Electrical Energy and Capacitance

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Hubert Wiggins
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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

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