ANSWERS TO TEST YOURSELF QUESTIONS 5 1physics for the iB Diploma © camBriDge University press 2015
answers to test yourself questionstopic 55.1 electric fields
1 a FkQQ
r= =
× × × × ××
− −1 22
9 6 68 99 10 2 0 10 4 0 10
5 0 10
. . .
( . −− = ≈2 2 28 8 29)
. N
b Theforcewouldbecome4timesassmall.
i ′ = =FkQQ
r
F1 222 4( )
ii ′ = =Fk QQ
r
F2
2 21 22( )
iii ′ =×
=Fk Q Q
rF
2 2
21 2
2( )
2 Themiddlechargeisattractedtotheleftbythechargeontheleftwithaforceof
FkqQ
r12
2
9 6 68 99 10 2 0 10 4 0 10
4 0 10= =
× × × × ××
− −. . .
( . −− =2 2 45)
N.Itisattractedtotherightbythechargeontheright
withaforceofFkqQ
r22
2
9 6 68 99 10 2 0 10 3 0 10
2 0 10= =
× × × × ××
− −. . .
( . −− =2 2 135)
N.Thenetforceisthus135 45 90− = N
directedtowardstheright.
3 Supposewecallthedistance(incm)fromtheleftchargex.Thenweneed8 99 10 2 0 10 4 0 10 8 99 10 2 0 19 6 6
2
9. . . . .× × × × ×=
× × ×− −
x
00 3 0 10
64 0 3 0
6
6 6
2
2 2
− −× ×−
=−
.
( ). .
( )
x
x x
Thismeansthat
4 0 6 3 0
4 36 12 3
48 144 0
2 2
2 2
2
. ( ) .
( )
− =
− + =
− + =
x x
x x x
x x
Thesolutionisx = 3 22. cm.
4 Theforcesareasshown.ThedistancebetweenthechargeQandthecharge2Qis5.0cm.
F1
F2
2 physics for the iB Diploma © camBriDge University press 2015ANSWERS TO TEST YOURSELF QUESTIONS 5
ThemagnitudesareFkQ Q
d1 2
9 6 62 8 99 10 3 0 10 6 0 10
5 0= =
× × × × ××
− −( ) . . .
( . 11064 72 2− =
). Nand
FkQQ
d2 2
9 6 2
2 2
8 99 10 3 0 10
3 0 1089= =
× × ××
=−
−
. ( . )
( . )..9 N.WeneedtofindthecomponentsofF1:
F x1 64 7 64 74
551 76= = × =. cos . .θ NandF y1 64 7 64 7
3
538 82= = × =. sin . .θ N.Thecomponentsof
thenetforceare:Fx = −51 76. NandFy = − = −38 82 89 9 51 08. . . N .Thenetforcehasmagnitude
F = + = ≈51 76 51 08 72 7 732 2. . . Nanddirection18051 08
51 76224 6 225° + = ° ≈ °arctan
.
.. .
5 a Adiagramisthefollowinginwhichtheangleθofeachstringtotheverticalisgivenbysin .θ θ= ⇒ =5
853 37.
F F
θ
WehavethatTcosθ=mgandT FkQ
dsinθ = =
2
2sothatdividingsidebysidegives
tantan . .θ θ
= ⇒ = =× × × ×−kQ
mgdQ
mgd
k
2
2
2 6 2100 10 9 8 0 1 ttan .
..
3 37
8 99 108 0 109
9°×
= × − C.
b Thiscorrespondsto8 0 10
1 6 105 0 10
9
1910.
..
××
= ×−
−electroniccharges.
6 a Sincethemolarmassofwateris18 gpermole,amassof60 kgcorrespondsto60 10
183333
3×= molesi.e.
3333 6 02 10 2 1023 27× × = ×. moleculesofwater.Amoleculeofwatercontains10electrons(2fromhydrogen
and8fromoxygen)andsowehave2 1028× electronsineachperson.
b TheelectricforceisthereforeFkQQ
r11 22
9 28 19 2
2
9 10 2 0 10 1 6 10
10= =
× × × × × −( . . )
( )== × ≈9 10 1026 27 N,an
enormousforce. c AssumptionsincludetheuseofCoulomb’slawforobjectsthatarenotpointcharges,assumingthesame
distancebetweenchargesetc. d Wehaveneglectedtheexistenceofprotonswhichgiveseachpersonazeroelectricchargeandhencezero
electricforce.
7 EFq
3.0 10
5.0 106.0 NC
5
61= = ×
×=
−
−−
8 ThemagnitudeofeachofthefieldsproducedatPis:EkQ
r= =
× × ×
+( )=
−
2
9 6
2 22
9 0 10 2 00 10
0 05 0 301 95
. .
. .. ×× −105 1N C .The
verticalcomponentsoftheelectricfieldswillcanceloutleavingonlythehorizontalcomponents.Thehorizontal
componentisE E Ed
da
x = =+
= × ×+
cos ..
..
θ2
2
5
22
4
1 95 100 30
0 104
0 30
== × −1 92 105 1. N C .Thenetfieldisthen
directedhorizontallytotherightandhasmagnitude2 1 92 10 3 84 105 5 1× × = × −. . N C .
ANSWERS TO TEST YOURSELF QUESTIONS 5 3physics for the iB Diploma © camBriDge University press 2015
9 ThetwoelectricfieldsareE E1 25 11 95 10= = × −. NC .AddingvectoriallybytakingcomponentsgivesEx = 0and
Ey = × × × × = × × ×+
2 1 95 10 2 1 95 100 05
0 104
0
5 5
2. .
.
.sinθ
..
.
30
6 4 102
5 1= × −NC .
10 I nqAv= = × × × × × ×− −8 5 10 1 6 10 0 90 10 3 628 19 3 2. . ( . ) .π ×× = ≈−10 12 45 124 . A
11 a Thecurrentwillbethesamebyconservationofcharge.
b SinceI nqAv= ,wehavethatA v A v1 1 2 2= andsovA v
A
r v
r21 1
2
12
1
22
2 4
2
1 0 2 2 10
2 05 5 10= = =
× ×= ×
−. .
.. −− −5 1m s .
12 FromI nqAv= wegetvI
nqA= =
× × × × × ×=− −
5 0
5 8 10 1 6 10 2 104
28 19 3 2
.
. . ( ).
π33 10 4 105 5 1× ≈ ×− − −ms .
13 a Onehouris1 60 60 3600× × = sandsoQ It= = × = ×10 3600 3 6 104. C.
bNQ
e= =
××
= × ≈ ×−
3 6 10
1 6 102 25 10 2 2 10
4
1923 23.
.. .
14 aandbThesepointsareinsidetheconductingspheresotheelectricfieldiszerothere.
c EkQ
R= =
× × ××
= ×−
−2
9 6
2 2
8 99 10 4 0 10
15 0 101 6 10
. .
( . ). 66 1NC−
d At20cm,E = × ×
= × −1 6 10
15
209 0 106
25 1. . NC
5.2 heating effect of electric currents
15 Electronsmakingupthecurrentcollidewithlatticeatomsandtransfersomeoftheirkineticenergytotheseatoms.Theaveragekineticenergyoftheatomsincreasesandsincetemperatureisproportionaltotheaveragekineticenergyoftheatomsthetemperatureofthewireincreases.Theelectricfieldkeepsacceleratingtheelectronsandsothisprocesscontinues.
16 Doublingthelengthofthewiredoublesthepotentialdifferenceacrossitsendswhilethecurrentstaysthesame.
SinceRV
I= theresistancedoubles.
17 a Yessincethegraphsarestraightlinesthroughtheorigin. b TheresistanceforwireAislowerandsothiswirecorrespondstothelowertemperature.
18 Sincetheresistanceisconstant,6 0
1 5 3 514
.
. .= ⇒ =V
V V.
19 ItobeysOhm’slawsotheresistanceisthesame,12 Ω.
20 RV
I= = =
220
1515 Ω
21 aV IR= = × =2 4 8 VacrossthefirstandV IR= = × =2 6 12 Vacrossthesecond. b ThereisnopotentialdifferencebetweenBandCsincethereisnoresistancebetweenthesepoints.
22 a TheresistanceisRV
P= = = ≈ ×
2 22220
120403 4 0 10. Ω.
b 4032 0 10
0 03 100 57
6
3 2=× ×
× ×⇒ =
−
−
.
( . ).
LL
πm
4 physics for the iB Diploma © camBriDge University press 2015ANSWERS TO TEST YOURSELF QUESTIONS 5
23 a Thetwo4.0 Ωresistorsareinseriesandareequivalentto8.0 Ω.Thelowertwo2.0 Ωareequivalentto4.0 Ω.
The8.0 Ωandthe4.0 Ωresistorsareinparallelandareequivalentto1 1
8 0
1
4 02 7
RR= + ⇒ =
. .. Ω.
b The6.0 Ωand4.0 Ωresistorsareinparallelandareequivalentto1 1
6 0
1
4 02 4
RR= + ⇒ =
. .. Ω.Thisandthe
othertwoareinseriesforantotalofR = + + =2 0 2 4 8 0 12 4. . . . Ω.
c Allthreeareinparallelforatotalof1 1
3 0
1
3 0
1
3 01 0
RR= + + ⇒ =
. . .. Ω.
24 Wehavethat12 1 2= +I R R( )andε = IR2whereR1istheresistanceofwireACandR2theresistanceofwireBC.
Thusε12
2
1 2
=+R
R R.Buttheresistancesareproportionaltothelengthsandso
ε ε12
54
1006 48= ⇒ = . V.
25 a ApplyingKirchoff ’slawstothetwoloopsgives:3 0 20 30. ( ) ( )= + + +x y x y and
2 0 20 30 10. ( ) ( )= + + + +x y x y y.Thesesimplifyto
3 0 50 50
2 0 50 60
.
.
= += +
x y
x y
Thesearesolvedtogive10 1 0 0 10y y= − ⇒ = −. . A,andx = 0 16. A.
b Thepotentialdifferencesare
20 20 20 0 16 0 10 1 2
30 30
ΩΩ: ( ) ( . . ) .
: (
V x y
V x
= + = × − ==
V
++ = × − == = × −
y
V y
) ( . . ) .
: ( .
30 0 16 0 10 1 8
10 10 10 0 10
V
Ω )) .= −1 0 V
26 ApplyingKitchhoff ’slaw:9 0 3 0 4 0. . .+ = xsorightawayx = 3 0. A.3 0 3 0 2 0 5 5. . ( ) . ( )= − − − − = − +x y x y x y.Hencey = 3 6. A.
27 ApplyingKitchhoff ’slaw:9 0 2 0 5 0 1 0 2 0. . . . .= + × ⇒ =x x A.Insecondloopε = × + × =3 0 1 67 5 0 1 0 10. . . . V.
28 a Fromthegraph,whenthepotentialdifferenceacrosseachresistoris1.5 VthecurrentinXisabout2.68 AandinY1.55 Aforatotalcurrentleavingthecellof4.2 A.
b Thishastobedonebytrialanderror.ThevoltageacrossXplusthatacrossYmustgive1.5 V.Thisisachievedforacurrentofabout1.1 Aforwhichthevoltagesare0.5 Vand1.0 Vaddingupto1.5 V.
29 Thetoploopgives:6 0 4 0 3 2 0 833. . . .= + ⇒ =x x x A.Thelowerloopgives:2 0 0 8333 2 4. . .= × ⇒ =R x R Ω.L
L1
2 4
40 60= ⇒ =
.. m.
5.2 electric cells
30 Chemicalenergyinthetopcellgetsconvertedintothermalenergyintheresistor,mechanicalenergy(andsomethermalenergy)inthemotor(whichinturngetsconvertedintogravitationalpotentialenergyastheloadisbeingraised)andfinallyelectricalenergythatchargesthelowerbattery.
ANSWERS TO TEST YOURSELF QUESTIONS 5 5physics for the iB Diploma © camBriDge University press 2015
31
0 5 10 15 20I / A
0
2
4
6
8
10
12
14P / W
R=2.0Ω.
32 a V
I
slope = −rvertical intercept = emf
ε
b SinceV Ir= −ε , i theslopeisthenegativeoftheinternalresistanceofthebattery iitheverticalinterceptistheemfofthebattery
33 a Theinternalresistanceistheslopeandsoequals− =−−
⇒ =r r2 4 9 6
8 0 2 01 2
. .
. .. Ω.
b Extendingthelinetofindtheverticalinterceptgivesanemfofabout12V.
34 a Thetwoparallelresistorsareequivalentto1 1
10
1
206 67
RR= + ⇒ = . Ω.Thetotalresistanceofthecircuitis
thenRT = 8 67. Ω .ThetotalcurrentisthenIT A= =12 0
8 671 38
.
.. .Thepotentialdifferenceacrossthe2.0Ω
resistorisV = × = ≈1 38 2 0 2 77 2 8. . . . V.Thepotentialdifferenceacrosstheparallelresistorsisthen
V = − =12 0 2 77 9 2. . . V.Soeachofthetworesistorsgetacurrentof9 2
100 92
..= Aand
9 2
200 46
..= A.
9.2 V0.46 A
9.2 V0.92 A
2.8 V1.38 A
b Thetwoparallelresistorshaveatotalof2.0Ωmakingatotalcircuitresistanceof6.0Ω.Thetotalcurrentis
thenIT A= =6 0
6 01 0
.
.. .Theinternalresistanceandthe2.0Ωresistorget1.0Aofcurrentandthepotential
differenceacrosseachis2.0V.Thepotentialdifferenceacrosstheparallelcombinationis2.0Vandsoeachgets0.50Aofcurrent.
2 V0.5 A
2 V0.5 A
2 V1 A
2 V1 A
6 physics for the iB Diploma © camBriDge University press 2015ANSWERS TO TEST YOURSELF QUESTIONS 5
35 aandbLetrbetheinternalresistanceandεtheemf.Thetotalresistancewheninparallelis2.0+randso
3 02 0
..
=+
εr.Wheninseriesthetotalresistanceis8.0+randso1 4
8 0.
.=
+εr.Wemustsolvethesystemof
equations3 02 0
1 48 0
..
..
=+
=+
ε
εr
r
Dividingsidebyside3 0
1 4
8 0
2 02 14
8 0
2 04 28 2 14
.
.
.
..
.
.. .=
++
⇒ =++
⇒ +r
r
r
rrr r r r= + ⇒ = ⇒ =8 0 1 14 3 17 3 25. . . . Ω
andsoε = × ≈5 25 3 0 16. . V.
36 a Thecurrenteverywhereisthesame,callitx.Then9 0 3 0 8 0 75. . .− = ⇒ =x x A. b Thepowerinthetopcellis9 0 0 75 6 8. . .× ≈ Wandintheloweritis− × ≈ −3 0 0 75 2 2. . . W. c Thepowerinthelowercellisnegativeimplyingthatitisbeingcharged.
5.4 magnetic fields
37
38 Wemustapplytherighthandruleforforce a Themagneticfieldisintothepage b Theforceisintothepage c Themagneticfieldisoutofthepage d Theforceiszerosincethevelocityisantiparalleltothefield e Theforceiszerosincethevelocityisparalleltothefield
ANSWERS TO TEST YOURSELF QUESTIONS 5 7physics for the iB Diploma © camBriDge University press 2015
39
40 Themagneticfieldisdirectedintothepage.Inatherighthandrule(foranegativecharge)givesaforcedownwardsawayfromthewire.Inbitgivesaforcetotheright.
41 a thefieldistotherightandsotheforceisintothepage b thevelocityisparalleltothefieldandtheforceiszero c theforceistowardsthemagnet(upthepage)
42 a eE evB BE
e= ⇒ = =
××
= × −2 4 10
2 0 101 2 10
3
52.
.. T.Theelectricforceisupwardssothemagneticforceis
downwards.Thereforethefieldmustbeintothepage. b Theconditioninaisindependentofchargeandmasssotheprotonwillbeundeflectedaswell. c Theelectricforcewillstaythesamebutthemagneticforcewilldouble.Thereforetheelectronwillbe
deflecteddownwards.
43 a Thereareequalandoppositeforcesatthepolesofthemagnetgivinganetforceofzero. b Theforcesareoppositesotheywillrotatethemagnetcounterclockwise.
44 TheforceisF BIL= = × × × × ° =−sin . . sin .θ 5 00 10 3000 30 0 30 2 255 N.
45 Note:Thisrequiresknowledgeofcircularmotion(Topic6).
a WehavethatevB mv
rv
eBr
m= =⇒ =
2
.Butv fr= 2π andso22
ππ
rfeBr
mf
eB
m= ⇒ = .
b Themassisdifferentandsotheanswerchanges.
46 a ThecombinedmagneticfieldfromthetwowiresatpointRmustpointdownwardssoastocanceltheuniformfield.SinceRisclosertoQ,thefieldofQislargerthanthefieldfromP.HencethecurrentinQmustgooutofthepage.
b Ifthecurrentincreases,thenetfieldfromPandQincreasesaswell,sothatthetotalfieldatRisnolongerzero.IfwemoveclosertoQthefieldfromQwillbemuchlargerthanthefieldfromPandsotheircombinedfieldwillbedownwardsandmuchlargerthanexternalfield.Hencethepointhastomovetotheleft.