnts /8 xsymmtc poblms a somtms fd to as adally symmtc poblms. hy a gomtcally th-dmnsonal but mathmatcally only two-dmnsonal n th physcs of th poblm. In oth
More informationPHYS Week 5. Reading Journals today from tables. WebAssign due Wed nite
PHYS 015 -- Week 5 Readng Jounals today fom tables WebAssgn due Wed nte Fo exclusve use n PHYS 015. Not fo e-dstbuton. Some mateals Copyght Unvesty of Coloado, Cengage,, Peason J. Maps. Fundamental Tools
More informationChapter 23: Magnetic Field Shielding
ELECTROMAGNETIC COMPATIBILITY ANDBOOK 1 Chapt : Magntc Fld Shldng.1 Usng th Bt-Savat law, vfy th magntc fld xpssn X (pvdd by yu nstuct) gvn n th cunt dstbutns and th magntc flds tabl n ths chapt.. Usng
More information8 - GRAVITATION Page 1
8 GAVITATION Pag 1 Intoduction Ptolmy, in scond cntuy, gav gocntic thoy of plantay motion in which th Eath is considd stationay at th cnt of th univs and all th stas and th plants including th Sun volving
More information4.4 Linear Dielectrics F
4.4 Lina Dilctics F stal F stal θ magntic dipol imag dipol supconducto 4.4.1 Suscptiility, mitivility, Dilctic Constant I is not too stong, th polaization is popotional to th ild. χ (sinc D, D is lctic
More informationGMm. 10a-0. Satellite Motion. GMm U (r) - U (r ) how high does it go? Escape velocity. Kepler s 2nd Law ::= Areas Angular Mom. Conservation!!!!
F Satllt Moton 10a-0 U () - U ( ) 0 f ow g dos t go? scap locty Kpl s nd Law ::= Aas Angula Mo. Consaton!!!! Nwton s Unsal Law of Gaty 10a-1 M F F 1) F acts along t ln connctng t cnts of objcts Cntal Foc
More informationa v2 r a' (4v) 2 16 v2 mg mg (2.4kg)(9.8m / s 2 ) 23.52N 23.52N N
Conceptual ewton s Law Applcaton Test Revew 1. What s the decton o centpetal acceleaton? see unom ccula moton notes 2. What aects the magntude o a ctonal oce? see cton notes 3. What s the deence between
More informationSignal Circuit and Transistor Small-Signal Model
Snal cut an anto Sall-Snal Mol Lctu not: Sc. 5 Sa & Sth 6 th E: Sc. 5.5 & 6.7 Sa & Sth 5 th E: Sc. 4.6 & 5.6 F. Najaba EE65 Wnt 0 anto pl lopnt Ba & Snal Ba Snal only Ba Snal - Ba? MOS... : : S... MOS...
More informationGrand Canonical Ensemble
Th nsmbl of systms mmrsd n a partcl-hat rsrvor at constant tmpratur T, prssur P, and chmcal potntal. Consdr an nsmbl of M dntcal systms (M =,, 3,...M).. Thy ar mutually sharng th total numbr of partcls
More informationThe angle between L and the z-axis is found from
Poblm 6 This is not a ifficult poblm but it is a al pain to tansf it fom pap into Mathca I won't giv it to you on th quiz, but know how to o it fo th xam Poblm 6 S Figu 6 Th magnitu of L is L an th z-componnt
More informationPhysics 2A Chapter 3 HW Solutions
Phscs A Chapter 3 HW Solutons Chapter 3 Conceptual Queston: 4, 6, 8, Problems: 5,, 8, 7, 3, 44, 46, 69, 70, 73 Q3.4. Reason: (a) C = A+ B onl A and B are n the same drecton. Sze does not matter. (b) C
More informationCollege Prep Physics I Multiple Choice Practice Final #2 Solutions Northville High School Mr. Palmer, Physics Teacher. Name: Hour: Score: /zero
Collg Pp Phsics Multipl Choic Pactic inal # Solutions Nothvill High School M. Palm, Phsics Tach Nam: Hou: Sco: /zo You inal Exam will hav 40 multipl choic qustions woth 5 points ach.. How is cunt actd
More information5.61 Fall 2007 Lecture #2 page 1. The DEMISE of CLASSICAL PHYSICS
5.61 Fall 2007 Lctu #2 pag 1 Th DEMISE of CLASSICAL PHYSICS (a) Discovy of th Elcton In 1897 J.J. Thomson discovs th lcton and masus ( m ) (and inadvtntly invnts th cathod ay (TV) tub) Faaday (1860 s 1870
More information24-2: Electric Potential Energy. 24-1: What is physics
D. Iyad SAADEDDIN Chapte 4: Electc Potental Electc potental Enegy and Electc potental Calculatng the E-potental fom E-feld fo dffeent chage dstbutons Calculatng the E-feld fom E-potental Potental of a
More informationChapter 23: Electric Potential
Chapte 23: Electc Potental Electc Potental Enegy It tuns out (won t show ths) that the tostatc foce, qq 1 2 F ˆ = k, s consevatve. 2 Recall, fo any consevatve foce, t s always possble to wte the wok done
More informationPhy213: General Physics III 4/10/2008 Chapter 22 Worksheet 1. d = 0.1 m
hy3: Gnral hyscs III 4/0/008 haptr Worksht lctrc Flds: onsdr a fxd pont charg of 0 µ (q ) q = 0 µ d = 0 a What s th agntud and drcton of th lctrc fld at a pont, a dstanc of 0? q = = 8x0 ˆ o d ˆ 6 N ( )
More informationCHAPTER 33: PARTICLE PHYSICS
Collg Physcs Studnt s Manual Chaptr 33 CHAPTER 33: PARTICLE PHYSICS 33. THE FOUR BASIC FORCES 4. (a) Fnd th rato of th strngths of th wak and lctromagntc forcs undr ordnary crcumstancs. (b) What dos that
More informationˆ (0.10 m) E ( N m /C ) 36 ˆj ( j C m)
7.. = = 3 = 4 = 5. The electrc feld s constant everywhere between the plates. Ths s ndcated by the electrc feld vectors, whch are all the same length and n the same drecton. 7.5. Model: The dstances to
More informationChapter 3 Basic Crystallography and Electron Diffraction from Crystals. Lecture 11. CHEM 793, 2008 Fall
Chapt 3 Basc Cystalloaphy and Elcton Dacton om Cystals Lctu CHEM 793 8 all Top o thn ol Cystal plan (hl) Bottom o thn ol Ba Law d snθ nλ hl CHEM 793 8 all Equons connctn th Cystal Paamts (h l) and d-spacn
More informationCOLLEGE OF FOUNDATION AND GENERAL STUDIES PUTRAJAYA CAMPUS FINAL EXAMINATION TRIMESTER /2017
COLLEGE OF FOUNDATION AND GENERAL STUDIES PUTRAJAYA CAMPUS FINAL EXAMINATION TRIMESTER 1 016/017 PROGRAMME SUBJECT CODE : Foundaton n Engneeng : PHYF115 SUBJECT : Phscs 1 DATE : Septembe 016 DURATION :
More informationMon. Tues. 6.2 Field of a Magnetized Object 6.3, 6.4 Auxiliary Field & Linear Media HW9
Fi. on. Tus. 6. Fild of a agntid Ojct 6.3, 6.4 uxiliay Fild & Lina dia HW9 Dipol t fo a loop Osvation location x y agntic Dipol ont Ia... ) ( 4 o I I... ) ( 4 I o... sin 4 I o Sa diction as cunt B 3 3
More informationPhysics 11b Lecture #2. Electric Field Electric Flux Gauss s Law
Physcs 11b Lectue # Electc Feld Electc Flux Gauss s Law What We Dd Last Tme Electc chage = How object esponds to electc foce Comes n postve and negatve flavos Conseved Electc foce Coulomb s Law F Same
More informationHydrogen atom. Energy levels and wave functions Orbital momentum, electron spin and nuclear spin Fine and hyperfine interaction Hydrogen orbitals
Hydogn atom Engy lvls and wav functions Obital momntum, lcton spin and nucla spin Fin and hypfin intaction Hydogn obitals Hydogn atom A finmnt of th Rydbg constant: R ~ 109 737.3156841 cm -1 A hydogn mas
More informationPhysics 240: Worksheet 15 Name
Physics 40: Woksht 15 Nam Each of ths poblms inol physics in an acclatd fam of fnc Althouh you mind wants to ty to foc you to wok ths poblms insid th acclatd fnc fam (i.. th so-calld "won way" by som popl),
More informationTEST-03 TOPIC: MAGNETISM AND MAGNETIC EFFECT OF CURRENT Q.1 Find the magnetic field intensity due to a thin wire carrying current I in the Fig.
TEST-03 TPC: MAGNETSM AND MAGNETC EFFECT F CURRENT Q. Fnd the magnetc feld ntensty due to a thn we cayng cuent n the Fg. - R 0 ( + tan) R () 0 ( ) R 0 ( + ) R 0 ( + tan ) R Q. Electons emtted wth neglgble
More informationAakash. For Class XII Studying / Passed Students. Physics, Chemistry & Mathematics
Aakash A UNIQUE PPRTUNITY T HELP YU FULFIL YUR DREAMS Fo Class XII Studying / Passd Studnts Physics, Chmisty & Mathmatics Rgistd ffic: Aakash Tow, 8, Pusa Road, Nw Dlhi-0005. Ph.: (0) 4763456 Fax: (0)
More informationJEE-2017 : Advanced Paper 2 Answers and Explanations
DE 9 JEE-07 : Advancd Papr Answrs and Explanatons Physcs hmstry Mathmatcs 0 A, B, 9 A 8 B, 7 B 6 B, D B 0 D 9, D 8 D 7 A, B, D A 0 A,, D 9 8 * A A, B A B, D 0 B 9 A, D 5 D A, B A,B,,D A 50 A, 6 5 A D B
More informationPhysics 2A Chapter 11 - Universal Gravitation Fall 2017
Physcs A Chapte - Unvesal Gavtaton Fall 07 hese notes ae ve pages. A quck summay: he text boxes n the notes contan the esults that wll compse the toolbox o Chapte. hee ae thee sectons: the law o gavtaton,
More informationBethe-Salpeter Equation Green s Function and the Bethe-Salpeter Equation for Effective Interaction in the Ladder Approximation
Bh-Salp Equaon n s Funcon and h Bh-Salp Equaon fo Effcv Inacon n h Ladd Appoxmaon Csa A. Z. Vasconcllos Insuo d Físca-UFRS - upo: Físca d Hadons Sngl-Pacl Popagao. Dagam xpanson of popagao. W consd as
More informationALL QUESTIONS ARE WORTH 20 POINTS. WORK OUT FIVE PROBLEMS.
GNRAL PHYSICS PH -3A (D. S. Mov) Test (/3/) key STUDNT NAM: STUDNT d #: -------------------------------------------------------------------------------------------------------------------------------------------
More informationChapter 3 Basic Crystallography and Electron Diffraction from Crystals. Lecture 12. CHEM 793, 2008 Fall
Chapt 3 Bac Cytalloaphy and Elcton Dacton om Cytal Lctu 1 CHEM 793, 008 all Announcmnt Mdtm Exam: Oct., Wdnday, :30 4:30 CHEM 793, 008 all Th xctaton o, Ba' Law and th Lau quaton pdct dacton at only pc
More informationEconomics 600: August, 2007 Dynamic Part: Problem Set 5. Problems on Differential Equations and Continuous Time Optimization
THE UNIVERSITY OF MARYLAND COLLEGE PARK, MARYLAND Economcs 600: August, 007 Dynamc Part: Problm St 5 Problms on Dffrntal Equatons and Contnuous Tm Optmzaton Quston Solv th followng two dffrntal quatons.
More informationSchool of Aerospace Engineering Origins of Quantum Theory. Measurements of emission of light (EM radiation) from (H) atoms found discrete lines
Ogs of Quatu Thoy Masuts of sso of lght (EM adato) fo (H) atos foud dsct ls 5 4 Abl to ft to followg ss psso ν R λ c λwavlgth, νfqucy, cspd lght RRydbg Costat (~09,7677.58c - ),,, +, +,..g.,,.6, 0.6, (Lya
More informationElectromagnetics: The Smith Chart (9-6)
Elctomagntcs: Th Smth Chat (9-6 Yoonchan Jong School of Elctcal Engnng, Soul Natonal Unvsty Tl: 8 (0 880 63, Fax: 8 (0 873 9953 Emal: yoonchan@snu.ac.k A Confomal Mappng ( Mappng btwn complx-valud vaabls:
More informationPhysics Exam 3
Physcs 114 1 Exam 3 The numbe of ponts fo each secton s noted n backets, []. Choose a total of 35 ponts that wll be gaded that s you may dop (not answe) a total of 5 ponts. Clealy mak on the cove of you
More informationSolid state physics. Lecture 3: chemical bonding. Prof. Dr. U. Pietsch
Solid stat physics Lctu 3: chmical bonding Pof. D. U. Pitsch Elcton chag dnsity distibution fom -ay diffaction data F kp ik dk h k l i Fi H p H; H hkl V a h k l Elctonic chag dnsity of silicon Valnc chag
More informationHeisenberg Model. Sayed Mohammad Mahdi Sadrnezhaad. Supervisor: Prof. Abdollah Langari
snbrg Modl Sad Mohammad Mahd Sadrnhaad Survsor: Prof. bdollah Langar bstract: n ths rsarch w tr to calculat analtcall gnvalus and gnvctors of fnt chan wth ½-sn artcls snbrg modl. W drov gnfuctons for closd
More informationPHYS 2421 Fields and Waves
PHYS 242 Felds nd Wves Instucto: Joge A. López Offce: PSCI 29 A, Phone: 747-7528 Textook: Unvesty Physcs e, Young nd Feedmn 23. Electc potentl enegy 23.2 Electc potentl 23.3 Clcultng electc potentl 23.4
More informationRigid Bodies: Equivalent Systems of Forces
Engneeng Statcs, ENGR 2301 Chapte 3 Rgd Bodes: Equvalent Sstems of oces Intoducton Teatment of a bod as a sngle patcle s not alwas possble. In geneal, the se of the bod and the specfc ponts of applcaton
More informationRectification and Depth Computation
Dpatmnt of Comput Engnng Unvst of Cafona at Santa Cuz Rctfcaton an Dpth Computaton CMPE 64: mag Anass an Comput Vson Spng 0 Ha ao 4/6/0 mag cosponncs Dpatmnt of Comput Engnng Unvst of Cafona at Santa Cuz
More informationStructure and Features
Thust l Roll ans Thust Roll ans Stutu an atus Thust ans onsst of a psly ma a an olls. Thy hav hh ty an hh loa apats an an b us n small spas. Thust l Roll ans nopoat nl olls, whl Thust Roll ans nopoat ylnal
More informationEngineering Mechanics. Force resultants, Torques, Scalar Products, Equivalent Force systems
Engneeng echancs oce esultants, Toques, Scala oducts, Equvalent oce sstems Tata cgaw-hll Companes, 008 Resultant of Two oces foce: acton of one bod on anothe; chaacteed b ts pont of applcaton, magntude,
More informationChapter-10. Ab initio methods I (Hartree-Fock Methods)
Chapt- Ab nto mthods I (Hat-Fock Mthods) Ky wods: Ab nto mthods, quantum chmsty, Schodng quaton, atomc obtals, wll bhavd functons, poduct wavfunctons, dtmnantal wavfunctons, Hat mthod, Hat Fock Mthod,
More informationΕρωτήσεις και ασκησεις Κεφ. 10 (για μόρια) ΠΑΡΑΔΟΣΗ 29/11/2016. (d)
Ερωτήσεις και ασκησεις Κεφ 0 (για μόρια ΠΑΡΑΔΟΣΗ 9//06 Th coffcnt A of th van r Waals ntracton s: (a A r r / ( r r ( (c a a a a A r r / ( r r ( a a a a A r r / ( r r a a a a A r r / ( r r 4 a a a a 0 Th
More informationPHY126 Summer Session I, 2008
PHY6 Summe Sesson I, 8 Most of nfomaton s avalable at: http://nngoup.phscs.sunsb.edu/~chak/phy6-8 ncludng the sllabus and lectue sldes. Read sllabus and watch fo mpotant announcements. Homewok assgnment
More informationCHAPTER 5 CIRCULAR MOTION AND GRAVITATION
84 CHAPTER 5 CIRCULAR MOTION AND GRAVITATION CHAPTER 5 CIRCULAR MOTION AND GRAVITATION 85 In th pious chapt w discussd Nwton's laws of motion and its application in simpl dynamics poblms. In this chapt
More informationANALYSIS: The mass rate balance for the one-inlet, one-exit control volume at steady state is
Problm 4.47 Fgur P4.47 provds stady stat opratng data for a pump drawng watr from a rsrvor and dlvrng t at a prssur of 3 bar to a storag tank prchd 5 m abov th rsrvor. Th powr nput to th pump s 0.5 kw.
More informationPhysics 256: Lecture 2. Physics
Physcs 56: Lctur Intro to Quantum Physcs Agnda for Today Complx Numbrs Intrfrnc of lght Intrfrnc Two slt ntrfrnc Dffracton Sngl slt dffracton Physcs 01: Lctur 1, Pg 1 Constructv Intrfrnc Ths wll occur
More informationGrid Transformations for CFD Calculations
Coll of Ennn an Comput Scnc Mchancal Ennn Dpatmnt ME 69 Computatonal lu Dnamcs Spn Tct: 5754 Instuct: La Catto Intoucton G Tansfmatons f CD Calculatons W want to ca out ou CD analss n altnatv conat sstms.
More informationChapter 8: Potential Energy and The Conservation of Total Energy
Chapter 8: Potental Energy and The Conservaton o Total Energy Work and knetc energy are energes o moton. K K K mv r v v F dr Potental energy s an energy that depends on locaton. -Dmenson F x d U( x) dx
More informationRemember: When an object falls due to gravity its potential energy decreases.
Chapte 5: lectc Potental As mentoned seveal tmes dung the uate Newton s law o gavty and Coulomb s law ae dentcal n the mathematcal om. So, most thngs that ae tue o gavty ae also tue o electostatcs! Hee
More informationThe Hyperelastic material is examined in this section.
4. Hyprlastcty h Hyprlastc matral s xad n ths scton. 4..1 Consttutv Equatons h rat of chang of ntrnal nrgy W pr unt rfrnc volum s gvn by th strss powr, whch can b xprssd n a numbr of dffrnt ways (s 3.7.6):
More informationEAcos θ, where θ is the angle between the electric field and
8.4. Modl: Th lctric flux flows out of a closd surfac around a rgion of spac containing a nt positiv charg and into a closd surfac surrounding a nt ngativ charg. Visualiz: Plas rfr to Figur EX8.4. Lt A
More informationk of the incident wave) will be greater t is too small to satisfy the required kinematics boundary condition, (19)
TOTAL INTRNAL RFLTION Kmacs pops Sc h vcos a coplaa, l s cosd h cd pla cocds wh h X pla; hc 0. y y y osd h cas whch h lgh s cd fom h mdum of hgh dx of faco >. Fo cd agls ga ha h ccal agl s 1 ( /, h hooal
More informationChapter 10 DIELECTRICS. Dielectrics
86 Dlctcs Chat DILCTRICS Dlctcs : Dlctcs a fct nsulatos. In dlctcs lctons a vy tghtly bound to th atoms so that at odnay tmatus thy do not conduct any lctc cunt. xamls: Solds: glass, ocln; gass: H, N ;
More informationAnalysis of a M/G/1/K Queue with Vacations Systems with Exhaustive Service, Multiple or Single Vacations
Analyss of a M/G// uu wth aatons Systms wth Ehaustv Sv, Multpl o Sngl aatons W onsd h th fnt apaty M/G// uu wth th vaaton that th sv gos fo vaatons whn t s dl. Ths sv modl s fd to as on povdng haustv sv,
More informationfor the magnetic induction at the point P with coordinate x produced by an increment of current
5. tatng wth th ffnta psson B fo th magntc nucton at th pont P wth coonat pouc by an ncmnt of cunt at, show pcty that fo a oop cayng a cunt th magntc nucton at P s B Ω wh Ω s th so ang subtn by th oop
More informationFolding of Regular CW-Complexes
Ald Mathmatcal Scncs, Vol. 6,, no. 83, 437-446 Foldng of Rgular CW-Comlxs E. M. El-Kholy and S N. Daoud,3. Dartmnt of Mathmatcs, Faculty of Scnc Tanta Unvrsty,Tanta,Egyt. Dartmnt of Mathmatcs, Faculty
More informationHomework 1: Solutions
Howo : Solutos No-a Fals supposto tst but passs scal tst lthouh -f th ta as slowss [S /V] vs t th appaac of laty alty th path alo whch slowss s to b tat to obta tavl ts ps o th ol paat S o V as a cosquc
More informationSoft k-means Clustering. Comp 135 Machine Learning Computer Science Tufts University. Mixture Models. Mixture of Normals in 1D
Comp 35 Machn Larnng Computr Scnc Tufts Unvrsty Fall 207 Ron Khardon Th EM Algorthm Mxtur Modls Sm-Suprvsd Larnng Soft k-mans Clustrng ck k clustr cntrs : Assocat xampls wth cntrs p,j ~~ smlarty b/w cntr
More informationPhysics 202, Lecture 2. Announcements
Physcs 202, Lectue 2 Today s Topcs Announcements Electc Felds Moe on the Electc Foce (Coulomb s Law The Electc Feld Moton of Chaged Patcles n an Electc Feld Announcements Homewok Assgnment #1: WebAssgn
More informationLINEAR MOMENTUM. product of the mass m and the velocity v r of an object r r
LINEAR MOMENTUM Imagne beng on a skateboad, at est that can move wthout cton on a smooth suace You catch a heavy, slow-movng ball that has been thown to you you begn to move Altenatvely you catch a lght,
More informationCOHORT 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 informationgravity r2,1 r2 r1 by m 2,1
Gavtaton Many of the foundatons of classcal echancs wee fst dscoveed when phlosophes (ealy scentsts and atheatcans) ted to explan the oton of planets and stas. Newton s ost faous fo unfyng the oton of
More information167 T componnt oftforc on atom B can b drvd as: F B =, E =,K (, ) (.2) wr w av usd 2 = ( ) =2 (.3) T scond drvatv: 2 E = K (, ) = K (1, ) + 3 (.4).2.2
166 ppnd Valnc Forc Flds.1 Introducton Valnc forc lds ar usd to dscrb ntra-molcular ntractons n trms of 2-body, 3-body, and 4-body (and gr) ntractons. W mplmntd many popular functonal forms n our program..2
More information2 dependence in the electrostatic force means that it is also
lectc Potental negy an lectc Potental A scala el, nvolvng magntues only, s oten ease to wo wth when compae to a vecto el. Fo electc els not havng to begn wth vecto ssues woul be nce. To aange ths a scala
More informationHall-Effect Thruster Simulations with 2-D Electron Transport and Hydrodynamic Ions
Hall-Effct Thust Smulatons wth -D Elcton Tanspot and Hydodynamc Ions IEPC-9-114 Psntd at th 31st Intnatonal Elctc Populson Confnc, Unvsty of Mchgan Ann Abo, Mchgan USA Sptmb 4, 9 Ioanns G. Mkllds, * Ia
More informationE F. and H v. or A r and F r are dual of each other.
A Duality Thom: Consid th following quations as an xampl = A = F μ ε H A E A = jωa j ωμε A + β A = μ J μ A x y, z = J, y, z 4π E F ( A = jω F j ( F j β H F ωμε F + β F = ε M jβ ε F x, y, z = M, y, z 4π
More information10/7/14. Mixture Models. Comp 135 Introduction to Machine Learning and Data Mining. Maximum likelihood estimation. Mixture of Normals in 1D
Comp 35 Introducton to Machn Larnng and Data Mnng Fall 204 rofssor: Ron Khardon Mxtur Modls Motvatd by soft k-mans w dvlopd a gnratv modl for clustrng. Assum thr ar k clustrs Clustrs ar not rqurd to hav
More informationFrom Structural Analysis to FEM. Dhiman Basu
From Structural Analyss to FEM Dhman Basu Acknowldgmnt Followng txt books wr consultd whl prparng ths lctur nots: Znkwcz, OC O.C. andtaylor Taylor, R.L. (000). Th FntElmnt Mthod, Vol. : Th Bass, Ffth dton,
More informationPHYS ,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 informationChapter 3 Binary Image Analysis. Comunicação Visual Interactiva
Chapt 3 Bnay Iag Analyss Counação Vsual Intatva Most oon nghbohoods Pxls and Nghbohoods Nghbohood Vznhança N 4 Nghbohood N 8 Us of ass Exapl: ogn nput output CVI - Bnay Iag Analyss Exapl 0 0 0 0 0 output
More informationMulti-linear Systems and Invariant Theory. in the Context of Computer Vision and Graphics. Class 4: Mutli-View 3D-from-2D. CS329 Stanford University
Mult-lna Sytm and Invaant hoy n th Contxt of Comut Von and Gahc Cla 4: Mutl-Vw 3D-fom-D CS39 Stanfod Unvty Amnon Shahua Cla 4 Matal W Wll Cov oday Eola Gomty and Fundamntal Matx h lan+aallax modl and latv
More informationNEWTON S THEORY OF GRAVITY
NEWTON S THEOY OF GAVITY 3 Concptual Qustions 3.. Nwton s thid law tlls us that th focs a qual. Thy a also claly qual whn Nwton s law of gavity is xamind: F / = Gm m has th sam valu whth m = Eath and m
More informationPHYS 1441 Section 002 Lecture #15
PHYS 1441 Secton 00 Lecture #15 Monday, March 18, 013 Work wth rcton Potental Energy Gravtatonal Potental Energy Elastc Potental Energy Mechancal Energy Conservaton Announcements Mdterm comprehensve exam
More informationElectrochemical Equilibrium Electromotive Force. Relation between chemical and electric driving forces
C465/865, 26-3, Lctur 7, 2 th Sp., 26 lctrochmcal qulbrum lctromotv Forc Rlaton btwn chmcal and lctrc drvng forcs lctrochmcal systm at constant T and p: consdr G Consdr lctrochmcal racton (nvolvng transfr
More informationMTH 263 Practice Test #1 Spring 1999
Pat Ross MTH 6 Practce Test # Sprng 999 Name. Fnd the area of the regon bounded by the graph r =acos (θ). Observe: Ths s a crcle of radus a, for r =acos (θ) r =a ³ x r r =ax x + y =ax x ax + y =0 x ax
More informationFourier transforms (Chapter 15) Fourier integrals are generalizations of Fourier series. The series representation
Pof. D. I. Nass Phys57 (T-3) Sptmb 8, 03 Foui_Tansf_phys57_T3 Foui tansfoms (Chapt 5) Foui intgals a gnalizations of Foui sis. Th sis psntation a0 nπx nπx f ( x) = + [ an cos + bn sin ] n = of a function
More informationYou will analyze the motion of the block at different moments using the law of conservation of energy.
Physcs 00A Homework 7 Chapter 8 Where s the Energy? In ths problem, we wll consder the ollowng stuaton as depcted n the dagram: A block o mass m sldes at a speed v along a horzontal smooth table. It next
More informationFree carriers in materials
Lctu / F cais in matials Mtals n ~ cm -3 Smiconductos n ~ 8... 9 cm -3 Insulatos n < 8 cm -3 φ isolatd atoms a >> a B a B.59-8 cm 3 ϕ ( Zq) q atom spacing a Lctu / "Two atoms two lvls" φ a T splitting
More informationEMU Physics Department
Physcs 0 Lecture 8 Potental Energy and Conservaton Assst. Pro. Dr. Al ÖVGÜN EMU Physcs Department www.aovgun.com Denton o Work W q The work, W, done by a constant orce on an object s dened as the product
More information