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# General Physics II PHYS 1420

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This 63 page Class Notes was uploaded by Alia Wilkinson on Sunday October 25, 2015. The Class Notes belongs to PHYS 1420 at University of North Texas taught by Bibhudutta Rout in Fall. Since its upload, it has received 16 views. For similar materials see /class/229203/phys-1420-university-of-north-texas in Physics 2 at University of North Texas.

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CHAPTER 18 ELECYRIC FORCES AND ELE CIR C FIELDS ANSWERS T 0F OCUS 0N CONCEPT S QUESTIONS N E 4 V39 0 gt1 19 x 1013 b Suppose that A is positive and B is negative Since C and A also attract each other C must be negative Thus B and C repel each other because they have like charges both negative Suppose however that A is negative and B is positive Since C and A also attract each other C must be positive Again we conclude that B and C repeal each other because they have like charges both positive a The ball is electrically neutral net charge equals zero However it is made from a conducting material so it contains electrons that are free to move The rod attracts some of these negative electrons to the side of the ball nearest the rod leaving the opposite side of the ball positively charged Since the negative side of the ball is closer to the positive rod than the positive side a net attractive force arises d The fact that the positive rod repels one object indicates that that object carries a net positive charge The fact that the rod repels the other object indicates that that object carries a net negative charge Since both objects are identical and made from conducting material they share the combined net charges equally after they are touched together Since the rod repels each object after they are touched each object must then ca1ry a net positive charge But the net electric charge of any isolated system is conserved so the total net charge initially must also have been positive This means that the initial positive charge had the greater magnitude c This distribution is not possible because of the law of conservation of electric charge The total charge on the three objects here is q whereas only qwas present initially c This is an example of charging by induction The negatively charged rod repels free electrons in the metal These electrons move through the point of contact and into the sphere farthest away from the rod giving it an induced charge of q The sphere nearest the rod acquires an induced charge of q As long as the rod is kept in place while the spheres are separated these induced charges cannot recombine and remain on the spheres lqlllqzl r2 the magnitude of each charge as in A would increase the numerator by a factor of four but this is offset by the change in separation which increases the denominator by a factor of b Coulomb s law states that the magnitude of the force is given by F k Doubling 22 4 Doubling the magnitude of only one charge as in D would increase the numerator by 8 e Coulomb s law states that the magnitude of the force is given by F k Chapter 18 Answers to Focus on Concepts Questions 943 a factor of two but this is offset by the change in separation which increases the 2 denominator by a factor of 2 lqlllqzl r2 is directed along the line between the charges and is an attraction for unlike charges and a repulsion for like charges Charge B is attracted by charge A with a force of magnitude 99 up The force and repelled by charge C with a force of the same magnitude Since both forces lqllql point to the left the net force acting on B has a magnitude of 2k d2 Charge A is attracted and also by charge C with a force of k lqllqlz d 26 lqllql d2 Since both by charge B with a force of k forces point to the right the net force acting on A has a magnitude of 125k Charge lqllql d2 C is pushed to the right by B with a force of k and pulled to the left by A with a force of k Mqu 26 Since these two forces have different directions the net force acting on C has a lqllql d2 39 magnitude of 075k 9 b According to Coulomb s law the magnitude of the force that any one of the point charges lqllql d2 the triangle The charge at B experiences a repulsive force from the charge at A and an attractive force from the charge at C Both forces have vertical components but one points in the ydirection and the other in thr ydirection These vertical components have equal magnitudes and cancel leaving a resultant that is parallel to the xaxis exerts on another point charge is given by F k where dis the length of each side of 85 uC e According to Equation 182 the force exerted on a charge by an electric field is proportional to the magnitude of the charge Since the negative charge has twice the magnitude of the positive charge the negative charge experiences twice the force Furthermore the direction of the force on the positive charge is in the same direction as the field so that we can conclude that the field points due west The force on the negative charge points opposite to the field and therefore points due east 944 ELECTRIC FORCES AND ELECTRIC FIELDS k 12 c The electric eld created by a point charge has a magnitude E and is inversely r proportional to the square of the distance r If r doubles the charge magnitude must increase by a factor of 22 4 to keep the eld the same LA b To the left of the positive charge the two contributions to the total eld have opposite directions There is a spot in this region at which the eld from the smaller but closer positive charge exactly offsets the eld from the greater but more distant negative charge 5 e Consider the charges on opposite comers In all of the arrangements these are like charges This means that the two eld contributions created at the center of the square point in opposite directions and therefore cancel Thus only the charge opposite the empty comer determines the magnitude of the net eld at the center of the square Since the point charges all have the same magnitude the net eld there has the same magnitude in each arrangement 18 x10 6Cm2 V39 9 c The tangent to the eld line gives the direction of the electric eld at a point At A the tangent points due south at B southeast and at C due east gt1 a The electric eld has a greater magnitude where the eld lines are closer together They are closest together at B and farthest apart at A Therefore the eld has the greatest magnitude at B and the smallest magnitude at A 9 d In a conductor electric charges can readily move in response to an electric eld In A B and C the electric charges experience an electric eld and hence a force from neighboring charges and will move outward away from each other They will rearrange themselves so that the electric eld within the metal is zero at equilibrium This means that they will reside on the outermost surface Thus only D could represent charges in equilibrium 19 13 NmzC 20 045 NmzC Chapter 18 Problems 945 CHAPTER 18 ELECYRIC FORCES AND ELE CIR C FIELDS PROBLEMS 1 REASOMNGAND SOLUYYON The total charge ofthe electrons is q7e60 gtlt 10137160 x 10 19 C q 7 96 X 10 6 C 796 uC The net charge on the sphere is therefore qnet 8390 796 Inc 2 REASONING The charge of a single proton is 6 and the charge of a single electron is 76 where e 160X10 19 C The net charge ofthe ionized atom is the sum ofthe charges of its constituent protons and electrons SOLUTION The ionized atom has 26 protons and 7 electrons so its net electric charge q is q 26e7 e l9e l9l60gtlt10 19 C 3 REASONING a Since the objects are metallic and identical the charges on each combine and produce a net charge that is shared equally by each object Thus each object ends up with onefourth of the net charge b The number of electrons or protons that make up the nal charge on each object is equal to the nal charge diVided by the charge of an electron or proton SOLUYYON a The net charge is the algebraic sum of the individual charges The charge q on each object after contact and separation is onefourth the net charge or ql6 uC62 uC748 uC794uC 946 ELECTRIC FORCES AND ELECTRIC FIELDS b Since the charge on each object is negative the charge is comprised of electrons The number of electrons on each object is the charge q divided by the charge e of a single electron 76 Number of electrons i LOIS 8 l60 X 10 C REASONING When N electrons each carrying a charge 7 e 716X10719 C are transferred from the plate to the rod the system consisting of the plate and the rod is isolated Therefore the total charge qli qu of the system is unchanged by the process where q1i is the initial charge of the plate and qu is the initial charge of the rod At the end the rod and the plate each have the same nal charge qlf qu Therefore each must have a charge equal to half the total charge of the system qlf qu qh qu SOLUTION The final charge q2f on the rod is equal to its initial charge qu plus the charge transferred to it which is equal to the product of the number N of electrons transferred and the charge 6 of each electron Therefore 92f2921NeZi Ne 1 Since the final charge on the rod is equal to half the total initial charge of the system we can substitute qu qh qugt into Equation 1 and solve for N qzi qli Ze q1iqziq2iNe 0r Neq2iqliq2iq2iqli or N Therefore the number of electrons that must be transferred to the rod is 121 qn 20x10 6 C 30x10 6 C N 2e 216x10 19 C REASONING Identical conducting spheres equalize their charge upon touching When spheres A and B touch an amount of charge q ows from A and instantaneously neutralizes the 7q charge on B leaving B momentarily neutral Then the remaining amount of charge equal to 4q is equally split between A and B leaving A and B each with equal amounts of charge 2q Sphere C is initially neutral so when A and C touch the 2q on A splits equally to give q on A and q on C When B and C touch the 2q on B and the q on C combine to give a total charge of 3q which is then equally divided between the spheres B and C thus B and C are each left with an amount of charge 15q Chapter 18 Problems 947 SOLUYYON Taking note of the initial values given in the problem statement and summarizing the nal results determined in the REASONING above we conclude the following a Sphere C ends up with an amount of charge equal to b The charges on the three spheres before they were touched are according to the problem statement 5q on sphere A 7q on sphere B and zero charge on sphere C Thus the total charge on the spheres is 5 q q 0 c The charges on the spheres after they are touched are q on sphere A 15 q on sphere B and 15q on sphere C Thus the total charge on the spheres is q 15 q l5q REASONING The conservation of electric charge states that during any process the net electric charge of an isolated system remains constant is conserved Therefore the net charge q1 q2 on the two spheres before they touch is the same as the net charge after they touch When the two identical metal spheres touch the net charge will spread out equally over both of them When the spheres are separated the charge on each is the same SOLUYYON a Since the nal charge on each sphere is 50 LC the nal net charge on both spheres is 250 LC 100 LC The initial net charge must also be 100 uC The only spheres whose net charge is 100 uC are B qB 20 uC and D qD 120 uC b Since the nal charge on each sphere is 30 uC the nal net charge on the three spheres is 330 LC 90 uC The initial net charge must also be 90 uC The only spheres whose net charge is 90 uC are AqA 80 uC C qC 50 uC and D qD 120 uC c Since the nal charge on a given sphere in part b is 30 uC we would have to add 30 LC to make it electrically neutral Since the charge on an electron is l6 X 10 19 C the number of electrons that would have to be added is 30gtlt10 6 C Number of electrons 19 X 1013 AMINO 948 ELECTRIC FORCES AND ELECTRIC FIELDS REASONING a The number N of electrons is 10 times the number of water molecules in 1 liter of water The number of water molecules is equal to the number 11 of moles of water molecules times Avogadro s number NA N 10nNA b The net charge of all the electrons is equal to the number of electrons times the change on one electron SOLUTYON a The number Nof water molecules is equal to 10 n M where 11 is the number of moles of water molecules and NA is Avogadro s number The number of moles is equal to the mass m of 1 liter of water divided by the mass per mole of water The mass of water is equal to its density p times the volume as expressed by Equation 111 Thus the number of electrons is N10n1 10 L M 10 L NA 180 gmol 180 gmol 1000g 0 1000 kgm3l00 gtlt1073 m3 1kg J 6022gtlt1023 molil 180 gmol 335 X 1026 electrons b The net charge Q of all the electrons is equal to the number of electrons times the change on one electron Q 335 X 1026 160 gtlt10719 C 536 gtlt107 C REASONING The magnitude F of the forces that point charges q1 and q2 exert on each IququI r2 other varies with the distance r separating them according to F k Equation 18 1 Where k 899X109 Nm2C2 We note that both charges are given in units of microcoulombs LC rather than the base SI units of coulombs C We will replace the pre x uwith 10 6 when calculating the distance rfrom Equation 181 SOLUTION Solving F k IqIIIZqZI Equation 181 for the distance r we obtain r 2zqulII92I or r IqulIquI F F Chapter 18 Problems 949 Therefore when the force magnitude F is 066 N the distance between the charges must be r lqu1q2J899X109 NmzCz84x106 C56gtlt10 6 C F 066N 080 m REASONING The number N of excess electrons on one of the objects is equal to the charge q on it divided by the charge of an electron e or N q e Since the charge on the object is negative we can write q lq where lql is the magnitude of the charge The magnitude of the charge can be found from Coulomb s law Equation 181 which states that the magnitude F of the electrostatic force exerted on each object is given by F qul lql r2 where r is the distance between them SOLULUON The number N of excess electrons on one of the objects is N 7 7 u 1 To nd the magnitude ofthe charge we solve Coulomb s law F ququrz for lql HzFr2 qVc Substituting this result into Equation 1 gives F 455x10 21N180x10 3 m2 l l k 899x109 Nm2C2 N77 e e 160 X 10 19 C O REASONING The gravitational force is an attractive force To neutralize this force the electrical force must be a repulsive force Therefore the charges must both be positive or both negative Newton s law of gravitation Equation 43 states that the gravitational force depends inversely on the square of the distance between the earth and the moon Coulomb s law Equation 181 states that the electrical force also depends inversely on the square of the distance When these two forces are added together to give a zero net force the distance can be algebraically eliminated Thus we do not need to know the distance between the two bodies 950 ELECTRIC FORCES AND ELECTRIC FIELDS SOLUHON Since the repulsive electrical force neutralizes the attractive gravitational force the magnitudes of the two forces are equal qullql 2 WM r2 r2 H Electrical Gravitational orce force Equation 181 Equation 43 Solving this equation for the magnitude lql of the charge on either body we nd 2 667 gtlt10711N7HZ1 598 gtlt1024 kg735 gtlt1022 kg GMeMm kg 13 qv k 9Nm2 57lgtlt10 C V 899gtlt10 7 REASONING Initially the two spheres are neutral Since negative charge is removed from the sphere which loses electrons it then carries a net positive charge Furthermore the neutral sphere to which the electrons are added is then negatively charged Once the charge is transferred there eXists an electrostatic force on each of the two spheres the magnitude of which is given by Coulomb s law Equation 181 Fkqlllq2r2 SOLUYYON a Since each electron carries a charge of l60gtlt10 19 C the amount of negative charge removed from the first sphere is 160 gtlt10719 C 30 gtlt1013 electrons l electron 48gtlt106 C Thus the first sphere carries a charge 48 x 104 C while the second sphere carries a charge 48 X 10 6 C The magnitude of the electrostatic force that acts on each sphere is therefore F M 83999X109 NmZCz48x1m6 CIZ r2 050 m2 b Since the spheres carry charges of opposite sign the force is Chapter 18 Problems 951 12 REASONING Let F2 and F1 represent the forces exerted on the 3 charge q at the origin by the point charges q1 and q2 respectively According to Equation 181 the magnitudes of these forces are q2 given by q 725 C lqlllql lqzllql 1 1171 15 2 and FZ k 2 1 1 r 2 Fl where r1 1s the distance between q1 and q r2 1s the d1stance q84 ILC between q2 and q and k 899X109 Nm2C2 The directions of the forces are determined by the signs of each chargepair The F2 sign of q1 is opposite that of q so F1 is an attractive force pointing in the positive ydirection The signs of q2 and q are both positive so F2 is a repulsive force pointing in the negative ydirection see the drawing Because the net force F F1 F2 acting on qpoints in the positive ydirection the force F1 must have a greater magnitude than the force F2 Therefore the magnitude F of the net electric force acting on q is equal to the magnitude of the attractive force F1 minus that of the repulsive force F2 F F1 F2 2 SOLUTION Substituting Equations 1 into Equation 2 yields FzFer zkl jz quZLI ql 3 Solving Equation 3 for qul we obtain qu22qqu12qF or Iq2r22 i r2 r1 1 Substituting the given values we find that 76 25gtlt10 C 27 N qul 034 m2 022 m2 899x109 NmzCZ84x10 6 cl 952 ELECTRIC FORCES AND ELECTRIC FIELDS 13 REASONING AND SOLUTION The net y I electrostatic force on charge 3 at x 30 m is the 1 vector sum of the forces on charge 3 due to the other 30 m39k HSIJC two charges 1 and 2 According to Coulomb s law I Equation 181 the magnitude of the force on charge 5 45 I 3 due to charge 1 is 712 PC I E45PC 4 k 2 3 E3 IqIIIq3I 0 x 13 Figure 1 where the distance between charges 1 and 3 is r13 According to the Pythagorean theorem r123 x2 y2 Therefore 899x109 Nm2 CZ18x10 6 C45x10 6 C F 0405 N 13 30 m2 30 m2 Charges 1 and 3 are equidistant from the origin so that H 45 see Figure 1 Since charges 1 and 3 are both positive the force on charge 3 due to charge 1 is repulsive and along the line that connects them as shown in Figure 2 The components of F13 are F13x F13 cos 45 0286 N and F13 7Fl3 sin 45 41286 N y The second force on charge 3 is the attractive force I opposite signs due to its interaction with charge 2 I located at the origin The magnitude of the force on 1 f charge 3 due to charge 2 is according to Coulomb s law I I k k F23 quIIqsI quIIqsI 2 3 0 x2 5 15 I F23 899x109 NmZCZ12x10 6 C45x10 6 C 30 m2 0539 N Since charges 2 and 3 have opposite signs they attract each other and charge 3 experiences a force to the left as shown in Figure 2 Taking up and to the right as the positive directions we have 0286 N 0539 N 0253 N F3x Fl3xF23x F3y Fl3y 0286 N Chapter 18 Problems 953 Using the Pythagorean theorem we nd the magnitude of F3 to be 0253 N 2 2 2 2 F3 JF3X F3y J 0253 N 0286 N 038 N 0286N F3 The direction of F3 relative to the 7X axis is speci ed by the angle see Figure 3 where Figure 3 tan71 49 below the x axis 4 REASONING The electrical force that each charge exerts on charge 2 is shown in the following drawings F21 is the force exerted on 2 by l and F23 is the force exerted on 2 by 3 Each force has the same magnitude because the charges have the same magnitude and the distances are equal q F21 W W W F23q F21 W W 1221 q Q 77777 I rrrrrrrrrrrrrrrrrrrr 77 lt gt SW 1 F23 2 3 1 2 3 1 2 F23 a b 31 OW c The net electric force F that acts on charge 2 is shown in the following diagrams F23 F21 F23 F21 4 L F F i 0 N a b It can be seen from the diagrams that the largest electric force occurs in a followed by c and then by b SOLUYYON The magnitude F21 of the force exerted on 2 by l is the same as the magnitude F23 of the force exerted on 2 by 3 since the magnitudes of the charges are the same and the distances are the same Coulomb s law gives the magnitudes as 954 ELECTRIC FORCES AND ELECTRIC FIELDS qu F21F23 I 899 x109 Nm2C286 X1076 C8396 X1076 C 4 6 104 N I X 38 X 10 3 m2 In part a of the drawing showing the net electric force acting on charge 2 both F21 and F23 point to the left so the net force has a magnitude of F2Flz246x104N In part b of the drawing showing the net electric force acting on charge 2 F21 and F23 point in opposite directions so the net force has a magnitude of In part c showing the net electric force acting on charge 2 the magnitude of the net force can be obtained from the Pythagorean theorem F1IF221F223 I46gtlt104 N2 46 x104 N2 REASONING AND SOLUHON a Since the gravitational force between the spheres is one of attraction and the electrostatic force must balance it the electric force must be one of repulsion Therefore the charges must have I the same algebraic signs both positive or both negative b There are two forces that act on each sphere they are the gravitational attraction FG of one sphere for the other and the repulsive electric force FE of one sphere on the other From the problem statement we know that these two forces balance each other so that FG FE The magnitude of FG is given by Newton39s law of gravitation Equation 43 FG Gmlmz r2 while the magnitude of FE is given by Coulomb39s law Equation 181 FE kq1q2r2 Therefore we have Gm m k 12 2 I91I2Iq2I or Gm2 qu2 r r Chapter 18 Problems 955 since the spheres have the same mass m and carry charges of the same magnitude Solving for lql we nd 711 2 2 lqlzm lgzao X104 kg 667gtlt10 9 N 1112ng k 899gtlt10 Nm C 17x10 16 C 16 REASONING AND SOLUHON The electrostatic forces decreases with the square of the distance separating the charges Ifthis distance is increased by a factor of 5 then the force will decrease by a factor of 25 The new force is then 35 N F 7 014 N 25 17 REASONING The electrons transferred increase the magnitudes of the positive and negative charges from 200 2C to a greater value We can calculate the number N of electrons by dividing the change in the magnitude of the charges by the magnitude 6 of the charge on an electron The greater charge that exists after the transfer can be obtained from Coulomb s law and the value given for the magnitude of the electrostatic force SOLUYYON The number N of electrons transferred is e where Iqa erl and lqbeforel are the magnitudes of the charges after and before the transfer of we apply Coulomb s law with a value of 680 N for the electrons occurs To obtain lqa er electrostatic force Iq I2 Fr2 a er Fk r2 or lqa er k Using this result in the expression for N we find that 2 680 N 00300 m 200x10 6 C Will I N k qbeforez 899gtlt109Nm2C2 e 160x10 19 C 956 ELECTRIC FORCES AND ELECTRIC FIELDS 18 REASONING AND SOLUYYON Calculate the magnitude of each force acting on the center charge Using Coulomb s law we can write F qu4IIq3 899 gtlt109 Nm2 C2400 gtlt10 6 C300 gtlt10 6 C 43 r423 0100 m2 108 N toward the south F qu5q3 899 x109 Nm2 C2500 X 10 6 C300 gtlt1076 C 53 r523 0100 m2 135 N toward the east Adding F 43 and F53 as vectors we have FaIFf3 17523 I108 N2 135 N2 173 N F 6 tan 1 i tanil M 387 s ofE F53 135 N 19 REASONING According to Newton s second law the centripetal acceleration experienced by the orbiting electron is equal to the centripetal force divided by the electron s mass Recall from Section 53 that the centripetal force F0 is the name given to the net force required to keep an object on a circular path of radius r For an electron orbiting about two protons the centripetal force is provided almost exclusively by the electrostatic force of attraction between the electron and the protons This force points toward the center of the circle and its magnitude is given by Coulomb s law SOLUYYON The magnitude ac of the centripetal acceleration is equal to the magnitude F c of the centripetal force divided by the electron s mass ac Fcm Equation 53 The centripetal force is provided almost entirely by the electrostatic force so Fc F where F is the magnitude of the electrostatic force of attraction between the electron and the two protons Thus ac F m The magnitude of the electrostatic force is given by Coulomb s law Fqu1Hq2Ir2 Equation 181 where q1I e and q2 I2e are the magnitudes of the charges r is the radius of the orbit and k 899gtlt109 Nm2 C2 Substituting this expression for Finto ac F m and using 121 911 X 10 31 kg for the mass ofthe electron we nd that Chapter 18 Problems 957 k l el l2el F r2 k l el l2el c m m mr2 9 2 2 719 719 899gtlt10 Nm C l 160 X 10 CH2x160 X 10 Cl 911x10 31 kg265x10 11 m2 20 REASONING The drawing at the right shows the forces that act on the charges at each corner For example FAB is the force exerted on the charge at corner A by the charge at corner B The directions of the forces are consistent with the fact that like charges repel and unlike charges attract Coulomb s law indicates that all of the forces shown have the same magnitude namely F qul2 L2 where lql is the magnitude of each of the charges and Lis the length of each side of the equilateral triangle The magnitude is the same for each force because lql and Lare the same for each pair of charges The net force acting at each corner is the sum of the two force vectors shown in the drawing and the net force is greatest at corner A This is because the angle between the two vectors at A is 60 With the angle less than 90 the two vectors partially reinforce one another In comparison the angles between the vectors at corners B and C are both 120 which means that the vectors at those corners partially offset one another The net forces acting at corners B and C have the same magnitude since the magnitudes of the individual vectors are the same and the angles between the vectors at both B and C are the same 120 Thus vector addition by either the tailtohead method see Section 16 or the component method see Section 18 will give resultant vectors that have different directions but the same magnitude The magnitude of the net force is the smallest at these two corners 958 ELECTRIC FORCES AND ELECTRIC FIELDS REASONING the magnitude of any individual force vector is F kq2 L2 With this in mind we apply the component method for vector addition to the forces at comer A which are shown in the drawing at the right together with the appropriate components The Xcomponent ZFX and the F A SOLUYYON As pointed out in the y 5 Bsin 6000 F AgX ycomponent 2F of the net force are I y F AB cos 600 ZFXA FAB cos 600 FAC Fcos600 1 ZFyA FAB sin600 Fsin600 where we have used the fact that F AB F AC F The Pythagorean theorem indicates that the magnitude of the net force at comer A is 2FA 1I2Fx2Fy F2cos600 12 Fsin600 2 2 F cos600 l2sin600 2 in I2 cos600 l2sin600 2 50x10 6 C2 2 2 899x109Nm2C2 cos600 1 sin600 0030 m 430 N We now apply the component method for vector addition to the forces at FBC comer B These forces together with the appropriate components are shown in the drawing at the right We note E 6000 immediately that the two vertical FBC COS 6000 lt components cancel since they have opposite directions The two horizontal components in contrast reinforce since FB they have the same direction Thus we have the following components for the net force at comer B X A cos 600 Chapter 18 Problems 959 ZFXB BC cos 600 FBA cos 600 2F cos 600 050 where we have used the fact that FBC FBA F The Pythagorean theorem indicates that the magnitude of the net force at comer B is 2FB 2F 2 2F 2 2Fcos600 202 2Fcos600 IqIZ 50x10 6 C2 2kL 2cos 600 2899 X 109 Nm2 C2 As discussed in the REASONING the magnitude of the net force acting on the charge at comer C is the same as that acting on the charge at comer B so ZFC 250 N 0 030 my cos 600 21 REASONING a There are two electrostatic forces that act on ql that due to q2 and that due to q3 The magnitudes of these forces can be found by using Coulomb s law The magnitude and direction of the net force that acts on q1 can be determined by using the method of vector components b According to Newton s second law Equation 42b the acceleration of q1 is equal to the net force divided by its mass However there is only one force acting on it so this force is the net force SOLUHON a The magnitude F12 of the force exerted on q1 by q2 is given by Coulomb s law Equation 181 where the distance is specified in the drawing Fl 2 qulllq2 899 X 109 Nm2C2800 gtlt10 6 C500 gtlt10 6 C 0213N r122 130 m2 960 ELECTRIC FORCES AND ELECTRIC FIELDS Since the magnitudes of the charges and the distances are the same the magnitude of F13 is the same as the magnitude ofF12 or F13 0213 N From the drawing it can be seen that the Xcomponents of the two forces cancel so we need only to calculate the ycomponents of the forces Force ycomponent F12 1712 sin 230 0213 N sin 230 00832 N F13 Fl3 sin 230 0213 N sin 230 00832 N F Fy 0166 N Thus the net force is F 0166 N directed along the y axis b According to Newton s second law Equation 42b the acceleration of q1 is equal to the net force diVided by its mass However there is only one force acting on it so this force is a the net force 3 111 ms2 m 150 X107 kg where the plus sign indicates that the acceleration is along the y axis F 0166N 22 REASONING We will use Coulomb s law to calculate the force that any one charge exerts on another charge Note that in such calculations there are three separations to consider Some of the charges are a distance d apart some a distance 2d and some a distance 3d The greater the distance the smaller the force The net force acting on any one charge is the vector sum of three forces In the following drawing we represent each of those forces by an arrow These arrows are not drawn to scale and are meant only to symbolize the three different force magnitudes that result from the three different distances used in Coulomb s law In the drawing the directions are determined by the facts that like charges repel and unlike charges attract By examining the drawing we will be able to identify the greatest and the smallest net force Mm d d d A B c D The greatest net force occurs for charge C because all three force contributions point in the same direction and two of the three have the greatest magnitude while the third has the next Chapter 18 Problems 961 greatest magnitude The smallest net force occurs for charge B because two of the three force contributions cancel SOLUYYON Using Coulomb s law for each contribution to the net force we calculate the ratio of the greatest to the smallest net force as follows larl2 2512 11i 14 5qu quV k lqlz 4 72 W 2 2 2 k k k 2FC d2 d2 2F REASONING The kinetic energy of the orbiting electron is KE Emv2 Equation 62 where m and vare its mass and speed respectively We can obtain the speed by noting that the electron experiences a centripetal force whose magnitude Fe is given by F0 mv2 r Equation 53 where r is the radius of the orbit The centripetal force is provided almost exclusively by the electrostatic force of attraction F between the electron and the protons so Fc F The electrostatic force points toward the center of the circle and its magnitude is given by Coulomb s law as F qulllqzlr2 Equation 181 where qll and q2 are the magnitudes of the charges SOLUYYON The kinetic energy of the electron is 62 Solving the centripetalforce expression Fc mv2 r Equation 53 for the speed V and substituting the result into Equation 62 gives rF KEmv2VFC 1 The centripetal force is provided almost entirely by the electrostatic force so Fc F where F is the magnitude of the electrostatic force of attraction between the electron and the three protons This force is given by Coulomb s law Fkqlllq2r2 Equation 181 Substituting Coulomb s law into Equation 1 yields rpirqulllqzl 171911qu 2 r2 2r 962 ELECTRIC FORCES AND ELECTRIC FIELDS Setting Iqu I eI and quI I3eI we have KE 2 kI eII3eI 2r 899gtlt109 Nm2 CZI 160 X 10 19 CII3gtlt 160 X 10 19 CI 196X1047 J 2176 X10711m 24 REASONING When the airplane and the other end of the guideline carry point charges q and7 q the airplane is subject to an attractive electric force of magnitude 2 F kw kw kq Z Equation 181 where k 899gtlt109 N m2 C2 and r is the r r r 2 sz length of the guideline This electric force provides part of the centripetal force F02 Equation 53 necessary to keep the airplane mass 12 ying along its circular path at the higher speed V2 which is associated with the greater kinetic energy KE2 The remainder of the centripetal force is provided by the maximum tension T in the guideline so we have max that 1 When the airplane is neutral and ying at the slower speed V1 which is associated with the 2 mv smaller kinetic energy KEI there is no electrical force so the centripetal force Fcl 1 acting on the airplane is due solely to the maximum tension Tmax in the guideline mvl2 T max F01 r 2 We note that the kinetic energy of the airplane is given by KE mv2 Equation 62 so that the quotquot in the of E 1 quot J energies KE2 and KE1 of the airplane 1 and 2 are proportional to the kinetic m 2KE2 and mv12 2KE1 3 2 SOLUTION Solving F kq 2 Equation 181 for q2 and taking the square root of both r sides we obtain Chapter 18 Problems 963 Fr2 Fr2 2 or 4 q k q k Substituting Equations 3 into Equations 1 and 2 yields 2 KB 2 KB FTmaX 2 and Tm 1 5 r r Substituting the second of Equations 5 into the first of Equations 5 we obtain 2 KB 2 KB 2 KB 2 KB 2 KB KE Flt1gtlt 2 or Flt 2gtlt1gtlt 2 1 6 r r r r r Substituting Equation 6 into Equation 4 we nd that Therefore the magnitude q of the charge on the airplane is 2518 J 500 J30 m 899x109 Nm2C2 25 REASONING Consider the drawing at the right It is given that the charges qA ql and q2 are each positive Therefore the charges q1 and q2 each exert a repulsive force on the charge qA As the drawing shows these forces have magnitudes F A1 vertically downward and F A2 horizontally to the left The unknown charge placed at the empty comer of the rectangle is qU and it exerts a force on qA that has a magnitude F AU In order that the net force acting on qA point in the vertical direction the horizontal component of F AU must cancel out the horizontal force F A2 Therefore F AU must point as shown in the drawing which means that it is an attractive force and qU must be negative since qA is positive 964 ELECTRIC FORCES AND ELECTRIC FIELDS SOLUYYON The basis for our solution is the fact that the horizontal component of F AU must cancel out the horizontal force F A2 The magnitudes of these forces can be expressed using Coulomb s law F qullq39lr2 where r is the distance between the charges qand q Thus we have quAIIqUI 4d2d2 zquAIquI F 4602 AU and F A2 where we have used the fact that the distance between the charges qA and qU is the diagonal of the rectangle which is I4d I2 612 according to the Pythagorean theorem and the fact that the distance between th harges qA and q2 is 4d The horizontal component of F AU is FAU cos 9 which must be equal to FM so that we have quAIIqUI cos6 quAIIq2I or mcos8M 4502 d2 4502 17 6 The drawing in the REASONING reveals that cos64d I4d2d241 7 Therefore we nd that ij or hi W lqzl 7aoxm c 17J 1 64 6 As discussed in the REASONING the algebraic sign of the charge qU is 2 ON REASONING AND SOLUYY ON a To nd the charge on each ball we rst need to determine the electric force acting on each ball This can be done by noting that each thread makes an angle of 180 with respect to the vertical 7 o 7 74 2 o 7 73 Fe 7mgtan 18 7 80 X 10 kg980 ms tan 18 7 2547 X 10 N We also know that k k 2 F 39q39239q239 where r 2025 m sin 180 01545 In Now 27 Chapter 18 Problems 965 2547x10 3 N F k 899x109N39m2C2 MT m b The tension is due to the combination of the weight of the ball and the electric force the two being perpendicular to one another The tension is therefore T mg2 Fe2 80gtlt10 4 kg980 MSZUZ 2547gtlt10 3 NY 82 gtlt1073 N 1 1 quot V REASONING The charged insulator experiences an electric force due to the presence of the charged sphere shown in the drawing in the text The forces acting on the insulator are the downward force of gravity ie its weight W mg the electrostatic force F qullquIrz see Coulomb s law Equation 181 pulling to the right and the tension T in the wire pulling up and to the left at an angle 6with respect to the vertical as shown in the drawing in the problem statement We can analyze the forces to determine the desired quantities Hand T SOLUHON a We can see from the diagram given with the problem statement that TX F which gives Tsint9kq1q2r2 and Ty W which gives T cos 6 mg Dividing the first equation by the second yields 7 k r2 smH tang lqlllqzl T cosH mg Solving for 6 we find that 639 tan 1kq132 mgr 1111 899x109 NmzC20600gtlt10 6 C0900x10 6 C 800 gtlt10 2 kg980 ms20 150 m2 966 ELECTRIC FORCES AND ELECTRIC FIELDS b Since T COS 6 mg the tension can be obtained as follows mg 800 X 10 2 kg 980 ms2 cos 154 T cos6 28 REASONING AND SOLUYYON Since the objects attract each other initially one object has a negative charge and the other object has a positive charge Assume that the negative charge has a magnitude of lqll and that the positive charge has a magnitude of q2 Assume also that lqzl is greater than lqll The magnitude F of the initial attractive force between the objects is k F IquI 12I r 1 After the objects are brought into contact and returned to their initial positions the charge on each object is the same and has a magnitude of lqzl lqll2 The magnitude F of the k lqzl WT r force between the objects is now F 2 It is given that F is the same in Equations 1 and 2 Equations 1 and 2 and rearrange the result to nd that Therefore we can equate I 11I2 6I 12II 11IJrI 12I2 0 3 The solutions to this quadratic equation are q15828q2 and q10l7l6q2 4 Since we have assumed that q2 is greater than lqll we must choose the solution Iq1I 01716Iq2I Substituting this result into Equation 1 and using the given values of F 120 N and r 0200 m we find that q10957 uC and q2 558 uC 5 Note that we need not have assumed that q2 is greater than lqll We could have assumed that lqzl is less than lqll Had we done so we would have found that Chapter 18 Problems 967 q1 558 uC and q20957 uC 6 Considering Equations 5 and 6 and remembering that q1 is the negative charge we conclude that the two possible solutions to this problem are q1 0957 uC and q 558 uCl or q1 558 uC and q 0957 uC 2 O SOLUHON Knowing the electric eld at a spot allows us to calculate the force that acts on a charge placed at that spot without knowing the nature of the object producing the eld This is possible because the electric eld is de ned as E Fqo according to Equation 182 This equation can be solved directly for the force F if the eld E and charge qo are known SOLUYYON Using Equation 182 we nd that the force has a magnitude of F Eq0 260 000 NC70 gtlt1076 C18 N If the charge were positive the direction of the force would be due west the same as the direction of the eld But the charge is negative so the force points in the opposite direction or due east Thus the force on the charge is 18 N due east REASONING a The magnitude of the electric eld is obtained by dividing the magnitude of the force obtained from the meter by the magnitude of the charge Since the charge is positive the direction of the electric eld is the same as the direction of the force b As in part a the magnitude of the electric eld is obtained by dividing the magnitude of the force by the magnitude of the charge Since the charge is negative however the direction of the force as indicated by the meter is opposite to the direction of the electric eld Thus the direction of the electric eld is opposite to that of the force SOLUYYON a According to Equation 182 the magnitude of the electric eld is M Elql 20010 As mentioned in the REASONING the direction of the electric eld is the same as the direction of the force or 968 ELECTRIC FORCES AND ELECTRIC FIELDS b The magnitude of the electric eld is F 200 uN E77 20 NC q 100 C Since the charge is negative the direction of the electric eld is opposite to the direction of the force or Thus the electric elds in parts a and b are the same LA I i REASONING The electric eld created by a point charge is inversely proportional to the square of the distance from the charge according to Equation 183 Therefore we expect the distance r2 to be greater than the distance r1 since the eld is smaller at r2 than at r1 The ratio rZrl then should be greater than one SOLU770N Applying Equation 183 to each position relative to the charge we have qul qul d E r12 an 2 r22 E1 Dividing the expression for E1 by the expression for E2 gives E1 quIr12 22 E2 quIrzz r12 Solving for the ratio r2r1 gives V2 IE1 248 NC 37 r1 E2 132 NC As expected this ratio is greater than one 32 REASONINGAND SOLUYYON The electric eld lines must originate on the 4q positive charges and terminate on the negative charges They cannot cross one another Furthermore the number of eld lines beginning or ending on any charge must be proportional to the magnitude of the charge If 10 electric eld lines leave the 5q charge then siX lines must originate from the 3q charge and eight lines must end on each Jiq charge The drawing shows the electric eld lines that meet these criteria 3q Chapter 18 Problems 969 LA REASONING Each charge creates an electric eld at the center of the square and the four elds must be added as vectors to obtain the net eld Since the charges all have the same magnitude and since each comer is equidistant from the center of the square the magnitudes qul of the four individual elds are identical Each is given by Equation 183 as E 2 The r directions of the various contributions are not the same however The eld created by a positive charge points away from the charge while the eld created by a negative charge points toward the charge SOLUHON The drawing at the right shows each of the eld contributions at the center of the square see black dot Each is directed along a diagonal of the square Note that ED and EB point in opposite directions and therefore cancel since they have the same magnitude In contrast E A and EC point in the same direction toward comer A and therefore combine to give a net eld that is twice the magnitude of E A or EC In other words the net eld at the center of the square is given by the following vector equation ZEEAEBECEDEAEBEC EBEAEC 21A Using Equation 183 we nd that the magnitude of the net eld is 2E2EA 970 ELECTRIC FORCES AND ELECTRIC FIELDS In this result r is the distance from a comer to the center of the square which is one half of the diagonal distance 039 Using Lfor the length of a side of the square and taking advantage of the Pythagorean theorem we have r d L2 L2 With this substitution for r the magnitude of the net eld becomes qul 4qu 4899x109 NmZCZ24x10 12 C 2 2 IL2L2 L EE2 0040 m2 REASONING Part a of the drawing given in the text The electric eld produced by a charge points away from a positive charge and toward a negative charge Therefore the electric eld E 2 produced by the 20 uC charge points away from it and the electric elds E73 and E75 produced by the 30 HC and 50 uC charges point toward them see the lefthand side of the following drawing The magnitude of the electric eld produced by a point charge is given by Equation 183 as E quI12 Since the distance from each charge to the origin is the same the magnitude of the electric eld is proportional only to the magnitude lql of the charge Thus the X component E X of the net electric eld is proportional to 50 uC 20 uC 30 uC Since only one ofthe charges produces an electric eld in the y direction the ycomponent E yof the net electric eld is proportional to the magnitude of this charge or 50 uC Thus the Xand ycomponents are equal as indicated at the righthand side of the following drawing where the net electric eld E is also shown 750 C A E75 E E 20 uC E2 730 HC 0 gt E 39 a E X Part b of the drawing given in the text Using the same arguments as earlier we nd that the electric elds produced by the four charges are shown at the lefthand side of the following drawing These elds also produce the same net electric eld E as before as indicated at the righthand side of the following drawing Chapter 18 Problems 971 AE l0pC E6 M E El E EON 710 MC a E4 EX E a 60uC SOLUHON Part a of the drawing given in the text The net electric eld in the Xdirection is 899 gtlt109 Nm2C220 gtlt1076 C 899 gtlt109 Nm2C230 gtlt1076 C X 0061 m2 0061 m2 12 gtlt107 NC The net electric eld in the ydirection is 899x109 Nm2C250gtlt10 6 C 7 Ey 2 l2gtlt10 NC 006m The magnitude of the net electric eld is 2 2 EJE E 12x107NC l2gtlt107NC l7gtlt107NC Part 17 of the drawing given in the text The magnitude of the net electric eld is the same as determined for part a E 17 gtlt107 NC REASONING AND SOLUHON a In order for the eld to be zero the point cannot be between the two charges Instead it must be located on the line between the two charges on the side of the positive charge and away from the negative charge If Xis the distance from the positive charge to the point in question then the negative charge is at a distance 30 m 1 meters from this point For the eld to be zero here we have kw or w 30mx2 x2 30mx2 x 972 ELECTRIC FORCES AND ELECTRIC FIELDS Solving for the ratio of the charge magnitudes gives qil q 160 uC30mx2 or 4030 mx2 40 uC x2 39 x2 Suppressing the units for convenience and rearranging this result gives 40x2 30x2 or 40x2 9060xx2 or 3x2 60x 900 Solving this quadratic equation for xwith the aid of the quadratic formula see Appendix C4 shows that X 30 m or X l0 m We choose the positive value for 2 since the negative value would locate the zero eld spot between the two charges where it can not be see above Thus we have b Since the eld is zero at this point the force acting on a charge at that point is 36 REASONING a The magnitude E of the electric eld is given by E 0 80 Equation 184 where O39is the charge density or charge per unit area and go is the permittivity of free space b The magnitude F of the electric force that would be exerted on a K ion placed inside the membrane is the product of the magnitude lqol of the charge and the magnitude E of the electric eld see Equation 182 or F IqOIE SOLUHON a The magnitude of the electric eld is 76 2 E 7391X10 mm 80x105 NC 0 885x10 12 CZNm2 b The magnitude Fofthe force exerted on a K4r ion q0 6 is Fq0EIeIEIL60 X 10 19 CI80gtlt105 NC Chapter 18 Problems 973 37 REASONING a The drawing shows the two point charges q1 and q2 Point A is located at X 0 cm and point B is at X 60 cm A E 4 0 gt q E ql 2 2 Since q1 is positive the electric eld points away from it At point A the electric eld E1 points to the left in the Xdirection Since q2 is negative the electric eld points toward it At point A the electric eld E2 points to the right in the Xdirection The net electric eld is E E1 E2 We can use Equation 183 Ekqr2 to nd the magnitude of the electric eld due to each point charge b The drawing shows the electric elds produced by the charges q1 and q2 at point B which is located at X 60 cm A B q2 30 cm 30 cm 30 cm 0 rrrrrrrrrrrrrrrrrrrrrrrrr E1 E2 Since q1 is positive the electric eld points away from it At point B the electric eld points to the right in the Xdirection Since q2 is negative the electric eld points toward it At point B the electric eld points to the right in the Xdirection The net electric eld is E E1 E2 SOLUHON a The net electric eld at the origin point A is E E1 E2 E E1E2 L2qllm I r l 2 899 gtlt109 Nm2C285 gtlt1076 C 899 gtlt109 Nm2C221gtlt10 6 C 30 gtlt1072 m2 90 gtlt1072 m2 62 gtlt107 NC 974 ELECTRIC FORCES AND ELECTRIC FIELDS The minus sign tells us that the net electric eld points along the XaXis b The net electric eld at X 60 cm point B is E E1 E2 kq kq my EE1E2 2 r1 V2 899 gtlt109 Nm2C285 X 10 6 C 899 gtlt109 Nm2C221gtlt10 6 C 30 X 10 2 m2 29 gtlt108 NC The plus sign tells us that the net electric eld points along the XaXis 30 gtlt10 2 m2 REASONING At every position in space the net electric eld E is the vector sum of the external electric eld Eext and the electric eld Epoint created by the point charge at the origin E Eext Epoim eld is uniform which means that it has the same magnitude Eext 4500 NC at all locations and that it always points in the positive Xdirection see the drawing The magnitude Ep of the electric eld due to The external electric 0th qul 0th r 2 the point charge at the origin is given by EF Equation 183 where r is the distance between the origin and the location where the electric eld is to be evaluated q is the charge at the origin and k 899X109 Nm2C2 All three locations given in the problem are a distance r 015 m from the origin so we have that qul 899x109 Nm2C2 80gtlt10 9 C E point z 2 2 3200 NC r 015 m The direction of the electric eld Epoint varies from location to location but because the charge q is negative Elwint is always directed towards the origin see the drawing Chapter 18 Problems 975 SOLUTION a At X 7015 m the electric eld of the point charge and the external electric eld both point in the positive Xdirection see the drawing so the magnitude E of the net electric eld is the sum of the magnitudes of the individual electric elds E Eext Epoint 4500 NC 3200 NC 7700 NC b At X 015 m the electric eld of the point charge is opposite the external electric eld see the drawing Therefore the magnitude E of the net electric eld is the difference between the magnitudes of the individual electric elds E Eext Epoint 4500 NC 3200 NC 1300 NC c At y 015 m the electric elds Epoint and Eext are perpendicular see the drawing This makes them in effect the Xcomponent Em and y component Elmm of the net electric eld E The magnitude E of the net electric eld then is given by the Pythagorean theorem Equation 17 E Jngt E oim J4500 Nc2 3200 me2 5500 NC 0 REASONING Since the charged droplet charge q is suspended motionless in the electric eld E the net force on the droplet must be zero There are two forces that act on the droplet the force of gravity W mg and the electric force F qE due to the electric eld Since the net force on the droplet is zero we conclude that mg We can use this reasoning to determine the sign and the magnitude of the charge on the droplet SOLUTION a Since the net force on the droplet is zero and the weight of magnitude W F points downward the electric force of magnitude F qE must point upward Since the electric eld points upward the excess charge on the dro let must be ositive in order for the force F to oint u ward p p p W b Using the expression mg lqlE we nd that the magnitude of the excess charge on the droplet is 79 2 lqlz z 350gtlt10 kg980 ms 2404X1042 C E 8480 NC 976 ELECTRIC FORCES AND ELECTRIC FIELDS The charge on aproton is 160 X 10 19 C so the excess number of protons is l proton 404x10 12 C 19 253x107 protons 160x10 C 4 O REASONING Since the proton and the electron have the same charge magnitude 6 the electric force that each experiences has the same magnitude The directions are different however The proton being positive experiences a force in the same direction as the electric eld due east The electron being negative experiences a force in the opposite direction due west Newton s second law indicates that the direction of the acceleration is the same as the direction of the net force which in this case is the electric force The proton s acceleration is in the same direction due east as the electric eld The electron s acceleration is in the opposite direction due west as the electric eld Newton s second law indicates that the magnitude of the acceleration is equal to the magnitude of the electric force divided by the mass Although the proton and electron experience the same force magnitude they have different masses Thus they have accelerations of different magnitudes SOLUYYON According to Newton s second law Equation 42 the acceleration a of an object is equal to the net force divided by the object s mass m In this situation there is only one force the electric force F so it is the net force According to Equation 182 the magnitude of the electric force is equal to the product of the magnitude of the charge and the magnitude of the electric eld or F lqOIE Thus the magnitude of the acceleration can be written as IquE m a E m The magnitude of the acceleration of the proton is 167x10 27 kg 77gtlt1012 ms2 m The magnitude of the acceleration of the electron is lqolE 160 gtlt10 19 C80 gtlt104 NC W a 14x1016ms2 m Chapter 18 Problems 977 41 REASONING AND SOLUYYON Figure 1 at the right shows q q the con guration given in text Figure 1820a The electric eld 1 2 at the center of the rectangle is the resultant of the electric elds at the center due to each of the four charges As discussed in Conceptual Example 11 the magnitudes of the electric eld at the center due to each of the four charges are equal However the elds produced by the charges in corners 1 and 3 are in opposite directions Since they have the same magnitudes they 4 3 combine to give zero resultant q q The elds produced by the charges in corners 2 and 4 point in the same direction toward corner 2 Thus EC EC2 EC4 Flgure 1 where EC is the magnitude of the electric eld at the center of the rectangle and EC2 and EC4 are the magnitudes of the electric eld at the center due to the charges in corners 2 and 4 respectively Since both EC2 and EC4 have the same magnitude we have EC 2 ECZ The distance r from any of the charges to the center 1 2 of the rectangle can be found using the Pythagorean theorem see Figure 2 at the right d 500 cm of 300 cm2500 cm2 583 cm 9 d 4 3 Therefore r 292 cm 292 gtlt1072 m 2 300 cm Figure 2 The electric eld at the center has a magnitude of 2k 9 2 2 712 EC 2EC2 I2612 2899gtlt10 N m C 28620gtlt10 C 181X102 MC V 292gtlt10 m Figure 3 at the right shows the con guration given in text q q Figure 18201 All four charges contribute a nonzero 1 T T 2 component to the electric eld at the center of the rectangle As discussed in Conceptual Example 11 the contribution E E from the charges in corners 2 and 4 point toward corner 2 and 13 24 the contribution from the charges in corners 1 and 3 point C toward corner 1 4 3 q q Notice also the magnitudes of E24 and E13 are equal and from the rst part of this problem we know that Figure3 2 EuiE13 7 181 X 10 NC 978 ELECTRIC FORCES AND ELECTRIC FIELDS The electric eld at the center of the rectangle is the vector sum of E24 and E13 The X components of E24 and E13 are equal in magnitude and opposite in direction hence E19 5724 0 Therefore EC E13y E24y 2 Elgy 2 E1 sin6 From Figure 2 we have that 500 cm 500 cm d 583 cm EC 2E13sin6 2181gtlt102 NC0858 sin6 0858 and 42 REASONINGANDSOLUYYON The magnitude of the force on q1 due to q2 is given by Coulomb s law k 91 92 F12 I II I 1 12 The magnitude of the force on q1 due to the electric eld of the capacitor is given by 039 F1c Iq1IEC 2 0 Equating the right hand sides of Equations 1 and 2 above gives qulIIq2I g r122 Iqu go Solving for r12 gives k r12 I go OIqZI 712 2 2 9 2 2 6 II885X10 C N m 899110 sz C 500810 C 553x1072m V 130x10 Cm 43 Chapter 18 Problems 979 REASONING The magnitude E of the electric eld is the magnitude F of the electric force exerted on a small test charge divided by the magnitude of the charge E F According to Newton s second law Equation 42 the net force acting on an object is equal to its mass mtimes its acceleration a Since there is only one force acting on the object it is the net force Thus the magnitude of the electric eld can be written as Eziz 9 9 The acceleration is related to the initial and nal velocities VO and V and the time tthrough V VO Equation 24 as a Substituting this expression for a into the one above for E gives V VO l r W lql larlt SOLUYYON The magnitude E of the electric eld is E mv V0 90 gtlt1075 kg20gtlt103 ms 0ms qt 75 gtlt1076 C096 s 44 REASONING The external electric eld E exerts a force FE qE Equation 182 on the sphere where q 66 uC is the net charge of the sphere The external electric eld E is directed upward so the force FE it exerts on the positively charged sphere is also directed upward see the freebody diagram Balancing this upward force are two downward forces the weight mg of the sphere where m is the mass of the sphere and g is the acceleration F due to gravity and the force FS exerted on the sphere by the spring see the freebody diagram We know that the spring exerts a downward force on the sphere because the equilibrium length L0059m of the spring is shorter than its unstrained length 10 0074 m The magnitude FS of the spring force is given by ES kx Equation 102 without the minus sign where k is the spring constant of the spring and X 10 7L 0074m 7 0059 m 0015 m is the distance by which the spring has been compressed Freebody diagram 980 ELECTRIC FORCES AND ELECTRIC FIELDS SOLUTION Solving FE qE Equation 182 for the magnitude E of the external electric eld we nd that E 1 Q lr The sphere is in equilibrium so the upward force FE must exactly balance the two downward forces mg and FS Therefore the magnitudes of the three forces are related by FE mg FS 2 Substituting Equation 2 into Equation 1 yields F E amp 3 q Lastly substituting FS kx Equation 102 without the minus sign into Equation 3 we obtain the desired electric field magnitude mgkx 51x10 3 kg980 ms224 Nm0015 m E 6 q 66x10 C 13x104 NC 4 UI REASONING The E2 E1 drawing shows the arrangement of the 111 T three charges Let Eq represent the djg Eq d electr1c field at the empty comer due to g 6 W2 i the 7q charge Furthermore let E1 and E2 be the electric fields at the empty comer due to charges q1 and q2 respectively lt 2d gt According to the Pythagorean theorem the distance from the charge 7q to the empty comer 201 2 0112 50112 015 The magnitude of each electric field is given by Equation 183 E qulrz Thus the magnitudes of each ofthe along the diagonal is given by electric fields at the empty comer are given as follows qul 4 r2 CAgr 5612 Chapter 18 Problems 981 qull 2602 kq kg and E2 The angle 6that the diagonal makes with the horizontal is 9 tan 1d2d 2657quot Since the net electric eld Enet at the empty comer is zero the horizontal component of the net eld must be zero and we have M7 qulcos 2657quot 0 E1 iEq cos 2657 0 or 4612 5612 Similarly the vertical component of the net eld must be zero and we have k 39 o EziEqsin 2657 0 or imw d 5512 These last two expressions can be solved for the charge magnitudes lqll and q2 SOLUYYON Solving the last two expressions for lqll and lq2l we nd that q1 25 cos 2657quot 46 REASONINGAND SOLUYYON From kinematics Vy2 0 ms The acceleration of the electron is given by 7 2 7 V0y 2a Since the electron starts from rest v0y a y 5 m m where e and m are the electrons charge magnitude and mass respectively and E is the magnitude of the electric eld The magnitude of the electric eld between the plates of a parallel plate capacitor is E 050 where O39is the magnitude of the charge per unit area on each plate Thus ay tea12150 Combining this expression for a with the kinematics V22 E y y mso equation we have 982 ELECTRIC FORCES AND ELECTRIC FIELDS Solving for V gives 3 2 V E y mgo 2160x10 19 C18x10 7 CmZ15x10 2 m I 911x10 31 kg885x10 12 CZNm2 10x107 ms 4 1 REASONING The fact that the net electric eld points upward along the vertical aXis holds the key to this problem The drawing at the right shows the elds from each charge together with the horizontal components of each The reason that the net eld points upward is that these horizontal components point in opposite directions and cancel Since they cancel they must have equal magnitudes a fact that will quickly lead us to a solution SOLUYYON Setting the magnitudes of the horizontal components of the elds equal gives E2 sin 600 E1 sin 300 The magnitude of the electric eld created by a point charge is given by Equation 183 Using this expression for E1 and E2 and noting that each point charge is the same distance r from the center of the circle we obtain qulI r2 kI 12I sin 600 sin300 r2 or lqzlsin 600 q1sin 300 Solving for the ratio of the charge magnitudes gives qul sin 300 q1 sin600 039577 48 REASONING a The drawing at the right shows the electric elds at point P due to the two charges in the case that the second charge is positive The presence of the 92 11 P E2 d d I E1 Chapter 18 Problems 983 second charge causes the magnitude of the net field at P to be twice as great as it is when only the rst charge is present Since both elds have the same direction the magnitude of E2 must then be the same as the magnitude of E1 But the second charge is further away from point P than is the first charge and more distant charges create weaker elds To offset the weakness that comes from the greater distance the second charge must have a greater magnitude than that of the rst charge b The drawing at the right shows the E electr1c elds at pomt P due to the two C3 Cq12 PE1 charges in the case that the second 039 039 charge is negative The presence of the second charge causes the magnitude of the net field at P to be twice as great as it is when only the rst charge is present Since the elds now have opposite directions the magnitude of E2 must be greater than the magnitude of E1 This is necessary so that E2 can offset E1 and still lead to a net eld with twice the magnitude as E1 To create this greater eld E2 the second charge must now have a greater magnitude than it did in question a SOLUHON a The magnitudes of the eld contributions of each charge are given according to k Equation 183 as E With q2 present the magnitude of the net eld at P is twice what r it is when only q1 is present Using Equation 183 we can express this fact as follows M kl hl 22M or M M 612 d2 d2 Solving for q2 gives lq2 4q1 4050 uC 20 uC Thus the second charge is q2 20 W b Now that the second charge is negative we have k92 M2M 0 qu2 3 d2 d2 d2 Solving for q2 gives q212qll12050 uC60 uC Thus the second charge is q2 60 w 984 49 ELECTRIC FORCES AND ELECTRIC FIELDS REASONING Since we know the initial velocity and time we can determine the particle s displacement from an equation of kinematics provided its acceleration can be determined The acceleration is given by Newton s second law as the net force acting on the particle divided by its mass The net force is the electrostatic force since the particle is moving in an electric eld The electrostatic force depends on the particle s charge and the electric eld both of which are known SOLUYYON To obtain the displacement Xof the particle we employ Equation 35a from the equations of kinematics x voxtaxt2 We use this equation because two of the variables the initial velocity VOXand the time L are known The initial velocity is zero since the particle is released from rest The acceleration ax can be found from Newton s second law as given by Equation 42a as the net force EFX acting on the particle divided by its mass m ax ZFX m Only the electrostatic force F X acts on the proton so it is the net force Setting ZFX Fx in Newton s second law gives ax Fx m Substituting this result into Equation 35a we have that F l 2 l x 2 x voxt axt v0xt 7m t 1 Since the particle is moving in a uniform electric eld EX it experiences an electrostatic force FX given by Fx q0 E x Equation 182 where go is the charge Substituting this expression for F X into Equation 1 gives JR 76 0 nus16x10 2 slzxmcmmr6x102 s2 38x10 5 kg F E x v0xtit2 v0xtq0 x m m REASONING The following drawing shows the two particles in the electric eld EX They are separated by a distance d Ifthe particles are to stay at the same distance from each other after being released they must have the same acceleration so axl ag 2 According to Newton s second law Equation 42a the acceleration aXof each particle is equal to the net force ZFXacting on it divided by its mass m or ax ZFX m Chapter 18 Problems 985 d 0 BE X q1 70 uC q2 18 uC m1 14 gtlt1075 kg m2 26 gtlt1075 kg SOLUYYON The net force acting on each particle and its resulting acceleration are 11 The charge q1 experiences a force qlEx due to the electric eld see Equation 182 The charge also experiences an attractive force in the Xdirection due to the presence of q2 This force is given by Coulomb s law as kqlllq2d2 see Equation 181 The net force acting on q1 is EFX1 qlEx k7 qufzi The acceleration of q1 is 2F 1 qlEx k iiquiiqui X am m1 m1 The charge q2 experiences a force qux due to the electric eld It also experiences an attractive force in the X direction due to the presence of ql This force is given by Coulomb s law as kq1q2d2 The net force acting on q2 is lqlll hl EFX 2 qux k 0112 The acceleration of q2 is Settlng axl axz g1ves qlEx k7lq16lllle qux ailquil m1 m2 986 ELECTRIC FORCES AND ELECTRIC FIELDS Solving this expression for d we nd that HqI i1 2 m1 m2 E Li X m2 m1 2 899x109 N 1211 70X1076 CHHMM C C 14x10 kg 26gtlt10 kg 65m 76 76 2500 NC WLOSC 26x10 kg 14x10 kg 51 REASONINGAM SOLUHON The net electric eld at point Pin Figure 1 is the vector sum of the elds E and E which are due respectively to the charges q and g These elds are shown in Figure 2 P 1 1 W P 2d I Figure 1 Figuiez According to Equation 183 the magnitudes of the elds E and E7 are the same since the triangle is an isosceles triangle with equal sides of length Z Therefore E E7 qulZZ The vertical components of these two elds cancel while the horizontal components reinforce leading to a total eld at point Pthat is horizontal and has a magnitude of E E cosa E cosa2Mcosa P 7 2 At point Min Figure 1 both E and E are horizontal and point to the right Again using Equation 183 we nd k k 2k EM EE7 I2qI d d d Chapter 18 Problems 987 Since EMF 90 we have 2 EM 2kqd 1 90 EP 2kqcos0 2 cos ad2 2 But from Figure l we can see that d cos 0c Thus it follows that 13 90 or cosa3l90 048 COS a The value for dis then a cos 1048 REASONING The net charge q carried by the particle determines the magnitude F of the electrical force that the external electric eld of magnitude E exerts on the particle via F qE Equation 182 The electric eld is horizontal so the electric force acting on the particle is also horizontal Taking east as the positive Xdirection the electric force gives the particle a horizontal component of acceleration aX Newton s second law holds that the F particle s horizontal acceleration is ax Equation 42a where m is the mass of the m particle Combining Equation 182 and Equation 42a we obtain ax 1 E m l r q We will use Equation 1 to determine q Note that the algebraic sign of q is the same as that of ax which we will obtain by using kinematics The particle s horizontal acceleration is related to its horizontal displacement x by x Voxtaxt2 Equation 35a where VOX is the horizontal component of its initial velocity v0 and t is the elapsed time The particle s initial velocity is entirely horizontal so that VOX v0 and Equation 35a becomes 1 2 x v0t axt 2 In order to determine the elapsed time t we consider the vertical component of the particle s projectile motion Because the particle is launched horizontally the vertical component voy of its initial velocity is zero and the vertical component ay of its acceleration is that due to gravity 657980 ms2 where we have taken upward as the positive y direction Therefore from y Voyt ayt2 Equation 35b we have that 988 ELECTRIC FORCES AND ELECTRIC FIELDS y0 mstayt2ayt2 3 where y 7671X1073 m is the particle s vertical displacement The value of yis negative because the particle moves downward SOLUTION Solving Equation 1 for q we obtain max 4 q E Solving Equation 2 for ax yields 2 x v t 2 0 axt x vot or ax t2 5 Substituting Equation 5 into Equation 4 we nd that 2m x v t q 2 0 I 6 E t Solving Equation 3 for tyields 2 671x10 3 m t22y or t 2y 200370s ay ay 980 ms Therefore from Equation 6 the charge carried by the particle is 2mxvot 2150x10 6 kg0160 m 880 ms00370 s q 2 Et 392x10 7 C 925 NC00370 s2 REASONING AND SOLUYYON Since the thread makes an angle of 3000 with the vertical it can be seen that the electric force on the ball Fe and the gravitational force mg are related by F8 mgtan 3000 The force Fe is due to the charged ball being in the electric eld of the parallel plate capacitor That is Chapter 18 Problems 989 Fe Elqball 1 where lqbaul is the magnitude of the ball39s charge and E is the magnitude of the eld due to the plates According to Equation 184 E is A 184 where q is the magnitude of the charge on each plate and A is the area of each plate Substituting Equation 184 into Equation 1 gives qlqballl F m tan300 e g 8014 Solving for q yields soAmg tan 300 Iqballl 885 gtlt10712 C2 N39m200150 m2650 gtlt1073 kg980 ms2 tan 300 0150x10 6 C 325 X10 C 54 REASONINGAND SOLUHON Gauss Law is given by text Equation 187 GDE 2 0 where Qis the net charge enclosed by the Gaussian surface 35 10 6 C an X 40gtlt105 Nm2C 885x10 12 C2Nm2 23 0 6 C b anTJ 726x10 Nm C 885gtlt10 C N m 35 10 6 C 23 10 6 C an W 14x105 Nm2C 885x10 12 C2Nm2 990 55 ELECTRIC FORCES AND ELECTRIC FIELDS REASONING As discussed in Section 189 the electric ux QDE through a surface is equal to the component of the electric eld that is normal to the surface multiplied by the area of the surface QDE ELA where E L is the component of E that is normal to the surface of area A We can use this expression and the gure in the text to determine the ux through the two surfaces SOLUYYON a The ux through surface 1 is QDE 1 E cos 35 A1 250 NCcos 35 l7 m2 350 N m2C b Similarly the ux through surface 2 is an2 Ecos 55 A2 250 NCcos 55 32 m2 460 Nm2C REASONING The electric ux QDE through the circular surface is determined by the angle between the electric field and the normal to the surface as well as the magnitude E of the electric field and the area Aof the surface DE Ecos A 186 We will use Equation 186 to determine the angle SOLUTION Solving Equation 186 for the angle we obtain d3 cosg E or cosilgj 1 EA EA The surface is circular with a radius r so its area is A 7272 Making this substitution in Equation 1 yields q cos71 cos 1 EA En39r 78Nm2C I 71 144x104 NC7r0057 m REASONING AND SOLUYYON Since the electric field is uniform its magnitude and direction are the same at each point on the wall The angle between the electric field and the normal to the wall is 35 Therefore the electric ux is an Ecos A 150 NCcos35 59 m25 m 18gtlt103 Nm2C Chapter 18 Problems 991 58 REASONING The charge Q inside the rectangular box is related to the electric ux GDE that passes through the surfaces of the box by Gauss law Q SOQDE Equation 187 where so is the permittivity of free space The electric ux is the algebraic sum of the ux through each ofthe six surfaces SOLUYYON The charge inside the box is QgoqgtE goqgt1qgt2 qgt3qgt4qgt5qgt6 2 2 2 2 885x10 12 C 21500N m 2200N m 4600N m N m 2 2 2 1800N m 3500N m 5400N m C C C 21x10 8 C 59 REASONING The electric ux through each y E face of the cube is given by QDE Ecos A see Section 189 where E is the magnitude of the electric field at the face Ais the area of the face and is the angle between the electric field and the outward normal of that face We can use this expression to x calculate the electric ux QDE through each of the six faces of the cube SOLUHON a On the bottom face of the cube the outward normal points parallel to the iyaxis in the opposite direction to the electric field and 180 Therefore anbottom 1500 NCcos1800020 m2 W On the top face of the cube the outward normal points parallel to the yaxis and I 000 The electric ux is therefore q 1500 NC cos 00 020m2 60x101N39m2C E top 992 ELECTRIC FORCES AND ELECTRIC FIELDS On each of the other four faces the outward normals are perpendicular to the direction of the electric eld so 90 So for each ofthe four side faces ansides 1500 NCcos 90 020 m2 0 Nm2 C b The total ux through the cube is QB total 2 QB top qDE bottom qDE side 1 qDE side 2 qDE side 3 qDE side 4 Therefore antotal 60x101N39m2C40 X101N39m2C0000 0 N m2 C 60 REASONING Gauss Law 2E cos AAg Equation 187 relates the electric field 8 0 magnitude E on a Gaussian surface to the net charge Q enclosed by that surface GDE 2E cos AA Equation 186 is the electric ux through the Gaussian surface divided into many tiny sections of area AA and so is the permittivity of free space We are to determine the magnitude E of the electric field due to electric charges that are spread uniformly over the surfaces of two concentric spherical shells The electric field due to these charges possesses spherical symmetry so we will choose Gaussian surfaces in the shape of spheres concentric with the shells The radius r of each Gaussian surface will be equal to the distance from the common center of the shells and will be the distance at which we are to evaluate the electric field Because the electric field has spherical symmetry the magnitude E of the electric field is constant at all points on any such spherical Gaussian surface Furthermore the electric field is directed either radially outward if the net charge within the Gaussian surface is positive or radially inward if the net charge within the Gaussian surface is negative This means that the angle between the electric field and the normal to any such spherical Gaussian surface is either 000 or 180 The quantity E cos therefore is constant and may be factored out of the summation in Equation 186 EZEcos AAEcos ZAl 1 The sum 2AA of all the tiny sections of area AA that compose a spherical Gaussian surface is the total surface area 2AA A 47rr 2 of a sphere of radius r Thus Equation 1 becomes QDE Ecos ZAA 47239r2E cos 2 Chapter 18 Problems 993 SOLUTION a The outer shell has a radius r2 015 m and we are to determine the electric eld at a distance r020m from the common center of the shells Therefore we will choose a spherical Gaussian surface radius r 020 m that encloses both shells and shares their common center According to Gauss law and Equation 2 we have that the net electric ux GDE through this sphere is Z Ecos AA47239r2Ecos g 3 80 Solving Equation 3 for E yields Q 4 f 4723980r cos Because the chosen Gaussian surface encloses both shells the net charge Q enclosed by the surface is Q q1 qz The positive charge q2 on the outer shell has a larger magnitude than the negative charge ql on the inner shell so that Q is a positive net charge Therefore the electric field is directed and the angle between the electric field and the normal to the surface of the spherical Gaussian surface is 00quot Therefore Equation 4 gives the electric field magnitude as Q q1q2 16gtlt10 6 C51gtlt10 6 c E 47r80r2cos 4723980r2 cos 4n885x10 12 CzNm2020 m2 cos00 79x105 NC b We again choose a spherical Gaussian surface concentric with the shells this time of radius r 010 m The radius of this sphere is greater than the radius r1 0050 m of the inner shell but less than the radius r2 015 m of the outer shell Therefore this Gaussian surface is located between the two shells and encloses only the charge on the inner shell Q ql This is a negative charge so that the electric field is directed radiall inward and the angle between the electric field and the normal to the surface of the Gaussian sphere is 180 From Equation 4 then we have that E ql 16x10 6 C 47r80r2 cos 47r80r2cos 47r885x10 12 CzNm2010 m2cos180 14x106 NC 994 ELECTRIC FORCES AND ELECTRIC FIELDS c Choosing a spherical Gaussian surface with a radius of r 0025 m we see that it is entirely inside the inner shell r1 0050 m Therefore the enclosed charge is zero Q 0 C Equation 4 shows that the electric eld at this distance from the common center is zero 0C E 2 2 47r 0r cos 47r 0r cos REASONING We use a Gaussian surface that is a sphere centered within the solid sphere The radius r of this surface is smaller than the radius R of the solid sphere Equation 187 gives Gauss law as follows 2Ecos AAg 50 187 Electric flux DE We will deal first with the left side of this equation and evaluate the electric ux DE Then we will evaluate the net charge Qwithin the Gaussian surface SOLUYYON The positive charge is Normal spread uniformly throughout the solid sphere and therefore is spherically Angle between E IE symmetric Consequently the electric and the normalis 00 field is directed radially outward and for each element of area AA is perpendicular to the surface This means that the angle between the normal to the surface and the field is 0 as the drawing shows Furthermore the electric field has the same magnitude everywhere on the Gaussian surface Because of these considerations we can write the electric ux as follows EEcos AAEEcos0 Al EEAA The term 2AA is the entire area of the spherical Gaussian surface or 47112 With this substitution the electric ux becomes ZEcos AAE2AAE4Irr2 1 Now consider the net charge Q within the Gaussian surface This charge is the charge density times the volume m encompassed by that surface 617 4 3 9 3 Q X 7239r 2 g REI 3 I R3 W H Charge Volume of denslty Gaussmn surface Chapter 18 Problems 995 Substituting Equations 1 and 2 into Equation 187 gives Eumz 50 Rearranging this result shows that M L 47rr2 50 47r50R3 REASONING Because the charge is distributed uniformly along the straight wire the electric eld is directed radially outward as the following end view of the wire illustrates Gausssian ng straight Wire Gaussian cylind 39 E cylinder 4 lt L gt End View And because of symmetry the magnitude of the electric eld is the same at all points equidistant from the wire In this situation we will use a Gaussian surface that is a cylinder concentric with the wire The drawing shows that this cylinder is composed of three parts the two at ends 1 and 3 and the curved wall 2 We will evaluate the electric ux for this threepart surface and then set it equal to 050 Gauss law to nd the magnitude of the electric eld SOLUYYON Surfaces l and 3 7 the at ends of the cylinder 7 are parallel to the electric eld so cos cos 90O 0 Thus there is no ux through these two surfaces 1433 0Nm2C Surface 2 7 the curved wall 7 is everywhere perpendicular to the electric eld E so cos cos 00 1 Furthermore the magnitude E of the electric eld is the same for all points on this surface so it can be factored outside the summation in Equation 186 q 2E cos 0 AAE2A The area 2A of this surface is just the circumference 211 r of the cylinder times its length L EA 211 rL The electric ux through the entire cylinder is then 996 ELECTRIC FORCES AND ELECTRIC FIELDS an qgt1qgt2 qgt3 0E27rrL0E27rrL Following Gauss law we set GDE equal to 050 where Q is the net charge inside the Gaussian cylinder E211 rL 050 The ratio QL is the charge per unit length of the wire and is known as the linear charge density 1 A Q L Solving for E we nd that Q L A r 27r 0 r 63 REASONING AND SOLUYYON The electric eld lines must originate on the positive charges and terminate on the negative charge They 3q cannot cross one another Furthermore the number of eld lines beginning or terminating on any charge must be proportional to the magnitude of the charge Thus for every eld line that leaves the charge q two eld lines must leave the charge 2q These three lines must terminate on the 3q charge If the sketch is to have six Jrq 2q eld lines two of them must originate on q and four of them must originate on the charge 2q 6 4 REASONING a The magnitude of the electrostatic force that acts on each sphere is given by Coulomb s law as F qullquIrz where lqll and lqzl are the magnitudes of the charges and r is the distance between the centers of the spheres b When the spheres are brought into contact the net charge after contact and separation must be equal to the net charge before contact Since the spheres are identical the charge on each after being separated is onehalf the net charge Coulomb s law can be applied again to determine the magnitude of the electrostatic force that each sphere experiences SOLUHON a The magnitude of the force that each sphere experiences is given by Coulomb s law as F kq12q2 899 X 109 Nm2C2200 X107 C500 X 10 6 C V 250 X 10 2 m Because the charges have opposite signs the force is Chapter 18 Problems 997 b The net charge on the spheres is 200 uC 500 uC 300 uC When the spheres are brought into contact the net charge after contact and separation must be equal to the net charge before contact or 300 uC Since the spheres are identical the charge on each after being separated is onehalf the net charge so q1 q2 150 uC The electrostatic force that acts on each sphere is now F kq12q2 899 gtlt109 Nm2C2150 x10 6 C150 gtlt1076 C 2 r 250 gtlt1072 m2 Since the charges now have the same signs the force is 6 5 REASONINGAND SOLUYYON The 2q of charge initially on the sphere lies entirely on the outer surface When the q charge is placed inside of the sphere then a a charge will still be induced on the interior of the sphere An additional q will appear on the outer surface giving a net charge of 6 ON REASONING Two forces act on the charged ball charge q they are the downward force of gravity mg and the electric force F due to the presence of the charge q in the electric field E In order for the ball to oat these two forces must be equal in magnitude and opposite in direction so that the net force on the ball is zero Newton s second law Therefore F must point upward which we will take as the positive direction According to Equation 182 F qE Since the charge q is negative the electric field E must point downward as the product qE in the expression F qE must be positive since the force F points upward The magnitudes of the two forces must be equal so that mg qE This expression can be solved for E SOLUYYON The magnitude of the electric field E is 0012 kg980 ms2 Ez 18x10 6 C 65 gtlt103 NC lql As discussed in the reasoning this electric field points 998 ELECTRIC FORCES AND ELECTRIC FIELDS 67 REASONING The drawing at the right shows the setup Here the electric eld E points along the y axis and applies a force of F to the q charge and a force of 7F to the g charge where q 80 uC denotes the magnitude of each charge Each force has the same magnitude of F E lql according to Equation 182 The torque is measured as discussed in Section 91 According to Equation 91 the torque produced by each force has a magnitude given by the magnitude of the force times the lever arm which is the perpendicular distance between the point of application of the force and the axis of rotation In the drawing the zaxis is the axis of rotation and is midway between the ends of the rod Thus the lever arm for each force is half the length Lof the rod or D2 and the magnitude of the torque produced by each force is E lql l 2 SOLUYYON The F and the 7F force each cause the rod to rotate in the same sense about the Z axis Therefore the torques from these forces reinforce one another Using the expression E q 2 for the magnitude of each torque we find that the magnitude of the net torque is Magnitude of EqlJElqlgjElqlL net torque 50 gtlt103 NC80 X104 C40 m 6 00 REASONING The magnitude of the electrostatic force that acts on particle 1 is given by Coulomb s law as F qulllqzlrz This equation can be used to find the magnitude lqzl of the charge SOLUYYON Solving Coulomb s law for the magnitude lqzl of the charge gives 2 l 899 gtlt109 Nm2C235 gtlt1076 C Since ql is positive and experiences an attractive force the charge q2 must be 6 O REASONING Each particle will experience an electrostatic force due to the presence of the other charge According to Coulomb s law Equation 181 the Chapter 18 Problems 999 magnitude of the force felt by each particle can be calculated from F qullquIrz where lqll and q2 are the respective charges on particles 1 and 2 and r is the distance between them According to Newton39s second law the magnitude of the force experienced by each particle is given by F ma where a is the acceleration of the particle and we have assumed that the electrostatic force is the only force acting SOLUYYON a Since the two particles have identical positive charges the data for particle l q1q2q and we have using 9392 r2 1 l we nd that 2 q ljj b Since each particle experiences a force of the same magnitude From Newton39s third law we can write F1 F2 or mlal 1212a2 Solving this expression for the mass 1212 ofparticle 2 Solving for lg 600x104quot kg 460x103 ms2 260x10 2 m2 l 899x109 Nm2C2 we have m mlal 600x10 6 kg460gtlt103 m2 2 850x103 ms2 325x104quot kg 02 7 O REASONINGAND SOLUHON The electric eld is de ned by Equation 182 E Fqo Thus the magnitude of the force exerted on a charge q in an electric eld of magnitude E is given by FME D The magnitude of the electric eld can be determined from its xand ycomponents by using the Pythagorean theorem 2 2 E lEf Ej 600gtlt103 NC 800gtlt103 NC l00gtlt104 NC a From Equation 1 above the magnitude of the force on the charge is F 750 gtlt1076 C100 X104NC 75 gtlt1072 N 1000 ELECTRIC FORCES AND ELECTRIC FIELDS b From the de ning equation for the electric eld it follows that the direction of the force on a charge is the same as the direction of the eld provided that the charge is positive Thus the angle that the force makes with the XaXis is given by E 3 g m4 z m4 X 600gtlt103 NC 71 REASONING The two charges lying on the XaXis produce no net electric eld at the coordinate origin This is because they have identical charges are located the same distance from the origin and produce electric elds that point in opposite directions The electric eld produced by q3 at the origin points away from the charge or along the y direction The electric eld produced by q4 at the origin points toward the charge or along the y direction The net electric eld is then E 7E3 E4 where E3 and E4 can be determined by using Equation 183 SOLUYYON The net electric eld at the origin is k k E E3 E4 r3 r4 899 gtlt109 Nm2C230 X 10 6 C 899 gtlt109 Nm2C280 X 10 6 C 72 2 72 2 50x10 m 70gtlt10 m 39gtlt106 NC The plus sign indicates that lthe net electric eld points along the y directionl 72 REASONING The unknown charges can be determined using Coulomb s law to express the electrostatic force that each unknown charge exerts on the 400 W charge In applying this law we will use the fact that the net force points downward in the drawing This tells us that the unknown charges are both negative and have the same magnitude as can be understood with the help of the freebody diagram for the 400 W charge that is shown at the right The diagram shows Chapter 18 Problems 1001 the attractive force F from each negative charge directed along the lines between the charges Only when each force has the same magnitude which is the case when both unknown charges have the same magnitude will the resultant force point vertically downward This occurs because the horizontal components of the forces cancel one pointing to the right and the other to the left see the diagram The vertical components reinforce to give the observed downward net force SOLUHON Since we know from the REASONING that the unknown charges have the same magnitude we can write Coulomb s law as follows k 400x106 CqA k 400x106 CqB Iquot Iquot The magnitude of the net force acting on the 400 w charge then is the sum of the magnitudes of the two vertical components F cos 300o shown in the freebody diagram 2Fk cos300 k os300 r r 400 10 6 C 2kX 2qAcos 300 r Solving for the magnitude of the charge gives 2Fr2 lqu 6 2k400 gtlt10 Ccos300 2 405 N00200 m 260X106 C 2899 gtlt109 N m2 C2400gtlt 10 6 Ccos300 Thus we have qA qB 260gtltlOT6 C 73 REASONING The electric field is given by Equation 182 as the force F that acts on a test charge qo divided by qo Although the force is not known the acceleration and mass of the charged object are given Therefore we can use Newton s second law to determine the force as the mass times the acceleration and then determine the magnitude of the field directly from Equation 182 The force has the same direction as the acceleration The direction of the field however is in the direction opposite to that of the acceleration and force This is 1002 ELECTRIC FORCES AND ELECTRIC FIELDS because the object carries a negative charge while the eld has the same direction as the force acting on a positive test charge SOLUYYON According to Equation 182 the magnitude of the electric eld is F E Iqu According to Newton s second law the net force acting on an object of mass m and acceleration a is 2F ma Here the net force is the electrostatic force F since that force alone acts on the object Thus the magnitude of the electric eld is F m 30x10 3kg25x103ms2 E lqol lqol 34x10 6 c 22le NC The direction of this eld is opposite to the direction of the acceleration Thus the eld points along the XaXis REASONING The magnitude of the electric eld between the plates of a parallel plate U capacitor is given by Equation 184 as E where 01s the charge dens1ty for each plate 8 0 and 50 is the permittivity of free space It is the charge density that contains information about the radii of the circular plates for charge density is the charge per unit area The area ofa circle is The second capacitor has a greater electric eld so its plates must have the greater charge density Since the charge on the plates is the same in each case the plate area and hence the plate radius must be smaller for the second capacitor As a result we expect that the ratio r2r1 is less than one SOLUYYON Using lql to denote the magnitude of the charge on the capacitor plates and A 7272 for the area of a circle we can use Equation 184 to express the magnitude of the eld between the plates of a parallel plate capacitor as follows E g Iql 0 807272 Applying this result to each capacitor gives E IqI 2 and E2 IqI 2 07171 507172 a First capacitor Second capacitor Chapter 18 Problems 1003 Dividing the expression for E1 by the expression for E2 gives 5 lql 507 12 i E2 lqlSOIL39VZ r12 Solving for the ratio rZr1 gives E 5 L2 1 M 1 E2 38gtlt105 NC As expected this ratio is less than one V39 REASOMNGAND SOLUYYON Before the spheres have been charged they exert no forces on each other After the spheres are charged each sphere experiences a repulsive force F due to the charge on the other sphere according to Coulomb s law Equation 181 Therefore since each sphere has the same charge the magnitude F of this force is F qulllqzl 899 gtlt109 Nm2C2160 X 10 6 C2 r2 0100 m2 230 N The repulsive force on each sphere compresses the spring to which it is attached The magnitude of this repulsive force is related to the amount of compression by Equation 101 FxApphed kx Setting F XAppled F and solving for k we find that F 230 N k 920 Nm x 00250 m 9 REASONINGAND SOLUHON In order for the net force on any charge to be directed inward toward the center of the square the charges must be placed with alternate and 7 signs on each successive comer The magnitude of the force on any charge due to an adjacent charge located at a distance r is 2 qul2 899 gtlt109 Nm2C220 X 10 6 C r2 030 m2 F 040 N The forces due to two adjacent charges are perpendicular to one another and produce a resultant force that has a magnitude of F adj acent le2 2040 N2 057 N 1004 ELECTRIC FORCES AND ELECTRIC FIELDS The magnitude of the force due to the diagonal charge that is located at a distance of rxE is quotIQII2 MM020N 2 Fdiagonal 52 2r2 since the diagonal distance is rxE The force Fdiagonal is directed opposite to Fadjacent the diagonal charges are of the same sign Therefore the net force acting on any of the charges is directed inward and has a magnitude Fnet Fadjacent T Fdiagonal 03957 N T 03920 N since

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