Instructor Solutions Manualfor
g g g
Physics g
by
Halliday, Resnick, and Krane
g g g
PaulgStanleygB
eloitg College
Volumeg 2
, AgNotegTogTheg Instructor...
Thegsolutionsgheregaregsomewhatgbrief,gasgtheygaregdesignedgforgtheginstructor,gnotgforgthegstuden
t.gCheckgwithgthegpublishersgbeforegelectronicallygpostingganygpartgofgthesegsolutions;gwebsite,gftp,gorg
serverg accessg mustg beg restrictedg tog yourg students.
Ighavegbeengsomewhatgcasualgaboutgsubscriptsgwhenevergitgisgobviousgthatgagproblemgisgonegdime
nsional,gorgthatgthegchoicegofgthegcoordinategsystemgisgirrelevantgtogthegnumericalgsolution.gAlthoughg
thisg doesg notg changeg theg validityg ofg theg answer,g itg willg sometimesg obfuscateg theg approachgifg viewe
dg byg ag novice.
Thereg areg someg traditionalg formula,g suchg as
v2g =gv2g +g2axx,
x 0x
whichg areg notg usedg ing theg text.g Theg workedg solutionsg useg onlyg materialg fromg theg text,g sog thereg ma
ygbegtimesgwhengthegsolutiongheregseemsgunnecessarilygconvolutedgandgdrawngout.gYes,gIgknowgangea
sierg approachg existed.gButg ifg itg wasg notg ing theg text,g Ig didg notg useg itg here.
Igalsogtriedgtogavoidgreinventinggthegwheel.gTheregaregsomegexercisesgandgproblemsgingthegtextgwh
ichgbuildgupongpreviousgexercisesgandgproblems.gInsteadgofgrederivinggexpressions,gIgsimplygrefergyo
ug tog theg previousg solution.
Igadoptgagdifferentgapproachgforgroundinggofgsignificantgfiguresgthangpreviousgauthors;gingpartic-
gular,g Igusuallygroundgintermediateganswers.g Asgsuch,g somegofgmyganswersgwillgdiffergfromgthosegingt
heg backg ofg theg book.
ExercisesgandgProblemsgwhichgaregenclosedgingagboxgalsogappeargingthegStudent’sgSolutiongManual
gwithgconsiderablygmoregdetailgand,gwhengappropriate,gincludegdiscussiongonganygphysicalgimplicatio
nsgofgtheganswer.g Thesegstudentgsolutionsgcarefullygdiscussgthegstepsgrequiredgforgsolvinggproblems,gp
ointgoutgthegrelevantgequationgnumbers,gorgevengspecifygwheregingthegtextgadditionalginformationgcang
begfound.gWhengtwogalmostgequivalentgmethodsgofgsolutiongexist,goftengbothgaregpresented.gYougarege
ncouragedgtogrefergstudentsgtogthegStudent’sgSolutiongManualgforgthesegexercisesgandgproblems.gHow
ever,g theg materialg fromg theg Student’sg Solutiong Manualg mustg notg beg copied.
PaulgStanley
BeloitgCollegegsta
g
1
,E25-1 Theg chargeg transferredg is
Qg=g(2.5g×g104gC/s)(20g×g10−6gs)g=g5.0g×g10−1gC.
E25-2gggg Useg Eq.g 25-4:
s
(8.99×109 N·m2 /C2 )(26.3×10−6 C)(47.1×10−6 C)
rg= =g 1.40gm
(5.66gN)
E25-3gggg UsegEq.g 25-
4:
(8.99×109N·m2/C2)(3.12×10−6C)(1.48×10−6C)g(0.
=g 2.74gN.
Fg = 123gm)2
E25-4gggg (a)g Theg forcesg areg equal,g sog m1a1g =g m2a2,g or
m2g =g (6.31×10−7kg)(7.22g m/s2)/(9.16g m/s2)g =g 4.97×10−7kg.
(b) Useg Eq.g25-4:
s
(6.31×10−7 kg)(7.22gm/s2 )(3.20×10−3 m)2 −11
qg = =g7.20×10 C
(8.99×109N·m2/C2)
E25-5 (a) UsegEq.g 25-4,
1gggg q1q2
ggg 1 (21.3gµC)(21.3gµC)
Fg = g = =g 1.77gN
4πϵ0 r12
2
(1.52gm)2
4π(8.85×10−12g C2/Ng ·g m2
)
(b) Ing partg (a)g weg foundg F12;g tog solveg partg (b)g weg needg tog firstg findg F13.g Sinceg q3g =g q2g and
r13g =gr12,g weg cang immediatelyg concludeg thatg F13g =gF12.
Weg mustg assessg theg directiong ofg theg forceg ofg q3g ong q1;g itg willg beg directedg alongg theg lineg whic
hgconnectsgthegtwogcharges,gandgwillgbegdirectedgawaygfromgq3.g Thegdiagramgbelowgshowsgthegdirectio
ns.
F
23
Fg12
F Fg12
23 Fgnet
Fromg thisg diagramg weg wantg tog findg theg magnitudeg ofg theg netg forceg ong q1.g Theg cosineg lawg isga
ppropriateg here:
Fgnet2 =gggg F12
g +gF g −g2F12F13gcosgθ,
2 2
13
=gggg (1.77gN)2g +g(1.77gN)2g −g2(1.77gN)(1.77gN)gcos(120◦),
=gggg 9.40gN2
2
, ,gFgnet =gggg 3.07gN.
3
g g g
Physics g
by
Halliday, Resnick, and Krane
g g g
PaulgStanleygB
eloitg College
Volumeg 2
, AgNotegTogTheg Instructor...
Thegsolutionsgheregaregsomewhatgbrief,gasgtheygaregdesignedgforgtheginstructor,gnotgforgthegstuden
t.gCheckgwithgthegpublishersgbeforegelectronicallygpostingganygpartgofgthesegsolutions;gwebsite,gftp,gorg
serverg accessg mustg beg restrictedg tog yourg students.
Ighavegbeengsomewhatgcasualgaboutgsubscriptsgwhenevergitgisgobviousgthatgagproblemgisgonegdime
nsional,gorgthatgthegchoicegofgthegcoordinategsystemgisgirrelevantgtogthegnumericalgsolution.gAlthoughg
thisg doesg notg changeg theg validityg ofg theg answer,g itg willg sometimesg obfuscateg theg approachgifg viewe
dg byg ag novice.
Thereg areg someg traditionalg formula,g suchg as
v2g =gv2g +g2axx,
x 0x
whichg areg notg usedg ing theg text.g Theg workedg solutionsg useg onlyg materialg fromg theg text,g sog thereg ma
ygbegtimesgwhengthegsolutiongheregseemsgunnecessarilygconvolutedgandgdrawngout.gYes,gIgknowgangea
sierg approachg existed.gButg ifg itg wasg notg ing theg text,g Ig didg notg useg itg here.
Igalsogtriedgtogavoidgreinventinggthegwheel.gTheregaregsomegexercisesgandgproblemsgingthegtextgwh
ichgbuildgupongpreviousgexercisesgandgproblems.gInsteadgofgrederivinggexpressions,gIgsimplygrefergyo
ug tog theg previousg solution.
Igadoptgagdifferentgapproachgforgroundinggofgsignificantgfiguresgthangpreviousgauthors;gingpartic-
gular,g Igusuallygroundgintermediateganswers.g Asgsuch,g somegofgmyganswersgwillgdiffergfromgthosegingt
heg backg ofg theg book.
ExercisesgandgProblemsgwhichgaregenclosedgingagboxgalsogappeargingthegStudent’sgSolutiongManual
gwithgconsiderablygmoregdetailgand,gwhengappropriate,gincludegdiscussiongonganygphysicalgimplicatio
nsgofgtheganswer.g Thesegstudentgsolutionsgcarefullygdiscussgthegstepsgrequiredgforgsolvinggproblems,gp
ointgoutgthegrelevantgequationgnumbers,gorgevengspecifygwheregingthegtextgadditionalginformationgcang
begfound.gWhengtwogalmostgequivalentgmethodsgofgsolutiongexist,goftengbothgaregpresented.gYougarege
ncouragedgtogrefergstudentsgtogthegStudent’sgSolutiongManualgforgthesegexercisesgandgproblems.gHow
ever,g theg materialg fromg theg Student’sg Solutiong Manualg mustg notg beg copied.
PaulgStanley
BeloitgCollegegsta
g
1
,E25-1 Theg chargeg transferredg is
Qg=g(2.5g×g104gC/s)(20g×g10−6gs)g=g5.0g×g10−1gC.
E25-2gggg Useg Eq.g 25-4:
s
(8.99×109 N·m2 /C2 )(26.3×10−6 C)(47.1×10−6 C)
rg= =g 1.40gm
(5.66gN)
E25-3gggg UsegEq.g 25-
4:
(8.99×109N·m2/C2)(3.12×10−6C)(1.48×10−6C)g(0.
=g 2.74gN.
Fg = 123gm)2
E25-4gggg (a)g Theg forcesg areg equal,g sog m1a1g =g m2a2,g or
m2g =g (6.31×10−7kg)(7.22g m/s2)/(9.16g m/s2)g =g 4.97×10−7kg.
(b) Useg Eq.g25-4:
s
(6.31×10−7 kg)(7.22gm/s2 )(3.20×10−3 m)2 −11
qg = =g7.20×10 C
(8.99×109N·m2/C2)
E25-5 (a) UsegEq.g 25-4,
1gggg q1q2
ggg 1 (21.3gµC)(21.3gµC)
Fg = g = =g 1.77gN
4πϵ0 r12
2
(1.52gm)2
4π(8.85×10−12g C2/Ng ·g m2
)
(b) Ing partg (a)g weg foundg F12;g tog solveg partg (b)g weg needg tog firstg findg F13.g Sinceg q3g =g q2g and
r13g =gr12,g weg cang immediatelyg concludeg thatg F13g =gF12.
Weg mustg assessg theg directiong ofg theg forceg ofg q3g ong q1;g itg willg beg directedg alongg theg lineg whic
hgconnectsgthegtwogcharges,gandgwillgbegdirectedgawaygfromgq3.g Thegdiagramgbelowgshowsgthegdirectio
ns.
F
23
Fg12
F Fg12
23 Fgnet
Fromg thisg diagramg weg wantg tog findg theg magnitudeg ofg theg netg forceg ong q1.g Theg cosineg lawg isga
ppropriateg here:
Fgnet2 =gggg F12
g +gF g −g2F12F13gcosgθ,
2 2
13
=gggg (1.77gN)2g +g(1.77gN)2g −g2(1.77gN)(1.77gN)gcos(120◦),
=gggg 9.40gN2
2
, ,gFgnet =gggg 3.07gN.
3
