SOLUTIONS FOR CHAPTER PROBLEMS 4,, 9, 5,, 4, 5, 75, 89, 96..4.IDENTIFY: Use the mss m of the ring nd the tomic mss M of gold to clculte the number of gold toms. Ech tom hs 79 protons nd n equl number of electrons. SET UP: N 6. toms/mol. A proton hs chrge +e. A EXECUTE: The mss of gold is 7.7 g nd the tomic weight of gold is 97 gmol. So the number of 7.7 g toms is NAn(6. toms/mol) 5.4 toms 97 g mol. The number of protons is 4 np (79 protons/tom)(5.4 toms) 4.7 protons. 9 5 Q( np )(.6 C/proton) 6.8 C. 4 (b) The number of electrons is ne np 4.7. EVALUATE: The totl mount of positive chrge in the ring is ver lrge, but there is n equl mount of negtive chrge... IDENTIFY: Appl Coulomb s lw. SET UP: Like chrges repel nd unlike chrges ttrct. 6 qq (.55 C) q EXECUTE: () F. This gives. N 4P r 4 P (.m) 6 q.64 C 6. The force is ttrctive nd q, so q.64 C. (b) F. N. The force is ttrctive, so is downwrd. EVALUATE: The forces between the two chrges obe Newton's third lw. nd.9. IDENTIFY: Appl Coulomb s lw to clculte the force ech of the two chrges eerts on the third chrge. Add these forces s vectors. SET UP: The three chrges re plced s shown in Figure.9. Figure.9 EXECUTE: Like chrges repel nd unlike ttrct, so the free-bod digrm for q is s shown in Figure.9b. Figure.9b F qq 4 P r F qq 4 P r
(.5 C)(5. C) (8.988 N m /C ).685 N (. m) 9 9 9 6 F 9 9 9 (. C)(5. C) 7 F (8.988 N m /C ) 8.988 N (.4 m) The resultnt force is R F F. R. R F F 6 7 6.685 N +8.988 N =.58 N. 6 The resultnt force hs mgnitude.58 N nd is in the -direction. EVALUATE: The force between q nd q is ttrctive nd the force between q nd q is replusive..5. IDENTIFY: F q E. Since the field is uniform, the force nd ccelertion re constnt nd we cn use constnt ccelertion eqution to find the finl speed. 7 SET UP: A proton hs chrge +e nd mss.67 kg. 9 6 EXECUTE: () F (.6 C)(.75 N/C) 4.4 N 6 F 4.4 N (b) 7.6 m/s m.67 kg 6 5 (c) v v t gives v (.6 m/s )(. s).6 m/s EVALUATE: The ccelertion is ver lrge nd the grvit force on the proton cn be ignored... IDENTIFY: Eq. (.) gives the force on the prticle in terms of its chrge nd the electric field between the pltes. The force is constnt nd produces constnt ccelertion. The motion is similr to projectile motion; use constnt ccelertion equtions for the horizontl nd verticl components of the motion. () SET UP: The motion is sketched in Figure.. For n electron q e. Figure. F qe nd q negtive gives tht F nd E re in opposite directions, so F is upwrd. The freebod digrm for the electron is given in Figure.b. Figure.b EXECUTE: F m ee m Solve the kinemtics to find the ccelertion of the electron: Just misses upper plte ss tht. cm when.5 cm. -component 6 v v.6 m/s,,. m, t? vt t. m 8 t.5 s 6 v.6 m/s In this sme time t the electron trvels.5 m verticll:
-component 8 t.5 s, v,.5 m,? v t t ( ) (.5 m) 6.4 m/s 8 t (.5 s) (This nlsis is ver similr to tht used in Chpter for projectile motion, ecept tht here the ccelertion is upwrd rther thn downwrd.) This ccelertion must be produced b the electric-field force: ee m m (9.9 kg)(6.4 m/s ) E 64 N/C 9 e.6 C Note tht the ccelertion produced b the electric field is much lrger thn g, the ccelertion produced b grvit, so it is perfectl ok to neglect the grvit force on the elctron in this problem. 9 ee (.6 C)(64 N/C) (b).49 m/s 7 mp.67 kg This is much less thn the ccelertion of the electron in prt () so the verticl deflection is less nd the proton won t hit the pltes. The proton hs the sme initil speed, so the proton tkes the sme time 8 t.5 s to trvel horizontll the length of the pltes. The force on the proton is downwrd (in the sme direction s E, since q is positive), so the ccelertion is downwrd nd.49 m/s. v t t (.49 m/s )(.5 s).7 m. The displcement is 8 6 6.7 m, downwrd. (c) EVALUATE: The displcements re in opposite directions becuse the electron hs negtive chrge nd the proton hs positive chrge. The electron nd proton hve the sme mgnitude of chrge, so the force the electric field eerts hs the sme mgnitude for ech chrge. But the proton hs mss lrger b fctor of 86 so its ccelertion nd its verticl displcement re smller b this fctor..4. IDENTIFY: The net force on ech chrge must be zero. SET UP: The force digrm for the 6.5 C chrge is given in Figure.4. F E is the force eerted on the chrge b the uniform electric field. The chrge is negtive nd the field is to the right, so the force eerted b the field is to the left. F q is the force eerted b the other point chrge. The two chrges hve opposite signs, so the force is ttrctive. Tke the + is to be to the right, s shown in the figure. 6 8 EXECUTE: () F q E (6.5 C)(.85 N/C). N 6 6 qq 9 (6.5 C)(8.75 C) Fq k (8.99 N m /C ) 8.8 N r (.5 m) F gives T Fq FE nd T FE Fq 8 N. (b) Now F q is to the left, since like chrges repel. F gives T Fq FE nd T FE Fq. N. EVALUATE: The tension is much lrger when both chrges hve the sme sign, so the force one chrge eerts on the other is repulsive. Figure.4
.5. IDENTIFY: Appl Eq.(.7) to clculte the field due to ech chrge nd then clculte the vector sum of those fields. SET UP: The fields due to q nd to q re sketched in Figure.5. 9 (6. C) EXECUTE: E ˆ ˆ ( i) 5 i N/C. 4 P (.6 m) E 9 ˆ ˆ ˆ ˆ (4. C) (.6) (.8) (.6 8.8 ) N C 4 (. m) i (. m) j i + j. P ˆ ˆ E =E +E ( 8.4 N/C) i+ (8.8 N/C) j. E (8.4 N/C) (8.8 N/C).6 N/C t 8.8 tn.6 bove the is nd therefore 96. counterclockwise from the + is. 8.4 EVALUATE: E is directed towrd q becuse q is negtive nd E is directed w from q becuse q is positive. Figure.5.75. IDENTIFY: Use Coulomb's lw for the force tht one sphere eerts on the other nd ppl the st condition of equilibrium to one of the spheres. () SET UP: The plcement of the spheres is sketched in Figure.75. Figure.75 The free-bod digrms for ech sphere re given in Figure.75b. Figure.75b F c is the repulsive Coulomb force eerted b one sphere on the other.
(b) EXECUTE: From either force digrm in prt (): F m mg Tcos5.mg nd T cos 5. F m T sin 5.F nd F Tsin 5. c c Use the first eqution to eliminte T in the second: F qq q q c 4P r 4P r 4 P [(. m)sin 5. ] Fc mg/ cos 5. sin 5. mg tn 5. q Combine this with Fc mgtn 5. nd get mg tn 5. 4 P [(. m)sin 5. ] q.4 m sin 5. mg tn 5. /4P 5. kg 9.8 m/s tn 5. 6 q.4 msin 5..8 C 9 8.988 N m /C (c) The seprtion between the two spheres is given b Lsin. q.8c s found in prt (b). / 4 / sin P nd tn. Thus P F q L F mg c c q sin tn 4 P 4L mg 6.8 C / 4 q / Lsin mgtn. 9 8.988 N m / C.8. 4.6 m 5. kg 9.8 m/s Solve this eqution b tril nd error. This will go quicker if we cn mke good estimte of the vlue of tht solves the eqution. For smll, tn sin. With this pproimtion the eqution becomes sin.8 nd sin.69, so 4.9. Now refine this guess: sin tn 45..5 4..467 9.6.6 9.5.5 9.4.9 so 9.5 EVALUATE: The epression in prt (c) ss s L nd 9 s L. When L is decresed from the vlue in prt (), increses..89. IDENTIFY: Divide the chrge distribution into infinitesiml segments of length d. Clculte E nd E u due to segment nd integrte to find the totl field. SET UP: The chrge dq of segment of length d is dq ( Q / ) d. The distnce between segment t nd the chrge q is r. ( ) when. dq Qd Q EXECUTE: () de so E 4 P ( r) 4. P ( r) 4P r r Q r, so E. E. 4 P qq (b) F= qe= ˆ i. 4 P
EVALUATE: (c) For, kqq (( ) ) kqq ( ) kqq qq F. 4 P r (Note tht for, r.) The chrge distribution looks like point chrge from fr w, so the force tkes the form of the force between pir of point chrges..96. IDENTIFY: Divide the semicircle into infinitesiml segments. Find the electric field de due to ech segment nd integrte over the semicircle to find the totl electric field. SET UP: The electric fields long the -direction from the left nd right hlves of the semicircle cncel. The remining -component points in the negtive -direction. The chrge per unit length of the semicircle is Q k dl kd nd de. k sin d EXECUTE: de de sin. Therefore, k k k kq E sin d [ cos ], in the -direction. EVALUATE: For full circle of chrge the electric field t the center would be zero. For qurtercircle of chrge, in the first qudrnt, the electric field t the center of curvture would hve nonzero nd components. The clcultion for the semicircle is prticulrl simple, becuse ll the chrge is the sme distnce from point P.