Electrical Energy and Capacitance
|
|
- Hubert Wiggins
- 6 years ago
- Views:
Transcription
1 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 into kintic nrgy of th lctron. Not that th lctron movs to a rgion of highr potntial, but bcaus th lctron has ngativ charg this corrsponds to a dcras in th potntial nrgy of th lctron.. (b), (d). hargd particls always tnd to mov toward positions of lowr potntial nrgy. Th lctrical potntial nrgy of a chargd particl is PE qv and, for positivlychargd particls, this incrass as V incrass. For a ngativly-chargd particl, th potntial nrgy dcrass as V incrass. Thus, a positivly-chargd particl locatd at x A would mov toward th lft. A ngativly-chargd particl would oscillat around x B which is a position of minimum potntial nrgy for ngativ chargs.. (d). If th potntial is zro at a point locatd a finit distanc from chargs, ngativ chargs must b prsnt in th rgion to mak ngativ contributions to th potntial and cancl positiv contributions mad by positiv chargs in th rgion. 4. (c). Both th lctric potntial and th magnitud of th lctric fild dcras as th distanc from th chargd particl incrass. Howvr, th lctric flux through th balloon dos not chang bcaus it is proportional to th total charg nclosd by th balloon, which dos not chang as th balloon incrass in siz. 5. (a). From th consrvation of nrgy, th final kintic nrgy of ithr particl will b givn by ( ) ( ) ( ) KE KE + PE PE + qv qv q V V q V f i i f i f f i For th lctron, q and V V giving + KEf ( )( ) + and V V ( )( ) For th proton, q, so lctron. + V + V. KE V + V, th sam as that of th 6. (c). Th battry movs ngativ charg from on plat and puts it on th othr. Th first plat is lft with xcss positiv charg whos magnitud quals that of th ngativ charg movd to th othr plat. f 5
2 6 HAPTER 6 7. (a) dcrass. (b) Q stays th sam. (c) E stays th sam. (d) V incrass. () Th nrgy stord incrass. Bcaus th capacitor is rmovd from th battry, chargs on th plats hav nowhr to go. Thus, th charg on th capacitor plats rmains th sam as th plats ar pulld QA apart. Bcaus E σ, th lctric fild is constant as th plats ar sparatd. Bcaus V Ed and E dos not chang, V incrass as d incrass. Bcaus th sam charg is stord at a highr potntial diffrnc, th capacitanc has dcrasd. Bcaus Enrgy stord Q and Q stays th sam whil dcrass, th nrgy stord incrass. Th xtra nrgy must hav bn transfrrd from somwhr, so work was don. This is consistnt with th fact that th plats attract on anothr, and work must b don to pull thm apart. 8. (a) incrass. (b) Q incrass. (c) E stays th sam. (d) V rmains th sam. () Th nrgy stord incrass. Th prsnc of a dilctric btwn th plats incrass th capacitanc by a factor qual to th dilctric constant. Sinc th battry holds th potntial diffrnc constant whil th Q V will incras. Bcaus th potntial capacitanc incrass, th charg stord ( ) diffrnc and th distanc btwn th plats ar both constant, th lctric fild E V d will stay th sam. Th battry maintains a constant potntial diffrnc. With ( ) V ( ) constant whil capacitanc incrass, th stord nrgy Enrgy stord ( V) incras. will 9. (a). Incrasd random motions associatd with an incras in tmpratur mak it mor difficult to maintain a high dgr of polarization of th dilctric matrial. This has th ffct of dcrasing th dilctric constant of th matrial, and in turn, dcrasing th capacitanc of th capacitor.
3 Elctrical Enrgy and apacitanc 7 Answrs to Evn Numbrd oncptual Qustions. hanging th ara will chang th capacitanc and maximum charg but not th maximum voltag. Th qustion dos not allow you to incras th plat sparation. You can incras th maximum oprating voltag by insrting a matrial with highr dilctric strngth btwn th plats. 4. Elctric potntial V is a masur of th potntial nrgy pr unit charg. Elctrical potntial nrgy, PE QV, givs th nrgy of th total charg Q. 6. A sharp point on a chargd conductor would produc a larg lctric fild in th rgion nar th point. An lctric discharg could most asily tak plac at th point. 8. Thr ar ight diffrnt combinations that us all thr capacitors in th circuit. Ths combinations and thir quivalnt capacitancs ar: All thr capacitors in sris - q + + All thr capacitors in paralll q On capacitor in sris with a paralll combination of th othr two: q + +, q + +, q + + On capacitor in paralll with a sris combination of th othr two: q + + +, q + +, q +. Nothing happns to th charg if th wirs ar disconnctd. If th wirs ar connctd to ach othr, th charg rapidly rcombins, laving th capacitor unchargd.. All connctions of capacitors ar not simpl combinations of sris and paralll circuits. As an xampl of such a complx circuit, considr th ntwork of fiv capacitors,,, 4, and 5 shown blow. 5 4 This combination cannot b rducd to a simpl quivalnt by th tchniqus of combining sris and paralll capacitors.
4 8 HAPTER 6 4. Th matrial of th dilctric may b abl to withstand a largr lctric fild than air can withstand bfor braking down to pass a spark btwn th capacitor plats. 6. (a) i (b) ii 8. (a) Th quation is only valid whn th points A and B ar locatd in a rgion whr th lctric fild is uniform (that is, constant in both magnitud and dirction). (b) No. Th fild du to a point charg is not a uniform fild. (c) Ys. Th fild in th rgion btwn a pair of paralll plats is uniform.
5 Elctrical Enrgy and apacitanc 9 Answrs to Evn Numbrd Problms 4. (a) 6. J (b) 5 V J 8. (a) 5.5 m s (b) m s. 4. kv.. V J J m s. (a) 48. µ (b) 6. µ 4. (a) 8 V (b) Q Q 6.. Å 8.. kv. (a) 8. µ F (b).78 µ F.. pf and 6. pf 4. (a). µ F (b) Q 4 44 µ, Q7. µ, Q4 Q86 µ f 6. Ys. onnct a paralll combination of two capacitors in sris with anothr paralll combination of two capacitors. V 45. V. 8.. µ F µ F 4..9 µ F 44. (a).5 J (b) 68 V kg i
6 4 HAPTER (a). V m (b) 4.5 n (c) 5. p 5..4 m 54. ±, 4 4 p p p s p p p s 56. (a) 4.8 V (b) 4.6 V (c) 4.8 V (d) 5.4 J 58. (a) ab k b a ( ) 6. κ µ on, 89 µ on 64. V 66. (a). m (b) at x 4.4 mm
7 Elctrical Enrgy and apacitanc 4 Problm Solutions 6. (a) Th work don is cosθ ( ) W F s qe scosθ, or ( )( )( ) N. m cos 6.4 J W (b) Th chang in th lctrical potntial nrgy is J PE W (c) Th chang in th lctrical potntial is PE V 4. V q J (a) W follow th path from (,) to ( cm,) to ( cm,5 cm). Th work don on th charg by th fild is ( ) cosθ ( ) W W + W qe s + qe s cosθ ( qe) ( ) ( ). m cos +.5 m cos9 ( )( ) ( ) V m. m 6. J + Thus, PE W 4 6. J (b) PE V q 4 6. J 5 J 5 V Th work don by th agnt moving th charg out of th cll is input fild ( ) ( ) W W PE + q V J ( ) J 6.4 PE q( V ) q( Vf Vi ), so 7 PE.9 J q V V +6. J f i 9.
8 4 HAPTER V 5 J E d.5 m 6.7 N 6.6 Sinc potntial diffrnc is work pr unit charg W V, th work don is q ( ) ( )( ) W q V J 4. J 6.7 (a) V 6 J E d 5. m 5. N (b) F q E ( )( ) N.8 N (c) W F scosθ ( ) ( ) N 5..9 m cos 4.8 J mv q V v 6.8 From consrvation of nrgy, ( ) f or f q ( V) m (a) For th proton, v f 9 ( )( ).6 V 7.67 kg 5.5 m s (b) For th lctron, v f 9 ( )( + ).6 V 9. kg m s
9 Elctrical Enrgy and apacitanc (a) Us consrvation of nrgy ( KE + PE + PE ) ( KE + PE + PE ) s f s i k Q ur E or ( KE) + ( PE ) + ( PE ) s x ( KE) sinc th block is at rst at both bginning and nd. ( PEs ) kxmax, whr x max is th maximum strtch of th spring. ( ) ( ) max PE W QE x + kxmax QE xmax, giving Thus, ( ) x max ( 6 )( 5. 5 V m) QE 5. k N m.5 m (b) At quilibrium, Σ F F + F, or kx + QE s q Thrfor, x q QE xmax k.5 m Not that whn th block is rlasd from rst, it ovrshoots th quilibrium position and oscillats with simpl harmonic motion in th lctric fild. 6. Using y v yt+ ayt for th full flight givs v yt+ ayt, or a y v y t Thn, using + a ( y) ( y) v v for th upward part of th flight givs y y y ( v y t) (. m s)( 4. s) vy vy vyt.6 m max a 4 4 y
10 44 HAPTER 6 From Nwton s scond law, a y ΣFy mg qe qe g+ m m m. Equating this to th arlir rsult givs a y qe g+ m t v y, so th lctric fild strngth is ( ) m vy. kg. ms E g 9.8 m s.95 N 6 q t s Thus, ( V) ( ymax ) E ( )( ) 4.6 m.95 N 4. V 4. kv max 6. (a) V kq ( N m )(.6 9 ) 7.44 V - r. m kq kq r r r r (b) V V V ( k q) 9 9 ( 8.99 N m )(.6 ). m. m V 6. q q V V + V k + r, and r r whr.6 m.6 m r.6 m. m. m. Thus, N m. 6. V V. 6 m. m
11 Elctrical Enrgy and apacitanc (a) alling th. µ charg q, V kq q i k + + i r i r r r + r q q N m m. m 6 (.6 ) (. ) + m 6 V.67 V 6 6 (b) Rplacing. by. in part (a) yilds V. 6 V 6.4 W q( V) q( Vf Vi ), and V f sinc th 8. µ is infinit distanc from othr chargs. q q Vi k + r r 6.5 V 6 6 Thus, W ( 8. )(. ) N m m (. ) (.6 ) + m 5 V 9.8 J 6.5 (a) V i k q r i i.75 m.75 m N m V
12 46 HAPTER 6 (b) kq PE iq r 9 9 ( 5. )(. ) N m.5 m J Th ngativ sign mans that positiv work must b don to sparat th chargs (that is, bring thm up to a stat of zro potntial nrgy) Th potntial at distanc r. m from a charg Q + 9. is V kq r ( )( N m m ) +7 V Thus, th work rquird to carry a charg is W qv. + 7 V ( )( ) q. 8.9 J from infinity to this location 6.7 Th Pythagoran thorm givs th distanc from th midpoint of th bas to th charg at th apx of th triangl as ( ) ( ) r 4. cm. cm 5 cm 5 m Thn, th potntial at th midpoint of th bas is V kqi ri, or i ( 9 ) ( ) ( + 7. ) 9 N m V m. m 5 m 4. V. kv
13 Elctrical Enrgy and apacitanc Outsid th sphrical charg distribution, th potntial is th sam as for a point charg at th cntr of th sphr, V k Q r, whr Q 9. Thus, ( ) ( ) PE q V kq r f r i and from consrvation of nrgy ( KE) ( ) PE, or mv kq r f r i This givs kq v, or m rf r i 9 N m (. )(.6 ) v 9. kg. m. m 6 v 7.5 m s 6.9 From consrvation of nrgy, ( KE PE ) ( KE PE ) + +, which givs f i kqq + + or mv α i rf r f ( )( ) k 79 kqq mv α i mv α i r f 9 N m ( 58)(.6 ) 7 7 ( 6.64 kg)(. m s) 4.74 m 6. By dfinition, th work rquird to mov a charg from on point to any othr point on W Fcosθ s, th work an quipotntial surfac is zro. From th dfinition of work, ( ) is zro only if s or F cosθ. Th displacmnt s cannot b assumd to b zro in all cass. Thus, on must rquir that F cosθ. Th forc F is givn by F qe and nithr th charg q nor th fild strngth E can b assumd to b zro in all cass. Thrfor, th only way th work can b zro in all cass is if cosθ. But if cosθ, thn θ 9 or th forc (and hnc th lctric fild) must b prpndicular to th displacmnt s (which is tangnt to th surfac). That is, th fild must b prpndicular to th quipotntial surfac at all points on that surfac.
14 48 HAPTER 6 kq 6. V so r 9 9 ( )( ) kq 8.99 N m V m r V V V For V V, 5. V, and 5. V, r.79 m,.44 m, and.88 m Th radii ar invrsly proportional to th potntial. Q V µ 6. (a) ( ) ( )( ) F. V (b) ( V) ( )( ) Q µ F.5 V (a) 6 (. m ) A F d N m ( 8 m) (b) Q ( V) ( E d) max max max 8 6 ( )( )( ). F. N 8 m For a paralll plat capacitor, Q Q Qd V A d A. ( ) (a) Doubling d whil holding Q and A constant doubls V to 8 V. (b) ( ) A V Q Thus, doubling d whil holding V and A constant will cut th d charg in half, or Q Q f i 6.5 (a) (b) V d plat. V.8 m 4 E. V m. kv m dirctd toward th ngativ - A d ( 8.85 N m )( m ) -.8 m.74 F.74 pf
15 Elctrical Enrgy and apacitanc 49 (c) ( V) ( )( ) Q.74 F. V p on on plat and 74.7 p on th othr plat. 6.6 A, so d d A ( 8.85 N m )(. m ) 9. m F 9 Å d (. m). Å - m 6.7 (a) (b) ( 4 )(. m) 4 ( 8.85 N m )( 5. m ) Q Q Qd V A d A V 9.4 V E - d. m V m 9.4 V mg 6.8 Σ Fy Tcos5. mg or T cos5. 5. T ur or Σ F qe Tsin5. mgtan5. x mg tan5. E q mgdtan5. V Ed q mg ur ur ur F qe 6 ( )( )( ) 5 kg 9.8 m s.4 m tan5. V. V. kv (a) For sris connction, + q ( ) Q q V V + (.5 µ F)(. µ F) ( ).5 µ F +. µ F q + 4 V. µ on ach
16 5 HAPTER 6 (b) Q ( V) ( µ )( ) µ.5 F 4 V. Q ( V ) ( µ )( ) µ. F 4 V (a) For paralll connction, + + ( + + ) µ µ F 8. F q (b) For sris connction, + + q + +, giving q.78 µ F 5. µ F 4. µ F 9. µ F q 6. (a) Using th ruls for combining capacitors in sris and in paralll, th circuit is rducd in stps as shown blow. Th quivalnt capacitor is shown to b a. µ F capacitor a b c a b c a c.. V. V. V Figur Figur Figur
17 Elctrical Enrgy and apacitanc 5 (b) From Figur : Q ( V ) ( µ )( ) µ. F. V 4. ac ac ac From Figur : Q Q Q 4. µ ab bc ac Thus, th charg on th. µ F capacitor is Q 4. µ ontinuing to us Figur, ( V ) bc 4. µ and ( V) ( V) bc µ Qab 4. µ 4. V ab 6. µ F Q 8. V. F From Figur, ( V) 4 ( V) ( V) 4. V and 4 4 bc ( V ) ( µ )( ) µ Q 4 4. F 4. V 6. ( ) ( )( ) Q V. µ F 4. V 8. µ ab ab 6. paralll + 9. pf 9. pf () + sris sris +. pf Thus, using quation (), sris ( 9. pf ) ( 9. pf ) +. pf which rducs to ( 9. pf) + 8. ( pf), or ( pf)(. ) 6. pf Thrfor, ithr 6. pf and, from quation (),. pf or. pf and 6. pf. W conclud that th two capacitancs ar. pf and 6. pf.
18 5 HAPTER 6 6. a c. b a.5 6. c. b a 8.5 c. b Figur Figur Figur (a) Th quivalnt capacitanc of th uppr branch btwn points a and c in Figur is s ac ( 5. µ F)(. µ F) 5. µ F +. µ F.5 µ F Thn, using Figur, th total capacitanc btwn points a and c is.5 µ F+6. µ F8.5 µ F From Figur, th total capacitanc is q µ F. µ F 5.96 µ F (b) Q Q ( ) Q V ab ac cb ab q ( )( µ ) µ 5. V 5.96 F 89.5 Thus, th charg on th. µ is Q Q 89.5 µ cb 89.5 µ 5. V.5 V ac ab bc. µ F ( V) ( V) ( V) ( ) ( µ ) µ Thn, Q6 V 6. F 6. ac and ( ) ( Q5 Q V.5 µ F 6. µ ac )
19 Elctrical Enrgy and apacitanc (a) Th combination rducs to an quivalnt capacitanc of. µ F in stags as shown blow. 6. V V V. Figur Figur Figur (b) From Figur, Q ( µ )( ) µ 4 4. F 6. V 44 ( µ )( ) µ Q. F 6. V 7. and ( µ )( ) µ Q. F 6. V Thn, from Figur, Q4 Q8 Q 6 6 µ 6.5 a a a 4. V. 8. b c V b c V c 4. Figur Figur Figur Th circuit may b rducd in stps as shown abov. Using th Figur, Q ( µ )( ) µ Thn, in Figur, ( V ) ac 4. F 4. V 96. Qac 96. µ ab 6. V 6. µ F and ( V) ( V) ( V) 4. V 6. V 8. V ab bc ac ab
20 54 HAPTER 6 Finally, using Figur, Q ( V ) ( µ )( ) µ. F 6. V 6. ab ( µ )( ) µ Q ( µ )( V) µ Q5 5. F V 8., ab 8 8. F 64. bc and Q ( µ )( V) µ 4 4. F. bc 6.6 Th tchnician combins two of th capacitors in paralll making a capacitor of capacitanc µ F. Thn sh dos it again with two mor of th capacitors. Thn th two rsulting µ F capacitors ar connctd in sris to yild an quivalnt capacitanc of µ F. Bcaus of th symmtry of th solution, vry capacitor in th combination has th sam voltag across it, ( V) ( 9. V) V 45. V ab A B 6.7 (a) From Q ( V), Q ( µ )( ) µ 5 5. F 5. V.5.5 m and Q 4 ( µ )( ) µ 4. F 5. V.. m (b) Whn th two capacitors ar connctd in paralll, th quivalnt capacitanc is q + 5. µ F+4. µ F 65. µ F. Sinc th ngativ plat of on was connctd to th positiv plat of th othr, th total charg stord in th paralll combination is Q Q Q µ µ 75 µ Th potntial diffrnc across ach capacitor of th paralll combination is Q 75 µ V 65. µ F q.5 V and th final charg stord in ach capacitor is Q ( V ) ( µ )( ) µ 5 5. F.5 V 88 and Q 4 Q Q 5 75 µ 88 µ 46 µ
21 Elctrical Enrgy and apacitanc From Q ( V), th initial charg of ach capacitor is ( µ )( ) µ and Q ( ) Q. F. V x Aftr th capacitors ar connctd in paralll, th potntial diffrnc across ach is V. V, and th total charg of Q Q + Q x µ is dividd btwn th two capacitors as x ( )( ) Q. µ F. V. µ and Q Q Q µ. µ 9. µ x Thus, x Q x 9. µ V. V. µ F 6.9 From Q ( V), th initial charg of ach capacitor is ( µ )( ) µ and ( µ )( ) Q. F. V. Q. F Aftr th capacitors ar connctd in paralll, th potntial diffrnc across on is th sam as that across th othr. This givs V Q Q. µ F. µ F or Q Q () From consrvation of charg, Q + Q Q + Q. µ. Thn, substituting from quation (), this bcoms Q + Q. µ, giving Q µ Finally, from quation (), Q µ
22 56 HAPTER Th original circuit rducs to a singl quivalnt capacitor in th stps shown blow. a a a a s s p q p b b b b s + +. µ F 5. µ F. µ F + + ( µ ) + µ µ. F. F 8.66 F p s s + ( µ ) µ p. F. F q + + p p 8.66 µ F. µ F 6.4 µ F 6.4 Rfr to th solution of Problm 6.4 givn abov. Th total charg stord btwn points a and b is Q ( V ) ( µ )( ) µ q q ab 6.4 F 6. V 6 Thn, looking at th third figur, obsrv that th chargs of th sris capacitors of that figur ar Qp Qp Qq 6 µ. Thus, th potntial diffrnc across th uppr paralll combination shown in th scond figur is Qp 6 µ p 4.8 V 8.66 µ F ( V ) p Finally, th charg on is ( ) ( µ )( ) µ Q V. F 4.8 V 8.6 p
23 Elctrical Enrgy and apacitanc Rcogniz that th 7. µ F and th 5. µ F of th cntr branch ar connctd in sris. Th total capacitanc of that branch is s +.9 µ F a b Thn rcogniz that this capacitor, th 4. µ F capacitor, and th 6. µ F capacitor ar all connctd in paralll btwn points a and b. Thus, th quivalnt capacitanc btwn points a and b is µ F +.9 µ F+6. µ F.9 µ F q 6.4 Th capacitanc is A ( 8.85 N m )(. 4 m ).54 F - and th stord nrgy is d 5. m.54 F. V.55 J ( ) ( )( ) W V 6.44 (a) Whn connctd in paralll, th nrgy stord is W ( V) + ( V) ( + )( V) 6 ( ) F ( V).5 J (b) Whn connctd in sris, th quivalnt capacitanc is q + µ F 4.7 µ F From W ( V) th potntial diffrnc rquird to stor th sam nrgy as q, in part (a) abov is W V q (.5 J) F 68 V
24 58 HAPTER Th capacitanc of this paralll plat capacitor is A 6 (. m ) F d N m ( 8 m ) 6 With an lctric fild strngth of E. N and a plat sparation of d 8 m, th potntial diffrnc btwn plats is ( )( ) V Ed 6 9. V m 8 m.4 V Thus, th nrgy availabl for rlas in a lightning strik is W V. 8 9 ( ) ( F)(.4 V). J 6.46 Th nrgy transfrrd to th watr is ( 5. )(. 8 V) W Q( V) Thus, if m is th mass of watr boild away, 7.5 J ( ) v W m c T + L bcoms m J kg ( ) J J kg + giving 7.5 J m.55 J kg 9.79 kg 6.47 Th initial capacitanc (with air btwn th plats) is i Q ( V) i capacitanc (with th dilctric insrtd) is Q ( V) quantity of charg stord on th plats. f f, and th final whr Q is th constant ( ) ( ) f V i V Thus, th dilctric constant is κ V 5 V i f (a) V 6. V E - d. m. V m
25 Elctrical Enrgy and apacitanc 59 (b) With air btwn th plats, th capacitanc is air 4 (. m ) ( ) A F d N m. m and with watr ( κ 8) ( )( κ air ( ) ( btwn th plats, th capacitanc is F 7.8 F Th stord charg whn watr is btwn th plats is ) )( ) Q V 7.8 F 6. V n (c) Whn air is th dilctric btwn th plats, th stord charg is air air ( ) ( )( ) Q V 8.85 F 6. V p 6.49 (a) Th dilctric constant for Tflon is κ., so th capacitanc is κ A d ( )( N m )( 75 4 m ) -.4 m 9 8. F 8. nf (b) For Tflon, th dilctric strngth is is E max 6 6. V m, so th maximum voltag max max 6 - ( 6. V m)(.4 m ) V E d V max.4 V.4 kv 6.5 Bfor th capacitor is rolld, th capacitanc of this paralll plat capacitor is ( ) A κ w L κ d d whr A is th surfac ara of on sid of a foil strip. Thus, th rquird lngth is 8 ( 9.5 F)(.5 m) (.7)( 8.85 N m )( 7. m) d L κ w.4 m
26 6 HAPTER (a). kg m V 9.9 m ρ kg m 6 4π r Sinc V, th radius is V r 4π, and th surfac ara is ( 6 m ) V 9.9 A 4πr 4π 4π 4.54 m 4π 4π (b) κ d A ( )( N m )( 4.54 m ) 9 m. F (c) Q ( V) ( )( ) - 4. F V. and th numbr of lctronic chargs is 4. Q n Sinc th capacitors ar in paralll, th quivalnt capacitanc is q A or q whr A A + A + A d ( A A A ) A A A + + d d d d 6.5 Sinc th capacitors ar in sris, th quivalnt capacitanc is givn by d d d d + d + d A A A A q A or q whr d d + d + d d
27 Elctrical Enrgy and apacitanc For th paralll combination: p + which givs p () For th sris combination: + s or s s s s Thus, w hav s and quating this to Equation () abov givs p s s or p p s + s s W writ this rsult as : + p p s and us th quadratic formula to obtain ± 4 p p p s Thn, Equation () givs 4 p p p s 6.55 Th charg stord on th capacitor by th battry is ( ) ( ) Q V V This is also th total charg stord in th paralll combination whn this chargd capacitor is connctd in paralll with an unchargd.- F th rsulting voltag across th paralll combination, ( ) V is µ capacitor. Thus, if ( ) Q p V givs ( V) ( +. µ F)( ) or ( 7. V) (. V)(. µ F). V. V. F 4.9 F 7. V and ( µ ) µ 6.56 (a) Th.-µ is locatd.5 m from point P, so its contribution to th potntial at P is 6 q 9. 4 V k ( 8.99 N m ).8 V r.5 m (b) Th potntial at P du to th.-µ charg locatd.5 m away is 6 q 9. 4 V k ( 8.99 N m ).6 V r.5 m
28 6 HAPTER 6 (c) Th total potntial at point P is V V V ( ) P V.8 V (d) Th work rquird to mov a charg q. µ to point P from infinity is 6 4 ( P ) ( )(.8 ) W q V q V V. V 5.4 J 6.57 Th stags for th rduction of this circuit ar shown blow V 48. V 48. V 48. V Thus, 6.5 µ F q 6.58 (a) Du to sphrical symmtry, th charg on ach of th concntric sphrical shlls will b uniformly distributd ovr that shll. Insid a sphrical surfac having a uniform charg distribution, th lctric fild du to th charg on that surfac is zro. Thus, in this rgion, th potntial du to th charg on that surfac is constant and qual to th potntial at th surfac. Outsid a sphrical surfac having a uniform charg kq distribution, th potntial du to th charg on that surfac is givn by V r whr r is th distanc from th cntr of that surfac and q is th charg on that surfac. In th rgion btwn a pair of concntric sphrical shlls, with th innr shll having charg + Q and th outr shll having radius b and charg Q, th total lctric potntial is givn by ( Q) kq k V Vdu to +Vdu to + kq innr shll outr shll r b r b
29 Elctrical Enrgy and apacitanc 6 Th potntial diffrnc btwn th two shlls is thrfor, b a V V V k r a r b Q kq kq a b b b ab Th capacitanc of this dvic is givn by Q ab V k b a ( ) (b) Whn b >> a, thn ba b. Thus, in th limit as b, th capacitanc found abov bcoms ab a 4π a k k ( b) W V. Hnc, 6.59 Th nrgy stord in a chargd capacitor is ( ) ( ) W J V 4.47 V 4.47 kv -6. F 6.6 From Q ( V), th capacitanc of th capacitor with air btwn th plats is Q V 5 µ V Aftr th dilctric is insrtd, th potntial diffrnc is hld to th original valu, but th charg changs to Q Q + µ 5 µ. Thus, th capacitanc with th dilctric slab in plac is Q 5 µ V V Th dilctric constant of th dilctric slab is thrfor κ 5 µ V 5 V 5 µ 5.
30 64 HAPTER Th chargs initially stord on th capacitors ar Q ( V ) ( µ )( ) µ 6. F 5 V.5 i and Q ( V ) ( µ )( ) µ. F 5 V 5. i Whn th capacitors ar connctd in paralll, with th ngativ plat of on connctd to th positiv plat of th othr, th nt stord charg is Q Q Q µ µ µ Th quivalnt capacitanc of th paralll combination is + 8. µ F. Thus, th final potntial diffrnc across ach of th capacitors is q Q µ 8. µ F ( V ) q. 5 V and th final charg on ach capacitor is 6. F 5 V m Q ( V ) ( µ )( ) µ and. F 5 V 5.5 m Q ( V ) ( µ )( ) µ 6.6 Whn connctd in sris, th quivalnt capacitanc is q µ F. µ F µ F and th charg stord on ach capacitor is 4 4 Q Q Qq q ( V) µ F ( V ) µ i Whn th capacitors ar rconnctd in paralll, with th positiv plat of on connctd to th positiv plat of th othr, th nw quivalnt capacitanc is + 6. µ F and th nt stord charg is Q Q + Q 8 µ diffrnc across ach of th capacitors is Q 8 µ 44.4 V 6. µ F ( V ) q. Thrfor, th final potntial q
31 Elctrical Enrgy and apacitanc 65 Th final charg on ach of th capacitors is 4. F 44.4 V.8 Q ( V ) ( µ )( ) µ and. F 44.4 V 89 Q ( V ) ( µ )( ) µ kq 6.6 (a) kq kq V V V V x + d x x d kq ( ) ( ( d ) ) ( ) x x d x d x x d + + x x kqd kqd which simplifis to V x x x xd (b) Whn x>> d, thn x d x ( d ) kqd and V bcoms V x x d ( ) kqd x 6.64 Th nrgy rquird to mlt th lad sampl is ( ) W m cpb T L + f ( ) ( )( ) 6 6. kg 8 J kg J kg +.8 J Th nrgy stord in a capacitor is ( ) W V, so th rquird potntial diffrnc is ( ) W.8 J V F V
32 66 HAPTER Th capacitanc of a paralll plat capacitor is κ d A Thus, κ A d, and th givn forc quation may b rwrittn as ( Q) ( V) Q Q F κ A d d d With th givn data valus, th forc is ( V) 6 ( F)( V) F d. m ( ) 5 N 6.66 Th lctric fild btwn th plats is dirctd downward with magnitud E y V V - d. m 4 5. N m Sinc th gravitational forc xprincd by th lctron is ngligibl in comparison to th lctrical forc acting on it, th vrtical acclration is a y (.6 9 )( 5. 4 N m) Fy qey m s m m 9. kg 5 (a) At th closst approach to th bottom plat, v y. Thus, th vrtical displacmnt from point O is found from v v a ( y) + as y y y sin in 45 y.89 mm 6 ( v ) ( 5.6 m s) s θ 5 a y 8.78 ( ms) Th minimum distanc abov th bottom plat is thn D d + y. mm.89 mm. mm
33 Elctrical Enrgy and apacitanc 67 (b) Th tim for th lctron to go from point O to th uppr plat is found from y v yt+ ay t as m +. m 5.6 sin 45 t s m t s 6 5 Solving for t givs a positiv solution of t displacmnt from point O at this tim is 9. s 6 9 ( ) ( ) x vxt 5.6 m s cos 45. s 4.4 mm. Th horizontal
34 68 HAPTER 6
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 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 informationPart 7: Capacitance And Capacitors
Part 7: apacitanc And apacitors 7. Elctric harg And Elctric Filds onsidr a pair of flat, conducting plats, arrangd paralll to ach othr (as in figur 7.) and sparatd by an insulator, which may simply b air.
More information7.4 Potential Difference and Electric Potential
7.4 Potntial Diffrnc and Elctric Potntial In th prvious sction, you larnd how two paralll chargd surfacs produc a uniform lctric fild. From th dfinition of an lctric fild as a forc acting on a charg, it
More information5 Chapter Capacitance and Dielectrics
5 Chaptr Capacitanc and Dilctrics 5.1 Introduction... 5-3 5. Calculation of Capacitanc... 5-4 Exampl 5.1: Paralll-Plat Capacitor... 5-4 Exampl 5.: Cylindrical Capacitor... 5-6 Exampl 5.3: Sphrical Capacitor...
More information1997 AP Calculus AB: Section I, Part A
997 AP Calculus AB: Sction I, Part A 50 Minuts No Calculator Not: Unlss othrwis spcifid, th domain of a function f is assumd to b th st of all ral numbrs for which f () is a ral numbr.. (4 6 ) d= 4 6 6
More informationSAFE HANDS & IIT-ian's PACE EDT-15 (JEE) SOLUTIONS
It is not possibl to find flu through biggr loop dirctly So w will find cofficint of mutual inductanc btwn two loops and thn find th flu through biggr loop Also rmmbr M = M ( ) ( ) EDT- (JEE) SOLUTIONS
More information1.2 Faraday s law A changing magnetic field induces an electric field. Their relation is given by:
Elctromagntic Induction. Lorntz forc on moving charg Point charg moving at vlocity v, F qv B () For a sction of lctric currnt I in a thin wir dl is Idl, th forc is df Idl B () Elctromotiv forc f s any
More informationA 1 A 2. a) Find the wavelength of the radio waves. Since c = f, then = c/f = (3x10 8 m/s) / (30x10 6 Hz) = 10m.
1. Young s doubl-slit xprint undrlis th instrunt landing syst at ost airports and is usd to guid aircraft to saf landings whn th visibility is poor. Suppos that a pilot is trying to align hr plan with
More informationVoltage, Current, Power, Series Resistance, Parallel Resistance, and Diodes
Lctur 1. oltag, Currnt, Powr, Sris sistanc, Paralll sistanc, and Diods Whn you start to dal with lctronics thr ar thr main concpts to start with: Nam Symbol Unit oltag volt Currnt ampr Powr W watt oltag
More informationThe pn junction: 2 Current vs Voltage (IV) characteristics
Th pn junction: Currnt vs Voltag (V) charactristics Considr a pn junction in quilibrium with no applid xtrnal voltag: o th V E F E F V p-typ Dpltion rgion n-typ Elctron movmnt across th junction: 1. n
More informationDIELECTRIC AND MAGNETIC PROPERTIES OF MATERIALS
DILCTRIC AD MAGTIC PROPRTIS OF MATRIALS Dilctric Proprtis: Dilctric matrial Dilctric constant Polarization of dilctric matrials, Typs of Polarization (Polarizability). quation of intrnal filds in liquid
More informationChapter 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 information22/ Breakdown of the Born-Oppenheimer approximation. Selection rules for rotational-vibrational transitions. P, R branches.
Subjct Chmistry Papr No and Titl Modul No and Titl Modul Tag 8/ Physical Spctroscopy / Brakdown of th Born-Oppnhimr approximation. Slction ruls for rotational-vibrational transitions. P, R branchs. CHE_P8_M
More informationHigh Energy Physics. Lecture 5 The Passage of Particles through Matter
High Enrgy Physics Lctur 5 Th Passag of Particls through Mattr 1 Introduction In prvious lcturs w hav sn xampls of tracks lft by chargd particls in passing through mattr. Such tracks provid som of th most
More informationPH2200 Practice Final Exam Spring 2004
PH2200 Practic Final Exam Spring 2004 Instructions 1. Writ your nam and studnt idntification numbr on th answr sht. 2. This a two-hour xam. 3. Plas covr your answr sht at all tims. 4. This is a closd book
More informationcycle 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 informationDifferentiation of Exponential Functions
Calculus Modul C Diffrntiation of Eponntial Functions Copyright This publication Th Northrn Albrta Institut of Tchnology 007. All Rights Rsrvd. LAST REVISED March, 009 Introduction to Diffrntiation of
More informationAP PHYSICS C: ELECTRICITY AND MAGNETISM 2015 SCORING GUIDELINES
AP PHYSICS C: ELECTRICITY AND MAGNETISM 2015 SCORING GUIDELINES Qustion 1 15 points total Distribution of points (a) i. For at last on arrow btwn th plats pointing downward from th positiv plats toward
More informationLast time. Resistors. Circuits. Question. Quick Quiz. Quick Quiz. ( V c. Which bulb is brighter? A. A B. B. C. Both the same
Last tim Bgin circuits Rsistors Circuits Today Rsistor circuits Start rsistor-capacitor circuits Physical layout Schmatic layout Tu. Oct. 13, 2009 Physics 208 Lctur 12 1 Tu. Oct. 13, 2009 Physics 208 Lctur
More informationPARTICLE 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 information2008 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 informationHydrogen Atom and One Electron Ions
Hydrogn Atom and On Elctron Ions Th Schrödingr quation for this two-body problm starts out th sam as th gnral two-body Schrödingr quation. First w sparat out th motion of th cntr of mass. Th intrnal potntial
More informationAlpha and beta decay equation practice
Alpha and bta dcay quation practic Introduction Alpha and bta particls may b rprsntd in quations in svral diffrnt ways. Diffrnt xam boards hav thir own prfrnc. For xampl: Alpha Bta α β alpha bta Dspit
More informationElectromagnetism Physics 15b
lctromagntism Physics 15b Lctur #8 lctric Currnts Purcll 4.1 4.3 Today s Goals Dfin lctric currnt I Rat of lctric charg flow Also dfin lctric currnt dnsity J Charg consrvation in a formula Ohm s Law vryon
More informationExam 1. It is important that you clearly show your work and mark the final answer clearly, closed book, closed notes, no calculator.
Exam N a m : _ S O L U T I O N P U I D : I n s t r u c t i o n s : It is important that you clarly show your work and mark th final answr clarly, closd book, closd nots, no calculator. T i m : h o u r
More informationAddition of angular momentum
Addition of angular momntum April, 0 Oftn w nd to combin diffrnt sourcs of angular momntum to charactriz th total angular momntum of a systm, or to divid th total angular momntum into parts to valuat th
More information4. Money cannot be neutral in the short-run the neutrality of money is exclusively a medium run phenomenon.
PART I TRUE/FALSE/UNCERTAIN (5 points ach) 1. Lik xpansionary montary policy, xpansionary fiscal policy rturns output in th mdium run to its natural lvl, and incrass prics. Thrfor, fiscal policy is also
More informationv = = ˆ ˆ ˆ B x y x y x y y x = C m s 0.15 T m s T
Chaptr 8 1. (a) Equation 8-3 lads to F v = = sinφ (b) Th kintic nrgy of th proton is 17 650. 10 N = 5 400. 10 ms. 3 160. 10 C. 60 10 T sin 3. 0 c hc h 1 1 1.67 10 7 kg 4.00 10 5 m s 1.34 10 16 mv J ( )(
More informationSolution: APPM 1360 Final (150 pts) Spring (60 pts total) The following parts are not related, justify your answers:
APPM 6 Final 5 pts) Spring 4. 6 pts total) Th following parts ar not rlatd, justify your answrs: a) Considr th curv rprsntd by th paramtric quations, t and y t + for t. i) 6 pts) Writ down th corrsponding
More informationAddition of angular momentum
Addition of angular momntum April, 07 Oftn w nd to combin diffrnt sourcs of angular momntum to charactriz th total angular momntum of a systm, or to divid th total angular momntum into parts to valuat
More informationCh. 24 Molecular Reaction Dynamics 1. Collision Theory
Ch. 4 Molcular Raction Dynamics 1. Collision Thory Lctur 16. Diffusion-Controlld Raction 3. Th Matrial Balanc Equation 4. Transition Stat Thory: Th Eyring Equation 5. Transition Stat Thory: Thrmodynamic
More informationElectrochemistry L E O
Rmmbr from CHM151 A rdox raction in on in which lctrons ar transfrrd lctrochmistry L O Rduction os lctrons xidation G R ain lctrons duction W can dtrmin which lmnt is oxidizd or rducd by assigning oxidation
More informationANSWERS C C =
107 CHAPTER E1 ANSWERS 1.2 No. of xcss lctrons -0.1 10-6 C -1.6 10-19 C 6.2 1011 6.2 10 11 lctrons hav bn transfrrd from your hair to th comb. 1.3 Elctrical: 1, 2, 3, 4, 6, 8, 12, 13 Non-lctrical: 5, 11
More informationExam 2 Thursday (7:30-9pm) It will cover material through HW 7, but no material that was on the 1 st exam.
Exam 2 Thursday (7:30-9pm) It will covr matrial through HW 7, but no matrial that was on th 1 st xam. What happns if w bash atoms with lctrons? In atomic discharg lamps, lots of lctrons ar givn kintic
More informationCalculus 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 informationPair (and Triplet) Production Effect:
Pair (and riplt Production Effct: In both Pair and riplt production, a positron (anti-lctron and an lctron (or ngatron ar producd spontanously as a photon intracts with a strong lctric fild from ithr a
More informationPreliminary Fundamentals
1.0 Introduction Prliminary Fundamntals In all of our prvious work, w assumd a vry simpl modl of th lctromagntic torqu T (or powr) that is rquird in th swing quation to obtain th acclrating torqu. This
More informationGradebook & Midterm & Office Hours
Your commnts So what do w do whn on of th r's is 0 in th quation GmM(1/r-1/r)? Do w nd to driv all of ths potntial nrgy formulas? I don't undrstand springs This was th first lctur I actually larnd somthing
More informationPHYS-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 informationProblem Set 6 Solutions
6.04/18.06J Mathmatics for Computr Scinc March 15, 005 Srini Dvadas and Eric Lhman Problm St 6 Solutions Du: Monday, March 8 at 9 PM in Room 3-044 Problm 1. Sammy th Shark is a financial srvic providr
More information2. Laser physics - basics
. Lasr physics - basics Spontanous and stimulatd procsss Einstin A and B cofficints Rat quation analysis Gain saturation What is a lasr? LASER: Light Amplification by Stimulatd Emission of Radiation "light"
More informationELECTROMAGNETIC INDUCTION CHAPTER - 38
. (a) CTOMAGNTIC INDUCTION CHAPT - 38 3 3.dl MT I M I T 3 (b) BI T MI T M I T (c) d / MI T M I T. at + bt + c s / t Volt (a) a t t Sc b t Volt c [] Wbr (b) d [a., b.4, c.6, t s] at + b. +.4. volt 3. (a)
More informationThe 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 informationSPH4U 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 informationCoulomb s Law. understand & solve problems using Coulomb s Law. Labs, Activities & Demonstrations:
Add Important oulomb s Law Pag: 431 NGSS Standards: HS-PS-4, HS-PS3-5 oulomb s Law MA urriculum ramworks (006): 5.4 AP Physics 1 Larning Objctivs: 3...1, 3..,, 5.A..1 Skills: undrstand & solv problms using
More informationPrecise Masses of particles
/1/15 Physics 1 April 1, 15 Ovrviw of topic Th constitunts and structur of nucli Radioactivity Half-lif and Radioactiv dating Nuclar Binding Enrgy Nuclar Fission Nuclar Fusion Practical Applications of
More informationSeptember 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 informationsurface of a dielectric-metal interface. It is commonly used today for discovering the ways in
Surfac plasmon rsonanc is snsitiv mchanism for obsrving slight changs nar th surfac of a dilctric-mtal intrfac. It is commonl usd toda for discovring th was in which protins intract with thir nvironmnt,
More informationElectrochemical Energy Systems Spring 2014 MIT, M. Z. Bazant. Midterm Exam
10.66 Elctrochmical Enrgy Systms Spring 014 MIT, M. Z. Bazant Midtrm Exam Instructions. This is a tak-hom, opn-book xam du in Lctur. Lat xams will not b accptd. You may consult any books, handouts, or
More informationJOHNSON COUNTY COMMUNITY COLLEGE Calculus I (MATH 241) Final Review Fall 2016
JOHNSON COUNTY COMMUNITY COLLEGE Calculus I (MATH ) Final Rviw Fall 06 Th Final Rviw is a starting point as you study for th final am. You should also study your ams and homwork. All topics listd in th
More informationDIFFERENTIAL EQUATION
MD DIFFERENTIAL EQUATION Sllabus : Ordinar diffrntial quations, thir ordr and dgr. Formation of diffrntial quations. Solution of diffrntial quations b th mthod of sparation of variabls, solution of homognous
More information10. Limits involving infinity
. Limits involving infinity It is known from th it ruls for fundamntal arithmtic oprations (+,-,, ) that if two functions hav finit its at a (finit or infinit) point, that is, thy ar convrgnt, th it of
More informationMATHEMATICS (B) 2 log (D) ( 1) = where z =
MATHEMATICS SECTION- I STRAIGHT OBJECTIVE TYPE This sction contains 9 multipl choic qustions numbrd to 9. Each qustion has choic (A), (B), (C) and (D), out of which ONLY-ONE is corrct. Lt I d + +, J +
More informationSupplementary Materials
6 Supplmntary Matrials APPENDIX A PHYSICAL INTERPRETATION OF FUEL-RATE-SPEED FUNCTION A truck running on a road with grad/slop θ positiv if moving up and ngativ if moving down facs thr rsistancs: arodynamic
More informationu x v x dx u x v x v x u x dx d u x v x u x v x dx u x v x dx Integration by Parts Formula
7. Intgration by Parts Each drivativ formula givs ris to a corrsponding intgral formula, as w v sn many tims. Th drivativ product rul yilds a vry usful intgration tchniqu calld intgration by parts. Starting
More informationDeepak Rajput
Q Prov: (a than an infinit point lattic is only capabl of showing,, 4, or 6-fold typ rotational symmtry; (b th Wiss zon law, i.. if [uvw] is a zon axis and (hkl is a fac in th zon, thn hu + kv + lw ; (c
More informationMath 34A. Final Review
Math A Final Rviw 1) Us th graph of y10 to find approimat valus: a) 50 0. b) y (0.65) solution for part a) first writ an quation: 50 0. now tak th logarithm of both sids: log() log(50 0. ) pand th right
More informationTitle: Vibrational structure of electronic transition
Titl: Vibrational structur of lctronic transition Pag- Th band spctrum sn in th Ultra-Violt (UV) and visibl (VIS) rgions of th lctromagntic spctrum can not intrprtd as vibrational and rotational spctrum
More informationQuasi-Classical States of the Simple Harmonic Oscillator
Quasi-Classical Stats of th Simpl Harmonic Oscillator (Draft Vrsion) Introduction: Why Look for Eignstats of th Annihilation Oprator? Excpt for th ground stat, th corrspondnc btwn th quantum nrgy ignstats
More informationChapter 6 Current and Resistance
Chaptr 6 Currnt and Rsistanc 6.1 Elctric Currnt... 1 6.1.1 Currnt Dnsity... 1 6. Ohm s Law... 3 6.3 Elctrical Enrgy and Powr... 6 6.4 Summary... 6.5 Solvd Problms... 8 6.5.1 Rsistivity of a Cabl... 8 6.5.
More informationare given in the table below. t (hours)
CALCULUS WORKSHEET ON INTEGRATION WITH DATA Work th following on notbook papr. Giv dcimal answrs corrct to thr dcimal placs.. A tank contains gallons of oil at tim t = hours. Oil is bing pumpd into th
More informationCalculus concepts derivatives
All rasonabl fforts hav bn mad to mak sur th nots ar accurat. Th author cannot b hld rsponsibl for any damags arising from th us of ths nots in any fashion. Calculus concpts drivativs Concpts involving
More informationStatus of LAr TPC R&D (2) 2014/Dec./23 Neutrino frontier workshop 2014 Ryosuke Sasaki (Iwate U.)
Status of LAr TPC R&D (2) 214/Dc./23 Nutrino frontir workshop 214 Ryosuk Sasaki (Iwat U.) Tabl of Contnts Dvlopmnt of gnrating lctric fild in LAr TPC Introduction - Gnrating strong lctric fild is on of
More informationCoupled Pendulums. Two normal modes.
Tim Dpndnt Two Stat Problm Coupld Pndulums Wak spring Two normal mods. No friction. No air rsistanc. Prfct Spring Start Swinging Som tim latr - swings with full amplitud. stationary M +n L M +m Elctron
More information5. B To determine all the holes and asymptotes of the equation: y = bdc dced f gbd
1. First you chck th domain of g x. For this function, x cannot qual zro. Thn w find th D domain of f g x D 3; D 3 0; x Q x x 1 3, x 0 2. Any cosin graph is going to b symmtric with th y-axis as long as
More informationExtraction of Doping Density Distributions from C-V Curves
Extraction of Doping Dnsity Distributions from C-V Curvs Hartmut F.-W. Sadrozinski SCIPP, Univ. California Santa Cruz, Santa Cruz, CA 9564 USA 1. Connction btwn C, N, V Start with Poisson quation d V =
More informationProbability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers Roy D. Yates and David J.
Probability and Stochastic Procsss: A Frindly Introduction for Elctrical and Computr Enginrs Roy D. Yats and David J. Goodman Problm Solutions : Yats and Goodman,4.3. 4.3.4 4.3. 4.4. 4.4.4 4.4.6 4.. 4..7
More informationMor Tutorial at www.dumblittldoctor.com Work th problms without a calculator, but us a calculator to chck rsults. And try diffrntiating your answrs in part III as a usful chck. I. Applications of Intgration
More information1997 AP Calculus AB: Section I, Part A
997 AP Calculus AB: Sction I, Part A 50 Minuts No Calculator Not: Unlss othrwis spcifid, th domain of a function f is assumd to b th st of all ral numbrs x for which f (x) is a ral numbr.. (4x 6 x) dx=
More informationLecture 37 (Schrödinger Equation) Physics Spring 2018 Douglas Fields
Lctur 37 (Schrödingr Equation) Physics 6-01 Spring 018 Douglas Filds Rducd Mass OK, so th Bohr modl of th atom givs nrgy lvls: E n 1 k m n 4 But, this has on problm it was dvlopd assuming th acclration
More informationUnit 7 Charge-to-mass ratio of the electron
Unit 7 Charg-to-ass ratio of th lctron Kywords: J. J. Thoson, Lorntz Forc, Magntic Filds Objctiv: Obsrv th rsults of lctron ba influncd by th agntic fild and calculat th charg-to-ass ratio of th lctron.
More informationThat is, we start with a general matrix: And end with a simpler matrix:
DIAGON ALIZATION OF THE STR ESS TEN SOR INTRO DUCTIO N By th us of Cauchy s thorm w ar abl to rduc th numbr of strss componnts in th strss tnsor to only nin valus. An additional simplification of th strss
More informationUniversity of Illinois at Chicago Department of Physics. Thermodynamics & Statistical Mechanics Qualifying Examination
Univrsity of Illinois at Chicago Dpartmnt of hysics hrmodynamics & tatistical Mchanics Qualifying Eamination January 9, 009 9.00 am 1:00 pm Full crdit can b achivd from compltly corrct answrs to 4 qustions.
More informationPrinciples of Humidity Dalton s law
Principls of Humidity Dalton s law Air is a mixtur of diffrnt gass. Th main gas componnts ar: Gas componnt volum [%] wight [%] Nitrogn N 2 78,03 75,47 Oxygn O 2 20,99 23,20 Argon Ar 0,93 1,28 Carbon dioxid
More informationDerivation of Electron-Electron Interaction Terms in the Multi-Electron Hamiltonian
Drivation of Elctron-Elctron Intraction Trms in th Multi-Elctron Hamiltonian Erica Smith Octobr 1, 010 1 Introduction Th Hamiltonian for a multi-lctron atom with n lctrons is drivd by Itoh (1965) by accounting
More informationdt d Chapter 30: 1-Faraday s Law of induction (induced EMF) Chapter 30: 1-Faraday s Law of induction (induced Electromotive Force)
Chaptr 3: 1-Faraday s aw of induction (inducd ctromotiv Forc) Variab (incrasing) Constant Variab (dcrasing) whn a magnt is movd nar a wir oop of ara A, currnt fows through that wir without any battris!
More informationde/dx Effectively all charged particles except electrons
de/dx Lt s nxt turn our attntion to how chargd particls los nrgy in mattr To start with w ll considr only havy chargd particls lik muons, pions, protons, alphas, havy ions, Effctivly all chargd particls
More information6. The Interaction of Light and Matter
6. Th Intraction of Light and Mattr - Th intraction of light and mattr is what maks lif intrsting. - Light causs mattr to vibrat. Mattr in turn mits light, which intrfrs with th original light. - Excitd
More informationME 321 Kinematics and Dynamics of Machines S. Lambert Winter 2002
3.4 Forc Analysis of Linkas An undrstandin of forc analysis of linkas is rquird to: Dtrmin th raction forcs on pins, tc. as a consqunc of a spcifid motion (don t undrstimat th sinificanc of dynamic or
More informationContemporary, atomic, nuclear, and particle physics
Contmporary, atomic, nuclar, and particl physics 1 Blackbody radiation as a thrmal quilibrium condition (in vacuum this is th only hat loss) Exampl-1 black plan surfac at a constant high tmpratur T h is
More information1973 AP Calculus AB: Section I
97 AP Calculus AB: Sction I 9 Minuts No Calculator Not: In this amination, ln dnots th natural logarithm of (that is, logarithm to th bas ).. ( ) d= + C 6 + C + C + C + C. If f ( ) = + + + and ( ), g=
More information1 Minimum Cut Problem
CS 6 Lctur 6 Min Cut and argr s Algorithm Scribs: Png Hui How (05), Virginia Dat: May 4, 06 Minimum Cut Problm Today, w introduc th minimum cut problm. This problm has many motivations, on of which coms
More informationSchrodinger Equation in 3-d
Schrodingr Equation in 3-d ψ( xyz,, ) ψ( xyz,, ) ψ( xyz,, ) + + + Vxyz (,, ) ψ( xyz,, ) = Eψ( xyz,, ) m x y z p p p x y + + z m m m + V = E p m + V = E E + k V = E Infinit Wll in 3-d V = x > L, y > L,
More informationBackground: We have discussed the PIB, HO, and the energy of the RR model. In this chapter, the H-atom, and atomic orbitals.
Chaptr 7 Th Hydrogn Atom Background: W hav discussd th PIB HO and th nrgy of th RR modl. In this chaptr th H-atom and atomic orbitals. * A singl particl moving undr a cntral forc adoptd from Scott Kirby
More informationCollisions between electrons and ions
DRAFT 1 Collisions btwn lctrons and ions Flix I. Parra Rudolf Pirls Cntr for Thortical Physics, Unirsity of Oxford, Oxford OX1 NP, UK This rsion is of 8 May 217 1. Introduction Th Fokkr-Planck collision
More informationThe Matrix Exponential
Th Matrix Exponntial (with xrciss) by Dan Klain Vrsion 28928 Corrctions and commnts ar wlcom Th Matrix Exponntial For ach n n complx matrix A, dfin th xponntial of A to b th matrix () A A k I + A + k!
More informationMolecular Orbitals in Inorganic Chemistry
Outlin olcular Orbitals in Inorganic Chmistry Dr. P. Hunt p.hunt@imprial.ac.uk Rm 167 (Chmistry) http://www.ch.ic.ac.uk/hunt/ octahdral complxs forming th O diagram for Oh colour, slction ruls Δoct, spctrochmical
More informationECE 2210 / 00 Phasor Examples
EE 0 / 00 Phasor Exampls. Add th sinusoidal voltags v ( t ) 4.5. cos( t 30. and v ( t ) 3.. cos( t 5. v ( t) using phasor notation, draw a phasor diagram of th thr phasors, thn convrt back to tim domain
More informationIYPT 2000 Problem No. 3 PLASMA
IYPT 000 Problm No. 3 PLASMA Tam Austria Invstigat th lctrical conducivity of th flam of a candl. Examin th influnc of rlvant paramtrs, in particular, th shap and polarity of th lctrods. Th xprimnts should
More informationSECTION where P (cos θ, sin θ) and Q(cos θ, sin θ) are polynomials in cos θ and sin θ, provided Q is never equal to zero.
SETION 6. 57 6. Evaluation of Dfinit Intgrals Exampl 6.6 W hav usd dfinit intgrals to valuat contour intgrals. It may com as a surpris to larn that contour intgrals and rsidus can b usd to valuat crtain
More informationForces. Quantum ElectroDynamics. α = = We have now:
W hav now: Forcs Considrd th gnral proprtis of forcs mdiatd by xchang (Yukawa potntial); Examind consrvation laws which ar obyd by (som) forcs. W will nxt look at thr forcs in mor dtail: Elctromagntic
More informationPhysics 2D Lecture Slides Lecture 14: Feb 1 st 2005
Physics D Lctur Slids Lctur 14: Fb 1 st 005 Vivk Sharma UCSD Physics Compton Effct: what should Happn Classically? Plan wav [f,λ] incidnt on a surfac with loosly bound lctrons intraction of E fild of EM
More informationThe following information relates to Questions 1 to 4:
Th following information rlats to Qustions 1 to 4: QUESTIN 1 Th lctrolyt usd in this ful cll is D watr carbonat ions hydrogn ions hydroxid ions QUESTIN 2 Th product formd in th ful cll is D hydrogn gas
More informationPipe flow friction, small vs. big pipes
Friction actor (t/0 t o pip) Friction small vs larg pips J. Chaurtt May 016 It is an intrsting act that riction is highr in small pips than largr pips or th sam vlocity o low and th sam lngth. Friction
More informationFirst derivative analysis
Robrto s Nots on Dirntial Calculus Chaptr 8: Graphical analysis Sction First drivativ analysis What you nd to know alrady: How to us drivativs to idntiy th critical valus o a unction and its trm points
More information4 x 4, and. where x is Town Square
Accumulation and Population Dnsity E. A city locatd along a straight highway has a population whos dnsity can b approimatd by th function p 5 4 th distanc from th town squar, masurd in mils, whr 4 4, and
More informationElements of Statistical Thermodynamics
24 Elmnts of Statistical Thrmodynamics Statistical thrmodynamics is a branch of knowldg that has its own postulats and tchniqus. W do not attmpt to giv hr vn an introduction to th fild. In this chaptr,
More informationIntroduction to Condensed Matter Physics
Introduction to Condnsd Mattr Physics pcific hat M.P. Vaughan Ovrviw Ovrviw of spcific hat Hat capacity Dulong-Ptit Law Einstin modl Dby modl h Hat Capacity Hat capacity h hat capacity of a systm hld at
More informationSolid State Theory Physics 545 Band Theory III
Solid Stat Thory Physics 545 Band Thory III ach atomic orbital lads to a band of allowd stats in th solid Band of allowd stats Gap: no allowd stats Band of allowd stats Gap: no allowd stats Band of allowd
More information