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.