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# Physics WO Calc II PHY 2054

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This 50 page Class Notes was uploaded by Yolanda Weimann on Monday October 12, 2015. The Class Notes belongs to PHY 2054 at Florida International University taught by Jorge Rodriguez in Fall. Since its upload, it has received 79 views. For similar materials see /class/221802/phy-2054-florida-international-university in Physics 2 at Florida International University.

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Date Created: 10/12/15

ELECTRIC CHARGE AND ELECTRIC FIELD Answers to MultipleC hoice Problems 1D 2C 3C 4BC 5E 6B 7C 8A 9B 10B 11D 12A 13B 14C 15A Solutions to Problems 171 Set Up Unlike charges attract and like charges repel In a conductor some ofthe negative charge is free to move In an insulator the charge can shift position only slig tly Solve 3 Aluminum is a conductor and negative charge in the sphere moves my from the rod The distribution of charge is sketched in Figure 171a b The charges in the nonconducting sphere displace slightly with negative charge moving army from the rod The distribution ofcharge is sketched in Figure 171b 4 te lt z 7 43 Z 1 a b Figure 171 172 Set Up Copper is a conductor so some of the electrons are free to move The positive charge on the rod 39 39 L 39 L L When 139 39 39 by aconducting wire charge can ow between the ball and the earth Solve a Electrons move tovmrd the rod The distribution of charge is sketched in Figure 172a b Electrons from the eanh are attracted by the region ofpositive charge on the ball and ow onto the ball giving it a net negative charge When the rod is removed this net charge distributes uniformly over the surface ofthe ball as sketched in Figure 172b A b Figure 172 173 Set Up For an isolated sphere the excess charge is uniformly distributed over the surface ofthe conductor Unlike charges attract and like charges repel and in a conductor the excess charge is free to move Solve a The uniform distribution ofcharge over the surface ofeach sphere is sketched in Figure 173a bWhen L rL I Lnlher L Ar L A A Lotheras shown in Figure 173b 172 Chaptzr 1 7 c When the spheres are close to each other the excess negative charges on each sphere repel as shown in Fig ure 173c r lt C a Figure 173 Re ect We will learn later in the chapter that the excess charge on a conductor is on the surface ofthe conductor 174 SetUp 391 quot39 39 quot aconductor L L 39 t Solve Ll 39 L 4 L 39 39 L 4 cloud as sketched in Figure 174 39 the earth r quot Figure 174 175 SetUp The charge ofone electron is 2 160 X 10 C 1pc 10 6 C 1nC 10 9 C Q 7 250 gtlt10 6 C 160 X 10 C Q 7 250 gtlt10 9 C 2 160 X10quot C Solve aN 156 X 10 electrons bN 156 X 10 electrons 176 SetUp The total charge is the number ifions times the charge ofeach 2 160 X 10 C Solve N 56 gtlt10 rn15 gtlt10 1rn 84 X 10 ions Q N2 84 x109160 x10quot C 13 gtlt10 9C 13nC 177 Set Up Charge conservation requires that the total charge Q ofthe reactants equals the total charge ofthe products qn 0 qp 2an q 2 He charge 22 xquot and vquot each have c arge So eactan Q 2 products Q 2 Could not occur b reactants Q 0 products Q 2 2 0 Could occur c reactants Q 22 22 42 products Q 22 2 0 32 Could not occur d reactants Q 2 22 32 products Q 22 2 2 Could not occur ereactantsQ 2 2 2 2productsQ 2 2 2 2 2 2Cou1doccur Re ect Just because the reaction obeys charge conservation doesn t mean it actually occurs There couldbe other reasons why the reaction does not occur 178 SetUp Like charges repel and unlike charges attract F Jr q 32 r 6 9 2 2 300 gtlt10quot5C2 Solve q qz 300 X 10 C The force is repulsive F 899 X 10 Nrn C W 202N rn 179 SetUp F k z An electron has charge 2 and aproton has charge 2 where 2 160 X 1049C k 899 gtlt109Nnn2 C2 160 x10quot C 1 Solve r k q fz 107 X 10 rn since an electron and a 9qu protonL L 39 39 c L L 439 39 L t uelectrons Electric Charge and Electric Field l73 1710 SetUp F qulzzl r Solve a The force is attractive so the unknown charge qZ is positive 7 r 7 0200N0300mZ 7 m 7 899 gtlt109NmZCZ0550 gtlt10 6 c b The unknown charge exerts a downward force of 0200 N on the other charge Like charges repel and unlike charges attract lqzl 364 gtlt10 6C 364 ac 1711 Set Up The electrical force is given by Coulomb s law with k 899 X 109 N mZCZ A proton has charge 9 and an electron has charge 9 lqlqzl 7 899 gtlt109NmZCZ160 gtlt10 19cZ Z 7 r Solve a F k 0 100 gtlt10 15mZ 230 N Yes this force is about 52 lbs 100 gtlt10 15m 2 b The force is smaller than in part a by a factor of 710 so it is 230 X 10 8N No this force is 100 X 10 In very small 1712 Set Up One mole of carbon contains NA 602 X 1023 atoms Each electron has charge 9 160 X 10 19 C lfcharge q is removed from the sphere the sphere is left with a charge of q 100 g 120 gmol 833 X 10 2 mol 602 X 1023 atomsmol 501 X 1022 atoms Each atom has 6 electrons so the number of electrons is Ne 6N 301 X 1023 electrons Each electron has charge 9 so the total negative charge of the elec trons is qe eNe 7160 gtlt10 19c301gtlt1023 7482 gtlt10 c b The sphere would be left with positive charge qsphere 482 X 104 C The force would have magnitude 482 gtlt104 C2 150 m2 charges are opposite sign so the force is attractive Solve a The number of moles is n 833 X 10 Zmol The number of atoms is N 1erA 928 X 1018N This is an immense force The F 7 k lqeqsihael r 899 gtlt109NmZC2 1713 Set Up One mole ofCa containsNA 602 X 1023 atoms Eachproton has charge 9 160 X 10 19 C Solve a The mass ofone hand is 0010 75 kg 075 kg 750 g The number ofmoles ofCais 750 g n 187 mol 4018 gmol The number of atoms is N nNA 187 mol602 gtlt 10Z3 atomsmol 112 gtlt 1025 atoms b Each Ca atom contains positive charge 209 The total positive charge in each hand is Ne 112 X102520160 gtlt10 19c 358 gtlt107c If 10 is unbalanced by negative charge the net positive charge of each hand is q 0010358 gtlt107 c 36 gtlt105 c c The repulsive force each hand exerts on the other would be 36 gtlt 105 C2 17m 40 gtlt IOZON 2 i 1 i F 7 kr Z 7 899 gtlt109NmZCZ This is an immense force39 our hands would y off Re ect Ordinary objects contain a very large amount of charge But negative and positive charge is present in almost equal amounts and the net charge of a charged object is always a very small fraction of the total magnitude of charge that the object contains l74 Chapter I 7 1714 Set Up F r 0220 N 742 gtlt1077 c 899 gtlt 109N mzCZ 2 F Solve aq1 qz qF kq Zandq r 0150 m r Z 7 7 4912 1 F 7 7 7 6 bq274q1F7 andqligr 7371X10 Cqzil48gtlt10 C V 1715 Set Up The mass of a proton is 167 X 10 27 kg and the charge of a proton is 160 X 10 19 C The distance from the earth to the moon is 384 X 108 m The electrical force has magnitude Fe k with r m m k 899 X 109 N mZCZ The gravitational force has magnitude Fgm G 12 2 r with G 667 gtlt 10 11 NmZkg2 10 gtlt 10 3 kg 167 gtlt 10 27 kg boxis q Ne 599 X1023l60 gtlt10 19c 958 gtlt104c gt 958 gtlt104cgtZ 384 gtlt108mZ The tension in the string must equal this repulsive electrical force The weight of the box on earth is w mg 98 X 10 3 N and the weight of the box on the moon is even less since g is less on the moon The gravitational forces exerted on the boxes by the earth and by the moon are much less than the electrical force and can be neglected 10 gtlt 10731652 384 gtlt108mZ Re ect Both the electrical force and the gravitational force are proportional to lrz But in SI units the coef cientk in the electrical force is much greater than the coef cient G in the gravitational force And a small mass of protons contains a large amount of charge It would be impossible to put 10 g of protons into a small box because of the very large repulsive electrical forces the protons would exert on each other Solve a The number of protons in each box is N 599 X 1023 The total charge of each 2 Fe kq Z 899 gtlt109NmZCZ 560N 1301b r m m bFgm G 2 Z 667 gtlt 10 11 NmZkgz 45 gtlt 10 34N lqiqzl r2 q2 f 457 gtlt10 N SolveFk 0200m 39 1426gtlt10 16c r2 M r k gt 899gtlt109NmZCZ lql 1426gtlt10 16c The number of excess electrons is 719 891 e 160 X 10 C 1716 Set Up F k The charge ofan electron is e l60 X 10 19 C 1717 Set Up An electron has mass 911 X 10 31 kg and charge q e l60 X 10 19 C The nucleus ofa hydrogen atom is aproton with charge q 9 F V e2 kez 899 gtlt109Nmz CZ160 gtlt10719cZ Solve mg k gives r 508 m r mg 911 gtlt 10 31kg980ms2 Re ect The electrical force is much stronger than the gravitational force The electrical binding force within the atom is much larger than the weight of the electron 1718 Set Up A proton has charge 9 and an electron has charge 9 with e 160 X 10 19 C The force lqiqzl 2 between them has magnitude F 72 62 andis attractive since the charges have opposite sign A proton has mass mp 167 X 10 27 kg and an electron has mass 911 X 10 31 kg The acceleration is related to the net force FbyF m3 Electric Charge and Electric Field l75 160 gtlt10 19cZ 575 gtlt 10 9N 20 gtlt10 1 mZ Z Solve F ke Z 899 gtlt 109N mZCZ r F 575 X 10 9N 18 2 proton up m 34 X 10 ms mp 167 X 10 kg F 575 X 10 9N electron ae i 7 731 63 gtlt 1021msZ me 911 gtlt 10 kg The proton has an initial acceleration of 34 X 1018 ms2 toward the electron and the electron has an initial accelera tion of63 X 1021 ms2 toward the proton Note that the force the electron exerts on the proton is equal in magnitude to the force the proton exerts on the electron but the accelerations produced by this force are different because the masses are different 1719 Set Up a 250g 245 msz F ma with F k Ly An electron has charge 9 i160 gtlt10 19 C r 2 Solve F ma 855 X 10 3 kg 245 msz 209 N The spheres have equal charges q soF 16in and r F 209N M r 0150m 229gtlt10 6c k 899 gtlt 109NmZCZ l 229 gtlt 10 6 c 13 N m 143 X 10 electrons The charges on the spheres have the same s1gn so the electn e cal force is repulsive and the spheres accelerate away from each other Re ect As the spheres move apart the repulsive force they exert on each other decreases and their acceleration decreases 1720 Set Up qe 160 X 10 19 C qp 160 X 10 19 C The net force is the vector sum of the forces exerted by each electron Each force is attractive so is directed toward the electron that exerts it Solve Each force has magnitude 2 8988 X109N Z c2 160 gtlt10 19 c 2 qu m gt gt 1023 gtlt10 8N r e F F k k Z r2 150 gtlt10 1 mZ The vector force diagram is shown in Figure 1720 J Figure 1720 l7 6 Chapter I 7 FX 1023 X10 8NF1y 0 FR cmos650 432 gtlt 10 9NFzy Fzsin650 927 gtlt 10 9N FX FIX F2 146 X10 8NFV F1y F2y 927 gtlt 10 9N F VFf Ff 173 X10 8N F 927 gtlt 10 9N tan6 V 78 06349 and 0 3240 The net force 1s 173 X 10 8N and 1s d1rected toward a FX 146 X 10 N point midway between the two electrons qu l r2 1721 Set Up F k Like charges repel and unlike charges attract The charges and their forces on q3 are shown in Figure 1721 Figure 1721 150 gtlt1079c 500 gtlt1079c Solve F1 k q 899 gtlt 109NmZCZ 169 gtlt 10 6N r1 0200 m 320 gtlt10 9c 500 gtlt10 9c FZ k qZ qji 899 X109NmZCZZgt 899 gtlt 10 7 N r2 0400 m F1 and F2 are in the same direction so F F1 F2 259 X 10 6 N and the net force is in the y direction Re ect The forces are vectors and must be added as vectors We add the forces by adding their magnitudes only because the two force vectors are in the same direction MM 72 1722 Set Up F k Like charges repel and unlike charges attract The charges and their forces on q3 are shown in Figure 1722 Figure 1722 Electric Charge and Electric Field 177 400 gtlt 10 9c 0600 gtlt 10 9c Solve F1 k q1 Z3 899 x109NmZCZ 5394 gtlt 10 7 N r1 0200 m 500 gtlt 10 9c 0600 gtlt 10 9c FZ k 899 X109NmZCZ 2997 gtlt 10 7 N r2 0300 m FX FIX FZX F1 7 F2 240 gtlt 10 7 N The net force has magnitude 240 gtlt 10 7 N and is in the x direction I 1723 Set Up F kiqqZ Like charges repel and unlike charges attract The charges and the forces on the r 150 cm charges q1 and q2 in the dipole are shownin Figure 1723 Use the coordinates shown sin6 Wand 6 4860 cm y FIX ql 150Cm 150cm FZX 12 F FZ D Figure 1723 500 X 1076C 100 X 1076C Solve F1 FZ 899 gtlt 109NmZCZ 1124 gtlt 103 N 00200 my 150 cm 200 cm FXF1X F3 0Fy Fly Fly 2F1sin0 721124 gtlt103N 7169 gtlt 103 N The net force has magnitude 169 X 103 N and is in the direction from the 500 LC to the i500 LC charge b FIX and F2C each produce clockwise torques and each have a moment arm of 150 cm Fly and Fly each have zero moment arm and produce no torque T 2F1X00150 m 2F1cos600150m 21124 gtlt103 Neos486 00150m 223 Nm The torque is clockwise Re ect The x components of the two forces are in opposite directions so they cancel and the net force has no x component But the torques of FIX and FZX are in the same direction and therefore produce a net torque l7 8 Chapter I 7 1724 Set Up In the OHN combination the O is 0170 nm from the ILIJr and 0280 nm from the N In the NHN combination the N7 is 0190 nm from the H1r and 0300 nm from the other N7 Like charges repel and unlike charges attract The net force is the vector sum of the individual forces 2 Solve aF cm 32 ke Z r r OHN 160 gtlt10 19c Z 01H F 899 gtlt 109Nm2CZW 796 gtlt 10 9N attractive m 160 gtlt 10 19c Z oiN F 899 gtlt 109Nm2CZW 294 gtlt 10 9N repulsive m NHN 160 gtlt10 19cZ 638 X 10 9 N attractive 0190 gtlt 10 m N H F 899 gtlt 109NmZCZ 160 gtlt10 19cZ 256 gtlt 10 9N re ulsive 0300 gtlt 10 9mZ p N7N F 899 gtlt 109 N mZCZ The total attractive force is 143 X 10 8 N and the total repulsive force is 550 X 10 9 N The net force is attractive and has magnitude 143 gtlt 10 8N e 550 gtlt 10 9N 880 gtlt 10 9 N 160 gtlt10 19cZ 00529 gtlt 10 9mZ The bonding force of the electron in the hydrogen atom is a factor of 10 larger than the bonding force of the adenine thymine molecules 2 bF kiz 899 gtlt 109NmZCZ 822 gtlt 10 8N V 1725 Set Up In the OHO combination the O is 0180 nm from the H1r and 0290 nm from the other 0 In the NHN combination the N is 0190 nm from the H4r and 0300 nm from the other N In the OHN combination the O is 0180 nm from the Hir and 0290 nm from the other N7 Like charges repel and unlike charges attract The net force is the vector sum of the individual forces 2 Solve F 6 ke Z The attractive forces are O H 710 X 10 9 N N H 637 X 10 9N O H r r 710 X 10 9 N The total attractive force is 206 X 10 8N The repulsive forces are 0707 274 X 10 9N NTN 256 X 10 9 N OTN 274 X 10 9 N The total repulsive force is 804 X 10 9 N The net force is attrac tive and has magnitude 126 X 10 8 N 1726 Set Up Like charges repel and unlike charges attract The force increases as the distance between the charges decreases Solve The forces on the dipole that is between the slanted dipoles are sketched in Figure 1726a The forces are attractive because the and charges of the two dipoles are closest The forces are toward the slanted dipoles so have a net upward component In Figure 1726b in adjacent dipoles charges of opposite sign are closer than charges of the same sign so the attractive forces are larger than the repulsive forces and the dipoles attract F1 Fly F2y F2 Mr 63 69 01le b Figure 1726 Electric Charge and Electric Field 179 1727 Set Up The central charge will be 185 mm from the charge on the left and 115 mm from the charge on the right Like charges repel The force diagram for the central charge is given in Figure 1727 els m x llSnm 11 F2 F1 qz Figure 1 72 7 4l60 gtlt10 19cZ 2690 gtlt10 1 N 185 gtlt10 9mZ 2 2 Solve F1 Hi 899 gtlt 109NmZCZ V2 4l60 gtlt10 19cZ 6961 gtlt 10 10N 115 gtlt10 9mZ 29 2 F2 k 899 X109NmZCZ V2 The net force is 427 X 10 10 N to the left Re ect The electrical forces are vectors and must be added as vectors At the initial position of the central charge the net force on it is zero If the central charge is displaced in either direction the net force on it is in the direction that pushes it back toward the equilibrium position 1728 Set Up F The new charges are q1 2q1 and q2 ZqZ 2 2 Solve The new force is F kw kw Him 32 4F r r r 7 lqiqzl 7 7 lqlqzl 1729 Set Up F 7 k 2 ln1t1ally r 7D and F 7 k DZ The new separat10n 1s r The new force 1s r F 3F 1 3 Solve F kw 317 so 2 Zandr D W D W D Re ect The force is proportional to lrz so a decrease in r increases the force 1730 Set Up If the new acceleration a equals a5 then the new force F equals F5 F k iqquZi Initially r r dandF kiq2i The new separation is r l Solve F kw k so r d5 r 5 d 1731 SetUp F LIEAproton has chargeq 9 160 X 10 19 C F 200 gtlt 10 9N Solve a F E 79 250 NC S1nce the charge 1s negat1ve the force and electr1c M 800 gtlt 10 c eld are in opposite directions and the electric eld is upward bF ME eE 160 gtlt 10 19c250 N 400 gtlt 10 19N The charge is positive so the force is in the same direction as the electric eld the force is upward l7lO Chapter 17 1732 Set Up F LIE A proton has charge q 9 160 gtlt 10 19C andrnass m 167 X 10 27 kg Solve a The gravity force is downward so the electrical force must be upward An upward force from a downward electric eld requires a negative charge mg lq H mg 145gtlt10 3kg980ms2 q 650 NC The chargeis q 7219 gtlt 10 5 c mg 7 167 gtlt 10 Z7kg980 msz e 7 160 gtlt10 19c 219 gtlt10 5c b mg eEE 102 gtlt10 7Nc 1733 Set Up Use the coordinates shown in Figure 1733 Since the electric eld is uniform the force is constant 7 7 7 and the acceleration is constant F qE and F m3 120 0 For an electron q e i160 gtlt 10 19C and m 911 gtlt 10 31 kg y Positive plate 32 cm Negative plate x 397 Figure 1733 2y 2320 gtlt10 Zm 7 l 7 7 7 Solve ay 7 voyt Zaytz a 7 t2 7 15 X 1085 7 284 gtlt1014ms2 F may 911 gtlt 10 31kg284 gtlt 1014ms2 259 gtlt10 16N Fy 259 gtlt 10 16N 3 3 E 19 162 gtlt 10 NC andE 162 gtlt 10 NC M 160 gtlt 10 c 11 no a 0 284 gtlt10 msz15 gtlt10 8s 426 gtlt106ms Re ect We could also use the workenergy theorem and set the work done by the force equal to the gain in kinetic energy of the electron 1734 Set Up For apoint charge E E is toward a negative charge and away from a positive charge r Solve a The eld is toward the negative charge so is downward 300 gtlt 10 9c E 899 gtlt 109NmZCZ Z 432 NC 0250m k 899 gtlt109N Z c2 300 gtlt10r9c 0 E W 150 E 120 NC 1735 Set Up Forapoint chargeE r E Z 650 gtlt103N c 0100 2 Solve lql f 723 gtlt10 9c 899 gtlt 109NmZCZ Electric Charge andElectric Field l7 11 1736 Set Up For apoint chargeE r qul 899 gtlt109NmZCZ500 gtlt10 9c Solve r1 335m E 400 NC 1737 Set Up The electric eld of a point charge has magnitude E A proton has charge q r e 160 X 10 19 C 160 gtlt 10 19c Solve aE 899 gtlt 109NmZCzm 58 gtlt 1019Nc 160 gtlt 10 19c bE 899 gtlt 109NmZCZ 58 gtlt 109Nc 50 gtlt10 1 mZ Re ect The electric elds inside atoms are very large 1738 Set Up The weight of an electron is w mg with m 911 X 10 31 kg The magnitude of the force on the electron due to the electric eld is Fe lq lE 9E In part b the electric eld of a proton at a distance r from 9 the proton isE k Z r 558 gtlt 10 11Nc 911 gtlt 10411 980 2 Solve aw Fe gives mg Ee andE g w e 160 gtlt 10 19 c e ke 899 gtlt109NmZ CZ160 gtlt10 19 c bEk Zsor n508m r E 558 gtlt 10 NC 1739 Set Up lfthe axon is modeled as a point charge its electric eld is E The electric eld ofa point charge is directed away from the charge if it is positive r Solve a 56 X 1011 Nair ions enter per meter so in a 010 mm 10 X 10 4 In section 56 X 107 Naions enter This number ofions has chargeq 56 gtlt 107160 gtlt 10 19c 90 gtlt 10 12c M 90 gtlt10 12c bEk 899gtlt109N ZCZ r2 m gt 500 gtlt10 ZmZ k 899gtlt109N ZcZ 90gtlt10 c cr i m gt280m 10 gtlt 10 6Nc 32 NC directed away from the axon 1740 Set Up The electric eld of a negative charge is directed toward the charge E The net eld is the r vector sum of the elds produced by each charge Point A is 0100 m from qZ and 0150 m from ql Point B is a a a 0100 m from q1 and 0350 m from qz A charge q in an electric eld E experiences a force F qE lt 0150m0100m 0100mag0250m E1 3 gt g B 3 3 qr E1 A E2 12 E 11 qz z a b Figure 1740 l712 Chapter 17 Solve a The electric elds due to the charges at pointA are shown in Figure l740a 625 gtlt 10 9c E1 k i j 899 X109NmZCZ Z 250 gtlt103 NC r 0150 rn 125 gtlt 10 9 c W 1124 gtlt10 Nc m 52 16 899 gtlt 109NmZCZ 7A2 Since the two elds are in opposite directions we subtract their magnitudes to nd the net eld E E2 7 E1 874 gtlt 103 NC to the right b The electric elds at points B are shown in Figure 1740b W 625 gtlt109c E k 899gtlt109N ZcZ 5619gtlt103Nc r512 m gt 0100rnZ 125 gtlt10 9c 0350 m 917 gtlt 10Z NC EZ 16 899 X109NmZC2 752 Since the elds are in the same direction we add their magnitudes to nd the net eld E E1 H2 654 X 103 NC to the right c At A E 874 X 103NC to the right The force on a proton placed at this point would be F qE 160 gtlt10 19c874 X103NC 140 gtlt 10 15Ntothenght 1741 Set Up For a point charge E E is directed toward a negative charge and away from a positive r charge Let the points a b and c be the locations where the eld is calculated in parts a b and c The three points and the electric elds produced at those points by each ofthe two charges are shown in Figure 1741 c qr E1 a 12 b E2 4 o y o lt o gt gt x E2 E1 E2 E1 Figure 1741 7 899 X109Nm2CZ400 gtlt10 9c iql E k 899N c a r12 0200 rnZ 899 X109N Z c2 600 gtlt10 9c EZ kwizz a 150NC r2 0600 rn E E 3 7899 NC 7150 Nc 71049 NC The eld is 1050 NC in the x direction 899 gtlt109NrnZ c2 400 gtlt10 9c bE1 k e 250 NC r1 120 rn 899 X109N Z c2 600 gtlt10 9c Ezkiq m gtZ gt337NC r2 0400 rn E 91 92 7250 NC 337 NC 312 NC The eldis 312 NC in the 9 direction 7 899 gtlt 109NmZCZ400 gtlt 10 9c iql E k 899N c c r12 0200 rnZ 899 gtlt 109N Z c2 600 gtlt 10 9c E2k q m M gt539Nc r2 100 rn E E1 E2 899 NC 7539 Nc 845 NC The eld is 845 NC in the x direction Re ect In each case the two electric elds must be added as vectors Electric Charge andElectric Field l7 l3 1742 Set Up For a point charge E E is toward a negative charge and away from a positive charge The r two charges and their elds at each point are shown in Figures 1742ad y y E1 6015quot 015m4gte015mgt E1 X X EZ ql qz ql EZ qz e015mgtlt015mgt a b y 12 015m 015m ql x 9 l l l 050m 040m C Figure 1742 Solve a E1 EZ and E1 E2 are in opposite directions so the resultant electric eld is zero EX By E 0 600 gtlt 10 9 c 0150mgtZ 2397Nc 1131 899 gtlt 109NmZCZ 600 gtlt 10 9c 0450 my 266 NC EZ 899 X109NmZCZ EX Eu EbC E1 H2 2660 NCEy 0 The resultant electric eld has magnitude 2660 NC and is in the 9 direction gt600 X 10 9 C 0400 m2 600 gtlt 10 9 c 0500 my cE1 899 gtlt 109NmZCZ 337 NCE1X 0Ely 7337 NC EZ 899 X109NmZCZ 216 NC 0 5310193 Ezcos 130 NC 32y agging 7173 NC EX 3 32x 0 130 NC 130NCEy Ely 32y 7337 NC 7173 NC 7510NC i E x tan 392 d 757O andi makes an angle 360 757O 284 counterclockwise from the 9 axis E Bf Bf 526 Nc 600 gtlt 10 9c 0250mZ EX Eu 32x 06 Ely 52y 251st 2863 NCsin53l 1380 NC Ehas magnitude 1380 NC and is in the y direction d 0 531031 52 899 X109NmZCZ 863 NC 1714 Chapter 17 1743 Set Up For a point charge E E is toward a negative charge and away from a positive charge Let q1 0500 nC and q2 800 nC For the net electric eld to be zero E1 and E2 must have equal magnitudes and opposite directions I II III I II 111 E1 11EZ E1 qz E1 22 q E1 qz E1 E2 E2 E1 E2 7 E2 lt7X7gt 172m XH712H gt a b Figure1743 Solve The two charges and the directions of their electric elds in three regions are shown in Figure l743a Only in region 11 are the two electric elds in opposite directions Consider a point a distance x from q1 so a distance 7 0500 HC 7 800 HC 2 7 Z 7 120m 7x from qZE17EZ glves k Z 7 k Z 16x 7 120m 7 x 4x 7 120m 7 x x 120 7 x and x 024 m is the positive solution The electric eld is zero at a point between the two charges 024 m from the 0500 nC charge b Let qZ i800 nC be the negative charge The two charges and the directions of their electric elds in three regions are shown in Figure l743b E1 and E2 are in opposite directions in regions I and 111 But for the magni tudes of the elds to be equal the point must be closer to the charge q1 that has smaller magnitude and that occurs only for region I Consider a point a distance x to the left of q1 so 120m x from qz E1 EZ gives 0500 HC 7 800 nC x2 120 ac tion The electric eld is zero at a point 040 m from q1 and 160 m from qz Re ect In each case there is only one point along the line connecting the two charges where the net electric eld is zero 162 120m x 4x l20m x and x 040m is the positive solu 1744 Set Up The electric eld of a negative charge is directed toward the charge E The net eld is the r vector sum of the elds due to each charge Label the charges q1 q2 and q as shown in Figure l744a This gure also shows additional distances and angles The electric elds atpointP are shown in Figure l744b This gure also shows the xy coordinates we will use and the x andy components of the elds E1 E2 and E3 a Figure 1744 Electric Charge andElectric Field l715 9 Z Z 500gtlt10 6c 6 ove 1 3 m S E E 899X1ON c 0100 2 449gtlt10Nc m 200gtlt10 6c E2 899 gtlt 109N mZCZ 499 gtlt10 NC gt 00600mZ E 31 32 33 0 andE 31 E2 33 52 251cos531 104 gtlt107Nc E 104 gtlt 107 NC toward the 7200 no charge 1745 Set Up The force on a charge q that is in an electric eld E is F qE Solve a The force on q is F4r qE to the right The force on q is F qE to the le The net force on the dipole is zero b The axis is at the midpoint of the line connecting the charges and perpendicular to the plane ofthe gure The torque is zero when each force Frr and F has zero moment arm This is the case for 6 0 and 6 180 as shown in Figure 1745a and b Egt Egt F Fi F F H q q q r q a b E q F gt 17 E gt W F T 1 F q C 01 Figure 1745 c For 6 slightly greater than zero as shown in Figure 1745c the torque on the dipole is clockwise and is directed so as to return the dipole to the equilibrium position For 6 slightly less than zero the torque is counterclockwise and again tends to rotate the dipole back to its equilibrium position The 6 0 position of Figure 1745a is a sta ble orientation For 6 slightly less than 180 Figure 1745d the net torque on the dipole rotates it away from the 6 180 equilibrium position The same is true for 6 slightly greater than 180 The 6 180 position ofFigure 1745b is an unstable orientation d The electric eld of the dipole is directed from the positive charge and toward the negative charge Thus in Fig ure 1745a the electric eld ofthe dipole is to the le opposite to the direction of the external eld Re ect For any orientation ofthe dipole the net force on the dipole is zero But only for 6 0 and 6 180 is the net torque zero 1716 Chapter 17 1746 Set Up 3 qi T Fl where l is the moment arm The forces and moment arms are shown in Fig ure 1746 The axis is at the midpoint of the line connecting the charges and perpendicular to the page E gt F gt 175 mm 1 axis 300 30 L 175 nm 4 F Figure 1746 Solve F F ME 250 gtlt 10 SC7800 Nc 195 gtlt10 ZN 1 1 175 nmsin30 875 gtlt io wm Each force produces a clockwise torque so the net torque is T 7 T 27 2195 gtlt IO ZN875 gtlt IOTIOm 34 gtlt10 Nm The net torque is 34 X 10 11 N m clockwise lql 1747 Set Up E k Z r M M 1 M S F 10 EEk 2Ek k E4 ove a orr1 cm 1 r12 V2 V1 2 2r1gtz 4 mg M 1 M bN 3 Ek k 9 0WV3 71 3 3rlgtz 9 r12 1748 Set Up E kg r k k k Solve ForrREkQE2kgEzE10sorzx Q Q10 QZER R r2 EZ EIO lelR 1749 Set Up Electric elds come out of positive charge and go into negative charge The electric eld is stronger where the electric eld Lines are closer together Solve a The eld lines go from B toA so B must be positive andA must be negative b The eld lines are closer together nearA so the electric eld has greater magnitude nearA mm Chargiz and mm mid l7el7 1750 sd Up 72 l l l s d eleeme elds duetu each charge 311115 and Care sketched m Flgure l7 5n gure 1750 nlve MALE bum elds are m the 7y dueth and the net eld ls m the 7y du39ectlun C At c as 214 the x cumpunmt 1me elds cancel and me net eld ls m the 7y diremun 1751 sd Up Eleeme eld ls my lm pusmve charge and Ward negative charge h z I Figure 1751 Salve 2 The eleeme elds 2 and Z and thequot vectur sum me net eld ii are shuwn fur each pmnt m g ure l7 512 The elemc eldlstuwdA atpulntsB and Cznd me eldlszem 314 A d The eldlszem 21A andC Re ect Cumpzre yuurresults m the eld llnes shdwn m Flgurel7 242 and b m the textbuuk l7 l8 Chapter 17 1752 Set Up Electric eld points toward negative charge and away from positive charge L E E Q Er P E Eiq P a b Figure 1752 Solve a Figure 1752a shows 39 and El at pointP 39 must have the direction shown to produce a resultant eld in the speci ed direction E9 is toward Q so Q is negative In order for the horizontal components of the two elds to cancel Q and q must have the same magnitude b No lfthe lower charge were negative its eld would be in theadirection shownin Figure 1752b The two possi ble directions for the eld ofthe upper charge when it is positive Err or negative E are shown In neither case is the resultant eld in the direction shown in the gure in the problem 1753 Set Up Electric eld is directed away from positive charge and toward negative charge By symmetry far from the edges ofthe sheets the eld lines are perpendicular to the sheets there is no reason to prefer to the le or to the right for a component of electric eld ill tit ill I a a l a a r s r l s s l i l i i l l l l l l a b C d 8 Figure 1753 Solve a The elds of each sheet are sketched in Figure 1753a The solid lines are the eld due to the upper posi tive sheet and the dashed lines are the eld due to the lower negative sheet Between the two sheets the elds are in the same direction and add Outside the two sheets the elds are in opposite directions and cancel The net electric eld is sketched in Figure 1753b This result is consistent with the eld in the similar case described in Example 175 b The elds of each sheet are sketched in Figure 1753c The solid lines are the eld due to the upper sheet and the dashed lines are the eld due to the lower sheet Since both sheets are positive the eld of each sheet is directed away from that sheet Between the two sheets the elds are in opposite directions and cancel Outside the two sheets the elds are in the same direction and add The net eld is sketched in Figure 1753d c The elds are the same as in part b except they are toward each sheet The net eld of the two sheets is sketched in Figure 1753e Re ect More advanced treatments show that the electric eld of an in nite sheet is uniform and does not depend on the distance from the sheet Electric Charge andElectric Field l7 l9 Qencl 6 1754 SetUp Gauss s law says DE 60 8854 X 10 12 CZNmz 250 gtlt 10 6c 8854 gtlt 10 12cZNm2 b Qeml 50cm 8854 gtlt10 12cZNm2140Nunic 124 gtlt10 c Solve altlgtE 282 gtlt 105NmZC 1755 Set Up Gauss s law says that the net electric ux DE through any closed surface equals Qemleo where Qeml is the net charge enclosed by the surface Solve S1 Qencl 0 so DE 0 sz Qend 50 uc 90 MC 770 c 70 MC and DE 70 Mce0 79 gtlt105NmZc 3 Qend 90 uc 10 MG 7100 c 0 and DE 0 S4 Qend 80 uc 770 c 10Mc and DE 10 Mce0 11 gtlt 105NmZC s5 Qend 80 MG 770 c 50 MG 90 MG 10 MG 7100 C 60 MC and DE 60 Mce0 68 gtlt 105NmZC Re ect The electric eld varies in a complicated way in both magnitude and direction for each of the surfaces and direct calculation of the ux as EELAA would be very dif cult Gauss s law gives us a very simple way of calcu lating the net ux 1756 Set Up The cube is a closed surface so we can apply Gauss s law to it DE Qemd By symmetry the ux is the same through each ofthe 6 faces of the cube 0 ch1 800 gtlt 1059c Z Solve a The total ux through the cube 1s DE 904 N m C 60 8854 gtlt 10 12cZNmz DE 904NmZC 2 b The ux through each face 1s 7 T 151 N m C 1757 Set Up Example 1710 shows that E 0 inside a uniform spherical shell and that E kg outside the shell r Solve aE 0 150 gtlt 10 6 c 0060 my 2 150 gtlt 10 6 c C 0110 m2 Re ect Outside the shell the electric eld is the same as if all the charge were concentrated at the center of the shell But inside the shell the eld is not the same as for a point charge at the center of the shell inside the shell the electric eld is zero by 0060mandE 899 X109NmZCZ 375 gtlt 107Nc cr 0llOmandE 899 gtlt109Nm2 111gtlt107Nc 1758 Set Up A spherically symmetric uniform sphere of charge can be modeled as a series of concentric shells so E kg outside the sphere A proton has charge 9 V 92160 x10 19c 74 gtlt 10 15mZ 74 gtlt 10 15m 2 1 13 0 NC Solve aE k 899 gtlt 109NmZCZ 24 gtlt 10Zl NC V b Forr 10 gtlt10 1 mE 24 gtlt10 NC c E 0 inside a spherical shell l7 20 Chapter 17 1759 Set Up The method of Example 1710 shows that the electric eld outside the sphere is the same as for a point charge of the same charge located at the center of the sphere The charge of an electron has magnitude 9 160 X 10 19C Solve aE Forr R 0150 mE 1150 NC so r 7 Erz 7 1150 Nc0150 m2 M 288 X10 9c 7 k 899 gtlt 109NmZCZ 288 gtlt 10 9c w The number of excess electrons 1s 180 X 10 electrons 160 X 10 19Celectron M 7 288 x10 9c b R0100m0250mEk 7 899gtlt109NmZCZ 414NC r y gt 025 1760 Set Up The charge distribution has spherical symmetry so the electric eld is radial and depends only on the distance from the center of the charged sphere The surface area of a sphere is A 477V Solve 2 Apply Gauss s law to a spherical surface of radius r 2 R concentric with the sphere of charge The elec tric eld is constant over the Gaussian surface and perpendicular to it so DE EA E477r2 The charge Q Q enclosed is Q so Gauss s law givesE47rrZ andE 6o 4 Z This is the sameE as for a point charge Q at the Weor center of the sphere b No direction is preferred so E 0 1761 Set Up The charge distribution has spherical symmetry so the electric eld if nonzero is radial and depends only on the distance from the center of the shell Solve 2 Apply Gauss s law to a sphere of radius r lt a and concentric with the shell The electric eld if nonzero is constant over the Gaussian surface and perpendicular to it so DE E4arrz But no charge is enclosed by the Gaussian surface so Q8161 0 Gauss s law givesE477rZ 0 andE 0 bApply Gauss s law to a sphere ofradius r gt b DE E4 Trzgt Qeml Q Gauss s law givesE47rrZ Egand 0 477esz c The thick shell can be constructed from a series of concentric thin shells E 0 inside each of these thin shells soE 0 inside the thick shell Re ect In using Gauss s law it is very helpful to select a Gaussian surface of appropriate symmetry so it is simple to express the ux in terms of the electric eld at the surface 1762 Set Up Section 179 shows that the net charge of a charged conductor is entirely on its outer surface Solve a q 0 on the surface of the inner cavity of the conducting car b All the net charge is on the outer surface i850 uC 1763 Set Up E 0 everywhere within the conductor Any net charge must be on the inner and outer surfaces of the conductor Gaussian surface Figure 1763 Electric Charge andElectric Field 1721 Solve a and b Apply Gauss s law to a surface that is within the conductor just outside the cavity as shown in Figure 1763 E 0 everywhere on the Gaussian surface so DE 0 for that surface Gauss s law then says that Qeml for this surface is zero The conductor is neutral so if the outer surface has charge 12 4C the inner surface must have charge 12 ac To make Q8161 0 there must be 12 4C within the hole Re ect The charge in the hole creates the charge separation in the conductor It pulls 12 MC to the inner surface and that leaves 12 4C on the outer surface 1764 Set Up E 0 everywhere within the conductor Any net charge must be on the inner and outer surfaces of the conductor Solve Apply Gauss s law to a surface that is within the conductor just outside the cavity as shown in Figure 1764 E 0 everywhere on the Gaussian surface so DE 0 for that surface Gaussian surface Figure 1764 2 Q8161 0 for the Gaussian surface means that q 0 on the inner surface so the 16 4C of net charge is on the outer surface of the conductor b Q8161 0 for the Gaussian surface means that there must be 11 HC on the inner surface of the conductor Since the total net charge of the conductor is 16 nC ifthere is 11 HC on the inner surface there must be 5 nC on the outer surface 1765 Set Up F k i Like charges repel and unlike charges attract Charges q1 and q2 and the forces they a a exert on q3 at the origin are sketched in Figure 1765a For the net force on q3 to be zero F1 and F2 from q1 and q2 must be equal in magnitude and opposite in direction y qz q F1 ql x F2 a 030m 020m a y l H III F F qz F1 F F 2 1 s qel 1 2 x 13 q i 7 13 F2 0 Figure 1765 1722 Chapter 17 Solve 2 Since F1 F2 and Fm are all in the 9 direction F F1 F2 This gives iqiqgi lqzqgl 400 gtlt 10 6N 450 gtlt 10 9c 250 gtlt 10 9c k Z k Z 13 Z 2 r1 r2 899 gtlt 109NmZCZ 0200 m 0300 m 400 gtlt 10 6N and q3 317 gtlt10 9 c 317 nC b Both F1 and F2 are in the 9 direction so Fm F1 F2 is in the 9 direction c The forces F1 and F2 on q3 in each ofthe three regions are sketched in Figure l765b Only in regions I to the left of qz and III to the right of ql are F1 and F2 in opposite directions But since qzl lt ql q3 must be closer to qZ than to q1 in order for F1 F2 and this is the case only in region I Let q3 be a distance d to the left of qz so it is a distance d 0500 In from q1F1 FZ gives 450 nC 250 nC k Z k 2 d 0500 m d The positive solution is d 146 In This point is at x 0300 rn 146 In i176 rn Re ect At the point found in part c the electric eld is zero The force on any charge placed at this point will be zero 180012 d 0500 m V180d d 0500 m lqiqzl r2 1766 Set Up Let q1 and q2 be the charges on the spheres F k and q1 qZ 600 nC Fr2 0270 gtlt10 3N0100mZ Solve qlqz 7 16 Z 7 9 k 899Xl09NmZCZ 7300gtlt10 Cqz7600gtlt10 c7q1so q1600 gtlt10 9c 7 11 300 gtlt10 16c2qlZ 7 600 gtlt10 9cq1 300 gtlt10 16cZ 0 The quadratic formula gives 117600 gtlt10 8 600 gtlt 10 s2 7 4300 gtlt 10 16c 055 gtlt 10 8 c or 545 gtlt 10 8 c For q1 055 gtlt 10 8c qz 545 gtlt 10 8c and for q1 545 gtlt 10 8 c qz 055 gtlt 10 One sphere has charge 055 X 10 8 C and the other has charge 545 X 10 8 C qu l r2 1767 Set Up F k Like charges repel and unlike charges attract The three charges and the forces on q3 are shown in Figure 1767 y Fl F1y 0 F q3 lX 5915 a 0030m F Z TV9 0040m 12 x ql Figure 1767 Electric Charge andElectric Field 1723 500 gtlt 10 9 c 600 gtlt 10 9c 899 x109NmZCZ 1079 gtlt10quot c rn Solve aF1 kiqlzgi 71 0 369 le F1 cos0 863 gtlt 10 5NF1y F1sin0 648 gtlt 10 5N 200 gtlt 10 9c 600 gtlt 10 9c FZ kw 899 X109NmZCZ 120 gtlt10 4 c r2 00300 m FR 0Fzy in 7120 gtlt10 4NFX FIX FR 863 gtlt 10 5 N Fy Fly Fly 648 gtlt10 5N 7120 gtlt10quot N 7552 gtlt10 5N F bF VizZ Ff 102 gtlt10 4Ntan Fy x Re ect The individual forces on q3 are computed rom Coulomb s law and then added as vectors using components 0640 d 3260 below the 9 axis 1768 Set Up F Like charges repel and unlike charges attract The positions of the three charges are sketched in Figure 1768a and each force on q3 is shown The distance between q1 and q3 is 500 cm Figure 1768 300 gtlt 10 9c 500 gtlt 10 9 c 899 x109NmZCZW 5394 gtlt10 5N Solve aF1 kiqlzg Vi FIX Flcos 75394 gtlt 10 5N0600 73236 gtlt 10 5N Fly iFlsin 75394 gtlt 10 5N0800 74315 gtlt10 5N 200 gtlt 10 500 gtlt 10 9c FZ k 899 X109NmZCZ 9989 gtlt10 5N r2 300 gtlt 10 m FR 9989 gtlt 10 5NFzy 0 E FIX F3 9989 gtlt10 5N 3236 gtlt10 5 N 675 gtlt IO SN39JfV Fly Fzy 7432 gtlt10 5 N l724 Chapter 17 F bF and its components are shown in Figure l768bF FXZ FyZ 801 X 10 5 N tan6 Fy 0640 and a 6 3260 Fis 327 counterclockwise from the 9 axis X qu l 1769 Set Up F k 2 Like charges repel and unlike charges attract r q1 010m q3 11 020m 13 F1 F3 y 020m 020m F1 F3 14 x F2 14 12 qz F2 a b Figure 1769 Solve a The charges and the forces on the i100 2C charge are shown in Figure 1769a The distance r14 between q1 and q4 is r 20200 In 01414 In F2 and F3 are equal in magnitude and opposite in direction so E F Omdfm F1 gt300 gtlt10 9c100 gtlt10 6c 135 gtlt 10 3 N 01414 my F1 k Ih q 899 gtlt109NmZCZ V14 The resultant force has magnitude 135 X 10 3 N and is directed away from the vacant comer b The charges and the forces on the i100 2C charge are shown in Figure 176 The distance r14 between q1 and q4 is 20200 In 02828 In Use coordinates as shown in the gure FZr F3X and FIX 0 so Fx 0 Fy Fly Fzy F3y F1 2FZcos45 300 gtlt10 9c100 gtlt10 6 c lq1q4l F k 899 gtlt109N Z c2 3372 gtlt10 4N r142 m gt 02828 my 300 gtlt10 9c 100 gtlt10 6 c FZ k 899 X109NmZCZ 6742 gtlt10quot N m 0200 m Fy 3372 gtlt10quot N 26742 gtlt10 4Ncos45 129 gtlt10 3 N The resultant force has magnitude 129 X 10 3 N and is directed toward the center of the square 1770 Set Up The force on the negatively charged electronis opposite to the direction ofthe eld Since the eld is unifOIm the force and acceleration are constant Use coordinates with y upward and 9 to the Iight The electron is at x 00200m when y 000500 m ax 0 120x 500 X 106 ms voy 0 For an electron q e 7160 X10719Candm 911 gtlt 10 31kgx 003 g 36 00200 In 9 t 400gtlt 10 s um 500 gtlt 106ms 2y 7 2000500m 1 2 7 y 7 ont gayt g1ves ay 7 22 7 400 gtlt 109sZ 63925 X lowmSz39 F ma 911 gtlt 10 31 k 625 gtlt 10mm sZ Fy may andgy 1 y 7 7356 gtlt103NC e e 160 gtlt 10 c The electIic eld has magnitude 356 X 103 NC Electric Charge andElectric Field l725 1771 Set Up The ball is in equilibrium so for it EFX 0 and 0 The force diagram for the ball is given in Figure 1771 FE is the force exerted by the electric eld 3 qE Since the electric eld is horizontal FE is horizontal Use the coordinates shown in the gure The tension in the string has been replaced by its x and y components y T Ty 0 FE X TX mg Figure 1771 7 7 7 7 mg Solve 7 0g1vesTy mg 7 0Tcos6 mg 7 OandT7m 2E OgivesFE 7 T 0FE 7 Tsin0 0 m FE g0sin0 mgtan0 123 gtlt 10 3kg980ms2tan174 378 gtlt 10 ZN COS FE 378 gtlt IO ZN 4 FE lqlEsoE 7 341 gtlt 10 NC lql 7111gtlt10 6c q is negative and E is to the right so E is to the left in the gure Re ect The larger the electric eldE the greater the angle the string makes with the wall 1772 Set Up E E is toward a negative charge and away from a positive charge At the origin E1 due to r the i500 nC charge is in the 9 direction toward the charge 500 gtlt10 9c 120 m2 E 31 323 450Nc soEZ E 7 E1 450Nc 7 312Nc 138 NC 3 7 lQl 79er 7 138Nc0600mZ E1s away from Q so Q 1s pos1t1veEZ 7 7g1ves lQl 7 T 7 109 7450 NC soEZ E 7 E1 7450Nc 7 312 NC 7762 NC i d 7 9er 7 762 Nc0600 my 1s towar QsoQ1snegat1ve lQl 7 k 7 899 X 109NmZCZ Solve 211 899 gtlt 109N mZCZ 312 NCE1X 312 Nc 553 X 10 10 C 305 X 10 9 C 1773 Set Up The gravity force weight has magnitude w mg and is downward The electric force is F LIE Solve a To balance the weight the electric force must be upward The electric eld is downward so for an upward force the charge q ofthe person must be negative w F gives mg lq 5 and 7 mg 7 60 kg 980 msz M 7 7 150 Nc 39 CY 899 X 109NmZCZW 14 X 107 N The repulsive force is immense and this is In 39 C qu l Z bFk not a feasible means of ight l726 Chapter 17 1774 Set Up The force diagram for the 650 1C charge is given in Figure 1774 FE is the force exerted on the charge by the uniform electric eld The charge is negative and the eld is to the right so the force exerted by the eld is to the left FL is the force exerted by the other point charge The two charges have opposite signs so the force is attractive Take the 9 axis to be to the right as shown in the gure y Figure 1774 Solve aF ME 650 gtlt 10 6C185 X108NC 120 gtlt103N 650 gtlt 10 6c 875 gtlt 10 6c F kw 899 X109NmZCZ gt Z gt 818 gtlt IOZN r 00250m 2F0givesrFq 7FE OandTFEFq382N b Now FL is to the left since like charges repel EFX 0givesTin 7FE OandTFEFq 202 gtlt103N 1775 Set Up The force on the electron is upward in the gure so the electric eld must be downward To produce a net electric eld that is downward it must be that q1 is positive qZ is negative and lql l lqzl The eld due to q1 and q2 at the location of the electron are sketched in Figure 1775 The electron is r 361 cm from each charge 300 cm tan6 and 6 5630 200 cm 11 0 r 20cm E1 M 30cm x E2417 0 0 7quot qz V Figure 1775 cI1 2k9q1 Zkeql Solve The net electric eld1sE 2E1y Zk Zcos6F eE Z cos6F ma so 2 cos6 ma and r r r 614 gtlt10 6 C rzma 7 361gtlt10 Zm2911gtlt10 31kg825 gtlt1018ms 2899 gtlt 109NmZCZ160 gtlt10 19ccos563 ql Zke cos 6 qz 7614 gtlt10 6c Re ect The force that an electric eld exerts on a negative charge is opposite to the direction of the electric eld We could also do this problem by considering the force that each charge exerts on the electron Electric Charge andElectric Field l7 27 1776 Set Up When the forces balance a 0 and the molecule moves with constant velocity Kl Solve aF FD so qE KRU and F E E ET b U qandis constant x vt qT KR E K R ET ET cx 1 where is constant 1 2 1 and 1 3 1 K R K R Z R 1 R 3 R 1 ET ET ET ET we ZHF mm 2 we K R 2 K R 1 K R 3 K R 1 122 2717 For a circular orbit a The period T is The mass of the sun is mS 7 r v mlmZ 1777 Set Up Fgmv G 2 r 199 X 1030 kg and the orbit radius ofthe earth is 150 X 1011m mSmE UZ Gms Z mE so U r r r Solve F ma gives G T 2 r 2 r32 150 gtlt 1011m3Z 273 X 1073 arr 77 quotn s Gms VGmS 90 gtlt 109NmZkg2l99 gtlt 103 kg A present yearis 3146 X 107 s so the year in the parallel universe would be 87 X 10 11 ofapresent year GmS 90 gtlt109NmZ kgzl99 gtlt103 kg M Re ect U n 35 X 10 ms This exceeds by a large r 150 X 10 In factor the speed of light in our universe so the parallel universe must have a much larger speed of light than ours in order for this speed to be possible ELECTRIC POTENTIAL AND CAPACITANCE Answers to Multiple Choice Problems 1AC 2C 3A 4AD 5C 6BC 7BD 8C 9B 10C 11D 12A 13A 14AC 15C Solutions to Problems 181 Set Up Since the charge is positive the force on it is in the same direction as the electric eld Since the eld is uniform the force is constant and W Fscos Solve a F is upward and 3 is to the right so 1 90 and W 0 b Fis upward and 3 is upward so 1 0 W Fs qu 280 gtlt10 9C400 gtlt 104 NC0670 m 750 gtlt 10 4 J c Fis upward and is at 4500 below the horizontal so 1 13500 W Fscosd qucos 280 gtlt 10 9 C 400 gtlt104 NC260 m cos 13500 7206 gtlt 10 3 J Re ect The work is positive when the displacement has a component in the direction of the force and it is negative when the displacement has a component opposite to the direction of the force When the displacement is perpendicular to the force the work done is zero 182 Set Up a The proton has charge 9 and mass 167 X 10 27 kg Let point a be at the negative plate and point b be at the positive plate The electric eld is directed from the positive plate toward the negative plate The force on the positively charged proton is in this direction The proton moves in a direction opposite to the electric force so the work done by the electric force is negative Solve WM 7qu 7eEd 7160 gtlt10 19 C280 gtlt106NC0100 m 7448 gtlt 10 14J 183 Set Up U km r 7 1amZ 7 899 gtlt 109NmZCZ7720 gtlt 10 6 C230 gtlt 10 6C 7 U Solve r 7 0372m 70400 184 Set Up WM Ua 7 Ub U km rd 0150 m rb 0250 m2 0250 my 0354m r qlqz 7 899 X109NmZCZ24O gtlt 10 6C7430 gtlt IO SC 7 Solve Ua k 70619 I rd 0150m 899 gtlt 109N m2 C2 240 gtlt 10 6C 7430 gtlt 10 6C Ub kqlq lt gtlt gtlt gt 70262 rb 0354m WM Ua 7 Ub 70619J 7 70262J 70357J 182 Chapter I 8 185 Set Up An electron has charge q e l60 X 10 19 C For a pair of point charges U k Let r q1 300 nC and q2 200 nC Let q3 be the electron The potential energy of the electron is the sum of U13 and U23 Solve a V1 rZ 0250 m 300 gtlt 10 9c7160 gtlt10 19c 0250m 200 gtlt 10 9c7160 gtlt10 19c 0250m q1q3 U13 k 899 X109NmZCZ 7173 gtlt 10 17J Vi UZ3 kw 899 X109NmZCZ 7115 gtlt 10 17J r2 U U13 UZ3 7288 gtlt 10 17J b r1 0100 m and r2 0400 m 300 gtlt 10 9c7160 gtlt 10 19c W 200 gtlt 10 9c7160 gtlt10 19c 0400m U13 k 7432 gtlt 10 17J M 899 X109NmZCZ V 1 UZ3 kw 899 X109NmZCZ 7719 gtlt10 18J r2 U U13 UZ3 7504 gtlt 10 17 J Re ect The potential energy is negative since the charge of the electron has opposite sign from the charge of each of the other two particles The magnitude of the potential energy increases when the electron moves toward the larger charge 186 Set Up For a pair of point charges the electrical potential energy is U k In the OHN combination the O is 0170 nm from the Hir and 0280 nm from the N In the NHN combination the N7 is 0190 nm from the Hir and 0300 nm from the other N7 Uis positive for like charges and negative for unlike charges Solve a OHN 160 gtlt10 19 C2 0170 gtlt 10 9 m 160 gtlt10 19cZ 0111 U 7899 X109NmZCZ 7135 gtlt10 18J OZN U 899 gtlt 109NmZCZ 822 gtlt 10 191 NHN 160 x10 19c Z 7121X10 18J 0190 gtlt 10 m 160 gtlt10 19cZ 0300 gtlt 10 9m N H U 7899 X109NmZCZ N N U 899 X109NmZCZ 767 gtlt 10 19J The total potential energy is ULot 7135 gtlt10 18J 822 X10491 7121X10 18J 767 gtlt10 19J 7 971 gtlt 10491 b In the hydrogen atom the electron is 00529 nm from the proton 160 gtlt 10 19c Z 7435 gtlt 10481 00529 X 10 m The magnitude of the potential energy in the hydrogen atom is about a factor of 4 larger than what it is for the adeninethymine bond U 7899 X109NmZCZ Electric Potential and Capacitance 183 7 187 Set Up For a pair of point charges the electrical potential energy is U kg In the OHO combination the O is 0180 nm from the H1r and 0290 nm from the other 0 In the NHN co mbination the N is 0190 nm from the Hir and 0300 nm from the other N7 Uis positive for like charges and negative for unlike charges In the OHN combination the O is 0180 nm from the Hir and 0290 nm from the NT Uis positive for like charges and negative for unlike charges Solve OHO oth U 7128 gtlt 10 18 J39 010 U 793 gtlt 10 19J NHN NiH U 7121 gtlt 10 18JN N U 767 gtlt 10 19J OHN O H U 7128 gtlt 10718 O N U 793 gtlt 10 19J The total potential energy is 7377 gtlt 10ng 235 gtlt 10 18J 7142 gtlt 10 18J Re ect For pairs of opposite sign the potential energy is negative and for pairs of the same sign the potential energy is positive The net electrical potential energy is the algebraic sum of the potential energy of each pair 188 Set Up The initial potential energy is converted to kinetic energy after the charges have moved far apart Apply U k to each pair of charges and add to get the total electrical potential energy In part b the distance 7 betweenthe 400 4C and 300 4C charges is 180 cm 899 gtlt109NmZ cZ Solve a U k m M m 7400 x10 9c7200 x10 9c V12 V13 V23 0100m 7200 gtlt 10 9C7300 gtlt 1079c 7300 x10 9c7400 gtlt 1079c 234 gtlt 10 400 gtlt10 6c 200 gtlt10 6c b U kM m m 899 gtlt109NmZCZ M r12 r13 r23 0100m 200 gtlt 10 6c300 gtlt 10 400 gtlt 10 6c300 gtlt 10 6cgt 7 16 0150 m 0180 m I The maximum total kinetic energy will be 168 J The kinetic energy increases as the electrical potential energy decreases as the charges move apart The maximum kinetic energy is achieved only after a long time 7 189 Set Up U1 U rZ 2r1 2d qZ qu and q2 2q 1qu k2q1gt2q1 gt 2kq1q1 2U r 2d d 2 Re ect For point charges the electrical potential energy is proportional to lr and the electrical force is propor tional to lrz The force between the charges would not change Solve UZ k 1810 SetUp U qu UZ 3U1r1R r Solve Ur qu which is constant So Ulr1 Uzrz rZ y R3 UZ 3U1 1811 Set Up For apair of oppositely charged parallel metal plates Val Ed F Val 7 360 v d 7 450 X10 3m bF ME 240 gtlt 10 9C8000 NC 192 gtlt10 5N Solve aE 8000 vm 8000 NC 1812 Set Up For a pair of oppositely charged metal parallel metal plates Vab Ed Val 475 gtlt 103 v FW l58gtlt 10 3m 158mm m Solve d 184 Chapter I 8 1813 Set Up For two oppositely charged parallel plates Val Ed where Val is the potential difference between the two plates E is the uniform electric eld between the plates and dis the separation ofthe plates An electric eld E exerts a force F q E on a charge placed in the eld a Fm An electron has charge 9 and mass 911 gtlt 10 31kg Val 25 v Solve aE 3 d 20 X 10 In F e 160 gtlt10 19c125 gtlt104vm bF lqlEeEa m m m 911 X 10 kg Re ect The electric eld is the same at all points between the plates away from the edges so the acceleration would be the same at all points between the plates as it is for a point midway between the plates 125 gtlt104vm 220 gtlt 1015msZ 1814 Set Up For two oppositely charged sheets of charge Val Ed The positively charged sheet is the one at higher potential Val 70 gtlt 10 3 v Solve aE 79 d 75 gtlt 10 m g the axon since the outer surface of the membrane has positive charge and E points away from positive charge and toward negative charge Section 189 explores the effects of a material other than air between the plates b The outer surface has positive charge so it is at higher potential than the inner surface 93 X 106 Vm The electric eld is directed inward toward the interior of 1815 Set Up Val Ed for parallel plates 1 V Solve d b 15 gtlt106m 15 gtlt103km E 7 10 gtlt10 6vm Wwb 1816 Set Up The workenergy theorem says Wadi Kb Kg Va Vb Point at is the starting and point b is the ending point Since the eld is uniform Wadi Fscos E q scosdi The eld is to the le so the force on the positive charge is to the le The particle moves to the le so 1 0 and the work Wadi is positive Solve 2 WM Kb 7 K 150 gtlt10 617 0 150 gtlt10 61 WM 7 150 gtlt10 61 b V i V 7 357 V Pointais athi her otential than oint b 0 a b q 420 gtlt10 9c g p p W11 V7Vb 357v CE sW soE 595gtlt103Vm 0 M M lqls s 600 gtlt10 Zm 1817 Set Up From Example 184 with Vb 0 and x rather than y as the distance from the negative plate VX Ex V 50 v Solve E x 0200 m Re ect VX Ex says the potential increases linearly with x and the graph of VX versus x should be a straight line in agreement with the gure in the problem 25 vm k 1818 Set Up For apoint charge V q r Vr 7 480 v 0250 m Solve a 7 133 gtlt10 9C 0 1 k 899 X109NmZCZ bV k tt V V V V 48025390cm 160v r q consan so 1r1 Zr2 Z 1 r2 750 cm kq 1819 Set Up For apo1nt charge V r kq 899 gtlt 109NmZCZ250 gtlt 10 11 c 3 Solve ar 250 gtlt 10 m 250mrn V 900v bV k t t V V V1 250 90390V 750 r COHSZHSO r rrr mm mm 1 1 1 2 2 2 1 V2 300V Electric Potential and Capacitance 185 1820 Set Up For apoint charge V k1 rZ 3r1 r Solve V1 k1 VZ k1 k1 V3 r1 r2 3r1 1821 Set Up From Chapter 17 we know that for a spherical shell the electric eld outside the shell is the same as for a point charge located at the center of the shell The electric eld determines the force on a test charge placed outside the shell and the work done on the test charge as it moves between two points determines the potential difference between those points Therefore outside the shell the potential is the same as for apoint charge 899 X109NmZCZ350 gtlt10 9c 7 0200 m 0240 m 7 0440 m 0240 m Re ect The potential is positive since the charge is positive and the potential increases at points closer to the sur face of the sphere Solve a V k1 715 V r b Nowr 0240 In so V 715 v 131 v k 1822 Set Up For a single point charge V q The total potential is the sum of the potentials due to the two point charges WBHA qV5 VA r k k 899 gtlt109NmZ cZ Solve a VA 1 7 240 gtlt10 9c 7650 gtlt10 9c 7737v rm r 0050m k 240 gtlt 10 9 c 7650 gtlt 10 9c 899 X109NmZCZ r25 0080m 0060m c WSW qV5 7 VA 250 gtlt10 9c7704 v 7 7737 v 82 gtlt10 8J k b V5 i i704 V V15 1823 Set Up U Conservation ofenergy says Ua Ka Ub Kb K mvz 120 gtlt 10 6c 460 gtlt10 6 c Solve a U 899 X109NmZCZ 0198J m b i Ua 0198 LKa 0 rb 0500m so Ub Ua 0099 7 7 7 7 210 7 20099J 7 Kb 7 Uai Ka Ub 70198 0099J 70099112137 7 W7266ms m g 11m 500mso Ub 00099JKb 0198J 7 00099J 0188 J vb 367m s rb 500 In so Ub 000099 J Kb 019817 000099 0197 J vb 375 ms Re ect As the charge q moves away from Q the repulsive force does positive work on q and its kinetic energy increases 1824 Set Up Leta be when they are 0750 mm apart and b when they are very far apart A proton has charge 9 and mass 167 X 10 27 kg As they move apart the protons have equal kinetic energies and speeds Solve a They have maximum speed when they are far apart and all their initial electrical potential energy has been converted to kinetic energy Ka Ua Kb Ub Ka 0 and Ub 0 so 160 gtlt10 19 C2 0750 gtlt 10 9 m K 1 Z 1 2 K 2 d x b x 33907 X10719 136 gtlt104 mv mv so my an U 7 m s b 2 b 2 b b b b m 167 X 10 27 kg b Their acceleration is largest when the force between them is largest and this occurs at r 0750 nm when they are closest 2 Kb Ua ke 899 X109NmZCZ 307 gtlt10 19J rd 160 X10719C Z 2 F ke Z 899 X109NmZCZ 9 409 gtlt10 1 N r 0750 gtlt 10 In F 409 gtlt 10 10N 1 Z a m245X 10 ms m 167 gtlt 10 kg 186 Chapter I 8 1825 Set Up Apply Ka Ua Kb Ub with Ka 0 and Ub U Solve 2 Kb Ua Ub qVa Vb eVa Vb An electron gains energy when it moves to higher potential so Vb a Vande eV Kb mUZ so 21ltb 2eV U m m 2 160 gtlt10 19c 95 v 11 1 3Igtgt 58 gtlt106ms 911 gtlt 10 kg Re ect A positive charge gains kinetic energy when it moves from high potential to low potential but a negative charge gains kinetic energy when it moves from low potential to high potential 1826 Set Up From Problem 1825 U 2eVm for a particle with charge of magnitude 9 The speed of light is c 300 X 108 ms An electronhas mass 911 X 10 31 kg and aproton has mass 167 X 10 27 kg 2 31 6 Z Solve a u 0010c 300 gtlt 106ms V E W 26 v 29 2160 gtlt 10 19c b From Problem 1825 the kinetic energy the particle gains is K eV so V 26 V gives electrons and protons the same kinetic energy But the protons must be accelerated by a potential decrease whereas the electrons are acceler ated by a potential increase 2 V 2 160 gtlt 10 19c 26V cU 1 IL 1 Zgtgt 70 gtlt104ms 0024c m 167 gtlt 10 kg 1827 Set Up K U Kb Ub and U qVK my Solve aKa 0 soKb Ua Ub qVa Vb E M m m U1 1 2 2 x q which is constant Let Va Vb V the accelerating voltage Then VZ 2V1 m Vivi V and U1 U Thus U ZU1 2U V1 U12 U22 U22 b uZ 3UsoVZ V1 2 9V1 900v V1 Vz U1 Re ect If the charge is positive it is accelerated by a decrease in potential If it is negative it is accelerated by an increase in potential 1828 Set Up Example 184 shows that V Ey where y is the distance from the negative plate V 0 at the AV negative plate Fory 60 cm V 120 V The potential gradient is 120 v V 60 v y Solve a E 20 Vcm y 30 cm39 V 60 V when y 30 cm The 60 V 60 cm E 20 Vcm equipotential surface is a at sheet parallel to the plates and midway between them V 20 V b y 10 cm The 20 V equipotential surface is a at sheet parallel to the plates and 10 cm E 20 Vcm from the negative plate and 50 cm from the positive plate AV 120v 200W C 7 m Ay 60gtlt10 Zm Electric Potential and Capacitance 187 1829 Set Up Example 184 shows that V Ey where y is the distance from the negative plate V 0 at the negative plate Fory 25 cm V 500 V 500 V Solve aE 25 cm 20 Vcm So V 20 Vcmy V 100V for y 5cm V 200V for y 10cm V 300V for y 15 cm V 400V for y 20 cm and V 500 V fory 25 cm The equipotential surfaces are drawn in Figure 1829 Positive Plate m Vo500V 77777777777777 W V0400V 77777777777777 W Vo300V 77777777777777 W Vo200V 77777777777777 W V0100V negative plate 3 V0 Figure 1829 b Yes they are separated by 5 cm c The equipotential surfaces are at sheets parallel to the plates Re ect The electric eld lines are straight lines perpendicular to the plates so are perpendicular to the equipoten tial surfaces as they must be The electric eld is uniform so the equipotential lines of constant potential difference are equally spaced W 1830 Set Up For apomt charge V V kq 899 gtlt109 N mZCZ 500 gtlt10 12 c 450 v cm Solve ar 7 V V V 100V atr 450 cm39 V 200V atr 225 cm39 V 300V atr 150cm39 V 400V atr 112cm and V 500 V at r 090 cm The equipotential surfaces are sketched in Figure 1830 Figure 1830 b These equipotential surfaces are not equally spaced in distance The V 100 V and V 200 V surfaces are farther apart than the V 400 V and V 500 V surfaces The equipotential surfaces are closer together whereE is larger c The equipotential surfaces are concentric spheres with the point charge at their common center 188 Chapter I 8 k 1831 Set Up For a point charge V q r Solve a and c The V 0 equipotential is the at surface consisting of points equidistant from the two charges The projection ofthis surface with a drawing in which the point charges are in the plane of the paper is a line along the perpendicular bisector of the line that connects the two charges as shown in Figure 1831 200 cm 200 cm Dena 100 uC 1 100 pC 1 1 1 1 1 1 1 1 1 Figure 1831 b Since the point charges have equal magnitudes and opposite signs V 0 at the point midway between them since this point is equidistant from the two charges 1832 Set Up The electric eld lines are perpendicular to the equipotential surfaces The electric eld lines are closer together where E is larger and where E is larger the equipotential surfaces for equal potential differences are closer together The projection of the equipotential surfaces into the plane of the gure are lines that satisfy these two requirements Solve For the electric eld given in the problem it is not possible to draw equipotential lines that satisfy both requirements The electric eld shown in the problem cannot be produced by an electrostatic distribution of charges 1833 Set Up and Solve The electric eld lines and intersections of the equipotential surfaces with the plane of the drawing are shown in Figure 18 1 lb in the textbook Re ect The gure shows that the electric eld lines are always perpendicular to the equipotential surfaces Also the electric eld lines are closer together whereE is larger and where E is larger the equipotential surfaces for equal potential differences are closer together 1834 Set Up The mass of a drop is m pV where V arr3 the volume of a sphere p 0820 gcm3 820 kgm3 The drops have r 500 X 10 7 m The electric eld and electric force magnitudes are related by F lq E where q is the net charge ofthe drop Solve aFw so lqlE mg andE lql Se 800 gtlt10 19c 1 m pV p m3 820 kgm3ar500 gtlt10 7 m3 429 gtlt10 16kg 7 429 gtlt 10 16kg980 mszgt E 526 gtlt 103 vm 800 gtlt10 19c bVEd 526 gtlt103vm225 gtlt 10 Zm 118V V 738V 429 gtlt 10 16k 980m sZ cE Z 328 gtlt103vmlql g M 128 gtlt10 18c d 225 gtlt 10 m E 328 gtlt 103vm lql 7128 gtlt 10 18c 7 8 The number of electrons isN 19 i e 160 X 10 C 1835 Set Up leV 160 gtlt 10 19J K 51 c 300 gtlt 108 ms An electron has mass 911 gtlt 10 31 kg and aproton has mass 167 X 10 27 kg Electric Potential and Capacitance 189 7 7 719 7 Solve aK 7100eV 7160 X 10 J U 7 m 2160 gtlt 107191 5 electron ve m 593 X 10 rns 911 X 10 kg 2 160 gtlt10 19J v quot1 proton Up 727 138 gtlt104ms e p V1836 167 X 10 kg Up me bK 100kev 160 gtlt 104 electron ve 187 X 107 ms proton Up 438 X 105 ms c U up 00100c 300 gtlt106ms Ke may 911 gtlt 10 31kg300 gtlt 106ms Z 410 gtlt10 18J 00256keV Kp mpup2 H167 gtlt 107271ltg300 gtlt106msZ 752 gtlt 10 15J 470 keV The proton energy is larger by a factor of mpm Re ect When we useK mvZ we should express all quantities in SI units 6 1836 Set Up C g V ab 125 gtlt10 6C 7 Solve aC 111 gtlt 10 F 0111 1F 113v b Q CVab 728 gtlt10 6F250V 182 gtlt10quot C 18211C 1837 SetUp 1F 1CZNmand1N 1kgmsz 1CZ 1N CZsZ Solve 1F Z Nm 1kgmsZ kg m 1838 Set Up For aparallelplate capacitor Val EdE iandC 50A Vab Solve a Val Ed 400 gtlt 106Vmgt250 gtlt 10 3m 100 gtlt 104v Q 800 gtlt 10 9C bA 6 712 2 2 E50 400 gtlt 10 vm8854 gtlt 10 C Nm Q 800 gtlt 10 9C c C 4 Val 100 gtlt 10 v 226 X 10 3 In2 226 cm2 800 gtlt 10 12F 800 pF 5 A 1839 Set Up For a parallelplate capacitor C C Val Ed The surface charge densityis 039 ab 51 V Q 0200gtlt10 6C 0 V9 a ab C 5000 gtlt 10 12 F Cd 5000 gtlt 10 F0600 gtlt 10 3m b A 6 8854 X10 12CZNmz 400V 00339 In2 E V 400V 667x105v c m d 0600 X 10 3 In Q 0200 gtlt 10 6C A 00339 In Re ect If the plates are square each side is about 18 cm in length We could also calculate 039 fromE TEo 590 gtlt10 6CmZ A 1840 Set Up C go when there is air between the plates 7 8854 gtlt 10 Fm30 gtlt 10 ZmZ Solve C 3 159 gtlt 10 12F 159 pF 50 gtlt 10 In 6 1841 Set Up C g C V01 1 Solve aQ CV01 100 gtlt10 6F120 v 120 gtlt10quot C 12011C b When dis doubled Cis halved so Q is halved Q 60 MC 1810 Chapter 18 c If r is doubled A increases by a factor of 4 C increases by a factor of 4 and Q increases by a factor of 4 Q 480 uC Re ect When the plates are moved apart less charge on the plates is required to produce the same potential differ ence With the separation of the plates constant the electric eld must remain constant to produce the same potential difference The electric eld depends on the surface charge density 039 To produce the same 039 more charge is required when the area increases 1842 Set Up C C Val d Solve a 120 V b i When I is doubled C is halved Val g and Q is constant so Vdoubles V 240 V ii When r is doubledA increases by a factor of4 Vdecreases by a factor of4 and V 30 V A 1843 Set Up C 50 We measure that the diameter ofa quarter is 245 mm so r 122 X 10 3 m C g d V 8854 gtlt10 Fmar122 gtlt10 3mZ 7 Solve a C 73 207 X 10 12F 207 pF 200 gtlt 10 m bQ CV 207 gtlt10 12 F12 v 248 gtlt 10 11 C 248pC 6 1844 Set Up C 2101 Ed C Val 1 Val 7 100 gtlt102v E 100 gtlt104NC Solve a d 100 X lO zm 100 cm 7 Cd 7 500 gtlt10 12 F100 gtlt10 Zm A 50 8854 gtlt 10 12CZNmz 565 X 10 3m2 A A arrzsor 424 gtlt10 Zm 424cm bQ CV01 500 gtlt10 F100 gtlt102v 500 gtlt10 1 C 500pC A e 7172 1845 Set Up C 60 0 d d 2 2 Z 2 goal re eo 39r 507139 3m soar r0 Solve C0 71 3r039d 3d0CT3 do 3 3CD Re ect C for a capacitor depends only on the dimensions of the capacitor A 1846 Set Up Since the capacitor is disconnected from the battery Q does not change C 50 C A A Solve a C 60E Cnew 60E 60E 1 Q Q0 0V0 gqnew Q0 g 2 2V0 C Cnew C2 C 1847 Set Up The capacitors between b and c are in parallel This combination is in series with the 15 pF capacitor C1 15 pF CZ 90 pF and C3 11 pF Solve a For capacitors in parallel Ceq C1 C2 so CZ3 C2 C3 20 pF 1 l b C1 15 pF is in series with C23 20 pF For capacitors in series so Ceq 01 C2 1 l l C C 15 F 20 F andcm 123 P P gt8I6PFI C123 C1 C23 C1 C23 l5 pF 20 pF Re ect For capacitors in parallel the equivalent capacitance is larger than any of the individual capacitors For capacitors in series the equivalent capacitance is smaller than any of the individual capacitors Electric Potential and Capacitance 18 11 1848 Set Up For capacitors in series the voltage across the combination equals the sum of the voltages in the individual capacitors For capacitors in parallel the voltage across the combination is the same as the voltage across each individual capacitor Solve a Connect the capacitors in series so their voltages will add b V V1 VZ V3 NVI where N is the number of capacitors in the series combination since the capacitors are identical V 500 V N 5000 V1 010 v l l l 1849 Set Up For capac1tors in series F 7 E For capac1tors mparallel Ceq C1 C2 r Z C eq 1 Solve The possible combinations and their equivalent capacitance are shown in Figure 1849 jg AHHF 33 iF 20 iF ii 57 F 15F 30 iF Figure 1849 l l l 1850 Set Up For capac1tors in series C F F ForN equivalent capac1tors in series Ceq CN eq 1 2 For capacitors in parallel Ceq C1 C2 ForN equivalent capacitors in parallel Ceq NC Solve There are many ways to achieve the required equivalent capacitance In each case one simple solution is shown in Figure 1850 t w a HHHF 39339 b EHHHJ Figure 1850 1812 Chapter 18 1851 Set Up In parallel the potential difference is the same for each capacitor and is equal to the applied potential difference C Val Solve a and b The potential difference across each capacitor is 480 V Q CVab For the 400 4F capacitor Q 192 LC For the 600 4F capacitor Q 288 LC 1852 Set Up In series the charges on the capacitors are the same and the sum of the potential differences across the capacitors is the applied potential difference C V21 Let C1 400 4F and C2 600 MF Q1 QZ Q a Solve a V1 VZ 480 v 2 g 480 v C1 C2 0102 400 gtlt 10 6ur600 gtlt 10 6ur 7 Q 480 v f 480V 115 gtlt 10 4c C1 C2 400 X 10 LF 600 X 10 LF The charge on each capacitor is 115 LC Q 115 gtlt10quot c Q 115 gtlt10quot c bV1 76288VVZ 76 192v Cl 400 X 10 F CZ 600 X 10 F The potential difference across the 400 4F capacitor is 288 V and the potential difference across the 600 4F capac itor is 192 V 1853 Set Up C For two capacitors in parallel Ceq C1 C2 For two capacitors in series 1 1 1 0102 andCe Ceq c1 cZ 1 C1 02 For capacitors in parallel the voltages are the same and the charges add For capacitors in series the charges are the same and the voltages add Let C1 300 MF CZ 500 4F and C3 600 MF C12 IO l uni C b O l 1 b J a C3 1 Figure 1853 Solve a The equivalent capacitance for C1 and C2 in parallel is C12 C1 C2 800 MF This gives the circuit shown in Figure 1853a In that circuit the equivalent capacitance is 7 01203 7 800 ur600 Mr 7 C12 C3 7 8001F 600 uF This gives the circuit shown in Figure 1853b In Figure 1853b Q CV 343 gtlt 10 6 F240 v 823 X 10 5 C In Figure 1853a each capacitor therefore has charge 823 X 10 5 C The potential differences are 7 823 gtlt 10 5c 12 823 gtlt 10 5c 6137Vauani 6103v C3 600gtlt10 F C12 800gtlt10 F 343 ur V3 Note that V3 V12 240 V Then in the original circuit V1 VZ V12 103 V Q1 VICI 103 v300 gtlt10 6F 309 gtlt10 5 c QZ VZCZ 103 v500 x10 6F 515 gtlt10 5 c Q1 309 MC QZ 515 MC and Q3 823 uc Note that Q1 QZ Q3 b V1 103 v VZ 103 v and V3 137 v Re ect Note that Q1 QZ Q3 V1 V2 and V1 V3 240 V Electric Potential and Capacitance 1813 1854 Set Up C For two capacitors in parallel Ceq C1 C2 For two capacitors in series V 1 1 1 CICZ arid Ceq C C1 C2 9 1 7C1CZ39 For capacitors in parallel the voltages are the same and the charges add For capacitors in series the charges are the same and the voltages add Label the capacitors as shown in Figure 1854a C1 C2 C12 H C3 C3 b b o l 4 C4 a 1 C123 Ceq b b C4 C d Figure 1854 Solve The equivalent capacitar1ce for C1 and C2 in series is CICZ C C12 7 C1 C2 2 This gives the circuit shown in Figure 1854b The equivalent capacitance for C12 and C3 in parallel is 3C C123 C12 C3 This gives the circuit shown in Figure 1854c The equivalent capacitar1ce for C123 and C4 in series is 012304 3CZgtCgt 9 1 C123 C4 3C2 C This gives the circuit shown in Figure 1854d In Figure 1854d Q Cquab 240 gtlt 10 6 F 280 v 672 gtlt 10 5 C This is the charge on each capacitor in Figure 1854c Q4 672 4C and 7 Q4 672 uC 672 uC 168VV C4 400 hF 3 32400 uF 0600C 240 gtlt 10 6F V4 112 v Note that V4 V123 Vab Ther1ir1 Figure 1854h V3 V12 112 v Q3 Cng 400 uF112 v 448 11C Q12 CIZV12 7400 uF112 v 224 11C Q1 QZ 224 11C 224 C 224 C 56vVZ I39L CI 400 hF CZ 400 hF In summaryQ1 224 hc QZ 224 hc Q3 448 hc Q4 672 hc V1 56 v VZ 56 v V3 112 v and V4 168 V VI 56V 1855 Set Up For capacitors in series the voltages add and the charges are the same L For C 1 eq C2 sl 9 capacitors in parallel the voltages are the same and the charges add Ceq C1 C2 i C 1814 Chapter 18 Solve a The equivalent capacitance of the 50 2F and 80 2F capacitors in parallel is 130 2F When these two capacitors are replaced by their equivalent we get the network sketched in Figure 1855 The equivalent capacitance ofthese three capacitors in series is 347 2F b QLot CWV 347 uF 500 v 174 MC c QLot is the same as Q for each ofthe capacitors in the series combination shown in Figure 1855 so Q for each of the capacitors is 174 11C 100 21 90 LF n l 1 1 H4 130 LF Figure 1855 Qtot C13 193 V V10 V13 V9 174 V 134 V 193 V 501 V The sum ofthe voltages equals the Re ect The voltages across each capacitor in Figure 1855 are V10 174 V V13 134 V and 10 Qtot 09 applied voltage apart from a small difference due to rounding V9 1 1856 Set Up For capacitors in series the voltages add and the charges are the same C C C eq 1 2 capacitors in parallel the voltages are the same and the charges add Ceq C1 C2 C Solve a The equivalent capacitance ofthe 180 nF 300 HF and 100 nF capacitors in series is 529 nF When these capacitors are replaced by their equivalent we get the network sketched in Figure 1856 The equivalent capacitance of these three capacitors in parallel is 193 HF and this is the equivalent capacitance of the original network 75 nF 529 nF b H 65 nF Figure 1856 b QLot 0qu 193 HF 25 v 482 nC c The potential across each capacitor in parallel network of Figure 1856 is 25 V Q65 C65V65 65 nF25 v162nc d 25 v 1857 Set Up The energy stored in a charged capacitor is CVZ Solve CVZ H450 gtlt10 6F295 vZ 196J V 400 V 1858 Set Up The energy density is u 0192 Val EdE id m 800 gtlt 104vm 0 gtlt u 0192 8854 gtlt 10 1ZcZNmz800 gtlt 104 vmZ 00283 Jm3 1859 Set Up C U CVZ Solve a Q CV 50 MF15 v 75 11CU CVZ 50LF15 vZ 562 uJ b U CVZ CQCZ QZZC Q VZCU 250 gtlt10 6F101 32 gtlt10 3 c 32gtlt10 3c Vg 640 v C 50 gtlt 10 6F Electric Potential and Capacitance 18 15 1860 SetUp C g V Q Q Q2 Solve aVEsoUQV Q E E b Q CVso U CVV CVZ 7 Q 39 1 7 Q2 7 1 2 1861 Set Up C 7 The stored energy 1s gQVi E 7 ECV Q2 333 gtlt10 9CZ in Solve a Energy E m 853 X 10 J 0853 p V7 Q 7 333 gtlt10 9C C 650 gtlt 10 6F 512 X10 4V 0512mV b Energy CVZ 127 gtlt10 9F185 v 2 217 gtlt10 6J 217 m Q CV 127 gtlt10 9F185 v 235 gtlt10 7 C 0235 11C Re ect There are three equivalent expressions for the stored energy and we can choose the one most appropriate to the information we are given 2 e A 1862 Set Up The energy stored in a capacitoris The charge ofan electron is 160 X 10 19 C C Solve aQ 2C250J 2500 gtlt 10 9F250J 500 gtlt 10 4 C The number ofelectrons with 500 X 10 4 C this ma nitude ofch eis 312 X 1015 electrons g erg 160 gtlt 10 19 Celectron b To double the stored energy halve the capacitance To do this either double the plate separation or halve the plate area 1863 Set Up The two capacitors are in series For capacitors in series the voltages add and the charges are the 1 1 1 same Cg UCVZ Ceq C1 C2 V 1 1 1 CICZ 150 nF120 nF Solve a so Ceq 667 nF Ceq C1 C2 C1 CZ 150nF 120 nF Q CV 667nF36v 24 gtlt10 6C 2411C b Q 24 4C for each capacitor c U CquZ 667 gtlt10 9 F 36 v2 432 d We know C and Q for each capacitor so rewrite Uin terms ofthese quantities U ECVZ 20QCgtZ QZZC 24 gtlt10 6CZ 24 gtlt10 6CZ 150nF U 9 192 m 120 HP U 9 2401 2150 gtlt 10 F 2120 gtlt 10 F Note that 192 M 240 MJ 432 11 the total stored energy calculated in part c Q 24 gtlt10 6C Q 24 gtlt10 6C 150nFV 7916V120nFV 20v C 150gtlt10 F C 120gtlt10 F Note that these two voltages sum to 36 V the voltage applied across the network Re ect Since Q is the same the capacitor with smaller C stores more energy U Q22C and has alarger voltage V Qcgt 1864 Set Up For capacitors in parallel the voltages are the same and the charges add Ceq C1 C2 C U CVZ Solve aCeq C1 CZ 35 nF 75 HF 110 HP QLot Cqu 110 gtlt10 9F220 v 242 uC b V 220 V for each capacitor 35 n17Q35 C35V 35 gtlt10 9F220 v 77 MC 75 nF Q75 C75V 75 gtlt 10 9 F220 v 165 11C 1816 Chapter 18 NOte that Q35 Q75 Qtot c ULot CquZ 110 gtlt10 9F220 vZ 266 mJ d 35 nF U35 c VZ H35 gtlt 10 9 F 220 v2 085 mJ 75 nF U75 C75VZ H75 x10 9F220 vZ 181 mJ Since Vis the same the capacitor with larger C stores more energy e 220 V for each capacitor 1865 Set Up C Let C1 200 4F and C2 100 MF The energy stored in a capacitor is ab Q2 QVab CVabZ i7 ab Solve a The initial charge on the 200 4F capacitor is Q C1800V 200 gtlt 10 6F800V 00160 c b In the nal circuit charge Q is distributed between the two capacitors and Q1 QZ Q The nal circuit con tains only the two capacitors so the voltage across each is the same V1 V2 Q Q1 Q2 C1 V soVV 1ves Q2 C 1 2 C1 CZ Q1 CZ 2 Q2 Using this in Q1 QZ 00160 c gives 3QZ 00160 c and Q2 533 gtlt1073 c Q 2QZ 1066 gtlt 1072c 1066 gtlt 1072c 533 gtlt 1073c 7 7 533v CI 200 gtlt 10 6F CZ 100 gtlt 10 6F Vi The potential differences across the capacitors are the same as they should be c Energy C1VZ 0sz C1 CZV2 200 gtlt 10 6F 100 gtlt 10 6F533 v2 426 J 1 The 200 4F capacitor initially has energy C1VZ 200 X 10 6 F 800 V2 640 J The decrease in stored energy that occurs when the capacitors are connected is 640 J i 426 J 214 J Re ect The decrease in stored energy is because of conversion of electrical energy to other forms during the motion of the charge when it becomes distributed between the two capacitors Thermal energy is generated by the current in the wires and energy is emitted in electromagnetic waves 1 l l 1866 Set Up For capacitors in series the voltages add and the charges are the same C E F For eq 1 2 Q capacitors in parallel the voltages are the same and the charges add Ceq C1 C2 t i C U CVZ Solve 2 Find Ceq for the network by replacing each series or parallel combination by its equivalent The succes sive simpli ed circuits are shown in Figure l866ac Um CquZ 219 gtlt10 6F120vZ 158 gtlt10quot J158MJ 406 iF a b 860 5 480 iF 350 ir a 860 iF 756 LF 219 5 1141 48101le a nb b C Figure 1866 Electric Potential and Capacitance 18 17 b From Figure 1866c QLot Cqu 219 gtlt 10 6F120 v 263 gtlt 10 5 C From Figure 1866b 263 gtlt 10 5 C 548 v C48 480 gtlt 10 6F U48 CVZ 480 gtlt10 6 F548 VZ 721 gtlt 10 5J 721 uJ QLot 263 gtlt 10 5 C V48 This one capacitor stores nearly half the total stored energy A 1867 Set Up U CVZ C For aparallelplate capacitor C 50 Solve a U CVZ CQCZ QZZC Uis proportional to Q2 so increase the charge by a factor ofE A to double the stored energy We didn t use C 60 so this result applies to any capacitor 5 A b Still connected to the battery means Vstays constant and Q changes U CVZ d gt dZ means U 2U0 c U CVZ V 25 Vgoes to V 75 v Vgt3Vso Ugt 9U 90 J Re ect In parts b and c the change causes the charge on the capacitor to increase and this increases the electric eld between the plates and therefore increases the stored energy 1868 Set Up The eld when there is air between the plates isE0 The eld when the dielectric is between E E the plates isE 0 K 27 0180 gtlt10 6C 50A 8854 gtlt 10 12 CZN m2 100 gtlt 10 4 m2 7 E0 203 gtlt 106 vm 615 E 0330 gtlt 106Vm Solve E0 203 gtlt 106Vm K A 1869 Set Up C KC0 Keog A 10 cm2 10 X 10 4 m2 V Ed for a parallel plate capacitor this equation applies whether or not a dielectric is present 885 gtlt10712 Fm 10 gtlt1074 m2 118 F ercmZ 75 X10 9m P Solve a C 10 V 85 mV 6 bE 7 9 113 gtlt 10 Vm Kd 10 75 gtlt 10 m Re ect The dielectric material increases the capacitance and decreases the electric eld that corresponds to a given potential difference KeOA 1870 Set Up The capacitance with the dielectric present is C KC0 Val Ed The dielectric strength is the maximum allowed eld between the plates V 400 gtlt 103 v Solve d i E 200 gtlt 106Vm Cd 7 150 gtlt10 9F200 x10 4m A 7 K60 3208854 gtlt 10 12CZNm2 200 X 10 4 m 00106 m2 1871 Set Up C KCO U CVZ Solve aWith the dielectric C 375125 uF 469 uF before U COVZ 125 gtlt 10 6 F 240 V2 360 m a er U CVZ 469 gtlt10 6F240 VZ 135 m b AU 135 m i 36 mJ 99 m The energy increased Re ect The power supply must put additional charge on the plates to maintain the same potential difference when the dielectric is inserted U QV so the stored energy increases 1818 Chapter 18 Ks 1872 Set Up The capacitance with the dielectric presentis C KC0 0A is the maximum allowed eld between the plates Cd 0200 gtlt 10 6 F00800 gtlt 10 3 m Solve aA K50 2508854 gtlt 10 12cZNmm b Vab 500 gtlt106Vm00800 gtlt10 3m 200 gtlt103 v Vab Ed The dielectric strength 0723 m2 30 1873 Set Up E K E0 320 gtlt 105 vm Solve K 128 E 250 gtlt 105 vm k 1874 Set Up The potential ofa point charge is V q r y q Figure 1874 Solve a The positions ofthe two charges are shown in Figure 1874 b Any point on the x axis is the same distance from each charge so k k 7 V q i m 0 V V The potential is zero at all points on the x axis k k 1875 Set Up For a point charge E i qzi and V q The electric eld is directed toward a negative charge V and away from a positive charge r V kqr kq r2 498 v Solve aVgt0soqgt0 r 0415m E qulrZ r kq 120 vm V 0415 m 498 V bq r 230 X 104 C k 899 gtlt109 N mZCZ c q gt 0 so the electric eld is directed away from the charge Re ect The ratio of Vto E due to a point charge increases as the distance r from the charge increases because E falls off as lrZ and Vfalls offas lr 1876 SetUp U k Ua Ka Ub Kb The charge ofthe alpha particle is 29 andthe charge ofthe radon nucleusis 869 r Solve a The nal energy of the alpha particle 479 MeV equals the electrical potential energy of the alpharadon combinationjust before the decay U 479 MeV 766 X l 13 J qu 7 899 gtlt109 N mZCZ 2 86160 gtlt10 19 C2 U 7 766 gtlt 10 13J by 517 gtlt 10 14m Electric Potential and Capacitance 18 19 1877 Set Up With air between the layers E0 E and V0 Bod The energy density in the electric eld 6 60 is u 019 Z The volume of a shell ofthickness t and average radius R is 477th The volume of a solid sphere of E V radius R is 7rR3 With the dielectric present E K0 and V a 050 gtlt 10 3 CmZ Solve aE0 56 gtlt107 Vm 50 8854 gtlt 10 c Nm b V0 Bod 56 X 107 Vm 50 X 10 9 m 028 V The outer wall ofthe cell is at higher potential since it has positive charge c For the cell Vceu 4371393 and 3V 13 3 1016m3 13 R e g 29gtlt10 6m 477 477 The volume ofthe cell wall is Vmu 47139th 47729 gtlt 10 6m250 gtlt 10 9m 53 X 10719 m3 uo 0302 8854 gtlt10 12 CZNmz56 gtlt107 Vm Z 139 gtlt10 vm3 The total electric eld in the cell wall is 139 X 104 Vm353 gtlt 10 19m3 7 X 10 15 V E0 56 gtlt 107vm 7 V0 028v 11 10 gtlt 10 VmandV K 54 K 0052 V 1878 Set Up Let a be the initial situation where the alpha particle is very far from the gold nucleus and has kinetic energy K 100 MeV 160 X 10 12 J At a the gold nucleus has zero kinetic energy Let b be at the distance of closest approach when the distance between the two particles is rb Conservation of energy says K U Kb Ub U km The alpha particle has charge q1 29 and the gold nucleus has charge qZ 79e r Solve Ka 160 X 10 12 J and Kb 0 since at the distance of closest approach the alpha particle has momentar ily come to rest Ua 0 since rd is very large 7 k29gt79egt Vb Ub Conservation of energy gives Kg Ub 279gtezk 279160 gtlt10 19CZ899 gtlt109NmZCZ Vb K 160 gtlt 10 12J 23927 X 107 a 1879 Set Up Use conservation of energy U K Ub Kb to nd the distance of closest approach rb Kb 0 Initially the two protons are far apart so Ua 0 The maximum force is at the distance of closest approach 1 11qu Vbz qqu 32 Solve Ka Ub 2mvaz k mudZ k and Vb Vb Fk A proton has mass 167 X 10 27 kg and charge q 9 160 gtlt10719 C 7 kel 7 899 gtlt109NmZCZ160 gtlt10 19cZ 138 gtlt10 13 Vb my 167 gtlt10 Z71ltg100 gtlt106msZ m 2 160 gtlt10 19c 2 F ke Z 899 gtlt 109N 1r12CZ BgtZ 0012 N rb 138 gtlt 10 c Re ect The acceleration a F m of each proton produced by this force is extremely large 1880 Set Up The proton has charge qp 9 and mass mp 167 X 10 27 kg The alpha particle has charge qa 49 and mass ma 4mp 668 X 10 27 kg We can apply both conservation of energy and conservation of F linear momentum to the system a where F m r 1820 Chapter 18 Solve acceleration The maximum force and hence the maximum acceleration occurs just after they are released whenr 0225 nm 2160 gtlt10 19cZ F 899 gtlt 109NmZCZ 07225 X 10971 909 gtlt 10 9N F 909 gtlt10 9N 8 2 F 909 gtlt10 9N 8 Z aP m544X 10 msaa m 136gtlt 10 ms mp 167 X 10 kg mm 668 X 10 kg The acceleration of the proton is larger by a factor ofmamp speed Conservation ofenergy says U K Uf Kf K 0 and Uf 0 so Kf U 2160 gtlt10 19 C2 9 205 X10481 0225 gtlt 10 m U k 899 X109NmZCZ P so the total kinetic energy ofthe two particles when they are far apart is K 205 X 10 18 J Conservation oflinear momentum says how this energy is divided between the proton and alpha particle p pf 0 mp1p may and U 7 vp mu m 2 m K mpvpZ mavaZ mpvpZ UPZ mpvpzl p ma ma 7 2Kf 7 2205 gtlt 105181 7 4 quot mpl mpmm 167 gtlt10 Z7kg1 43943 X10 mS m u m pup H443 gtlt104ms111gtlt104ms The maximum acceleration occurs just after they are released The maximum speed occurs after a long time A 2 7 5 7 Q 7 7 Q 1 7i 2 7 1Q 1881 Set Up C 7 C 7 Vab 7 Ed andE 7 The stored energy 1s ZQVHb 7 ZCVab 7 Z d Val 50A 0 g 8854 gtlt 10 12cZ NmZ 0200m Z Solve a C 7 443 gtlt 10 11 F d 0800 gtlt 10 m bQ CV01 443 gtlt 10 11 F120 v 532 gtlt10 9c V 120 v cE 7391 7 150 gtlt104vm d 7 0800 gtlt10 Zm d Energy QV 532 x10 9c120 v 319 gtlt10 7J e Since the battery is disconnected the charge Q on the capacitor stays constant 6 60 so C 443 gtlt10 F 222 gtlt 10 11 F b The charge can t change so Q 532 X 10 9 C cE Since Q doesn t changeE doesn t change andE 150 X 104 Vm 6 Q2 d E nergy ZC Q doesn t change and C changes by a factor ofl so the stored energy doubles and becomes 638 X 10 7 J Re ect Since the stored energy increases work must be done by the force that pulls the plates apart 1882 Set Up Since the battery remains connected Vremains constant and Q changes 50A 7 8854 gtlt10 12 CZN m2 0200 m2 Solve a C 72 221 X 10 11 F d 160 gtlt 10 m bQ CV01 221 gtlt 10 11 F120v 265 gtlt10 9c V01 120 v 3 cE Z 750 gtlt 10 vm d 160 gtlt 10 m 1 Energy 50W H221 gtlt 10 11 F120 vZ 159 gtlt 10571 Electric Potential and Capacitance 1821 KeoA 1883 Set Up C T Q lt22 0 7 V7 Energy 7 QVb 7 mg 7 s 8854 gtlt 10 12cZNmz160 gtlt10quot m2 Solve aC 500 2 354 gtlt 10 H F 0200 gtlt 10 m bQ CVab 354 gtlt 10 F300V 106 gtlt10 8c c Energy CVZ 354 gtlt10 F300VZ 159 gtlt10 6J 18 84 7 Q 7 60A 7 39 1 Set Up C 7 C 7 V01 7 Ed The stored energy is EQV ab 00180 gtlt 10 6c 7n Solve aC 900 X 10 F 900 pF 200 v 5 Cd 900 gtlt 10 F 150 gtlt10 3m bC soA 00152m2 d 50 8854gtlt10 CNm c VEd 30 gtlt106vm150 gtlt10 3m 45 gtlt103v d Energy QV 00180 gtlt10 6c200v 180 gtlt10 6 7180 1 For l 1885 Set Up For capacitors in series the equivalent resistance Ceq is given by Ceq C1 C2 capacitors in parallel the equivalent capacitance Ceq is given by Ceq C1 C2 C1 C1 01 01 me W c2 G2 23 pf 02 69 LF b o l j D b o l ll C1 C1 01 C1 a b HQ i g 01 69 LF 0 Figure 1885 Solve a Using the rules for combining capacitors in series and in parallel gives the sequence of equivalent net works shown in Figure 1885 The equivalent capacitance of the network is 23 MF b In Figure 1885d the three capacitors in series have the same capacitance so the voltage across each is 420 v3 140 v Q1 ClVl 69 uF 140 v 966 MC The voltage across C2 in Figure 1885cis 140 v so QZ CZVZ 46 MF140V 644 MC Re ect We could continue to analyze the networks in Figure 1885 and nd V and Q for each capacitor in the network 1822 Chapter 18 1886 Set Up C When the two capacitors are connected to each other V1 V2 and Q1 QZ ql 7 qZ where q1 and q2 are the charges on the two capacitors when they are connected to the line in parallel Let C1 400 11F and C2 600 uF Solve a In parallel the voltage across each capacitor equals the line voltage q1 01V 400 MF660 v 264 gtlt10 3 c and qz CZV 60011F660 v 396 gtlt10 3 c Q1 Q2 0 b Q1 QZ 396 gtlt10 3 c 7 264 gtlt10 3 c 132 gtlt10 3 011 VZ so 3 7 3 Q2 2Q1150Q1 1 Z 1 Q1 150Q1 132 gtlt 10 3Cand Q1 528 gtlt10quot CQZ 150Q1 792 gtlt10 4c Q 528 X 10 4C Q2 792 X 10 4C V1 7 132VVZ 7 132V Cl 400 X 10 F CZ 600 X 10 F 1 1 1 1887 Set Up For capac1tors 1n series the equivalent capac1tance Ceq 1s g1ven by 7 7 7 7 In series Ce 01 C2 03 the charge on each capacitor equals the charge on the equivalent capacitor The energy stored in the three capacitors is equal to the energy stored in the equivalent capacitor 1 1 1 84 11F 4 gives Ceq 21 X 10 6 F 1 Solve a Ceq 84 11F 84 11F 42 11F Q Cqu 21 gtlt 10 6F36 v 756 gtlt10 5 c 756 uc b The total stored energy isQV 756 gtlt 10 5 c36 v 136 gtlt 10 3 J Re ect The three capacitors each have the same Q so the 42 11F capacitor has twice the voltage across it than is across each of the 84 11F capacitors We could also nd the total stored energy by nding the energy stored in each of the three capacitors 1888 Set Up The capacitor is equivalent to two capacitors in parallel as shown in Figure 1888 Each of these two capacitors have plates that are 120 cm by 60 cm For a parallelplate capacitor with dielectric lling the volume A between the plates C K60 Figure 1888 A 8854 gtlt10 Fm0120m0060m Solve a C C1 C2 C2 60 7 d 450 gtlt 10 In C1KCZ 340142 gtlt10 F 483 gtlt 10 11 F C 01 CZ 625 gtlt 10 11 F 625 pF b U CVZ 625 gtlt10 F180VZ 101 gtlt10 8J cNowC1 CzandC 2142 gtlt10 F 284 gtlt 10 11 F U CV2 284 gtlt10 F180vZ 460 gtlt10 9J 142 X 10 11 F The plexiglass increases the capacitance and that increases the energy stored for the same voltage across the capacitor Electric Potential and Capacitance 1823 1889 Set Up For a single capacitor with dielectric of dielectric constantK completely lling the volume between K6 its plates C 0A The capacitor is sketched in Figure 1889a It is equivalent to two capacitors in series as shown 1 l in Figure 188 For two capacitors C1 and C2 in series the equivalent capacitance C is given by E E 1 Z K 11 I I T a b Figure 1889 1 1 1 c c K 2K K 2K A Solve acl LOAI 0AandCZL0AZ 0 c c1 c2 0ch dZ d dZ d 2K150A2K250A d d 7 250A 1le 2K150A 2K250A 7 1 K1 K2 1 1 250A K2 7 KeOA TE 7 d 250A 1 50A d H b For K1 K2 C gt which is the correct expression c ForK K 10 C gt which is the correct ex ression 1 2 2 d P Re ect Our result applies only when each dielectric lls half the space between the plates When K1 K2 the elec tric eld has a different value in each dielectric

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