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23 Electric Fields CHAPTER OUTLINE 23.1 Properties of Electric Charges 23.2 Charging Objects by Induction 23.3 Coulomb’s Law 23.4 The Electric Field 23.5 Electric Field of a Continuous Charge Distribution 23.6 Electric Field Lines 23.7 Motion of a Charged Particle in a Uniform Electric Field ANSWERS TO QUESTIONS Q23.1 A neutral atom is one that has no net charge. This means that it has the same number of electrons orbiting the nucleus as it has protons in the nucleus. A negatively charged atom has one or more excess electrons. *Q23.2 (i) Suppose the positive charge has the large value 1 m C. The object has lost some of its conduction electrons, in number 10 6 C (1 e/1.60 × 10 19 C) = 6.25 × 10 12 and in mass 6.25 × 10 12 (9.11 × 10 31 kg) = 5.69 × 10 18 kg. This is on the order of 10 14 times smaller than the ~1g mass of the coin, so it is an immeasurably small change. Answer (d). (ii) The coin gains extra electrons, gaining mass on the order of 10 14 times its original mass for the charge 1 m C. Answer (b). Q23.3 All of the constituents of air are nonpolar except for water. The polar water molecules in the air quite readily “steal” charge from a charged object, as any physics teacher trying to perform electrostatics demonstrations in the summer well knows. As a result—it is difficult to accumulate large amounts of excess charge on an object in a humid climate. During a North American winter, the cold, dry air allows accumulation of significant excess charge, giving the potential (pun intended) for a shocking (pun also intended) introduction to static electricity sparks. Q23.4 Similarities: A force of gravity is proportional to the product of the intrinsic properties (masses) of two particles, and inversely proportional to the square of the separation distance. An electrical force exhibits the same proportionalities, with charge as the intrinsic property. Differences: The electrical force can either attract or repel, while the gravitational force as described by Newton’s law can only attract. The electrical force between elementary particles is vastly stronger than the gravitational force. Q23.5 No. The balloon induces polarization of the molecules in the wall, so that a layer of positive charge exists near the balloon. This is just like the situation in Figure 23.4a, except that the signs of the charges are reversed. The attraction between these charges and the negative charges on the balloon is stronger than the repulsion between the negative charges on the balloon and the negative charges in the polarized molecules (because they are farther from the balloon), so that there is a net attractive force toward the wall. Ionization processes in the air surrounding the balloon provide ions to which excess electrons in the balloon can transfer, reducing the charge on the balloon and eventually causing the attractive force to be insufficient to support the weight of the balloon. 1

Solucionario Fisica de serway Septima Edicion , Tomo II - Completo, ESPERO QUE LES SEA DE MUCHA AYUDA

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Solucionario Fisica de serway septima edicion II - Completo

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  • 1. 23Electric FieldsCHAPTER OUTLINE23.1 Properties of ElectricCharges23.2 Charging Objects byInduction23.3 Coulombs Law23.4 The Electric Field23.5 Electric Field of aContinuous Charge Distribution23.6 Electric Field Lines23.7 Motion of a Charged Particlein a Uniform Electric FieldANSWERS TO QUESTIONSQ23.1 A neutral atom is one that has no net charge. Thismeans that it has the same number of electronsorbiting the nucleus as it has protons in the nucleus.A negatively charged atom has one or more excesselectrons.*Q23.2 (i) Suppose the positive charge has the large value 1 mC.The object has lost some of its conduction electrons,in number 106 C (1 e/1.60 1019 C) = 6.25 1012 andin mass 6.25 1012 (9.11 1031 kg) = 5.69 1018 kg.This is on the order of 1014 times smaller than the~1g mass of the coin, so it is an immeasurably smallchange. Answer (d).(ii) The coin gains extra electrons, gaining mass onthe order of 1014 times its original mass for the charge1 m C. Answer (b).Q23.3 All of the constituents of air are nonpolar except for water. The polar water molecules in theair quite readily steal charge from a charged object, as any physics teacher trying to performelectrostatics demonstrations in the summer well knows. As a resultit is diffi cult to accumulatelarge amounts of excess charge on an object in a humid climate. During a North Americanwinter, the cold, dry air allows accumulation of signifi cant excess charge, giving the potential(pun intended) for a shocking (pun also intended) introduction to static electricity sparks.Q23.4 Similarities: A force of gravity is proportional to the product of the intrinsic properties (masses)of two particles, and inversely proportional to the square of the separation distance. An electricalforce exhibits the same proportionalities, with charge as the intrinsic property.Differences: The electrical force can either attract or repel, while the gravitational force asdescribed by Newtons law can only attract. The electrical force between elementary particles isvastly stronger than the gravitational force.Q23.5 No. The balloon induces polarization of the molecules in the wall, so that a layer of positivecharge exists near the balloon. This is just like the situation in Figure 23.4a, except that the signsof the charges are reversed. The attraction between these charges and the negative charges on theballoon is stronger than the repulsion between the negative charges on the balloon and thenegative charges in the polarized molecules (because they are farther from the balloon), so thatthere is a net attractive force toward the wall. Ionization processes in the air surrounding theballoon provide ions to which excess electrons in the balloon can transfer, reducing the chargeon the balloon and eventually causing the attractive force to be insuffi cient to support the weightof the balloon.1

2. 2 Chapter 23*Q23.6 Answer (c). Each charge produces fi eld as if it were alone in the Universe.*Q23.7 (i) According to the inverse square law, the fi eld is one-fourth as large at twice the distance. Theanswer is (c), 2 36 cm = 72 cm.(ii) The fi eld is four times stronger at half the distance away from the charge. Answer (b).Q23.8 An electric fi eld created by a positive or negative charge extends in all directions from the charge.Thus, it exists in empty space if that is what surrounds the charge. There is no material at point Ain Figure 23.21(a), so there is no charge, nor is there a force. There would be a force if a chargewere present at point A, however. A fi eld does exist at point A.*Q23.9 (i) We compute qAqB/r2 in each case. In (a) it is 400/4 = 100 (nC/cm)2. In (b) and (c), 300/4 =75 (nC/cm)2. In (d) 600/9 = 67 (nC/cm)2. In (e) 900/9 = 100 (nC/cm)2. The ranking is then a = e >b = c > d.(ii) We compute qA/r2 in each case. In (a) it is 20/4 = 5 nC/cm2. In (b) 30/4 = 7.5 nC/cm2. In (c)10/4 = 2.5 nC/cm2. In (d) 30/9 = 3.3 nC/cm2. In (e) 45/9 = 5 nC/cm2. The ranking is then b > a =e > d > c.*Q23.10 The charge at the upper left creates at the fi eld point electric fi eld to the left, with magnitude wecall E1. The charge at lower right creates downward electric fi eld with an equal magnitude E1.These two charges together create fi eld 2 1 E downward and to the left at 45. The positivecharge is 2 times farther from the fi eld point so it creates fi eld 2E1/( 2 )2 = E1 upward andto the right. The net fi eld is then ( 2 1)E1 downward and to the left. The answer to question(i) is (d).(ii) With the positive charge removed, the magnitude of the fi eld becomes 2 1 E , larger thanbefore, so the answer is (a).*Q23.11 The certain point must be on the same line as A and B, for otherwise the fi eld componentsperpendicular to this line would not add to zero. If the certain point is between A and B, it ismidway between them, and Bs charge is also +40 nC. If the certain point is 4 cm from Aand 12 cm from B, then Bs charge must be 9(40 nC) = 360 nC. These are the only twopossibilities. The answers are (a), (f), and (j).Q23.12 The direction of the electric fi eld is the direction in which a positive test charge would feel a forcewhen placed in the fi eld. A charge will not experience two electrical forces at the same time, butthe vector sum of the two. If electric fi eld lines crossed, then a test charge placed at the point atwhich they cross would feel a force in two directions. Furthermore, the path that the test chargewould follow if released at the point where the fi eld lines cross would be indeterminate.Q23.13 Both fi gures are drawn correctly.E1 andE2 are the electric fi elds separately created by the pointcharges q1 and q2 in Figure 23.12 or q and q in Figure 23.13, respectively. The net electric fi eldis the vector sum ofE1 andE2, shown as E. Figure 23.19 shows only one electric fi eld line ateach point away from the charge. At the point location of an object modeled as a point charge, thedirection of the fi eld is undefi ned, and so is its magnitude.*Q23.14 Answer (a). The equal-magnitude radially directed fi eld contributions add to zero.*Q23.15 Answer (c). Contributions to the total fi eld from bits of charge in the disk lie closer together indirection than for the ring. 3. Electric Fields 3*Q23.16 (i) Answer (c). Electron and proton have equal-magnitude charges.(ii) Answer (b). The protons mass is 1836 times larger than the electrons.*Q23.17 Answer (b).Q23.18 Linear charge density, l, is charge per unit length. It is used when trying to determine theelectric fi eld created by a charged rod.Surface charge density, s, is charge per unit area. It is used when determining the electric fi eldabove a charged sheet or disk.Volume charge density, r, is charge per unit volume. It is used when determining the electric fi elddue to a uniformly charged sphere made of insulating material.Q23.19 No. Life would be no different if electrons were + charged and protons were charged. Oppositecharges would still attract, and like charges would repel. The naming of + and charge is merelya convention.Q23.20 In special orientations the force between two dipoles can be zero or a force of repulsion. Ingeneral each dipole will exert a torque on the other, tending to align its axis with the fi eld createdby the fi rst dipole. After this alignment, each dipole exerts a force of attraction on the other.SOLUTIONS TO PROBLEMSSection 23.1 Properties of Electric ChargesP23.1 (a) The mass of an average neutral hydrogen atom is 1.007 9u. Losing one electron reduces itsmass by a negligible amount, to1.007 9(1.660 1027 kg) 9.111031 kg = 1.67 1027 kg .Its charge, due to loss of one electron, is0 1(1.60 1019 C) = +1.60 1019 C .(b) By similar logic, charge = +1.60 1019 Cmass = 22.99(1.66 1027 kg) 9.111031 kg = 3.82 1026 kg(c) charge of Cl = 1.60 1019 Cmass = 35.453(1.66 1027 kg)+ 9.111031 kg = 5.89 1026 kg(d) charge of Ca++ = 2(1.60 1019 C) = +3.20 1019 Cmass = 40.078(1.66 1027 kg) 2(9.111031 kg) = 6.65 1026 kg(e) charge of N3 = 3(1.60 1019 C) = 4.80 1019 Cmass = 14.007(1.66 1027 kg)+ 3(9.111031 kg) = 2.33 1026 kgcontinued on next page 4. 4 Chapter 23(f ) charge of N4+ = 4(1.60 1019 C) = +6.40 1019 Cmass = 14.007(1.66 1027 kg) 4(9.111031 kg) = 2.32 1026 kg(g) We think of a nitrogen nucleus as a seven-times ionized nitrogen atom.charge = 7(1.60 1019 C) = 1.12 1018 Cmass = 14.007(1.66 1027 kg) 7(9.111031 kg) = 2.32 1026 kg(h) charge = 1.60 1019 Cmass = ( )+ 2 1.007 9 15.999 1.66 1027 kg + 9.111031 kg = 2.99 1026 kgP23.2 (a) N = 10 06 02 1023 . .grams107.87 grams molatomsmolelectronsatom 47 = 2.62 1024(b) #= = .electrons addedC1.60 10Qe1 00 10 319 C electron= 6.25 1015or 2.38 electrons for every109 already present .Section 23.2 Charging Objects by InductionSection 23.3 Coulombs LawP23.3 If each person has a mass of 70 kg and is (almost) composed of water, then each personcontainsN 70 0006 02 1023 grams 18 grams molmol.eculesmolprotonsmolecule10 2.3 1028 protonsWith an excess of 1% electrons over protons, each person has a chargeq = 0.01(1.6 1019 C)(2.3 1028 ) = 3.7 107 Cq qr e = = ( ) ( ) 1 2 = So F k23 7 109 10 9257 220 64 10..N N ~1026 NThis force is almost enough to lift a weight equal to that of the Earth:Mg = 6 1024 kg(9.8 m s2 ) = 6 1025 N ~ 1026 N 5. Electric Fields 5*P23.4 In the fi rst situation,k q qF e A B i A Br on ,1 =21. In the second situation, qA and qB are the same. F F i B A A B= = e A B( )on ,2 on ek q qrFFk q=2221A B2e A Bqrrk q qFF rr22122= 1 1= 2 622 2N13.7 mmN . 17.7 mm =21.57 ThenFB on A,2 = 1.57 N to the left .k q qr eP23.5 (a) F= e =( )( )1 228.99 109 N m2 C2 1.60 10 19 C 2( ) = 1 .59 109N ( repulsion) 3 .80 1010 m2Gm mr g= =(b) F( ) 1 226.67 10 11 N m2 C2 (1.67 10 27 kg)( ) = 210 2453 80 101 29 10..mNThe electric force islarger by 1.24 1036 times .(c) If km mr eq qrG1 221 22 = with q q q 1 2 = = and m m m 1 2 = = , thenqm= = Gke =6 67 108 99 108119...2 2N m kgN m C2 2 611011 C kgP23.6 We fi nd the equal-magnitude charges on both spheres:qr e e = 1 2 =F kq qrk22= =( ) 2 so q rFke1 00 10N8.99 109 N m2 C2 05 103 C1 00 = 14..m .The number of electron transferred is thenN= e .C15 .C1 05 101 60 10e xfer= 6 59 10319. lectronsThe whole number of electrons in each sphere isNtot 10 0.g6 02 1023 = atoms m107.87 g mol ( . ol)(47 e atom) = 2.62 1024 eThe fraction transferred is thenf= = NNxfer = tot 6 59 102 62 102 51 11524... 09 = 2.51charges in every billion 6. 6 Chapter 23P23.7 F kq q1 221 e r8 99 109 7 00 10 6 2= =( . Nm2 C2 )( . C) .00 100 5000 503 6C2( )( ) =mN..q qr 2 eF k1 228 99 109 7 00 10 6 4= =( . Nm2 C2 )( . C) .00 100 5001 01 6C2( )( ) =mN..FFxy= + ==0 . 503 cos 60 . 0 1 . 01 cos 60 . 0 0 .7550 .5 N03 sin 60 . 0 1 . 01 sin 60 . 0 0 .4360 .755 NN ==F( )i (0.436 N)j = 0.872 N at an angle of 330P23.8 Let the third bead have charge Q and be located distance x from the left end of the rod. This beadwill experience a net force given byF = i i ( ) + ( )k 3q Q k q Qe e ( )x2 ( d x) 2The net force will be zero if 3 1=( )2 2 x d x, orx =d x3.This gives an equilibrium position of the third bead of x = 0.634d .The equilibrium is stable if the third bead has positive charge .P23.9 (a) The force is one of attraction . The distance r in Coulombs law is the distance betweencenters. The magnitude of the force isF= e = ( ) ( )k q qr1 22998 99 105 ( )12 0 10 C 180 102 2 .. .N m C( ) = 0 3002 16 1092 CmN..(b) The net charge of 6.00 109 C will be equally split between the two spheres, or3.00 109 C on each. The force is one of repulsion , and its magnitude isF= e = ( ) ( )k q qr1 22998 99 107 ( )3 00 10 C 3 00 102 2 .. .N m C( ) = 0 3008 99 1092 CmN..P23.10 The top charge exerts a force on the negative charge k qQe( d /2 )2 + x2which is directed upward and tothe left, at an angle of tan dx12to the x-axis. The two positive charges together exert force2k qQ2 4 2 2 4 2 1 2d xxd xe( + ) ( )( + ) i= maor for xd 2,a 2 xk qQemd3 8(a) The acceleration is equal to a negative constant times the excursion from equilibrium, as ina = 2x, so we have Simple Harmonic Motion with 216 = k qQe .3mdT 3mdk qQ e= = 22, where m is the mass of the object with charge Q.(b) vmax= A = ak qQmd4 e 3FIG. P23.7 7. Electric Fields 7P23.11 (a) F= e = ( ) 2 ( )k er2919 28 99 101 60 100 5...C 2 2N m C29 108 22 1010 28( )= m. N toward the otherparticle(b) We have F= v2mrfrom whichv= = 8 ( 10)Frm8 22 10 109 11 1031..N 0.529 mkg= 2.19 106 m sSection 23.4 The Electric FieldP23.12 The point is designated in the sketch. The magnitudes of theelectric fi elds, E1 (due to the 2.50 106 C charge) and E2(due to the 6.00 106 C charge), areE8 99 10 2 50 10k qer d1 29 62= =( . Nm2 C2 )( . C)(1)E. . N m2 C2 Ck qer d2 28 99 109 6 00 10 6( 1 00 m)2 = =( )( )+.(2)Equate the right sides of (1) and (2)to get (d +1 00 ) = 2 4 d . m 2 . 0 2or d +1.00 m = 1.55dwhich yields d = 1.82 mor d = 0.392 mThe negative value for d is unsatisfactory because that locates a point between the charges whereboth fi elds are in the same direction.Thus, d = 1.82 m to the left of the 2.50 C chargeFIG. P23.12 8. 8 Chapter 23P23.13 For equilibrium, F F e g = E = (j)or q mgThus,E = j mgq(a) E = j =( 9 11 10 kg )( 9 80m s2)( ) j = ( . N C )j mgq . .1 .60 1031915 58 10 11C(b) E = j =( 1 67 10 kg )( 9 80m s2)( ) j = (1.02 107 N C )jmgq . .1 .60 102719Cq qr e = =P23.14 F k( )( )1 229 19 2 8.99 10 N m2 C2 1.60 10 C 6.02 102 6 37 1051423 26 2( ) ( ) =. mkN*P23.15 The fi rst charge creates at the origin fi eld k Qe2 to the right.aBoth charges are on the x axis, so the total fi eld cannot have avertical comonent, but it can be either to the right or to the left.If the total fi eld at the origin is to the right, then q must be negative:k Qa k qi + ( i )= 2 k Q ( ) i q = 9Qaae e e2 2 2 3In the alternative, if the total fi eld at the origin is to the left,k Qa 2 i + (i)= (i) q = +27Qk qak Qae e e2 9 2 2P23.16 (a) Ek qr= e =( 8 99 10 )( 2 00 10)( ) =29 621 1214 400. ..N CEx = 0 and Ey = 2(14 400)sin 26.6 = 1.29 104 N CsoE = 1.29 104 j N C(b) F = qE = (3.00 106 )(1.29 104 j)= 3.86 102 j NP23.17 (a)E = r + r + r = k q ( )rk qrk qrk qae 1 e e e2 1122 2232 332 4 i + i cos . jsin . ( ) ( + )+ k q (k q e e )32 22 245 0 45 0aa2jE = 3 06 i + 5 06 j = 5 91 2 2 2 . . .k qak qak qae e e at 58.8(b) F = qE =k qa5 91 e22 . at 58.8x+Q x = 0 qFIG. P23.15FIG. P23.16 9. Electric Fields 9P23.18 The electric fi eld at any point x has the x-componentEk qx ak qk q ( 4ax)( )) = x ( ax a= e e e( ) +( ) 2 2 2 2 2When x is much, much greater than a, we fi nd E e 4 ( )a k qx3 .P23.19 (a) One of the charges creates at P a fi eldE=k Q nR xe2 2 at an angle q+to the x-axis as shown.When all the charges produce fi eld, for n1, the componentsperpendicular to the x-axis add to zero.The total fi eld ise e ( )+nk Q nR xk QxR x=( + )cosi i2 2 2 2 3 2 .(b) A circle of charge corresponds to letting n grow beyond all bounds, but the result doesnot depend on n. Smearing the charge around the circle does not change its amount or itsdistance from the fi eld point, so it does not change the field .Section 23.5 Electric Field of a Continuous Charge DistributionP23.20 Ek dqx= e 2 , where dq = dx 0 e = = E k1 The direction is or left for . i0 0dxkkexxx xex =0 2 000 0 0P23.21 E= e e e( + ) = ( )( 9 )( 6 )( )( + ) kd dk Qd dk Qd d( + ) =( + ) =. 8.99 10 22 0 100 . 290 0 . 140 0 .290E= 1.59 106 N C, directed toward the rod .P23.22 Ek Qxx a= e( 2 + 2 )3 2For a maximum, dEdxQk= 1 3x axx ae =( + ) ( + )02 2 3 222 2 5 2x2 + a2 3x2 = 0 or xa =2Substituting into the expression for E givesE23 3 6 3 32k Qaak Qak QaQa2= e e e 2 ( = = =2 ) 3 2 332 220FIG. P23.19FIG. P23.21 10. 10 Chapter 23P23.23 Ek xQx ax8 . 99 10 75 .0 10( x) = 2 3 2 = e( + ) =( )( )+2 2 3 29 62.10006 74 105.( + )2 3 2 0 .010 0xxWe choose the x axis alongthe axis of the ring.(a) At x = 0.010 0 m,E = 6.64 106 i N C = 6.64i MN C(b) At x = 0.050 0 m,E = 2.41107 i N C = 24.1i MN C(c) At x = 0.300 m,E = 6.40 106 i N C = 6.40i MN C(d) At x = 1.00 m,E = 6.64 105 i N C = 0.664i MN Cxx R e = 2 2 P23.24 E k+ 2 1Exx= ( )( ) + ( )2 8 99 10 7 90 10 1 0 3509 32 2 . .. = + 4 46 10 10 12382..xx(a) At x = 0.050 0 m , E = 3.83 108 N C = 383 MN C(b) At x = 0.100 m, E = 3 24 10 = 324 . 8 N C MN C(c) At x = 0.500 m, E = 8 07 10 = 80 7 . 7 N C . MN C(d) At x = 2.000 m, E = 6 68 10 = 6 68 . 8 N C . MN Cxx R e = 2 2 P23.25 (a) From the Example in the chapter text, E k+ 2 1Q R= = = ( )( ) =E231 84 108.C m1 . 04 10 N C0 .90029.36 107 N C = 93.6 MN Capproximation: E ke = 2 = 104 MN C (about 11% high)(b)Qr e = = ( ) approximation: E k5 .20 100 .30( ) =2962 8 99 10. MN C (about 0.6% high). 0 519E = ( ) + 1 04 10 130 03 0082...N Ccm30.02 cm = (1.04108 N C)(0.004 96) = 0.516 MN C 11. Electric Fields 11P23.26 The electric fi eld at a distance x is E kxx Rx e = + 2 12 2 This is equivalent to E kR xx e = + 2 111 2 2 For large x,Rx221and 1 12222+ R + Rxx2 = so E kR xk( R( ) )x ee+ ( ) =+2 111 221 22 22 x2+ R 2 ( x2) 11 2Substitute = QR2 , E( 2)= e k Q xR xk Q xRxe+ ( ) = + 11 2 2 2222But for xR, 121 , so Ex2 + R2 x2k Qx x e 2 for a disk at large distancessin dqr x e = sin = P23.27 Due to symmetry E dE y y = = 0, and E dE k2where dq = ds = rd ,so that E= e = e ( ) = e kkkdxrrr sin cos 002where = qLand rL =Thus, E( ) 2 2 8 99 10 7 50 10k qL x= e =29 6 . N m2 C2 . C( )( )0 140 2 . mSolving, Ex = 2.16 107 N CSince the rod has a negative charge,E = (2.16 107 i) N C = 21.6i MN C .FIG. P23.27 12. 12 Chapter 23P23.28 (a) We defi ne x = 0 at the point where we are to fi nd the fi eld. One ring, with thickness dx, hascharge Qdxhand produces, at the chosen point, a fi eldd= k xQdxE e ( x 2 + R2 )i3 2hThe total fi eld is+E = E = ik ed h d ( =+ ) k Qxdxh x Rdeall charge2 23 2 Qhx R xdxk Qhx R+d hx deiEi2222 2 3 22 2( + )=( += )1 2 +d h( ) =( + ) i1 2 ( )1 2 2 2 1 2 2( + ) +=1 1x de k Qh d R d h R2(b) Think of the cylinder as a stack of disks, each with thickness dx, charge Qdxh, and charge-per-area = QdxR2h . One disk produces a fi eldde E = ik QdxR hx( x + R) 21 2 2 21 2So,+ d hE = E = d =( +k QdxR hxx Rex dall charge2 1 2 2 2) 1 2iE d hi d +h= ( + ) k Q 2 2 1 2R h+ 2 1e dx x R xdx2 2=2dx d ( + )x R e +k QR hd h= +2 12 21 2x2 d2 1 2d hdi= + (( + ) + ) + +E2 ik Q1 2e d h d d h 2 R 2d2 R 2hRi2 2 2 1 2 2k QR he h d R d h2 1 22( ) = + ( + ) ( + ) E+ ( ) R21 2P23.29 (a) The electric fi eld at point P due to each element of length dx isdEk dqx y= e2 + 2 and is directed along the line joining the element topoint P. By symmetry,E dE x x = = 0 and since dq = dx,E E dE dE y y = = = cos where cos =yx2 y2+ 2 sin0 ( + ) 2 =2= eTherefore, E k ydxx yke2 2 3 2y 0(b) For a bar of infi nite length, 0 = 90 and E= e 2 ky yFIG. P23.29 13. Electric Fields 13P23.30 (a) The whole surface area of the cylinder is A = 2 r2 + 2 rL = 2 r (r + L).Q = A = (15.0 109 C m2 )2 (0.025 0 m)[0.025 0 m + 0.060 0 m] = 2.00 1010 C(b) For the curved lateral surface only, A = 2 rL.Q = A = (15.0 109 C m2 )2 (0.025 0 m)(0.060 0 m) = 1.41 1010 C(c) Q = V = r2L = ( 9 ) ( )2 500 10 0 025 0 0 060 0 C m m 3 . . m C ( ) = 5.89 1011P23.31 (a) Every object has the same volume, V = 8(0 030 ) = 2 16 10 . 0 m 3 . 4 m3 .For each, Q = V = (400 109 C m3 )(2.16 104 m3 ) = 8.64 1011 C(b) We must count the 9.00 cm2 squares painted with charge:(i) 6 4 = 24 squaresQ = A = (15.0 109 C m2 )24.0(9.00 104 m2 ) = 3.24 1010 C(ii) 34 squares exposedQ = A = (15.0 109 C m2 )24.0(9.00 104 m2 ) = 3.24 1010 C(iii) 34 squaresQ = A = (15.0 109 C m2 )34.0(9.00 104 m2 ) = 4.59 1010 C(iv) 32 squaresQ = A = (15.0 109 C m2 )32.0(9.00 104 m2 ) = 4.32 1010 C(c) (i) total edge length:= 24 (0.030 0 m)Q = = (80.0 1012 C m)24 (0.030 0 m) = 5.76 1011 C(ii) Q = = (80.0 1012 C m)44 (0.030 0 m) = 1.06 1010 C(iii) Q = = ( ) ( ) = 80.0 1012 C m 64 0.030 0 m 1.54 1010 C(iv) Q = = (80.0 1012 C m)40 (0.030 0 m) = 0.960 1010 CSection 23.6 Electric Field LinesP23.32 P23.33FIG. P23.32 FIG. P23.33 14. 14 Chapter 23P23.34 (a) q1q261813= = (b) q q 1 2 is negative, is positiveP23.35 (a) The electric fi eld has the general appearance shown. It iszero at the center , where (by symmetry) one can see thatthe three charges individually produce fi elds that cancel out.In addition to the center of the triangle, the electric fi eld lines inthe second fi gure to the right indicate three other points near themiddle of each leg of the triangle where E = 0, but they aremore diffi cult to fi nd mathematically.(b) You may need to review vector addition in Chapter Three.The electric fi eld at point P can be found by adding theelectric fi eld vectors due to each of the two lower pointcharges:E = E + E 1 2.The electric fi eld from a point charge isE = k rqr e 2 .As shown in the solution fi gure at right,Eqa e to the right and upward at 601 2 = kEqa e to the left and upward at 602 2 = kE = E + E = ( i + j)+ 1 2 2 k 60 60 60qa e cos sin cos sin sin .i j j + ( ) = ( ) =qa e60 2 601 72 k qa e3 2 kjSection 23.7 Motion of a Charged Particle in a Uniform Electric FieldP23.36 (a) aqEm= =( 1 . 602 10 )( 6 .00 10)( =1 67 10) 519 527..76 1013 m s so a = 5.76 1013 i m s2(b) v v f i f i 2 = 2 + 2a(x x )0 = v2 + 2(5 76 1013 )(0 070 0) i . .v i i= 2.84 106 m s(c) v v f i = + at0 = 2.84 106 + (5.76 1013 )tt = 4.93 108 sFIG. P23.35 15. Electric Fields 15P23.37 (a) a= = ( )qEm1 602 10 64010 .= 1 67 106 14 101927.. ms2(b) vf = vi + at 1.20 106 = (6.14 1010 )t t = 1.95 105 s(c)x x t f i i f = 1 ( + )2v vxf= 1 ( )( ) =21.20 106 1.95 10 5 11.7 m(d) K = m = ( )( ) = 12122 1 67 10 27 1 20 106 2 1 2 v . kg . m s . 0 1015 JP23.38 The particle feels a constant force: F = qE = (1 106 C)(2 000 N C)(j) = 2 103 N(j)and moves with acceleration: a2 F j= m = =( )( )2 10kg m s2 101 10316kg( 13 m s2 )(j)Note that the gravitational force is on the order of a trillion times smaller than the electrical forceexerted on the particle. The particles x-component of velocity is constant at(1.00 105 m s)cos37 = 7.99 104 m s. Thus it moves in a parabola opening downward. Themaximum height it attains above the bottom plate is described byv v yf yi y f i 2 = 2 + 2a (y y ): 0 6 02 104 2 10 0 = ( . m s)2 ( 13 m s2 )( ) yfyf= 1.81104 mSince this is less than 10 mm, the particle does not strike the top plate, but moves in a symmetricparabola and strikes the bottom plate after a time given by12y y t at f i yi y = + v +2 0 0 602 1012= + ( . 4 m s)t + (1 1013 m s2 )t 2since t0, t = 1.20 108 sThe particles range is x x t f i x = + v = 0 + (7.99 104 m s)(1.20 108 s) = 9.61104 m.In sum,The particle strikes the negative plate after moving in a parabola with a height of 0.181 mmand a width of 0.961 mm.P23.39 The required electric fi eld will be in the direction of motion .Work done = K122 v (since the fi nal velocity = 0)so, Fd = m iwhich becomes eEd = Kand EKed= 16. 16 Chapter 23P23.40vi = 9.55 103 m s(a) aeEm y= =( 1 60 10 )( 720)( = 1 67 10) 6 90 11927... 010m s2From the large magnitude of this vertical acceleration,we can note that the gravitational force on the particle isnegligible by comparison to the electrical force.R= v = 2aiy3 21 27 10sin.m so that9 55 10 23 . sin6 90 101 27 103 210..( )= sin 2 = 0.961 = 36.9 90.0 = 53.1(b) tR R= =v vcosix iFIG. P23.40If = 36 9 . , t = 167 ns . If = 53 1 . , t = 221 ns .P23.41 (a) tx= =x7 ..= =v0 050 04 50 101 11 10 111 5. s ns(b) aqEm y= =( 1 . 602 10 19 )( 9 .60 103)( =1 .67 1027) 9.211011 m s212y y t at f i yi y = v +2: yf= 1 ( )( ) = =29.21 1011 1.11 10 7 2 5.68 10 3 m 5.68 mm(c) vx= 4.50 105 m s v v yf yi y = + a t = (9.211011 )(1.11107 ) = 1.02 105 m sAdditional ProblemsP23.42 The two given charges exert equal-size forces of attractionon each other. If a third charge, positive or negative, wereplaced between them they could not be in equilibrium. If thethird charge were at a point x15 cm, it would exert astronger force on the 45 m C than on the 12 m C, and couldnot produce equilibrium for both. Thus the third charge mustbe at x = d0. Its equilibrium requiresk q( ) = ( )dC k qC( + d)e e 12 452 15 cm215 45= = d123 752 cm + d.x = 0 +12 mC 45 mCFIG. P23.42dq15 cm + d = 1.94d d = 16.0 cmThe third charge is at x = 16 0. cm . The equilibrium of the 12 m C requiresk q k e e 1216 0( C) ( 45 C )12C( = . cm) 2 ( 15 cm)2q = 51.3 Cx15 cmAll six individual forces are now equal in magnitude, so we have equilibrium as required, and thisis the only solution. 17. Electric Fields 17P23.43 The proton moves with acceleration aqEm p= =( 1 60 10 19)( 640)=1 673 1027..C NCkg6.13 1010 m s2while the e has acceleration ae =( 1 60 10 )( 640)=9 110 101 121931...C NCkg1014 m s2 = 1836ap(a) We want to fi nd the distance traveled by the proton (i.e., d atp = 122), knowing:4 0012121 837122 2 2 . cm = + = a t a t a t p e pThus, d atp = = = 124 002 21 8 ..cm1 837m(b) The distance from the positive plate to where the meeting occurs equals the distance thesodium ion travels (i.e., d at Na Na = 122). This is found from:4 001212. cm 2 2: Na Cl = a t + a t 4 0012 229912 3545. 2 2. .cmu u= + eEteEtThis may be written as 4 0012120 649 1 6512. cm 2 . 2 . 2 Na Na Na = a t + ( a )t = a t = = = cm 12so d at Na Na4 .00cm2 2 .43 1.65P23.44 (a) The fi eld, E1, due to the 4.00 109 C charge is in thex direction.E r 1 2k q( 8 99 109 )( 4 00 10 9) r= e =. N m2 C2 . C2 505 752 .. mN C( )= iiLikewise, E2 and E3 , due to the 5.00 109 C charge and the 3.00 109 C charge, areE r 2 2k q( 8 99 109 )( 5 00 10 9) r( ) i = N C i2= e =. N m2 C2 . C. . 0011 2 2 mE3. 2 2. 9 98 99 10 3 00 10N m C C21 20=( )( )( .m)i = 18.7 N C i E E E E R= + + = 1 2 3 24 2. NC in +x direction.FIG. P23.44 (a)continued on next page 18. 18 Chapter 23(b)= k qE1 e r = (8 . 46 N C )(0 . 243i + 0 . 970j) 2 r= = ( )(+ )= =E r j2 2E r3 211 2( 5 81)( )k qerk qer . .N CN C + 0.928N= + =0 3714 21 1 3. . i jE E E i x x x C E E E E N C y y y y = + + = 1 2 3 8.43jER = 9.42 N C = 63.4 above x axisP23.45 (a) Let us sum force components to fi ndF qE T x x = sin = 0, andF qE T mg y y = + cos = 0Combining these two equations, we getqmg=( + ) =E E x y( )( )cot1 00 10 9 80. . 3 . 00 cot 37.30 5 00 101 09 1010 958CnC( + ) = =...(b) From the two equations for Fx and Fy we also fi ndTqEx = = =sin .. .37 05 44 10 3 5 44N mNFIG. P23.44 (b)FIG. P23.45P23.46 This is the general version of the preceding problem. The known quantities are A, B, m, g, and .The unknowns are q and T.The approach to this problem should be the same as for the last problem, but without numbers tosubstitute for the variables. Likewise, we can use the free body diagram given in the solution toproblem 45.Again, Newtons second law: F = T sin + qA = 0 (1)x and F = + T cos + qB mg = 0(2)y (a) Substituting TqA =sininto Eq. (2), qAqB mgcossin+ =Isolating q on the left, qmg=( A cot + B)(b) Substituting this value into Eq. (1), TmgA=( A cos + Bsin )If we had solved this general problem fi rst, we would only need to substitute the appropriatevalues in the equations for q and T to fi nd the numerical results needed for problem 45. If youfi nd this problem more diffi cult than problem 45, the little list at the fi rst step is useful. It showswhat symbols to think of as known data, and what to consider unknown. The list is a guide fordeciding what to solve for in the analysis step, and for recognizing when we have an answer. 19. Electric Fields 19P23.47 Fk q qr= e 1 22 : tan = 15 0..60 0 = 14 0 . F1F( )( )9 62= 40 0( ) =238 99 10 10 0 100 150=. . .. N( 8 99 10 )( 10 0 10)0 6002 50( 8 )( )9 6222. ....( ) ==NF99 10 10 0 100 6192 359 6223( ) == . .. N2 cos14.0 = 2.50 2.35cos14.0 = 4.78 NF F x FF F y= 1 2 F sin14.0 = 40.0 2.35sin14.0 = 40.6 NF F F= + = ( ) + ( ) = Nnet x yFFy=2 2 2 2 4.78 40.6 40.9tanx= =40 64 78263.. P23.48 From Figure (a) we have d cos30.0 = 15.0 cmor d = 15 0.cos .cm30 0From Figure (b) we have = sin150 .0dcm1 . 15 0sin = = ( ) .cos .30 020 3cm50.0 cm Fmgq = tanor F mg q = tan 20.3 (1)From Figure (c) we have F F q = 2 cos30.0Fk qq= e ( ) 20 30030 022 .cos .m (2)Combining equations (1) and (2),2k qe mg.( )30 0 20 3 0 30022cos . tan .m =q( )= mg. tan .kqe2220 300 20 32 300( 2 00 13 )( )( )2=cos ..m kg . m s2 . m tan . . 10 30 00 9 80 0 300 20 32 8 99( 9 )= 14= N m C 4 20 10 C2 05 12 22cos .q . . 07 C = 0.205 CFIG. P23.47Figure (a)Figure (b)Figure (c)FIG. P23.48 20. 20 Chapter 23P23.49 ChargeQ2resides on each of the blocks, which repel as point charges:Fk ( Q 2 )( = Q2e )= k ( L L) 2i LSolving for Q, Q Lk L Lkie=( ) 2P23.50 If we place one more charge q at the 29th vertex, the total force on the central charge will add upto zero:F28 charges + k qQe2 away from vertex 29 = 0aF28 charges = toward vertex 29 k qQe2aP23.51 According to the result of an Example in the chapter text, the left-handrod creates this fi eld at a distance d from its right-hand end:Ek Qd a ddFk QQadxd d aFk Qadxa xx xx x b aeee=( + )=( + )=+22 222k Qa a2 212222ba 2beb a( )= + = lnFk Qaa bbbb ak Qabb= + e + + eln ln ln=22222422 4( )( + ) = b 2k Qaa b a 22 222 2 4 4b ae ln*P23.52 We model the spheres as particles. They have different charges. They exert on each other forcesof equal magnitude. They have equal masses, so their strings make equal angles q with thevertical. The distance r between them is described by sin q = (r/2)/40 cm,so r = 80 cm sin qLet T represent the string tension. We haveFx = 0: keq1q2/r2 = T sinqFy = 0: mg = T cosqDivide to eliminate T.k q qr mg r/e 1 22 2r240 4= =tan(cm)2 /(80 cm)2 2 = 3Cleared of fractions, k q q r mgr e 1 28.99 109(N m2/C2) 300 10 9C (200 109C) (0.8 m)2 r2 = 2.4 103 (9.8) N r3(0.8 m)2 r 2 = 1901 r 6We home in on a solution by trying values.r 0.64 r 2 1901 r 60 +0.640.5 29.30.2 +0.480.3 0.840.24 +0.220.27 0.170.258 +0.0130.259 0.001Thus the distance to three digits is 0.259 m .FIG. P23.51 21. Electric Fields 21*P23.5390 0Q = d = Rd = R9 cos sin 090 00 90 0. .. 0 00 001 1 212 0 2 0 600. . .= [ ( )] == =( )( R RQ C ) = ==( )m C so C m C( ) Rd0 12 0 10 0143 000. .. dFyd 2 14R =( )2003 00cos . C ( cos ) C 2 2 )( )= ( ) ( RFy290 0 .. C m 9 d68 99 103 00 10..N m C( )10 0 100 6006290 0.cos. m= ( ) ( ) + 8 99 30 00 60023 2 . ..F 10 d y1212N cos = ( ) + =2220 4501214F 2 0 y . N sin .707 N downwardcos q0 360cos2 q0 360FIG. P23.5310110Since the leftward and rightward forces due to the two halves of the semicircle cancel out, Fx = 0.*P23.54 (a) The two charges create fi elds of equal magnitude, both with outward components along thex axis and with upward and downward y components that add to zero. The net fi eld is thenk qx + k qx 109 m2 = 2 ( 8 . 99 N )52 10 9 Ce i ei r2 rr2rC2 m)2xx(( . ) /i0 25 + 2 3 2= 9352 3 2N mC(0.0625 m22xx+i) /(b) At x = 0.36 m,E= 935 N m 0 36mi +C (0.0625 m 0 36m22. ( . ). / = ikN/C 2 32 4 00)(c) We solve 1 000 = 935 x (0.0625 + x2)3/2 by tabulating values for the fi eld function:x 935 x (0.0625 + x2)3/20 00.01 5970.02 1 1850.1 4 7890.2 5 6980.36 4 0000.9 1 0321 854 0We see that there are two points where E = 1 000. We home in on them to determine theircoordinates as (to three digits) x = 0.016 8 m and x = 0.916 m.(d) The table in part (c) shows that the fi eld is nowhere so large as 16 000 N/C.(e) The fi eld of a single charge in Question 7 takes on all values from zero to infi nity, eachat just one point along the positive x axis. The vector sum of the fi eld of two charges,in this problem, is zero at the origin, rises to a maximum at 17.7 cm, and then decreasesasymptotically to zero. In the question and the problem, the fi elds at x = 36 cm happento take similar values. For large x the fi eld of the two charges in this problem showsjust the same inverse proportionality to x2 as the fi eld in the question, being largerby the factor 2(52 nC)/(57.7 nC) = 1.80 times. 22. 22 Chapter 23P23.55 (a) From the 2Q charge we have Fe T2 sin2 = 0 and mg T = 2 2 cos 0Combining these we fi ndFmgTTe = 2 2 =2 22sincostanFrom the Q charge we have F T e= = 1 1 sin 0 and mg T = 1 1 cos 0Combining these we fi ndFmgTTe = 1 1 =1 11sincostan or 2 1 == e = e 2 2k Qr e(b) Fk QQr222If we assume q is small then tan r 2FIG. P23.55Substitute expressions for Fand tan into either equation found in part (a) and solve for r.e Fe = tan , then 2 1r e mg22k Qr 2mg and solving for r we fi ndr4 2 1 3.k Qmg e P23.56 The bowl exerts a normal force on each bead, directed alongthe radius line or at 60.0 above the horizontal. Consider thefree-body diagram shown for the bead on the left side of thebowl:F n mg y = sin60.0 = 0,or nmg =sin60.0Also, F F n x e = + cos60.0 = 0,ork qRmg mg e= ncos . = =22 60 0tan .60 0 3Thus, q Rmgke= 31 2P23.57 (a) The total non-contact force on the cork ball is:F qE mg m gnFe 60.0mgFIG. P23.56qEm= + = + ,which is constant and directed downward. Therefore, it behaves like a simple pendulum inthe presence of a modifi ed uniform gravitational fi eld with a period given by:TLg qE m=+=+ ( )2 20 5002 00 10 6 1 0 .. .m9.80 m s2 C 0 10 1 00 100 3075 3 ( ) =N C kgs..(b) Yes . Without gravity in part (a), we get TLqE m= 2T =0 500 .m2 00 10 6 1 00 1 00 10 C 105 N C = 3 0 3142( . )( . ) . kg. s (a 2.28% difference). 23. Electric Fields 23*P23.58 (a) At A the top charge makes the dominant contribution to thefi eld and the net fi eld is downward. At B the total electricfi eld is zero. Between B and C the main change is weakeningof the downward electric fi eld of the top charge, so the net fi eldat C is upward. At E the fi elds of the two bottom charges cancelout and the total fi eld is downward. At F the total fi eld isdownward.(b) The fi eld is zero at B as it changes from downward at A toupward at C. As a continuous function, the fi eld must passthrough the value zero near D as it changes from upwardat C to downward at E and F.(c) Let y represent the distance from E up to the zero-fi eld point.The distance from P to E is (32 1.52)1/2 cm = 2.60 cm. Thenthe requirement that the fi eld be zero isk q= 2+ ++ ( ) + 2 2 3 2cm y2 cm) 2 cm2 k qyyye e(2.60 ) (1.5 1.51 5++ k qyye( ./cm)2k q=2 60y/ k qyye e( . ) ( .21 5 2 2 3 2 cm cm)2+ (1.52 + y2)3/2 2 y (2.60 y)2 = 0PABCDEF+FIG. P23.58As a check on our algebra, we note that y = (1/3)2.60 cm = 0.866 cm should be a solution,corresponding to point B. Substituting 0.866 gives 5.20 5.20 = 0 as it should. We home inon the smaller answer:y (1.52 + y2)3/2 2y(2.60 y)20 +3.3750.3 +0.4110.4 0.1240.37 +0.0140.373 0.000 6To three digits the answer is 0.373 cm .P23.59 (a) There are 7 terms which contribute:3 are s away (along sides)3 are 2s away (face diagonals) and sin = = cos 121 is 3s away (body diagonal) andsin = 13The component in each direction is the same by symmetry.F = + + i j k( + + ) = k qs (1.90)(i + j + k )k qs222 213 3e e222 1k qs x y z= 2 + 2 + 2 = e(b) F F F F22 3.29 away from the originFIG. P23.59 24. 24 Chapter 23P23.60 (a) Zero contribution from the same face due to symmetry, oppositeface contributese sin k qr4 2 where rs s s = + s s + = =2 21 5 1 222 22 . .sin = srEk qsrk qsk qs41 22= e = e e ( ) 4 =2 .18 3 .3 2 2 (b) The direction is the direction. kP23.61 The fi eld on the axis of the ring is calculated in an Example in the chapter text as= = eE Ek xQx ax( 2 + 2 )3 2The force experienced by a charge q placed along the axis of the ring isF kQqxx ae = ( + ) 2 2 3 2and when xa, this becomes Fk Qqax e = 3 .This expression for the force is in the form of Hookes law, with an effective springconstant of kk Qqa= e 3 .Since = 2 f =km, we have f= e 1k Qqma2 3.P23.62 d . k dq +2 2+ xxxe Ei j =+ ( )0 150 20 150. m 0 .150mme . i mj( ) =( + )+2( 3 220 m ) 2 0 1500 15k x dxx. E Ei j= =( + )x dxx0 400 d ke+ all chargem 0.150 2 0..150 2 3 20mm( ) = xE0 400+ ( ) ki j= ++ ( )xxe.. .2 200 1500 150mmm0 150 0 150 2 2 20 4000. ..m mm( ) + ( )xEE i = ( ) ( ) 8 99 10 35 0 10 2 34 6 67 9 9 . . . . N m C C m 2 2 ( ) + ( ) = ( +)103 N C = (1.36i +1.96j)kN Cm1 6 24 0 m11 36 1 96 .. . jE i jFIG. P23.60FIG. P23.62 25. Electric Fields 25P23.63 The electrostatic forces exerted on the two charges result in a net torque = 2Fa sin = 2Eqa sin .For small , sin and using p = 2qa, we have = Ep2The torque produces an angular acceleration given by = I = I ddt2Combining these two expressions for torque, we have d + dtEpI2 =2 0This equation can be written in the form d2 = which is the standard2dt2equation characterizing simple harmonic motion, with 2 = EpIThen the frequency of oscillation is f = w/2p, or fpEIqaEI= = 12122 P23.64= k q= k qE r (i)+ k q( i )+ ra( a) ( k qa e e e e2 2 2 2 3) ( )+ = + + + e e = 2 2 2 2211213 6i k qi ak qa2iP23.65 E Ei = = = ( )k x edk xdxxxe 0 03 0 00 = 12 2kx xi x dx k x i e3 =xex 0 0 200 00 ( ) iANSWERS TO EVEN PROBLEMSP23.2 (a) 2.621024 (b) 2.38 electrons for every 109 presentP23.4 1.57 N to the leftP23.6 2.51109P23.8 x = 0.634d. The equilibrium is stable if the third bead has positive charge.P23.10 (a) period = 2md3k qQ ek qQmdwhere m is the mass of the object with charge Q (b) 4 ae3 P23.12 1.82 m to the left of the negative chargeP23.14 514 kNP23.16 (a) 12 9 . j kN C (b) 38 6 . j mNP23.18 See the solution.P23.20e0kx0 ( ) iP23.22 See the solution.P23.24 (a) 383 MN/C away (b) 324 MN/C away (c) 80.7 MN/C away (d) 6.68 MN/C awayP23.26 See the solution.FIG. P23.63 26. 26 Chapter 23P23.28 (a)k Qhi2 2 1 2 2 2e d R d h R1 2+ ( ) + ( ) + ( ) (b)2i+ + ( ) + ( ) + ( ) k Q e 2 21 2 1 2h d R d h 2 R2R 2hP23.30 (a) 200 pC (b) 141 pC (c) 58.9 pCP23.32 See the solution.P23.34 (a) 13(b) q1 is negative and q2 is positiveP23.36 (a) 57 6 . iTm s2 (b) 2 84 . iMm s (c) 49.3 nsP23.38 The particle strikes the negative plate after moving in a parabola 0.181 mm high and 0.961 mm wide.P23.40 (a) 36.9, 53.1 (b) 167 ns, 221 nsP23.42 It is possible in just one way: at x = 16.0 cm place a charge of +51.3 C.P23.44 (a) 24.2 N/C at 0 (b) 9.42 N/C at 117P23.46 (a) mgAcot + B(b) mgAAcos + BsinP23.48 0.205 CP23.50 k qQe2atoward the 29th vertexP23.52 25.9 cmP23.54 (a)E= 935i ( )2 3 2N m22xxC 0.0625 m +/ (b) 4.00 ikN/C (c) At x = 0.016 8 m and at x = 0.916 m(d) Nowhere is the fi eld so large. (e) The fi eld of a single charge in Question 7 takes on allvalues from zero to infi nity, each at just one point along the positive x-axis. The vector sum ofthe fi eld of two charges, in this problem, is zero at the origin, rises to a maximum at 17.7 cm, andthen decreases asymptotically to zero. In the question and the problem, the fi elds at x = 36 cmhappen to take similar values. For large x the fi eld of the two charges in this problem showsjust the same inverse proportionality to x2 as the fi eld in the question, being larger by thefactor 2(52 nC)/(57.7 nC) = 1.8 times.P23.56 R1 2 mgke 3 P23.58 (a) At A downward. At B zero. At C upward. At E downward. At F downward. (b) See thesolution. (c) 0.373 cmP23.60 (a) See the solution. (b) k.P23.62 (1.36i +1.96j) kN CP23.64 2k qae i6 2 27. 24Gausss LawCHAPTER OUTLINE24.1 Electric Flux24.2 Gausss Law24.3 Application of Gausss Law toVarious Charge Distributions24.4 Conductors in ElectrostaticEquilibriumANSWERS TO QUESTIONSQ24.1 The luminous fl ux on a given area is less when the sun islow in the sky, because the angle between the rays of thesun and the local area vector, dA, is greater than zero.The cosine of this angle is reduced. The decreased fl uxresults, on the average, in colder weather.Q24.2 The surface must enclose a positive total charge.Q24.3 The net fl ux through any gaussian surface is zero. Wecan argue it two ways. Any surface contains zero charge,so Gausss law says the total fl ux is zero. The fi eld isuniform, so the fi eld lines entering one side of the closedsurface come out the other side and the net fl ux is zero.*Q24.4 (i) Equal amounts of fl ux pass through each of the six faces of the cube. Answer (e).(ii) Move the charge to very close below the center of one face, through which the fl ux is thenq/20. Answer (c).(iii) Move the charge onto one of the cube faces. Then the fi eld has no component perpendicularto this face and the fl ux is zero. Answer (a).*Q24.5 (i) Answer (a).(ii) the fl ux is zero through the two faces pierced by the fi lament. Answer (b).*Q24.6 (i) Answer (a).(ii) The fl ux is nonzero through the top and bottom faces, and zero through the other four faces.Answer (c).*Q24.7 (i) Both spheres create equal fi elds at exterior points, like particles at the centers of the spheres.Answer (c).(ii) The fi eld within the conductor is zero. The fi eld within the insulator is 4/5 of its surfacevalue. Answer (f).Q24.8 Gausss law cannot tell the different values of the electric fi eld at different points on the surface.When E is an unknown number, then we can say E cos dA = E cos dA. When E(x, y, z)is an unknown function, then there is no such simplifi cation.Q24.9 The electric fl ux through a sphere around a point charge is independent of the size of the sphere.A sphere of larger radius has a larger area, but a smaller fi eld at its surface, so that the productof fi eld strength and area is independent of radius. If the surface is not spherical, some parts arecloser to the charge than others. In this case as well, smaller projected areas go with strongerfi elds, so that the net fl ux is unaffected.27 28. 28 Chapter 24Q24.10 Inject some charge at arbitrary places within a conducting object. Every bit of the charge repelsevery other bit, so each bit runs away as far as it can, stopping only when it reaches the outersurface of the conductor.*Q24.11 (a) Let q represent the charge of the insulating sphere. The fi eld at A is (4/5)3q/[4p (4 cm)20].The fi eld at B is q/[4p (8 cm)20]. The fi eld at C is zero. The fi eld at D is q/[4p (16 cm)20].The ranking is ABDC.(b) The fl ux through the 4-cm sphere is (4/5)3q/0. The fl ux through the 8-cm sphere andthrough the 16-cm sphere is q/0. The fl ux through the 12-cm sphere is 0. The ranking isB = DAC.*Q24.12 The outer wall of the conducting shell will become polarized to cancel out the external fi eld. Theinterior fi eld is the same as before. Answer (c).Q24.13 If the person is uncharged, the electric fi eld inside the sphere is zero. The interior wall of the shellcarries no charge. The person is not harmed by touching this wall. If the person carries a (small)charge q, the electric fi eld inside the sphere is no longer zero. Charge q is induced on the innerwall of the sphere. The person will get a (small) shock when touching the sphere, as all the chargeon his body jumps to the metal.*Q24.14 (i) The shell becomes polarized. Answer (e).(ii) The net charge on the shells inner and outer surfaces is zero. Answer (a).(iii) Answer (c).(iv) Answer (c).(v) Answer (a).Q24.15 There is zero force. The huge charged sheet creates a uniform fi eld. The fi eld can polarize theneutral sheet, creating in effect a fi lm of opposite charge on the near face and a fi lm with an equalamount of like charge on the far face of the neutral sheet. Since the fi eld is uniform, the fi lms ofcharge feel equal-magnitude forces of attraction and repulsion to the charged sheet. The forcesadd to zero.SOLUTIONS TO PROBLEMSSection 24.1 Electric FluxP24.1 E = EAcos A = r2 = ( )2 = 0.200 0.126 m25.20 105 = E(0.126)cos0 E = 4.14 106 N C = 4.14MN CP24.2 = EAcos = (2.00 104 N C)(18.0 m2 )cos10.0 = 355 kNm2 CEP24.3 (a) = E A = (a i + b j ) A i= aAE (b) E = (ai + bj) Aj = bA(c) E = (ai + bj) Ak = 0 29. Gausss Law 29P24.4 (a) A = (10.0 cm)(30.0 cm)= =cm2 . m2cos.= =AEA E AE A300 0 030 07 80,,( )( )= 10 0 030 0 1802 344 . cos. , kN m2 C E A(b) E A EA A , = cos = (7.80 104 )( )cos60.0A w = ( )( ) = ( ) 10 030 0 30 0 60 0. ..cos .cm cmcm = == ( )(600 0 060 07 80 104 0 060 0FIG. P24.4cm2 . m2. . , E A )cos60.0 = +2.34 kNm2 C(c) The bottom and the two triangular sides all lie parallel toE, so E = 0 for each of these.Thus,E, . . total= 2 34 kNm2 C+ 2 34 kNm2 C+ 0 + 0 + 0 = 0P24.5 E = EAcos through the baseE = (52.0)(36.0)cos180 = 1.87 kNm2 CNote that the same number of electric fi eld lines go through the base as gothrough the pyramids surface (not counting the base).For the slanting surfaces, E = +1.87 kNm2 C .Section 24.2 Gausss LawP24.6 (a) Ek Qr= e2 ; 8 90 10( 8 99 10)( 0 750) 2=9..2 .QBut Q is negative sinceEpoints inward. Q = 5.57 108 C = 55.7 nC(b) The negative charge has a spherically symmetric charge distribution, concentric withthe spherical shell.P24.7 (a) E(+ + ) in C C C Cq ==05.00 9.00 27.0 84.0 8 85 10= 6 89 10 6 12C N m..N m C 2 2 2 2E = 6.89 MNm2 C(b) Since the net electric fl ux is negative, more lines enter than leave the surface.FIG. P24.5 30. 30 Chapter 24P24.8 (a) One-half of the total fl ux created by the charge q goes through the plane. Thus, E Eq q,plane ,total = = =1212 2 0 0(b) The square looks like an infi nite plane to a charge very close to the surface. Hence, E Eq,square ,plane =20(c) The plane and the square look the same to the charge.P24.9 Eq =in0Through S1 EQ Q Q = += 20 0Through S2 EQ Q =+ =00Through S3 E= 2 Q + Q Q 2Q = 0 0Through S4 E = 0P24.10 (a) Eq,= ..= in. shell= 061212 0 108 85 101 36 106 Nm2 C = 1.36MNm2 C(b) E,half shell= ( Nm2 C) = Nm 121.36 106 6.78 105 2 C = 678 kNm2 C(c) No ,the same number of fi eld lines will pass through each surface, no matter how theradius changes.P24.11 (a) With d very small, all points on the hemisphere arenearly at a distance R from the charge, so the fi eldeverywhere on the curved surface is k Qe2 radiallyRoutward (normal to the surface). Therefore, the fl uxis this fi eld strength times the area of half a sphere:curved local hemispherecurved= ==E Ad E AkeQRR QQ220 01241422=( ) = +(b) The closed surface encloses zero charge so Gausss law gives curved flat + =0 or flat curved = = Q2 0P24.12 Consider as a gaussian surface a box with horizontal area A, lying between 500 and 600 m elevation.E A =q d0: (+ ) + ( ) = ( )120 1001000N C N CmA AA =( )( )= 20 8 85 101001 77 101212 N C C N mmC. 2 2. m3The charge is positive , to produce the net outward fl ux of electric fi eld.Q 0FIG. P24.11 31. Gausss Law 31P24.13 The total charge is Q 6 q . The total outward fl ux from the cube is Q 6q0, of whichone-sixth goes through each face:E66 0Q q ( ) =one face .( ) Q q =E= ( ) one face6 5 00 6 00 10C N m60. . 6 22= . 18 8 12 C2kN m C6 8 85 10.P24.14 The total charge is Q 6 q . The total outward fl ux from the cube is Q 6q0, of whichone-sixth goes through each face:66 0( ) Q q =Eone face P24.15 If R d,, the sphere encloses no charge and Eq =in =00 .If Rd , the length of line falling within the sphere is 2 R2 d2so ER d = 2 2 20P24.16 EE A 2e k QRr ,.hole holeN= = ( 2) =8 99 109( 2 2 )( 10 0 10)( ) m C Cm0 1001 0062.. ( . 103 m)2E, . hole= 28 2 Nm2 CP24.17 Eq == C6170 10 = in2 2128.85 10 C N m 01.92 107 Nm2/C(a) E E ( ) = = one face. 7 E ( ) = one face1 1 92 10N m2 C663.20 MN m2 C(b) E= 19.2 MNm2 C(c) The answer to (a) would change because the flux through each face of the cube wouldnot be equal with an asymmetric charge distribution. The sides of the cube nearer thecharge would have more flux and the ones further away would have less. The answerto (b) would remain the same, since the overall flux would remain the same. 32. 32 Chapter 24Section 24.3 Application of Gausss Law to Various Charge DistributionsP24.18 (a) Ek Qra= e = 3 0(b) Ek Qra= e =( . )( . )( .)(38 99 109 26 0 10 6 0 100. ) = 3 365 kN C0 400(c) Ek Qr= e =( 8 99 10 )( 26 0 10)( ) =29 620 4001 46. ... MN C(d) Ek Qr= e =( 8 99 10 )( 26 0 10)( ) =29 620 600649. ..kN CThe direction for each electric fi eld is radially outward .P24.19 The charge distributed through the nucleus creates a fi eld at the surface equal to that of a pointcharge at its center: Ek qr= e2 .E =( 8 . 99 10 9 Nm2 C2 )( 82 1 .60 1019C)( 208 )1 32 1.20 10 15 mE = 2.33 1021 N C away from the nucleusP24.20 Note that the electric fi eld in each case is directed radially inward, toward the fi lament.(a) E( )( ) 2 2 8 99 10 90 0 10kr9 6 . .= e =N m2 C2 C m0 100.m= 16.2MN C(b) E( )( ) 2 2 8 99 10 90 0 10kr9 6 . .= e =N m2 C2 C m0 200.m= 8.09MN C(c) E( )( ) 2 2 8 99 10 90 0 10kr9 6 . .= e =2 2N m C C m1 00.m= 1.62MN CP24.21 E == ( ) =29 00 102 8 85 105080612..C mC N mk22 2 N C, upwardP24.22 (a) E= e 2 kr3 60 102 8 99 10 2 400 19049.. .. =( )(Q )Q = +9.13 107 C = +913 nC(b)E= 0Q A = =P24.23 mg qE q q = 2 2 0 0QA. . .2 2 8 85 10 0 01 9 8mgq==( )( )( ) 0 7 100126.= 2.48 C m2*P24.24 (a) A long cylindrical plastic rod 2.00 cm in radius carries charge uniformly distributedthroughout its volume, with density 5.00 C/m3. Find the magnitude of the electric fi eldit creates at a point P, 3.00 cm from its axis. As a gaussian surface choose a concentriccylinder with its curved surface passing through the point P and with length 8.00 cm.(b) We solve forE = ( . . )m) m (5 10 C/mC( .0 02 0 088 85 10 122 6 32 /N m2 m) 0.08 mkN/C=) ( ..2 0 033 77 33. Gausss Law 33P24.25 The volume of the spherical shell is430 25 0 20 3 19 10 3 3 2 . . . m m m3 ( ) ( ) = Its charge isV = (1.33 106 C m3 )(3.19 102 m3 ) = 4.25 108 CThe net charge inside a sphere containing the protons path as its equator is60 109 C 4.25 108 C = 1.02 107 CThe electric fi eld is radially inward with magnitudek qqer2r029 27Nm 1.02 CC8 99 10 104 2( ) = N C 0 252== ( )..m1.47 104For the protonF = ma eE= v2mrv = ( ) eErm1 2 1.60 10 19 C 1.47 104 N C 0.25 m== 105 94 10 271.67 kgm s 1 2. 5*P24.26 s = ( ) 1002cmm8 60 10 = . 6 C cm 8 . 60 1022 C m2E == ( )= 28 60 102 8 85 10.4 86 109 .0212. NC away from the wallSo long as the distance from the wall is small compared to the width and height of the wall, thedistance does not affect the fi eld.P24.27 If r is positive, the fi eld must be radially outward. Choose as thegaussian surface a cylinder of length L and radius r, containedinside the charged rod. Its volume is r2L and it enclosescharge r2L. Because the charge distribution is long, no electricfl ux passes through the circular end caps; EdA = EdAcos90.0 = 0.The curved surface has EdA = EdAcos0, and E must be the samestrength everywhere over the curved surface.Gausss law,E A =q d0 =, becomes E dA 2r LCurvedSurface0.Now the lateral surface area of the cylinder is 2 rL:E rLr L220( ) = Thus,E=r2 0FIG. P24.27radially away from the cylinder axis .P24.28 The distance between centers is 2 5.90 1015 m. Each produces a fi eld as if it were a pointcharge at its center, and each feels a force as if all its charge were a point at its center.FN kN ( )( ) = = = e = ( )( ) k q qr1 2292 198 99 1046 1 60 10..N m2 C2 Cm215 232 5 90 103 50 10 3 50.. . 34. 34 Chapter 24P24.29 (a)E= 0(b) Ek Qr= e =( 8 99 10 )( 32 0 10)( ) =29 620 2007 19. ... MN CE= 7.19 MN C radially outwardP24.30 Consider two balloons of diameter 0.2 m, each with mass 1 g, hanging apart witha 0.05 m separation on the ends of strings making angles of 10 with the vertical.(a) F T mg Tmgy = cos = =cos10 010 F = T sin10 F = 0 F = T sin10, sox e e mgFmg e == =( )cossin tan . .1010 10 0 001 9 kg 8 102 103 3m s N ~10 Nor1mN( 2 ) tanFek qr e(b) F= e222 10( 8 . 99 10 2 2)0 25 3N1 2 19 22 ( ) N m Cm..qq 07 C ~107 C or 100 nC(c) Ek qr= e ( . N m 2 C 2)( . C)(28 99 109 1 2 10 70 .25m) 21.7 104 N C ~10 kN C(d) Eq = 4 1 2 10N 2 2 m2 C ~10 kNm2 C = 07121 4 10..C8.85 10 C N mP24.31 (a) E( ) ( ) 2 2 8 99 109 2 00 10 6 7 00 . N m2 C2 . C . mkr= e =m0.100E = 51.4 kN C, radially outward(b) E = EAcos = E(2 r)cos0E= (5.14 104 N C)2 (0.100 m)(0.020 0 m)(1.00) = 646 Nm2 CSection 24.4 Conductors in Electrostatic EquilibriumP24.32 The fi elds are equal. The equation E = conductor0suggested in the chapter for the fi eld outside thealuminum looks different from the equation E = insulator2 0for the fi eld around glass. But its chargewill spread out to cover both sides of the aluminum plate, so the density is conductor = . Q2AThe glass carries charge only on area A, with insulator = QA. The two fi elds areQ2A 0 , the samein magnitude, and both are perpendicular to the plates, vertically upward if Q is positive.FIG. P24.30 35. Gausss Law 35q = ( ) = in EP24.33 EdA E rl20q lin2 2 0 0 r r==(a) r = 3.00 cmE= 0(b) r = 10 0. cmE= 30 0 10( )( ) =2 8 85 10 0 1005 400912. . .N C, outward(c) r = 100 cmE= 30 0 10( )( ) =2 8 85 10 1 00540912. . .N C, outward*P24.34 (a) All of the charge sits on the surface of the copper sphere at radius 15 cm. The fi eld insideis zero .(b) The charged sphere creates fi eld at exterior points as if it were a point charge at the center:E( )( ) k qr= =e228 99 109 40 10 9( ) outward = N C outward 0awayNm CC2...171 24 10 24m(c)E=( 8 99 10 9 )( 40 109)Nm CC m( 0 75)2..ou22 tward = 639 N C outward(d) All three answers would be the same. The solid copper sphere carries charge only on itsouter surface.P24.35 (a) E =0 = (8.00 104 )(8.85 1012 ) = 7.08 107 C m2 = 708 nC m2 , positive on one face and negative on the other.(b) = QAQ = A = (7 08 107 )(0 500)2 . . CQ = 1.77 107 C = 177 nC , positive on one face and negative on the other.*P24.36 Let the fl at box have face area A perpendicular to its thickness dx. The fl ux at x = 0.3 m is intothe box EA = (6 000 N/C m2)(0.3 m)2 A = (540 N/C) AThe fl ux out of the box at x = 0.3 m + dx+EA = (6 000 N/C m2)(0.3 m + dx)2 A = +(540 N/C) A + (3 600 N/C m) dx A(The term in (dx)2 is negligible.)The charge in the box is rA dx where r is the unknown. Gausss law is(540 N/C) A + (540 N/C) A + (3 600 N/C m) dx A = rA dx/0Then r = (3600 N/C m)0 = (3600 N/C m)(8.85 1012 C2/N m2) = 31.9 nC/m3P24.37 The charge divides equally between the identical spheres, with chargeQ2on each. Then theyrepel like point charges at their centers:F= e e ( )( )( + + ) =( + ) =2 2 k Q QL R Rk QL R4 28 99 10222 ( )( ) =. 9 N m CC mN2260 0 104 2012 006 22... 36. 36 Chapter 24*P24.38 The surface area is A = 4pa2. The fi eld is thenEk QaQaQA= e ===2 4200 0 It is not equal to s /20. At a point just outside, the uniformly charged surface looks just like auniform fl at sheet of charge. The distance to the fi eld point is negligible compared to the radiusof curvature of the surface.P24.39 (a) Inside surface: consider a cylindrical surface within the metal. Since E inside theconducting shell is zero, the total charge inside the gaussian surface must be zero, so theinside charge/length = .0 = + qin soqin= Outside surface: The total charge on the metal cylinder is 2 = q + q in outqout= 2 + so the outside charge/length is 3(b) E= e e ( )= = 2 3 6 3krkr r2 0radially outwardP24.40 An approximate sketch is given at the right. Note that the electric fi eldlines should be perpendicular to the conductor both inside and outside.P24.41 (a) The charge density on each of the surfaces (upper and lower) of the plate is: = =( )( ) = 11 224 00 100 5008 00 182qA...Cm08 C m2 = 80.0 nC m2(b)E = k = 8 00 108 85 10 0812 ..C mC N22 mkN C 2 k = (9.04 )k(c)E = (9.04 kN C)kAdditional ProblemsP24.42 In general,E = ayi + bzj + cxkIn the xy plane, z = 0 andE = ayi + cxkEE E A i k k= = ( + )= ==xwd ay cx dAch xdx c0h2 wx chwx02=2 2 =FIG. P24.40xyzx = 0x = wy = 0 y = hdA = hdxFIG. P24.42 37. Gausss Law 37P24.43 (a) UniformE, pointing radially outward, so E = EA. The arc lengthis ds = Rd , and the circumference is 2 r = 2 Rsin. = =( ) = 2 2 2 A rds R Rd R dR=02202sin sin( cos ) = ( cos ) 02 R2 1E1 ( )4Q =QR Rcos cos ( ) =2 1210220 [independent of R!](b) For = 90 0 . (hemisphere): EQ Q =( 1 90) =cos .2 2 0 0(c) For = 180 (entire sphere): EQ Q =( 1 cos 180) =2 0 0[Gausss Law].*P24.44 (a) qin = +3 C1 C = +2.00 C(b) The charge distribution is spherically symmetric and qin0. Thus, the fi eld is directedradially outward or to the right at point D.(c) E= e in = =k qr2.00 10 6 N/C2928 99 100 1670.( . )2 kN/C(d) Since all points within this region are located inside conducting material, E = 0 .(e) E E = d = q = =E A0 0 in 0(f) qin = +3.00 C(g) E= e = =k qrin3.00 10 69MN 228 99 100 084 21.( . ). /C to the right(radially outward).= = C + = +(h) q V in3C 5 434 1544333 .(i) E= e = =k qrin1.54 10 69MN 228 99 100 048 63.( . ). /C to the right(radially outward)(j) As in part (d), E = 0 for 10 cmr15 cm.Thus, for a spherical gaussian surface with10 cmr15 cm, q q in inner = +3 C+ = 0where qinner is the charge on the inner surfaceof the conducting shell. This yieldsqinner = 3.00 C .(k) Since the total charge on the conductingshell is q q q net outer inner = + = 1 C, we haveq q outer inner = 1 C = 1 C (3 C) = +2.00 C(l) This is shown in the fi gure to the right.FIG. P24.43Ea b c rFIG. P24.44(l) 38. 38 Chapter 24*P24.45 (a) The fi eld is zero within the metal of the shell. The exterior electric fi eld lines end at equallyspaced points on the outer surface. The charge on the outer surface is distributed uniformly.Its amount is given byEA = Q/0Q = (890 N/C) 4p (0.75 m)2 8.85 1012 C2/N m2 = 55.7 nC(b) and (c) For the net charge of the shell to be zero, the shell must carry +55.7 nC on its innersurface, induced there by 55.7 nC in the cavity within the shell. The charge in thecavity could have any distribution and give any corresponding distribution to thecharge on the inner surface of the shell. For example, a large positive charge mightbe within the cavity close to its topmost point, and a slightly larger negative chargenear its easternmost point. The inner surface of the shell would then have plentyof negative charge near the top and even more positive charge centered on theeastern side.P24.46 The sphere with large charge creates a strong fi eld to polarize the other sphere. That means itpushes the excess charge over to the far side, leaving charge of the opposite sign on the near side.This patch of opposite charge is smaller in amount but located in a stronger external fi eld, so itcan feel a force of attraction that is larger than the repelling force felt by the larger charge in theweaker fi eld on the other side.P24.47 (a)E A = ( ) =q d E r in4 20 4For ra, q r in = 33so E3 0r =For arb and cr, q Q in =So EQr=4 20 For b r c, E = 0, since E = 0 inside a conductor.(b) Let q1 = induced charge on the inner surface of the hollow sphere. Since E = 0 inside theconductor, the total charge enclosed by a spherical surface of radius b r c must be zero.Therefore, q Q 1 + = 0 and = q = 14 2 4 21 b QbLet q2 = induced charge on the outside surface of the hollow sphere. Since the hollowsphere is uncharged, we requireq q 1 2 + = 0 and = = q14 2 4 22 c QcFIG. P24.47 39. Gausss Law 39P24.48 First, consider the fi eld at distance rR from the center of a uniform sphere of positive charge(Q = +e) with radius R.( ) = r4 20 04334330 r Eq V eR== + inso EeR= 3 4 0r directed outward(a) The force exerted on a point charge q = e located at distance r from the center is thenF qE eeRreR= = r Kr = = 3 34 4 020(b) KeRk eR= e=20324 3e = = (c) F mak eR 23 , so ar r e rk eem Rr r re= = 23 2Thus, the motion is simple harmonic with frequency f= =2 k eem Re1223(d)( )8 99 10 1 60 10 15f= =( ) 2 47 10129 1.. .HzN m2 C29 2( . )R9 11 10 31 3Ckgwhich yields R3 = 1.05 1030 m3, or R = 1.02 1010 m = 102 pmP24.49 The vertical velocity component of the moving chargeincreases according tomv= mddtF yyddxdxdtqE yyv=Now dxdt x = v has the nearly constant value v. Sodqm= v vE dx y y vvqmvvy y y dE dxy= = 0 v0The radially outward component of the electric fi eld varies along the x axis, but is described by E dA = E ( d ) dx=Qy y 20Qd y =So E dx 2 0 and vv yqQm d=2 0 . The angle of defl ection is described bytanvy qQv v= =2 dm 02=qQdmvtan12 20FIG. P24.49vxvyxyq dQ 40. 40 Chapter 24P24.50 Consider the fi eld due to a single sheet and let E+ and E representthe fi elds due to the positive and negative sheets. The fi eld at anydistance from each sheet has a magnitude given by the textbook equationE+ E = =2 0(a) To the left of the positive sheet, E+ is directed toward theleft and E toward the right and the net fi eld over thisregion isE= 0 .(b) In the region between the sheets, E+ and E are both directedtoward the right and the net fi eld isE=0to the right(c) To the right of the negative sheet, E+ and E are againoppositely directed andE= 0 .P24.51 The magnitude of the fi eld due to each sheet given byEquation 24.8 isE =2 0directed perpendicular to the sheet(a) In the region to the left of the pair of sheets, both fi elds aredirected toward the left and the net fi eld isE=0to the left(b) In the region between the sheets, the fi elds due to the individual sheets are oppositelydirected and the net fi eld isE= 0(c) In the region to the right of the pair of sheets, both fi elds are directed toward the right andthe net fi eld isE=0to the rightP24.52 The resultant fi eld within the cavity is the superposition oftwo fi elds, oneE+ due to a uniform sphere of positive chargeof radius 2a, and the otherE due to a sphere of negativecharge of radius a centered within the cavity.43 r3 4 2 0r E = +so E rr r3 3 0 0+ == r3 2 r E so = 431 401E r r = r 11 3 3( ) = 010continued on next pageFIG. P24.50FIG. P24.51FIG. P24.52 41. Gausss Law 41Since r = a + r1,E3 0r a = ( )E E E 3 3 3 3r r a ai= + == + + +=00 0 0 0 a3 0 jThus, Ex = 0and E3 0ay =at all points within the cavityP24.53 Consider the charge distribution to be an unbroken charged spherical shell with uniform chargedensity s and a circular disk with charge per area . The total fi eld is that due to the wholesphere,44 0QRR= 20 24 R02 =outward plus the fi eld of the disk =2 2 0 0radiallyinward. The total fi eld is = 0 0 0 2 2outward .P24.54 The electric fi eld throughout the region is directed along x; therefore, Ewill be perpendicular to dA over the four faces of the surface which areperpendicular to the yz plane, and E will be parallel to dA over the two faceswhich are parallel to the yz plane. Therefore,E xx a x x a c = (E )A + (E )A = ( + a )ab + + (a + c) = =+ 3 2 2 3 2 22 2( )= ( + )ababc a cSubstituting the given values for a, b, and c, we fi nd E= 0.269 Nm2 C .Q E = = =0 2.38 10 12 C 2.38 pCP24.55E A = ( ) =q d E r in4 20AR Rin = ( ) = 2 2 (a) For rR, q Ar r dr054 45and EAR5 r2=50Ar r in = ( ) = 2 2(b) For rR, q Ar r dr05445and EAr =30 5FIG. P24.54 42. 42 Chapter 24P24.56 The total fl ux through a surface enclosing the charge Q isQ0.The fl ux through the disk is = E dAdisk where the integration covers the area of the disk. We must evaluate thisintegral and set it equal toQ140to fi nd how b and R are related. In the fi gure,take dAto be the area of an annular ring of radius s and width ds. The fl uxthrough dAis EdA = EdAcos = E(2 sds)cos.FIG. P24.56The magnitude of the electric fi eld has the same value at all points within the annular ring,EQrQs b== +1414 0 2 2 20and cos= =( + )brbs2 b2 1 2 .Integrate from s = 0 to s = R to get the fl ux through the entire disk.ER Qb sds ( + ) 2 2 0s bQb1 s b , disk = ( + ) =2 2 3 20 02 21 22 21 2 20 0=( + )RQ bR bThe fl ux through the disk equalsQ4 0 provided that bR2 b2 1 21( + ) 2 = .This is satisfi ed if R = 3b .P24.57E A =q a d= rr drrin0 02014 = E rardra rEar44 42220 0 020===constant magnitude(The direction is radially outward from center for positive a; radially inward for negative a.)P24.58 In this case the charge density is not uniform, and Gausss law is written asE A =1 d dV0 .We use a gaussian surface which is a cylinder of radius r, length , and is coaxial with the chargedistribution.(a) When r R, this becomes E r a rbdVr2 00 0( ) = . The element of volume is acylindrical shell of radius r, length , and thickness dr so that dV = 2 rdr. ( ) = E rr a rb222 3200 so inside the cylinder, Erarb= 00 223(b) When rR , Gausss law becomes( ) = E r a rbr drR2 0 20 0 ( ) or outside the cylinder, ERraRb= 020 223 43. Gausss Law 43P24.59 (a) Consider a cylindrical shaped gaussian surfaceperpendicular to the yz plane with one end in the yz planeand the other end containing the point x:Use Gausss law:E A =qin d0By symmetry, the electric fi eld is zero in the yz planeand is perpendicular to dAover the wall of the gaussiancylinder. Therefore, the only contribution to the integralis over the end cap containing the point x:E A =qin d0or EA( Ax )= 0so that at distance x from the mid-line of the slab, Ex =0.(b) aFm( ) = e Ememx= =e e e 0The acceleration of the electron is of the form a = 2x with = eme 0Thus, the motion is simple harmonic with frequency f eme= =212 0P24.60 Consider the gaussian surface described in the solution to problem 59.(a) For xd 2, dq = dV = Adx = CAx2dx E A =1d dq=2CA d d= EACAx dx130020038 ECd =30 24or E = i E i Cd for ; for =xd Cdx3030 24 2 24d2d(b) For x d2 2 E A =CAx x 1 d dq= = CA2x dx3 0 0030 E = i E i Cx for ; for 0 =xCxx3030330 xyz xgaussiansurfaceFIG. P24.59 44. 44 Chapter 24P24.61 (a) A point mass m creates a gravitational acceleration g = r Gmr2 at a distance rThe fl ux of this fi eld through a sphere isGmr g dA = ( ) = 4 2 4r Gm 2Since the r has divided out, we can visualize the fi eld as unbroken fi eld lines. The samefl ux would go through any other closed surface around the mass. If there are several or nomasses inside a closed surface, each creates fi eld to make its own contribution to the netfl ux according to g dA = 4 Gmin(b) Take a spherical gaussian surface of radius r. The fi eld is inward so g dA = g4 r2 cos180 = g4 r2and 4 = 443 Gm G r3 in43Then, g4 r = 4 G rand g = r G 4 2 33 Or, since = MERE433 , g= 3 or gM GrERE= M GrER3 inwardEP24.62 The charge density is determined by Q = a 43 3 = 3Q4 a3(a) The fl ux is that created by the enclosed charge within radius r:E q r rQaQra====in030303303434 33 4(b) EQ =0. Note that the answers to parts (a) and (b) agree at r = a .(c)arEQ000FIG. P24.62(c) 45. Gausss Law 45P24.63E A = ( ) =q d E r in4 20C2 N Q(a) ( ) ( ) =3 60 10 4 0 1008 85 10 3 212 . ..N C mm2 (arb)Q = 4.00 109 C = 4.00 nC(b) We take Q to be the net charge on the hollow sphere. Outside c,(+ ) ( ) = Q + Q2 00 10 4 0 5008 85 10 2 212 . ..N C mC2 Nm2(rc)Q +Q = +5.56 109 C , so Q = +9.56 109 C = +9.56 nC(c) For brc : E = 0 and q Q Q in= + = 1 0 where Q1 is the total charge on the innersurface of the hollow sphere. Thus, Q Q 1 = = +4.00 nC .Then, if Q2 is the total charge on the outer surface of the hollow sphere,Q Q Q 2 1 = = 9.56 nC 4.0 nC = +5.56 nC .P24.64 The fi eld direction is radially outward perpendicular to the axis. The fi eld strength depends onr but not on the other cylindrical coordinates q or z. Choose a gaussian cylinder of radius r andlength L. If ra,Eq =in0( ) = and E rLL20E2 r0=orE = r ( ra)2 r0 ( ) =If arb, E rLL r a L22 20+ ( )+ ( )( ) E = rr ara r b2 20 2 ( ) =If rb, E rLL b a L22 20+ ( )+ ( )() E = rb arr b2 20 2*P24.65 (a) Consider a gaussian surface in the shape of a rectangularbox with two faces perpendicular to the direction of thefi eld. It encloses some charge, so the net fl ux out of thebox is nonzero. The fi eld must be stronger on one sidethan on the other. The fi eld cannot be uniform in magnitude.(b) Now the volume contains no charge. The net fl ux out of thebox is zero. The fl ux entering is equal to the fl ux exiting.The fi eld magnitude is uniform at points along one fi eldline. The fi eld magnitude can vary over the faces of thebox perpendicular to the fi eld.+++++++FIG. P24.65(a) 46. 46 Chapter 24ANSWERS TO EVEN PROBLEMSP24.2 355 kNm2 CP24.4 (a) 2.34 kNm2 C (b) +2.34 kNm2 C (c) 0P24.6 (a) 55 7 . nC (b) The negative charge has a spherically symmetric distribution concentric withthe shell.P24.8 (a)q2 0 (b)q2 0 (c) Plane and square both subtend a solid angle of a hemisphere at thecharge.P24.10 (a) 1.36 MNm2 C (b) 678 kNm2 C (c) No; see the solution.P24.12 1.77 pC m3 positiveP24.14Q q66 0P24.16 28.2 Nm2 CP24.18 (a) 0 (b) 365 kN C (c) 1.46 MN C (d) 649 kN CP24.20 (a) 16.2 MN C toward the fi lament (b) 8.09 MN C toward the fi lament (c) 1.62 MN Ctoward the fi lamentP24.22 (a) 913 nC (b) 0P24.24 (a) A long cylindrical plastic rod 2.00 cm in radius carries charge uniformly distributed through-outits volume, with density 5.00 m C/m3. Find the magnitude of the electric fi eld it creates at apoint P, 3.00 cm from its axis. As a gaussian surface choose a concentric cylinder with its curvedsurface passing through the point P and with length 8.00 cm. (b) 3.77 kN/CP24.26 4.86 GN C away from the wall. It is constant close to the wall.P24.28 3.50 kNP24.30 (a) ~1 mN (b) ~100 nC (c) ~10 kN/C (d) ~10 kN m2/CP24.32E= Q/20 A vertically upward in each case if Q0P24.34 (a) 0 (b) 12 4 . kN C radially outward (c) 639 N Cradially outward (d) No answer changes.The solid copper sphere carries charge only on its outer surface.P24.36 31.9 nC/m3P24.38 The electric fi eld just outside the surface is given by s /0. At this point the uniformly chargedsurface of the sphere looks just like a uniform fl at sheet of charge.P24.40 See the solution.P24.42chw22 47. Gausss Law 47P24.44 (a) 2.00 m C (b) to the right (c) 702 kN/C (d) 0 (e) 0 (f ) 3.00 m C (g) 4.21 MN/Cradially outward (h) 1.54 m C (i) 8.63 MN/C radially outward ( j) 3.00 m C(k) 2.00 m C (l) See the solution.P24.46 See the solution.P24.48 (a, b) See the solution. (c) 122k eem R 3e(d) 102 pmP24.50 (a) 0 (b)0to the right (c) 0P24.52 See the solution.P24.54 0.269 Nm2 C; 2.38 pCP24.56 See the solution.P24.58 (a)00 223r a r b (b)020 223Rr a R b P24.60 (a)E = i3 d Cd for x;0 24 2E = i3 d Cd for x(b)0 24 2E = i Cxx30 3 for 0;E = i Cxx30 3 for 0P24.62 (a) Qr3 a30(b)Q0(c) See the solutionP24.64 For ra,E= l/2p0r radially outward.For arb,E= [l + rp(r2a2)]/2p0r radially outward.For rb,E= [l + rp(b2a2)]/2p0r radially outward. 48. 25Electric PotentialCHAPTER OUTLINE25.1 Electric Potential and PotentialDifference25.2 Potential Difference in a UniformElectric Field25.3 Electric Potential and PotentialEnergy Due to Point Charges25.4 Obtaining the Value of theElectric Field from the ElectricPotential25.5 Electric Potential Due toContinuous Charge Distributions25.6 Electric Potential Due to aCharged Conductor25.7 The Millikan Oil Drop Experiment25.8 Application of ElectrostaticsANSWERS TO QUESTIONSQ25.1 When one object B with electric charge is immersedin the electric fi eld of another charge or charges A, thesystem possesses electric potential energy. The energycan be measured by seeing how much work the fi eld doeson the charge B as it moves to a reference location. Wechoose not to visualize As effect on B as an action-at-a-distance,but as the result of a two-step process: ChargeA creates electric potential throughout the surroundingspace. Then the potential acts on B to inject the systemwith energy.*Q25.2 (i) The particle feels an electric force in the negative xdirection. An outside agent pushes it uphill against thisforce, increasing the potential energy. Answer (a).(ii) The potential decreases in the direction of the electricfi eld. Answer (c).*Q25.3 The potential is decreasing toward the bottom of the page, so the electric fi eld is downward.Answer (f).*Q25.4 (i) At points off the x axis the electric fi eld has a nonzero y component. At points on the negativex axis the fi eld is to the right and positive. At points to the right of x = 500 mm the fi eld is to theleft and nonzero. The fi eld is zero at one point between x = 250 mm and x = 500 mm. Answer (b).(ii) The electric potential is negative at this and at all points. Answer (c). (iii) Answer (d).(iv) Answer (d).Q25.5 To move like charges together from an infi nite separation, at which the potential energy of thesystem of two charges is zero, requires work to be done on the system by an outside agent. Henceenergy is stored, and potential energy is positive. As charges with opposite signs move togetherfrom an infi nite separation, energy is released, and the potential energy of the set of chargesbecomes negative.Q25.6 (a) The equipotential surfaces are nesting coaxial cylinders around an infi nite line of charge.(b) The equipotential surfaces are nesting concentric spheres around a uniformly charged sphere.*Q25.7 Answer (b). The potential could have any value.*Q25.8 The same charges at the same distance away create the same contribution to the total potential.Answer (b).49 49. 50 Chapter 25*Q25.9 The change in kinetic energy is the negative of the change in electric potential energy, so wework out qV = q(Vf Vi) in each case.(a) (e)(60 V 40 V) = +20 eV (b) (e)(20 V 40 V) = 20 eV(c) (e)(20 V 40 V) = +20 eV (d) (e)(10 V 40 V) = +30 eV(e) (2e)(50 V 40 V) = +20 eV (f) (2e)(60 V 40 V) = +40 eVWith also (g) 0 and (h) +10 eV, the ranking is fdc = e = ahgb.Q25.10 The main factor is the radius of the dome. One often overlooked aspect is also the humidity of theairdrier air has a larger dielectric breakdown strength, resulting in a higher attainable electricpotential. If other grounded objects are nearby, the maximum potential might be reduced.*Q25.11 (i) The two spheres come to the same potential, so q/R is the same for both. With twice theradius, B has twice the charge. Answer (d).(ii) All the charge runs away from itself to the outer surface of B. Answer (a).Q25.12 The grounding wire can be touched equally well to any point on the sphere. Electrons will drainaway into the ground and the sphere will be left positively charged. The ground, wire, and sphereare all conducting. They together form an equipotential volume at zero volts during the con-tact.However close the grounding wire is to the negative charge, electrons have no diffi culty inmoving within the metal through the grounding wire to ground. The ground can act as an infi nitesource or sink of electrons. In this case, it is an electron sink.SOLUTIONS TO PROBLEMSSection 25.1 Electric Potential and Potential DifferenceP25.1 (a) Energy of the proton-fi eld system is conserved as the proton moves from high to low poten-tial,which can be defi ned for this problem as moving from 120 V down to 0 V.Ki +Ui + E = Kf +Uf mech 0 012+ qV + = m 2 + 0 p vJ( 1 . 60 10 19 )( 120) 1.67 101 12 C V = 1 V C( 27 kg)v2pvp= 1.52 105 m s(b) The electron will gain speed in moving the other way,from Vi = 0 to Vf = 120 V: K U E K U i i f f + + = + mech0 0 012+ + = m 2 + qV e v012= (9.111031 kg)v2 + (1.60 1019 C) 120 J e ( C)ve= 6.49 106 m sP25.2 V = 14 0. V and Q NeA = = (6.02 1023 )(1.60 1019 ) = 9.63 104 CVWQ= , so W = QV = (9.63 104 C)(14.0 J C) = 1.35 MJ 50. Electric Potential 51Section 25.2 Potential Difference in a Uniform Electric FieldP25.3 E= = Vd6 ..= = 25 0 101 50 101 67 10 132. .J CmN C 67 MN CBCB = = P25.4 V V d d dB A =V VAACB A E s E s E s0 500( ) ( )0 400E cos dy E cos . dx ...180 90 00 3000 200325 0 800 260..V V = ( )( ) = + B A VP25.5 U m f i = 1 ( ) = ( ) 2122 2 9 11 10 31 1 40 105 v v . kg . m s m sJ( ) ( ) = 2 6 2183 70 106 23 10..U = qV: +6.23 1018 = (1.60 1019 )VV = 38 9 . V. The origin is at highest potential.P25.6 Assume the opposite. Then at some point A on some equipotential surface the electric fi eld has anonzero component Ep in the plane of the surface. Let a test charge start from point A and movesome distance on the surface in the direction of the fi eld component. Then V dB = E s isAnonzero. The electric potential charges across the surface and it is not an equipotential surface. Thecontradiction shows that our assumption is false, that Ep = 0, and that the fi eld is perpendicular to theequipotential surface.P25.7 (a) Arbitrarily choose V = 0 at 0. Then at other pointsV = Ex and U QV QEx e= =Between the endpoints of the motion,K U U K U U s ei s e f ( + + ) = ( + + )0 0 0 012+ + = + kx2 QEx max maxso xQEk max = 2(b) At equilibrium,F F F x s e = + = 0or kx = QESo the equilibrium position is at xQEk= .d xdt x = + =(c) The blocks equation of motion is F kx QE m22 .Let x = x QEk, or x xQEk= + ,FIG. P25.4FIG. P25.7continued on next page 51. 52 Chapter 25so the equation of motion becomes: + ( + ) k x+ =QEkQE md x QE kdt22 , ord xdtkmx22 = This is the equation for simple harmonic motion a x x = 2 with = kmThe period of the motion is then Tmk= = 22(d) K U U E K U U s ei s e f ( + + ) + = ( + + ) mech+ + = + 2 0 0 0 0122=mgx kx QEx( )kkxQEmax max maxmaxmgkP25.8 Arbitrarily take V = 0 at point P. Then the potential at the original position of the charge is E s =EL cos. At the fi nal point a, V = EL. Because the table is frictionless we have( K + U ) = ( K + U )i f 012 = v2 qEL m qEL= ( ) =2 qEL1 2 2 00 10mcoscos .v( 6 C )(300 NC )(1 50 m )(1 60 0 )0 010 0k. cos .. g= 0.300 m sP25.9 Arbitrarily take V = 0 at the initial point. Then at distance d downfi eld, where L is the rod length,V = Ed and U LEd e = .(a) K U K U i f ( + ) = ( + )0 01 222 2 40 0 10 6100+ = = =( )L LEdEdvv. Cm N C mkg mm s( )( 2 00)( ) =0 1000 400...(b) The same. Each bit of the rod feels a force of the same size as before.Section 25.3 Electric Potential and Potential Energy Due to Point ChargesP25.10 (a) Since the charges are equal and placed symmetrically, F = 0 .(b) Since F = qE = 0, E = 0 .qr e = = ( ) (c) V k2 2899 102 00 10 96..N m CC0.8002 2m V = 4.50 104 V = 45.0 kVFIG. P25.10 52. Electric Potential 53P25.11 (a) The potential at 1.00 cm isV kq1 e rN m2 C2 C8 99 109 1 60 10 191 00= =( . )( . ).= 101 44 10 27m. V(b) The potential at 2.00 cm isqr 2 eV kN m2 C2 C8 99 109 1 60 10 192 00= =( . )( . ).= 100 719 10 27m. VThus, the difference in potential between the two points is V = V V = 2 17.19 10 8 V .(c) The approach is the same as above except the charge is 1.60 1019 C. This changes thesign of each answer, with its magnitude remaining the same.That is, the potential at 1.00 cm is 1.44 107 V .The potential at 2.00 cm is 0.719 107 V, so V = V V = 2 17.19 10 8 VP25.12 (a) E= k q+ k qe exxx ( ) 1 =222 2 000.becomes E kqxqx x e = + + = 2 2( ) 22 000.Dividing by ke , 2 2 2 00 2 qx = q(x . ) x2 + 4.00x 4.00 = 0Therefore E = 0 when x = + = 4 00 16 0 16 024 83. . .. m(Note that the positive root does not correspond to a physically valid situation.)(b) Vk qxk q1 2 =x= e + e2 000.x e = + or V kqx2q= 2 000.Again solving for x, 2qx = q(2.00 x)For 0 x 2.00 V = 0 when x = 0.667 mand qxqx= 22For x0 x = 2.00 mP25.13 (a) EQr=4 0 2VQrr4VE=3 000= = =6 000 V500 V m. m(b) VQ = = ( ) 3 0004 600 0V . mQ = ( )( )= 3 0006 . 00 m 2 . 00C 9 V8.99 10 V m C 53. 54 Chapter 25P25.14 (a) UqQr==( )( ) 45 00 10 3 00 10 8 99 109 9. C . C . 00 3503 86 1097 V m CmJ( )( ) = . .The minus sign means it takes 3.86 107 J to pull the two charges apart from 35 cm to amuch larger separation.(b) VQ . C ( . V ) +4 r45 00 10 8 99 10Qr=1+2=0 10 2( ) 9 9m C ( )( )mC Vm C0 1753 00 10 9 8 99 109.. .0 175103. mV = VP25.15 V kqrVii i== ( )( 10) 18 99 10 7 00 0 010 010. 9 . 6. . .. .010 010 038 71 10 107 11 0 + V = V = MVP25.16 (a) Vk qrk qrk qre e e = + = 11222V =( . 9 2 2)( . 6)( ) +28 99 10 2 00 101 002.N m C Cm( 0 500)3 22 10 32 224.. .mV kV V = =(b) U = qV = (3.00 106 C)(3.22 104 J C) = 9.65 102 JP25.17 U qV qV qV q1qqq3 4 rrr3e= + + =+ + 4 1 4 2 4 3 4 01122 112 2 Ue= (10 0 10 ) (8 99 10 ) +0 600. 6 2 . 9.C Nm Cm0 1501+( 0 600 ) + ( 0 150)8. . 2 . 2.m m m U = e 95 JFIG. P25.15FIG. P25.16 54. Electric Potential 55*P25.18 (a) The fi rst expression, with distances squared, describes an electric fi eld. The second expressiondescribes an electric potential. Then a positive 7 nC charge is 7 cm from the origin. To createfi eld that is to the left and downward, it must be in the fi rst quadrant, with positionvector 7 cm at 70 . A negative 8 nC charge 3 cm from the origin creates an upwardelectric fi eld at the origin, so it must be at 3 cm at 90 . We evaluate the given expressions:E = 4 . 39 kN C i + 67 . 8kN Cj= 1 .50V kV(b) F = qE = 16 109 C(4.39i + 67.8j)103 N C = (7.03i 109j)105 N(c) U qV e= = 16 109 C(1.50 103 J C) = +2.40 105 J= + + += + + ( + )+ ( + +P25.19 U U U U U1 2 3 412 13 23 14 24 34 0 )= + + + U U U U U U U + + + 12U k Qsk Qsk Qs0 e e e1 11212 2 2 2 24= + k Qs22U =k Qse e5.41We can visualize the term 422+ as arising directly from the 4 side pairs and 2 face diagonalpairs.P25.20 Each charge creates equal potential at the center. The total potential is:V ( )k qRk qR= 5 e = e 5P25.21 (a) Each charge separately creates positive potential everywhere. The total potential producedby the three charges together is then the sum of three positive terms. There is no point ,at a fi nite distance from the charges, at which this total potential is zero.(b) V= e + e = e 2k qak qak qaFIG. P25.19 55. 56 Chapter 25P25.22 (a) V xk Qrk Qr(+ )+k Qx ak Qx a( ) = e + e = e +e(+ )+ (11222 2 2 )2( ) =V xk Qx ak Qa x aV xk Q ae ee+=( ) + ( )2 21 2 2 2( ) =2( ) +1 2 x ae e e e ( ) = + = +(b) V yk Qrk Qrk Qy ak Qy a( )+( )+1122( ) =V yk Qea ya yaV yk Q a y ae+ ( )( ) =1111111y a +1 P25.23 Consider the two spheres as a system.(a) Conservation of momentum: 0 1 1 2 2 = m v i + m v (i)FIG. P25.22(a)FIG. P25.22(b)or vv21 12= mmBy conservation of energy, 012121 221 12 1 22 21 2=( ) = + + ( )+k q qdm mk q qr re e v vandk q qr rk q qd = mv +mme 1 2 e1 21 22 11 12212121+ 2vv1v2 1 21 1 2 1 212 1 12 0 7=( + ) + =m k q qem m m r r d( . 00 kg)(8.99 109 Nm2 C2 )(2 106 C)(3 106 C)1 ( )( ) 0 100 0 800 8 101. kg . kg 3 m 1.00 m=10 .80 100( )= 1 1= 0 700 22..m skg 10.8 m svm vm kg= 1.55 m s(b) If the spheres are metal, electrons will move around on them with negligible energy loss toplace the centers of excess charge on the insides of the spheres. Then just before they touch,the effective distance between charges will be less than r r 1 2 + and the spheres will reallybe moving faster than calculated in (a) . 56. Electric Potential 57P25.24 Consider the two spheres as a system.(a) Conservation of momentum: 0 = m1v1i + m2v2 (i)or vv21 12= mm.By conservation of energy, 012121 221 12 1 22 21 2=( ) = + + ( )+k q qdm mk q qr re e v vandk q qr rk q qd = mv +mme 1 2 e1 21 22 11 12212121+ 2v.v1v=2 1 1 2 1 21 1 2 1 2212( + ) + =m k q qm m m r r dmme =2m k q q 1 1em m m r r d( + ) + v11 1 22 1 2 1 2(b) If the spheres are metal, electrons will move around on them with negligible energy loss toplace the centers of excess charge on the insides of the spheres. Then just before they touch,the effective distance between charges will be less than r r 1 2 + and the spheres will reallybe moving faster than calculated in (a) .P25.25 The original electrical potential energy is= = k qeed U qV qIn the fi nal confi guration we have mechanical equilibrium. The spring and electrostatic forces2on each charge are ( k qk 2 d ) + q e( ) == k q0 . Then ke. In the fi nal confi guration the total3d2 18 d3potential energy is1212 18234922322kx qVk qdd qk qdk qd+ = e ( ) + e = e . The missing energy must havebecome internal energy, as the system is isolated: k qd2 4 2k qd= +int .e e E9Ek qedint = 592P25.26 Using conservation of energy for the alpha particle-nucleus system,we have K U K U f f i i + = +But Uk q q= e gold and r Thus, U= 0i i r iiAlso Kf = 0 (vf = 0at turning point),so U K f i =ork q qrm e goldmin= 122 vr2 2(8 99 10 )(2) 79k q qemmin.gold= =29 N m2 C2v( )( . )( )( )1 60 106 64 10 2 00 1019 227 7. .Ckg m s 22 74 10 1427 4= =. .mfm 57. 58 Chapter 25P25.27 Each charge moves off on its diagonal line. All charges have equal speeds.( + ) = ( + )+ + = K U K Uk qLi fk qLme e+ 4 2024122 22 v4222 221221182 222k qLk qLk qLme ee++ == +vv k qemL2P25.28 A cube has 12 edges and 6 faces. Consequently, there are 12 edge pairs separated by s, 2 6 = 12face diagonal pairs separated by 2s, and 4 interior diagonal pairs separated by 3s.U2 212= e + + ek qsk qs =1224322 8 .Section 25.4 Obtaining the Value of the Electric Field from the Electric PotentialP25.29 V = a + bx = 10.0 V + (7.00V m)x(a) At x = 0, V = 10 0. VAt x = 3.00 m, V = 11.0 VAt x = 6.00 m, V = 32.0 V(b) EdVdx= = b = (7.00V m) = 7.00 N C in the + x directionP25.30 (a) For rR Vk QR= eEdVdr r = = 0(b) For r R Vk Qr= eE= dV= k Qe = k Qe rdrr2 r 2P25.31 V = 5x 3x2y + 2yz2Evaluate E at (1, 0, 2).EVx5 6 xy5 6 1 0 5EVyxxy= = + = + ( )( )= = = + 2 2 2 z3 2 2 31 2 2 54 40 2EVzyz z= ( ) ( ) = = = = ( )( ) =0= + + = ( ) + ( ) + =E E2 E2 E2 5 2 5 2 02 7 07 x y z . NC 58. Electric Potential 59Vs AB since =P25.32 (a) E E E(b) EVs B = = ( ) =6 2200V2 cmN Cdown(c) The fi gure is shown to the right, with sample fi eld linessketched in.P25.33 E= eVy yk Q yy y= + + ln2 2Ek Qyy2= e ey yk Q1y y y+ + + = +2 2 2 2 2 2Section 25.5 Electric Potential Due to Continuous Charge Distributionsk QR RP25.34 V V Vk QR Rk QR= = e e e+ ( ) = 15= 2 0 2 2 21 0.553k QeRP25.35 (a) [ ] = = =xCm mCm21(b) V k= + dqdxxdxLkkk L de re red xd Le = = =+0ln 1 P25.36 V= k dqxdxe=kre b + ( L x) 2 2 2Let zL= x2.Then xL= z2, and dx = dz + e =V kL z dzb zk L dzb zkzdzee( )( )+= +22 2 2 2 2 b zk Lz z b k z bVek Lee2 22 2 2 222+= ( + + )+ += lnlnLxL22x b kLxLe 2 2 20 + + + += + ( ) ++ ( ) +2202 22 22 22 2bVk L L L L bL L bLeln222222 2+ + + kLL bLb e = + ( ) + ( ) +Vk L b L Lb L Le24 24 22 22 2lnFIG. P25.32FIG. P25.35 59. 60 Chapter 25P25.37 V dVdqr= = 14 0 All bits of charge are at the same distance from O.So V1 Q= ( ) 4R=8 99 107 50 10C 2 2096..N m C0.140 mMV = 1.51 .P25.38 V kRR 3 dq +dse reR = =ke dxxkR + k all charge semi3 circledxx eR ln ln= ( ) + +V k xkRR k xV ke RRRR eRe Re=ln333 + k + k = k ( + ) e e e ln3 2ln3Section 25.6 Electric Potential Due to a Charged ConductorP25.39 (a) E = 0 ;Vk qR= e =( 8 99 10 )( 26 0 10) =0 1401 67. 9 . 6.. MV(b) Ek qr= e =( 8 99 10 )( 26 0 10)( ) =29 620 2005 84. ... MN C awayVk qR= e =( 8 99 10 )( 26 0 10) =0 2001 17. 9 . 6.. MV(c) Ek qR= e =( 8 99 10 )( 26 0 10)( ) =29 620 14011 9. ... MN C awayVk qR= e = 1.67 MVP25.40 Substituting given values into Vk qr= e7 50 10( 8 99 10 ) V0 30039... =2 2 qN m CmSubstituting q = 2.50 107C,N12 .e= 2 50 10= 1 56 10719.C1.60 10 Celectrons 60. Electric Potential 61P25.41 The electric fi eld on the surface of a conductor varies inversely with the radius of curvature of thesurface. Thus, the fi eld is most intense where the radius of curvature is smallest and vice-versa.The local charge density and the electric fi eld intensity are related byE =0or = 0 E(a) Where the radius of curvature is the greatest, = = ( )( ) =0E 8 85 10 12 2 80 104 2 min. C2 N m2 . N C 48 nC m2(b) Where the radius of curvature is the smallest, = = ( )( ) =0E 8 85 10 12 5 60 104 4 max . C2 N m2 . N C 96 nC m2P25.42 (a) Both spheres must be at the same potential according tok qrk qre 1 e122=+ = 1.20 106 Cwhere also q q 1 2q rr 1Then q2 12=q rr+ q= q2 122621 20 106= 1 20 101+..CC6 cm 2 cm= C on the smaller sphere=0 300 10161.q .20 10 6 0 300 10 6 0 900 10 61 = = C . C . CVk qer19 68 99 10 0 900 10. N m C . C26 10=( )( )=m2 21.35 105 V(b) Outside the larger sphere,E r r r 1= k q= = 1= 2 2 21111 35 105rVre . .V0.06 m5 106 V m awayOutside the smaller sphere,E r 2= 1 .35 105V = 6 .74 10 6 0.02 mV m awayThe smaller sphere carries less charge but creates a much stronger electric fi eld than thelarger sphere. 61. 62 Chapter 25Section 25.7 The Millikan Oil Drop ExperimentSection 25.8 Application of ElectrostaticsP25.43 (a) E1 1 6k Qrk Qr r= . = = Vmax max re e= 3 00 10 2 V m V E r max max = = 3.00 106 (0.150) = 450 kV(b)k Qr2 max =e max E{ k Qore max = V}rmax Q= = . ( )E rkemaxmax . .3 00 10 0 150.=2 6 298 99 107 51 CB =E sP25.44 (a) V V d B AAand the fi eld at distance r from a uniformlycharged rod (where rradius of charged rod) isE =rkr= e220In this case, the fi eld between the central wire and the coaxialcylinder is directed perpendicular to the line of charge so thatr ln = =rB Ar V Vkrdr kebrea b a 22or V krr eab= 2 ln(b) From part (a), when the outer cylinder is considered to be at zero potential, the potential ata distance r from the axis isV krr ea = 2 lnThe fi eld at r is given byE= a eVrk2 rrkerrr a= 2 =2But, from part (a), 2klnVr r e = ( )a b.Therefore, EVr r r a b= ( )ln.1FIG. P25.44 62. Electric Potential 63P25.45 (a) From the previous problem,ElnVr r r a b= ( )1We require just outside the central wire = 5 50 101 650 0 100 8503..( )ln . V mVmr r b b( ) mor 1100 850r =m 1 1 r bbln.We solve by homing in on the required valuerb (m) 0.0100 0.00100 0.00150 0.00145 0.00143 0.00142( 0 850 m110m 1) rr bbln.4.89 0.740 1.05 1.017 1.005 0.999Thus, to three signifi cant fi gures,rb = 1.42 mm(b) At ra ,50 0kV= E = ( )0 85010 8509.ln . .m 0.001 42 m m.20 kV m 63. 64 Chapter 25Additional Problems*P25.46 (a) The two particles exert forces of repulsion on each other. As the projectile approaches thetarget particle, the projectile slows. The target starts to move in the x direction. As long asthe projectile is moving faster than the second particle, the two will be approaching. Kineticenergy will be being converted into electric potential energy. When both particles movewith equal speeds, the distance between them will momentarily not be changing: this isthe instant of closest approach. Thereafter, the target particle, still feeling a forward force,will move faster than the projectile. The particles will separate again. The particles exertforces on each other but never touch. The particles will eventually be very far apart, withzero electric potential energy. All of the Ue they had at closest approach is converted backinto kinetic energy. The whole process is an elastic collision. Compare this problem withProblem 9.49 in Chapter 9.(b) Momentum is constant throughout the process. We equate it at the large-separation initialpoint and the point b of closest approach.v + v = v +vim1 1i m2 2i m1 1b m2 2b( 2 g)( 21 m s)+ 0= ( + )=2 5g g6 00m svv ibb . (c) Energy conservation between the same two points:1212k q qr i i bm m m m 2 1 20121221 122 21 2ebv + v + = ( + )v +0 002 21120 005 0 0120 007 2 2 . kg( m s) + . kg( ) + = . kg m s( )9 6 6+ . Nm CC22 C .526899 10 15 10 8 10 = J JJ mmrrrbbb0 441 0 1261 151 150 3. ....=15= 3.64 m(d) The overall elastic collision is described by conservation of momentum:v + v = v +vim m m m 1 1i 2 2i 1 1d 2 2d( g)(