23 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 Charged Particles in a Uniform Electric Field Electric Fields 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 When the comb is nearby, molecules in the paper are polarized, similar to the molecules in the wall in Figure 23.5a, and the paper is attracted. During contact, charge from the comb is transferred to the paper by conduction. Then the paper has the same charge as the comb, and is repelled. Q23.3 The clothes dryer rubs dissimilar materials together as it tumbles the clothes. Electrons are transferred from one kind of molecule to another. The charges on pieces of cloth, or on nearby objects charged by induction, can produce strong electric fields that promote the ionization process in the surrounding air that is necessary for a spark to occur. Then you hear or see the sparks. Q23.4 To avoid making a spark. Rubber-soled shoes acquire a charge by friction with the floor and could discharge with a spark, possibly causing an explosion of any flammable material in the oxygenenriched atmosphere. Q23.5 Electrons are less massive and more mobile than protons. Also, they are more easily detached from atoms than protons. Q23.6 The electric field due to the charged rod induces charges on near and far sides of the sphere. The attractive Coulomb force of the rod on the dissimilar charge on the close side of the sphere is larger than the repulsive Coulomb force of the rod on the like charge on the far side of the sphere. The result is a net attraction of the sphere to the rod. When the sphere touches the rod, charge is conducted between the rod and the sphere, leaving both the rod and the sphere like-charged. This results in a repulsive Coulomb force. Q23.7 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. 12 Electric Fields Q23.8 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. Q23.9 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 23.5a, 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. Q23.10 The electric field due to the charged rod induces a charge in the aluminum foil. If the rod is brought towards the aluminum from above, the top of the aluminum will have a negative charge induced on it, while the parts draping over the pencil can have a positive charge induced on them. These positive induced charges on the two parts give rise to a repulsive Coulomb force. If the pencil is a good insulator, the net charge on the aluminum can be zero. Q23.11 So the electric field created by the test charge does not distort the electric field you are trying to measure, by moving the charges that create it. Q23.12 With a very high budget, you could send first a proton and then an electron into an evacuated region in which the field exists. If the field is gravitational, both particles will experience a force in the same direction, while they will experience forces in opposite directions if the field is electric. On a more practical scale, stick identical pith balls on each end of a toothpick. Charge one pith ball + and the other –, creating a large-scale dipole. Carefully suspend this dipole about its center of mass so that it can rotate freely. When suspended in the field in question, the dipole will rotate to align itself with an electric field, while it will not for a gravitational field. If the test device does not rotate, be sure to insert it into the field in more than one orientation in case it was aligned with the electric field when you inserted it on the first trial. Q23.13 The student standing on the insulating platform is held at the same electrical potential as the generator sphere. Charge will only flow when there is a difference in potential. The student who unwisely touches the charged sphere is near zero electrical potential when compared to the charged sphere. When the student comes in contact with the sphere, charge will flow from the sphere to him or her until they are at the same electrical potential. Q23.14 An electric field once established by a positive or negative charge extends in all directions from the charge. Thus, it can exist in empty space if that is what surrounds the charge. There is no material at point A in Figure 23.23(a), so there is no charge, nor is there a force. There would be a force if a charge were present at point A, however. A field does exist at point A. Q23.15 If a charge distribution is small compared to the distance of a field point from it, the charge distribution can be modeled as a single particle with charge equal to the net charge of the distribution. Further, if a charge distribution is spherically symmetric, it will create a field at exterior points just as if all of its charge were a point charge at its center.Chapter 23 3 Q23.16 The direction of the electric field is the direction in which a positive test charge would feel a force when placed in the field. A charge will not experience two electrical forces at the same time, but the vector sum of the two. If electric field lines crossed, then a test charge placed at the point at which they cross would feel a force in two directions. Furthermore, the path that the test charge would follow if released at the point where the field lines cross would be indeterminate. Q23.17 Both figures are drawn correctly. E1 and E2 are the electric fields separately created by the point charges q1 and q2 in Figure 23.14 or q and –q in Figure 23.15, respectively. The net electric field is the vector sum of E1 and E2 , shown as E. Figure 23.21 shows only one electric field line at each point away from the charge. At the point location of an object modeled as a point charge, the direction of the field is undefined, and so is its magnitude. Q23.18 The electric forces on the particles have the same magnitude, but are in opposite directions. The electron will have a much larger acceleration (by a factor of about 2 000) than the proton, due to its much smaller mass. Q23.19 The electric field around a point charge approaches infinity as r approaches zero. Q23.20 Vertically downward. Q23.21 Four times as many electric field lines start at the surface of the larger charge as end at the smaller charge. The extra lines extend away from the pair of charges. They may never end, or they may terminate on more distant negative charges. Figure 23.24 shows the situation for charges +2q and –q. Q23.22 At a point exactly midway between the two changes. Q23.23 Linear charge density, λ, is charge per unit length. It is used when trying to determine the electric field created by a charged rod. Surface charge density, σ, is charge per unit area. It is used when determining the electric field above a charged sheet or disk. Volume charge density, ρ, is charge per unit volume. It is used when determining the electric field due to a uniformly charged sphere made of insulating material. Q23.24 Yes, the path would still be parabolic. The electrical force on the electron is in the downward direction. This is similar to throwing a ball from the roof of a building horizontally or at some angle with the vertical. In both cases, the acceleration due to gravity is downward, giving a parabolic trajectory. Q23.25 No. Life would be no different if electrons were + charged and protons were – charged. Opposite charges would still attract, and like charges would repel. The naming of + and – charge is merely a convention. Q23.26 If the antenna were not grounded, electric charges in the atmosphere during a storm could place the antenna at a high positive or negative potential. The antenna would then place the television set inside the house at the high voltage, to make it a shock hazard. The wire to the ground keeps the antenna, the television set, and even the air around the antenna at close to zero potential. Q23.27 People are all attracted to the Earth. If the force were electrostatic, people would all carry charge with the same sign and would repel each other. This repulsion is not observed. When we changed the charge on a person, as in the chapter-opener photograph, the person’s weight would change greatly in magnitude or direction. We could levitate an airplane simply by draining away its electric charge. The failure of such experiments gives evidence that the attraction to the Earth is not due to electrical forces.4 Electric Fields Q23.28 In special orientations the force between two dipoles can be zero or a force of repulsion. In general each dipole will exert a torque on the other, tending to align its axis with the field created by the first dipole. After this alignment, each dipole exerts a force of attraction on the other. SOLUTIONS TO PROBLEMS Section 23.1 Properties of Electric Charges *P23.1 (a) The mass of an average neutral hydrogen atom is 1.007 9u. Losing one electron reduces its mass by a negligible amount, to 1.007 9e1.660 × 10−27 kgj− 9.11 × 10−31 kg = 1.67 × 10−27 kg . Its charge, due to loss of one electron, is 0 − 1e−1.60 × 10−19 Cj = +1.60 × 10−19 C . (b) By similar logic, charge = +1.60 × 10−19 C mass = 22.99e1.66 × 10−27 kgj− 9.11 × 10−31 kg = 3.82 × 10−26 kg (c) charge of Cl− = −1.60 × 10−19 C mass = 35.453e1.66 × 10−27 kgj+ 9.11 × 10−31 kg = 5.89 × 10−26 kg (d) charge of Ca++ = −2e−1.60 × 10−19 Cj = +3.20 × 10−19 C mass = 40.078e1.66 × 10−27 kgj− 2e9.11 × 10−31 kgj = 6.65 × 10−26 kg (e) charge of N3− = 3e−1.60 × 10−19 Cj = −4.80 ×10−19 C mass = 14.007e1.66 × 10−27 kgj+ 3e9.11 ×10−31 kgj = 2.33 ×10−26 kg (f) charge of N4+ = 4e1.60 ×10−19 Cj = +6.40 × 10−19 C mass = 14.007e1.66 × 10−27 kgj− 4e9.11 × 10−31 kgj = 2.32 × 10−26 kg (g) We think of a nitrogen nucleus as a seven-times ionized nitrogen atom. charge = 7e1.60 × 10−19 Cj = 1.12 × 10−18 C mass = 14.007e1.66 × 10−27 kgj− 7e9.11 × 10−31 kgj = 2.32 × 10−26 kg (h) charge = −1.60 × 10−19 C mass = 2b1.007 9g+ 15.999 1.66 × 10−27 kg + 9.11 × 10−31 kg = 2.99 × 10−26 kgChapter 23 5 P23.2 (a) N =FH G IK J× FH G IK J FH G IK J= × 10 0 6 02 1023 47 2 62 1024 . . . grams 107.87 grams mol atoms mol electrons atom (b) # . electrons added . C 1.60 10 C electron = = × × = × − − Qe 1 00 10 6 25 10 3 19 15 or 2.38 electrons for every 109 already present . Section 23.2 Charging Objects by Induction Section 23.3 Coulomb’s Law P23.3 If each person has a mass of ≈ 70 kg and is (almost) composed of water, then each person contains N ≅ FH GIK J× FH GIK JFH GIK J≅ × 70 000 6 02 1023 10 2 3 1028 grams 18 grams mol molecules mol protons molecule . . protons . With an excess of 1% electrons over protons, each person has a charge q = 0.01e1.6 × 10−19 Cje2.3 × 1028 j = 3.7 × 107 C . So F k q q r = e = × × 1 2 = × 2 9 7 2 2 9 10 25 3 7 10 0 6 e je j 4 10 . . N N ~1026 N . This force is almost enough to lift a weight equal to that of the Earth: Mg = 6 × 1024 kge9.8 m s2 j = 6 × 1025 N~1026 N. *P23.4 The force on one proton is F = k q q r e 1 2 2 away from the other proton. Its magnitude is 8 99 10 1 6 10 2 10 9 57 5 19 15 2 . . × ⋅ . × × FH GIK J= − − N m C C m e 2 j N . P23.5 (a) F k q q r e = e = × ⋅ × × = × − − 1 2 − 2 9 19 2 10 2 9 8 99 10 1 60 10 3 80 10 1 59 10 . . . . N m C C m N repulsion e 2 2 je j e j b g (b) F Gm m r g = = × ⋅ × × = × − − − 1 2 − 2 11 27 2 10 2 45 6 67 10 1 67 10 3 80 10 1 29 10 . . . . N m C kg m N e 2 2 je j e j The electric force is larger by 1.24 × 1036 times . (c) If k q q r Gm m r e 1 2 2 1 2 2 = with q1 = q2 = q and m1 = m2 = m , then q m Gke = = × ⋅ × ⋅ = × − 6 67 10 − 8 99 10 8 61 10 11 9 . 11 . . N m kg N m C C kg 2 2 2 2 .6 Electric Fields P23.6 We find the equal-magnitude charges on both spheres: F k q q r k qr = e = e 1 2 2 22 so q r F ke = = × × ⋅ 1 00 = × − 1 00 10 1 05 10 4 . 3 . m . N 8.99 10 N m C C 9 2 2 a f . The number of electron transferred is then N e xfer C C = electrons × × = × − − − 1 05 10 1 60 10 6 59 10 3 19 . 15 . . . The whole number of electrons in each sphere is Ntot e e g 107.87 g mol = atoms mol atom FH GIK J× − = × − 10 0 6 02 1023 47 2 62 1024 . e . je j . . The fraction transferred is then f NN = = ×× FH GIK Jxfer = × − = tot 6 59 10 2 62 10 2 51 10 2 51 15 24 . 9 . . . charges in every billion. P23.7 F k q q r 1 e 1 2 2 9 6 6 2 8 99 10 7 00 10 2 00 10 0 500 = = 0 503 × ⋅ × × = . . − . − . . N m C C C m N e 2 2 je je j a f F k q q r 2 e 1 2 2 9 6 6 2 8 99 10 7 00 10 4 00 10 0 500 = = 1 01 × ⋅ × × = . . − . − . . N m C C C m N e 2 2 je je j a f FFxy= °+ ° = = °− ° = − = − = ° 0 503 60 0 1 01 60 0 0 755 0 503 60 0 1 01 60 0 0 436 0 755 0 436 0 872 . cos . . cos . . . sin . . sin . . . . . N N F a Nfi a Nfj N at an angle of 330 FIG. P23.7 P23.8 F k q q r = e = × ⋅ × × × = − 1 2 2 9 19 2 23 2 6 2 8 99 10 1 60 10 6 02 10 2 6 37 10 514 . . . . N m C C m kN e 2 2 je j e j e j P23.9 (a) The force is one of attraction . The distance r in Coulomb’s law is the distance between centers. The magnitude of the force is F k q q r = e = × ⋅ × × = × − − 1 2 − 2 9 9 9 2 8 99 10 5 12 0 10 18 0 10 0 300 . 2 16 10 . . . N m C . C C m e 2 2 j e je j N a f . (b) The net charge of −6.00 ×10−9 C will be equally split between the two spheres, or −3.00 ×10−9 C on each. The force is one of repulsion , and its magnitude is F k q q r = e = × ⋅ × × = × − − 1 2 − 2 9 9 9 2 8 99 10 7 3 00 10 3 00 10 0 300 . 8 99 10 . . . N m C . C C m e 2 2 j e je j N a f .Chapter 23 7 P23.10 Let the third bead have charge Q and be located distance x from the left end of the rod. This bead will experience a net force given by F = i + i − − k qQ x k qQ d x e 3 e 2 2 b g b g a f e j. The net force will be zero if 3 1 x2 d x 2 = a − f , or d x − = 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 . P23.11 (a) F k e r= e = × ⋅ ×× = × −− − 2 2 9 19 2 10 2 8 99 10 8 1 60 10 0 529 10 . 8 22 10 . . N m C . C m e 2 2 j e j N e j (b) We have F mvr = 2 from which v Fr m = = × × × = × − − − 8 22 10 10 9 11 10 2 19 10 8 10 31 6 . . . N 0.529 m kg m s e j . P23.12 The top charge exerts a force on the negative charge k qQx e d2 c h2 + 2 which is directed upward and to the left, at an angle of tan− FH GIK J1 2dx to the x-axis. The two positive charges together exert force 22 2 4 2 4 2 1 2 k qQx xx e m d + d FH GGIK JJ− + FH GGG IK JJJ= e j a f e j i a or for x d << 2 , a ≈ x −2 3 8 k qQ mde . (a) The acceleration is equal to a negative constant times the excursion from equilibrium, as in a = −ω 2x , so we have Simple Harmonic Motion with ω 2 3 16 = k qQ mde . T md keqQ = 2 = 2 π 3 ω π , where m is the mass of the object with charge −Q. (b) v A ak qQ md e max =ω = 4 38 Electric Fields Section 23.4 The Electric Field P23.13 For equilibrium, Fe = −Fg or qE = −mge−jj . Thus, E = j mg q . (a) E = j = j j ×− × = − × − − mg − q . . . . 9 11 10 9 80 1 60 10 5 58 10 31 19 11 kg m s C N C e je 2 j e j e j (b) E = j = j j × × = × − − mg − q . . . . 1 67 10 9 80 1 60 10 1 02 10 27 19 7 kg m s C N C e je 2 j e j e j P23.14 ΣFy = 0 : QEj +mge−jj = 0 ∴ = = × = − m QE g 24 0 10 610 9 80 1 49 . 6 . . C NC m s grams 2 e jb g P23.15 The point is designated in the sketch. The magnitudes of the electric fields, E1 , (due to the −2.50 ×10−6 C charge) and E2 (due to the 6.00 × 10−6 C charge) are E k qr d e 1 2 9 6 2 8 99 10 2 50 10 = = × ⋅ ×−e . N m2 C2 je . Cj (1) E k q r d e 2 2 9 6 2 8 99 10 6 00 10 1 00 = = × ⋅ × + . . − . N m C C m e 2 2 je j a f (2) FIG. P23.15 Equate the right sides of (1) and (2) to get ad + 1.00 mf2 = 2.40d2 or d + 1.00 m= ±1.55d which yields d = 1.82 m or d = −0.392 m. The negative value for d is unsatisfactory because that locates a point between the charges where both fields are in the same direction. Thus, d = 1.82 m to the left of the − 2.50 μC charge .Chapter 23 9 P23.16 If we treat the concentrations as point charges, E j j E j j E E E +− + − = = × ⋅ − = × − = = × ⋅ − = × − = + = × k q r k q r ee 2 9 2 5 2 9 2 5 5 8 99 10 40 0 1 000 3 60 10 8 99 10 40 0 1 000 3 60 10 7 20 10 . . . . . . . N m C C m N C downward N m C C m N C downward N C downward 2 2 2 2 e ja f b ge j e ja f e ja f b ge j e ja f *P23.17 The first charge creates at the origin field k Q ae2 to the right. Suppose the total field at the origin is to the right. Then q must be negative: k Q a k q a k Q a e e e 2 2 2 3 2 i + −i = i a f e j q = −9Q . In the alternative, the total field at the origin is to the left: k Q a k q a k Q a e e e 2 9 2 2 2 i + e−ij = e−ij q = +27Q . x +Q x = 0 q FIG. P23.17 P23.18 (a) E k q re 1 2 9 6 2 5 8 99 10 7 00 10 0 500 = = 2 52 10 × × = × . . − . . e je j a f N C E k q r E E E E Ee xy2 2 9 6 2 5 2 1 5 5 3 1 5 3 3 8 99 10 4 00 10 0 500 1 44 10 60 1 44 10 2 52 10 60 0 18 0 10 60 0 2 52 10 60 0 218 10 18 0 218 10 18 0 218 = = × × = × = − °= × − × °= × = − ° = − × ° = − × = − × = − . . − . . cos . . cos . . sin . . sin . . . e je j a f N C N C N C E i j N C i j kN C FIG. P23.18 (b) F = qE = e2.00 × 10−6 Cje18.0i − 218jj× 103 N C = e36.0i − 436jj× 10−3 N = e36.0i − 436jj mN P23.19 (a) E1 j j j 1 12 9 9 2 3 8 99 10 3 00 10 0 100 = − = 2 70 10 × × − = − × − k q re . . . e j e je j . a f e j e N Cj E i i i E E E i j 2 2 22 9 9 2 2 2 1 2 3 8 99 10 6 00 10 0 300 5 99 10 5 99 10 2 70 10 = − = × × − = − × = + = − × − × − k q re . . . . . . e je je j a f e j e j e j e j N C N C N C FIG. P23.19 (b) F = qE = e5.00 × 10−9 Cje−599i − 2 700jj N C F = e−3.00 × 10−6 i − 13.5 × 10−6 jj N = e−3.00i − 13.5jj μN10 Electric Fields P23.20 (a) E k q r = e = × × = − 2 9 6 2 8 99 10 2 00 10 1 12 14 400 . . . e je j a f N C Ex = 0 and Ey = 2b14 400gsin26.6°= 1.29 × 104 N C so E = 1.29 × 104 j N C . FIG. P23.20 (b) F = qE = e−3.00 × 10−6 je1.29 × 104 jj = −3.86 × 10−2 j N P23.21 (a) E = r + r + r = i + i °+j ° + j k q r k q r k q r k q a k q a k q a e 1 e e e e e 12 1 2 22 2 3 32 3 2 2 2 2 32 45 0 45 0 4 cos . sin . b g b ge j b g E = 3 06 i + 5 06 j = 5 91 ° 2 2 2 . . . k q a k q a k q a e e e at 58.8 (b) F = qE = ° k q a5 91 e 2 2 . at 58.8 P23.22 The electric field at any point x is E k q x a k q x a k q ax x a = e e e − − − − = a f c a fh −a f e j 2 2 2 2 2 4 . When x is much, much greater than a, we find E a k q x ≅ e 4 3 b g . P23.23 (a) One of the charges creates at P a field E = + k Q n R x e2 2 at an angle θ to the x-axis as shown. When all the charges produce field, for n > 1 , the components perpendicular to the x-axis add to zero. The total field is nk Q n R x k Qx R x e e b g e j cos i i 2 + 2 2 2 3 2 = + θ . FIG. P23.23 (b) A circle of charge corresponds to letting n grow beyond all bounds, but the result does not depend on n. Smearing the charge around the circle does not change its amount or its distance from the field point, so it does not change the field . P23.24 E r i i i i i = = − + − + − + = − + + + FH GIK JΣ = − k q r k q a k q a k q a k q a k q a e e e e e e 2 2 2 2 2 2 2 2 2 2 3 1 1 2 13 6 e j a f e j a f e j … … πChapter 23 11 Section 23.5 Electric Field of a Continuous Charge Distribution P23.25 E k d d k Q d d k Q d d = e e e + = + = + = × × + − λ a f b g a f a f e je j a fa f 8 99 10 22 0 10 0 290 0 140 0 290 . 9 . 6 . . . E = 1.59 × 106 N C, directed toward the rod. FIG. P23.25 P23.26 E k dq x= z e 2 , where dq = λ 0dx E k dx x k x ke x x e x = = − e FH GIK J= ∞ ∞ λ z λ λ 0 2 0 0 0 0 0 1 The direction is or left for − > iλ 0 0 P23.27 E k xQ x a x x x x = e + = × × + = × + − 2 2 3 2 9 6 2 2 3 2 5 2 3 2 8 99 10 75 0 10 0 100 6 74 10 e j 0 010 0 e je j e j e j . . . . . (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 = 2.41 × 107 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 C P23.28 E E i = = i i i L − N MMM OQ PPP= − = − − FH GIK Jz z z = − ∞ − ∞ ∞ d k xdx x k x x dx k x x k x e x e x e x e λ λ λ 0 0 λ 3 0 0 3 0 0 2 0 0 0 0 0 1 2 2 e j e j P23.29 E k Qx x a= e 2 + 2 3 2 e j For a maximum, dE dx Qk x a x x a = e + − + LNMMM OQPPP1 3 = 0 2 2 3 2 2 2 2 5 2 e j e j x2 + a2 − 3x2 = 0 or x = a2 . Substituting into the expression for E gives E k Qa a k Qa k Qa Q a = e = e = e = 2 3 ∈ 2 3 3 3 6 3 2 2 3 2 3 2 2 2 0 2 e j π .12 Electric Fields P23.30 E k x x R = e − + FH GIK J2 1 2 2 π σ E x x x x = × × −+ FH GGIK JJ= × − + FH GIK J2 8 99 10 7 90 10− 1 0 350 4 46 10 1 0 123 9 3 2 2 8 2 π . . . . . e je j a f (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 × 108 N C = 324 MN C (c) At x = 0.500 m, E = 8.07 × 107 N C = 80.7 MN C (d) At x = 2.00 m, E = 6.68 × 108 N C = 6.68 MN C P23.31 (a) From Example 23.9: E k x x R = e − + FH GIK J2 1 2 2 π σ σ π = = × = × = × = Q − R E 2 3 8 7 1 84 10 1 04 10 0 900 9 36 10 93 6 . . . . . C m N C N C MN C 2 e ja f appx: E = 2π keσ = 104 MN C babout 11% highg (b) E= × − + FH GIK J1 04 10 1 = × = 30 03 00 8 1 04 10 0 004 96 0 516 2 . 8 . . N C . . . cm 30.0 cm N C MN C 2 e j e jb g appx: E k Qr = e = × × = − 2 9 6 2 8 99 10 5 20 10 0 30 . 0 519 . . e j a f . MN C babout 0.6% highg P23.32 The electric field at a distance x is E k x x R x = e − + LN MMOQ PP2 1 2 2 π σ This is equivalent to E k R x x = e − + LN MMOQ PP2 1 1 1 2 2 π σ For large x, Rx 22 << 1 and 1 1 2 22 22 + R ≈ + x Rx so E k R x k R x R x x = e − e + FH GGIK JJ= + − + 2 1 1 1 2 2 1 2 1 2 2 1 2 2 2 2 2 π σ π σ e j e e j j e j Substitute σ π = QR2 , E k Q x R x k Q x R x e = e + = + FH GIK J1 1 2 2 2 2 2 2 e j 2 e j But for x >> R , 1 2 1 x2 + R2 x2 ≈ , so E k Q x x ≈ e2 for a disk at large distancesChapter 23 13 P23.33 Due to symmetry Ey = zdEy = 0 , and E dE k dqr x = z sin = e z sin θ θ 2 where dq = λds = λrdθ , so that, E kr d kr kx r = eλ z = e − = e θ θ λ θ π λ π sin cos 0 0 a f 2 where λ = qL and r L = π . FIG. P23.33 Thus, E k q L x = e = × ⋅ × − 2 2 8 99 10 7 50 10 0 140 2 9 6 2 π . . π . N m C C m e 2 2 je j a f . Solving, Ex = 2.16 × 107 N C . Since the rod has a negative charge, E = e−2.16 × 107 ij N C = −21.6i MN C . P23.34 (a) We define x = 0 at the point where we are to find the field. One ring, with thickness dx, has charge Qdx h and produces, at the chosen point, a field d k x x R Qdx h E = e i 2 + 2 3 2 e j . The total field is E E i i E i i = = + = + = +− = + − + + LNMMM OQPPP z z z + − =+ − =+ d k Qxdx h x R k Qh x R xdx k Qh x R k Qh d R d h R e d d h e x d d h e x d d h e all charge 2 23 2 2 2 3 2 2 2 1 2 2 21 2 2 2 1 2 2 2 2 1 2 1 1 e j e j e j b g e j ea f j (b) Think of the cylinder as a stack of disks, each with thickness dx, charge Qdx h , and chargeper-area σ π = Qdx R2h . One disk produces a field d k Qdx R h x x R E = e − i + FH GGIK JJ2 1 2 2 21 2 ππ e j . So, E E i = = −+ FH GGIK JJz z=+ d k Qdx R h x x R e x d d h all charge 2 1 2 2 21 2 e j E i i E i E i = − + LN MMOQ PP= − + LNMMM OQPPP = + − − + + + + LNMOQP = + + − + + LNMOQP + − =+ + + 2 z 1 z 2 2 2 12 12 22 2 2 2 1 2 2 2 21 2 2 2 2 1 2 2 21 2 2 2 21 2 2 2 1 2 k Q R h dx x R xdx k Q R h x x R k Q R h d h d d h R d R k Q R h h d R d h R e d d h x d d h e dd h dd h e e e j e j ea f j e j e j ea f j14 Electric Fields P23.35 (a) The electric field at point P due to each element of length dx, is dE k dq x y = e2 + 2 and is directed along the line joining the element to point P. By symmetry, Ex = zdEx = 0 and since dq = λdx , E = Ey = zdEy = zdEcosθ where cosθ = + y x2 y2 . Therefore, E k y dx x y k e y = e + 2 z = 2 2 2 3 2 02 λ 0 λ θ e j sin . FIG. P23.35 (b) For a bar of infinite length, θ 0 = 90° and E ky y= e 2 λ . P23.36 (a) The whole surface area of the cylinder is A = 2π r 2 + 2π rL = 2π rar + Lf . Q =σA = e15.0 × 10−9 C m2 j2π b0.025 0 mg 0.025 0 m+ 0.060 0 m = 2.00 × 10−10 C (b) For the curved lateral surface only, A = 2π rL . Q =σA = e15.0 × 10−9 C m2 j 2π b0.025 0 mga0.060 0 mf = 1.41 × 10−10 C (c) Q = ρV = ρπ r 2L = e500 × 10−9 C m3 jπ b0.025 0 mg2 b0.060 0 mg = 5.89 × 10−11 C P23.37 (a) Every object has the same volume, V = 8a0.030 0 mf3 = 2.16 × 10−4 m3 . For each, Q = ρV = e400 × 10−9 C m3 je2.16 × 10−4 m3 j = 8.64 × 10−11 C (b) We must count the 9.00 cm2 squares painted with charge: (i) 6 × 4 = 24 squares Q =σA = e15.0 × 10−9 C m2 j24.0e9.00 × 10−4 m2 j = 3.24 × 10−10 C (ii) 34 squares exposed Q =σA = e15.0 × 10−9 C m2 j34.0e9.00 × 10−4 m2 j = 4.59 × 10−10 C (iii) 34 squares Q =σA = e15.0 × 10−9 C m2 j34.0e9.00 × 10−4 m2 j = 4.59 × 10−10 C (iv) 32 squares Q =σA = e15.0 × 10−9 C m2 j32.0e9.00 × 10−4 m2 j = 4.32 × 10−10 C (c) (i) total edge length: = 24 × b0.030 0 mg Q = λ = e80.0 × 10−12 C mj24 × b0.030 0 mg = 5.76 × 10−11 C (ii) Q = λ = e80.0 × 10−12 C mj44 × b0.030 0 mg = 1.06 × 10−10 C continued on next pageChapter 23 15 (iii) Q = λ = e80.0 × 10−12 C mj64 × b0.030 0 mg = 1.54 × 10−10 C (iv) Q = λ = e80.0 × 10−12 C mj40 × b0.030 0 mg = 0.960 × 10−10 C Section 23.6 Electric Field Lines P23.38 FIG. P23.38 P23.39 FIG. P23.39 P23.40 (a) qq12 6 18 13 = − = − (b) q1 is negative, q2 is positive P23.41 (a) The electric field has the general appearance shown. It is zero at the center , where (by symmetry) one can see that the three charges individually produce fields that cancel out. In addition to the center of the triangle, the electric field lines in the second figure to the right indicate three other points near the middle of each leg of the triangle where E = 0 , but they are more difficult to find mathematically. (b) You may need to review vector addition in Chapter Three. The electric field at point P can be found by adding the electric field vectors due to each of the two lower point charges: E = E1 + E2 . The electric field from a point charge is E = k r q r e 2 . As shown in the solution figure at right, E1 2 = k q a e to the right and upward at 60° E2 2 = k q a e to the left and upward at 60° FIG. P23.41 E E E i j i j j j = + = ° + ° + − ° + ° = ° = 1 2 2 2 2 60 60 60 60 2 60 1 73 k q a k q a k q ae e e cos sin cos sin sin . e je j e j16 Electric Fields Section 23.7 Motion of Charged Particles in a Uniform Electric Field P23.42 F = qE = ma a qE m = v f = vi + at v qEt f = m electron: ve = × × × = × − − − 1 602 10 520 48 0 10 9 11 10 4 39 10 19 9 31 6 . . . . e ja fe j m s in a direction opposite to the field proton: vp = × × × = × − − − 1 602 10 520 48 0 10 1 67 10 2 39 10 19 9 27 3 . . . . e ja fe j m s in the same direction as the field P23.43 (a) a qE m = = × × = × − − 1 602 10 640 1 67 10 6 14 10 19 27 . 10 . . a f m s2 (b) v f = vi + at 1.20 × 106 = e6.14 × 1010 jt t = 1.95 × 10−5 s (c) x f − xi = vi + v f t 12d i x f = 1 × × − = 2 e1.20 106 je1.95 10 5 j 11.7 m (d) K = mv = × − × = × − 12 12 2 e1.67 10 27 kgje1.20 106 m sj2 1.20 10 15 J P23.44 (a) a qE m = = × × × = × − − 1 602 10 6 00 10 1 67 10 5 76 10 19 5 27 13 . . . . e je j e j m s so a = −5.76 × 1013 i m s2 (b) v f = vi + 2adx f − xi i 0 = vi2 + 2e−5.76 × 1013 jb0.070 0g vi = 2.84 × 106 i m s (c) v f = vi + at 0 = 2.84 × 106 + e−5.76 × 1013 jt t = 4.93 ×10−8 s P23.45 The required electric field will be in the direction of motion . Work done = ΔK so, −Fd = − mvi 12 2 (since the final velocity = 0 ) which becomes eEd = K and E Ked = .Chapter 23 17 P23.46 The acceleration is given by v f vi a x f xi 2 = 2 + 2 d − i or v a h f 2 = 0 + 2 a− f . Solving a vhf = − 2 2 . Now ΣF = ma : −mg + q = − mvh f j E 2 j 2 . Therefore q mvh mg f E = − + j FH GIK J2 2 . (a) Gravity alone would give the bead downward impact velocity 2e9.80 m s2 ja5.00 mf = 9.90 m s . To change this to 21.0 m/s down, a downward electric field must exert a downward electric force. (b) q mE vh g = f − FH GIK J= × × ⋅⋅ FH GIK J− LN MMOQ PP= 2 −3 2 2 1 00 10 21 0 2 5 00 9 80 3 43 . . . . . kg 1.00 10 N C N s kg m m s m m s C 4 2 2 b g a f μ P23.47 (a) t x vx = = × = × − = 0 050 0 4 50 10 111 10 111 5 . 7 . . s ns (b) a qE y = m = × × × = × − − 1 602 10 9 60 10 1 67 10 9 21 10 19 3 27 11 . . . . e je j e j m s2 y y v t a t f i yi y − = + 1 2 2 : y f = 1 × × − = × − = 2 e9.21 1011 je1.11 10 7 j2 5.68 10 3 m 5.68 mm (c) vx = 4.50 ×105 m s vyf = vyi + ayt = e9.21 × 1011 je1.11 × 10−7 j = 1.02 ×105 m s *P23.48 The particle feels a constant force: F = qE = e1 × 10−6 Cjb2 000 N Cge−jj = 2 × 10−3 Ne−jj and moves with acceleration: a F j = = j × ⋅ − × = × − Σ − m − 2 102 10 1 10 3 16 13 kg m s kg m s 2 2 e je je jej . Its x-component of velocity is constant at e1.00 × 105 m sjcos37°= 7.99 × 104 m s. Thus it moves in a parabola opening downward. The maximum height it attains above the bottom plate is described by vyf vyi ay y f yi 2 2 2 = + − d i: 0 6 02 10 2 10 0 4 2 13 = × − × − . m s m s2e j e jdy f i y f = 1.81 × 10−4 m. continued on next page18 Electric Fields Since this is less than 10 mm, the particle does not strike the top plate, but moves in a symmetric parabola and strikes the bottom plate after a time given by y y v t a t f i yi y = + + 1 2 2 0 0 602 10 12 1 10 4 13 2 = + × + − × . m s m s2e jt e jt since t > 0 , t = 1.20 × 10−8 s . The particle’s range is x f = xi + vxt = 0 + e7.99 × 104 m sje1.20 × 10−8 sj = 9.61 × 10−4 m. In sum, The particle strikes the negative plate after moving in a parabola with a height of 0.181 mm and a width of 0.961 mm. P23.49 vi = 9.55 × 103 m s (a) a eE y = m = × × = × − − 1 60 10 720 1 67 10 6 90 10 19 27 10 . . . e ja f e j m s2 R v a i y = = × − 2 2 3 1 27 10 sin . θ m so that 9 55 10 2 6 90 10 1 27 10 3 2 10 3 . sin . . × × = × − e j θ sin2θ = 0.961 θ = 36.9° 90.0°−θ = 53.1° ^ FIG. P23.49 (b) t R v R ix vi = = cosθ If θ = 36.9° , t = 167 ns . If θ = 53.1°, t = 221 ns . Additional Problems *P23.50 The two given charges exert equal-size forces of attraction on each other. If a third charge, positive or negative, were placed between them they could not be in equilibrium. If the third charge were at a point x > 15 cm, it would exert a stronger force on the 45 μ C than on the −12 μ C , and could not produce equilibrium for both. Thus the third charge must be at x = −d < 0 . Its equilibrium requires d x 15 cm q x = 0 – + –12 μ C 45 μ C FIG. P23.50 k q d k q d e 12 e 45 15 2 2 C C cm b μ g b μ g a f = + 15 45 12 3 75 2 cm+ FH GIK Jd = = d . 15 cm+ d = 1.94d d = 16.0 cm. The third charge is at x = −16.0 cm . The equilibrium of the −12 μ C requires keq 12 ke 16 0 45 12 2 2 C cm C C 15 cm b μ g μ μ a f b g . a f = q = 51.3 μ C . All six individual forces are now equal in magnitude, so we have equilibrium as required, and this is the only solution.Chapter 23 19 P23.51 The proton moves with acceleration a qE p = m = × × = × − − 1 60 10 640 1 673 10 6 13 10 19 27 10 . . . C NC kg m s2 e jb g while the e− has acceleration ae = ap × × = × = − − 1 60 10 640 9 110 10 112 10 1 836 19 31 14 . . . C NC kg m s2 e jb g . (a) We want to find the distance traveled by the proton (i.e., d = 1 apt 2 2 ), knowing: 4 00 12 12 1 837 12 2 2 2 . cm= + = FH GIK Japt aet apt . Thus, d = apt = = 12 4 00 2 . 21 8 . cm 1 837 μm . (b) The distance from the positive plate to where the meeting occurs equals the distance the sodium ion travels (i.e., dNa = 1 aNat 2 2 ). This is found from: 4 00 12 12 . cm= aNat2 + aClt2 : 4 00 12 2299 12 3545 . 2 2 . . cm u u = FH GIK J+ FH GIK JeE t eE t . This may be written as 4 00 12 12 0 649 1 65 12 2 2 2 . . . cm Na Na Na = + = FH GIK Ja t b a gt a t so dNa aNat cm 1.65 = = = cm 12 4 00 2 . 2 43. . P23.52 (a) The field, E1 , due to the 4.00 × 10−9 C charge is in the –x direction. E r i i 1 2 9 9 2 8 99 10 4 00 10 2 50 5 75 = = × ⋅ − × = − − k q re . . . . N m C C m N C e 2 2 je j a f FIG. P23.52(a) Likewise, E2 and E3 , due to the 5.00 × 10−9 C charge and the 3.00 × 10−9 C charge are E2 2 r i i 9 9 2 8 99 10 5 00 10 2 00 = = 11 2 × ⋅ × = − k q re . . . . N m C C m N C e 2 2 je j a f E3 i i 9 9 2 8 99 10 3 00 10 1 20 = 18 7 × ⋅ × = . . − . . N m C C m N C e 2 2 je j a f ER = E1 + E2 + E3 = 24.2 N C in +x direction. (b) E1 2 r i j = = −8 46 0 243 + 0 970 k q re b . N Cge . . j E r j E r i j i j 2 2 3 21 3 1 2 3 11 2 5 81 0 371 4 21 8 43 = = + = = − = + =− = + + = k q rk q r E E E E E E E ee x x x y y y y . . . . . N C N C +0.928 N C N C b ge j b ge j ER = 9.42 N C θ = 63.4° above − x axis FIG. P23.52(b)20 Electric Fields *P23.53 (a) Each ion moves in a quarter circle. The electric force causes the centripetal acceleration. ΣF = ma qE mv R = 2 E mv qR = 2 (b) For the x-motion, vxf vxi ax x f xi 2 = 2 + 2 d − i 0 = v2 + 2axR a vR Fm qEx m = − = x = x 2 2 E mvx = − qR2 2 . Similarly for the y-motion, v ayR 2 = 0 + 2 a vR qEy my = + = 2 2 E mvy = qR2 2 The magnitude of the field is E E mvqR x y x 2 2 2 2 + = at 135° counterclockwise from the -axis . P23.54 From the free-body diagram shown, ΣFy = 0 : T cos15.0°= 1.96 × 10−2 N . So T = 2.03 × 10−2 N . From ΣFx = 0 , we have qE = T sin15.0° or q T E = ° = × ° × = × = − sin . . sin . − . . . 15 0 2 03 10 15 0 1 00 10 5 25 10 5 25 2 3 6 N N C C C e j μ . FIG. P23.54 P23.55 (a) Let us sum force components to find ΣFx = qEx −T sinθ = 0 , and ΣFy = qEy + T cosθ −mg = 0 . Combining these two equations, we get q mg Ex Ey = + = × °+ × = × = − − cot . . . cot . . . . e θ j e ja f a f 1 00 10 9 80 3 00 37 0 5 00 10 1 09 10 10 9 3 5 8 C nC (b) From the two equations for ΣFx and ΣFy we also find T qEx = °= × − = sin . . . 37 0 5 44 10 3 N 5 44 mN . Free Body Diagram FIG. P23.55Chapter 23 21 P23.56 This is the general version of the preceding problem. The known quantities are A, B, m, g, and θ. The unknowns are q and T. The approach to this problem should be the same as for the last problem, but without numbers to substitute for the variables. Likewise, we can use the free body diagram given in the solution to problem 55. Again, Newton’s second law: ΣFx = −T sinθ + qA = 0 (1) and ΣFy = +T cosθ + qB −mg = 0 (2) (a) Substituting T qA = sinθ , into Eq. (2), qA qB mg cos sin θ θ + = . Isolating q on the left, q mg A B = a cotθ + f . (b) Substituting this value into Eq. (1), T mgA A B = a cosθ + sinθ f . If we had solved this general problem first, we would only need to substitute the appropriate values in the equations for q and T to find the numerical results needed for problem 55. If you find this problem more difficult than problem 55, the little list at the first step is useful. It shows what symbols to think of as known data, and what to consider unknown. The list is a guide for deciding what to solve for in the analysis step, and for recognizing when we have an answer. P23.57 F k q q r = e 1 2 2 : tan .. θ = 15 0 60 0 θ = 14.0° FFFF F F F F F xy1 9 62 2 3 9 62 2 2 9 62 2 3 2 1 2 8 99 10 10 0 10 0 150 40 0 8 99 10 10 0 10 0 600 2 50 8 99 10 10 0 10 0 619 2 35 14 0 2 50 2 35 14 0 4 78 14 0 40 0 2 35 14 0 40 6 = × × = = × × = = × × = = − − ° = − − ° = − = − − ° = − − ° = − −−− . . . . . . . . . . . . cos . . . cos . . sin . . . sin . . e je j a f e je j a f e je j a f N N N N N FIG. P23.57 F F F FFx y yx net = + = + = N = = −− = °2 2 2 2 4 78 40 6 40 9 40 6 4 78 263 . . . tan . . a f a f φ φ22 Electric Fields P23.58 From Figure A: d cos 30.0°= 15.0 cm, or d = ° 15 030 0 . cos . cm From Figure B: θ = FH GIK Jsin− . 1 50 0d cm θ = ° FH GIK Jsin− = ° . cos . 1 15 0 . 30 0 20 3 cm 50.0 cma f F mgq = tanθ or Fq = mg tan 20.3° (1) From Figure C: Fq = 2F cos30.0° F k q q = e LN MMOQ PP2 ° 0 300 30 0 2 2 . cos . a mf (2) Combining equations (1) and (2), 2 0 300 30 0 20 3 2 2 k q e mg . cos . tan . a mf LN MMOQ PP° = ° q mg k qq e 2 2 2 3 2 9 14 7 0 300 20 3 2 300 2 00 10 9 80 0 300 20 3 2 8 99 10 30 0 4 20 10 2 05 10 0 205 = ° ° = × ° × ⋅ ° = × = × = −− − . tan . cos . . . . tan . . cos . . . . m kg m s m N m C C C C 22 2 2 a f e je ja f e j μ Figure A Figure B Figure C FIG. P23.58 P23.59 Charge Q2 resides on each block, which repel as point charges: F k Q Q L k L L e = = − i 2 2 2 b gb g b g. Solving for Q, Q L k L L k i e = − 2 b g . *P23.60 If we place one more charge q at the 29th vertex, the total force on the central charge will add up to zero: F28 charges + k qQ ae 2 away from vertex 29 = 0 F28 charges ke = toward vertex 29 qQ a2 . P23.61 According to the result of Example 23.7, the left-hand rod creates this field at a distance d from its right-hand end: E k Q d a d dF k QQ a dx d d a F k Qa dx x x a k Qa a a x x e e e x b a b e b a b = + = + = + = − + FH GIK J = − − z 22 2 2 2 2 1 2 2 2 2 2 2 a fa f a f ln FIG. P23.61 F k Q a a b b b b a k Qa b b a b a k Qa b b a = e e e + − + + − FH GIK J= − + = FH GIK J− FH GIK J2 2 2 2 2 2 2 2 4 2 2 2 2 4 2 2 4 4 ln ln ln a fa f lnChapter 23 23 P23.62 At equilibrium, the distance between the charges is r = 2b0.100mgsin10.0° = 3.47 × 10−2 m Now consider the forces on the sphere with charge +q , and use ΣFy = 0 : ΣFy = 0 : T cos10.0°= mg , or T mg = cos10.0° (1) ΣFx = 0 : Fnet = F2 − F1 = T sin10.0° (2) Fnet is the net electrical force on the charged sphere. Eliminate T from (2) by use of (1). F mg net mg = kg m s2 N ° ° = °= × − °= × − sin . cos . tan . . . tan . . 10 0 10 0 10 0 e2 00 10 3 je9 80 j 10 0 3 46 10 3 Fnet is the resultant of two forces, F1 and F2 . F1 is the attractive force on +q exerted by −q , and F2 is the force exerted on +q by the external electric field. r L–q +q θ θ FIG. P23.62 Fnet = F2 − F1 or F2 = Fnet + F1 F1 9 8 8 3 2 8 99 10 2 5 00 10 5 00 10 3 47 10 = × ⋅ 1 87 10 × × × = × − − − . − . . . N m C . C C m e 2 2 j e je j N e j Thus, F2 = Fnet + F1 yields F2 = 3.46 × 10−3 N+ 1.87 × 10−2 N = 2.21 × 10−2 N and F2 = qE , or E Fq = = ×× = × = −− 2 28 2 21 10 5 4 43 10 443 . . N 5.00 10 C N C kN C . P23.63 Q d Rd R R R QdF d R Rd R F y y= = = = − − = = = = = = ∈ FH GIK J= ∈ FH GGIK JJ = × ⋅ × z z− °° − ° ° − λ λ θ θ λ θ λ λ μ λ μ λ μ π μ λ θ π μ λ θ θ 0 90 0 90 0 0 90 0 90 0 0 0 0 0 2 0 0 2 2 9 1 1 2 12 0 2 0 600 12 0 10 0 1 4 3 00 1 4 3 00 8 99 10 3 00 10 cos sin . . . . . cos . cos . . . . . . a f b ga f b gb g b ge j e j C m C so C m C C N m C 0 2 2 6 6 2 90 0 90 0 3 2 2 2 2 10 0 10 0 600 8 99 30 0 0 600 10 12 12 2 0 450 12 14 2 0707 C C m m N N N Downward. e je j a f a fe j a f . . cos . . . cos . sin . . . × = + FH GIK J = + FH GIK J= − − ° ° − − − z z θ θ θ θ θ θ π π π π d F d Fyy 10 –1 cosθ 0° 360° 0° 360° 10 cos2θ FIG. P23.63 Since the leftward and rightward forces due to the two halves of the semicircle cancel out, Fx = 0 . P23.64 At an equilibrium position, the net force on the charge Q is zero. The equilibrium position can be located by determining the angle θ corresponding to equilibrium. In terms of lengths s, 12 a 3 , and r, shown in Figure P23.64, the charge at the origin exerts an attractive force k Qq s ae + 12 2 e 3 j continued on next page24 Electric Fields The other two charges exert equal repulsive forces of magnitude k Qq re 2 . The horizontal components of the two repulsive forces add, balancing the attractive force, F kQqrs a net = e − + LNMMM OQPPP2 1 = 3 0 2 12 2 cosθ e j From Figure P23.64 r a = 12 sinθ s = 1 a 2 cotθ The equilibrium condition, in terms of θ, is F a net = keQq FH GIK J− + FH GGIK JJ4 = 2 1 3 0 2 2 2 cos sin cot θ θ e θ j . Thus the equilibrium value of θ satisfies 2 2 3 2 1 cosθ sin θ e + cotθ j = . One method for solving for θ is to tabulate the left side. To three significant figures a value of θ corresponding to equilibrium is 81.7°. The distance from the vertical side of the triangle to the equilibrium position is s = a °= a 12 cot 81.7 0.072 9 . FIG. P23.64 θ 2 θ θ 3 θ 60 70 80 90 81 81 5 81 7 2 2 cos sin cot .. + °°°°°°° e j 4 2.654 1.226 0 1.091 1.024 0.997 A second zero-field point is on the negative side of the x-axis, where θ = −9.16° and s = −3.10a . P23.65 (a) From the 2Q charge we have Fe − T2 sinθ 2 = 0 and mg − T2 cosθ 2 = 0 . Combining these we find F mg TT e = 2 2 = 2 2 2 sin cos tan θθ θ . From the Q charge we have Fe = T1 sinθ 1 = 0 and mg − T1 cosθ 1 = 0 . Combining these we find F mg TT e = 1 1 = 1 1 1 sin cos tan θθ θ or θ 2 =θ 1 . (b) F k QQ r k Q r e = e = e 2 2 2 2 2 FIG. P23.65 If we assume θ is small then tanθ ≈ r 2 . Substitute expressions for Fe and tanθ into either equation found in part (a) and solve for r. F mge = tanθ then 2 1 2 2 2 k Q r mg e r FH GIK J≈ and solving for r we find r k Q mg ≈ e FH GIK J4 2 1 3 .Chapter 23 25 P23.66 (a) The distance from each corner to the center of the square is L L L 2 2 2 2 2 FH GIK J+ FH GIK J= . The distance from each positive charge to −Q is then z 2 L2 2 + . Each positive charge exerts a force directed +q +q +q z +q –Q L/2 L/2 x FIG. P23.66 along the line joining q and −Q, of magnitude k Qq z L e 2 + 2 2 . The line of force makes an angle with the z-axis whose cosine is z z2 + L2 2 The four charges together exert forces whose x and y components add to zero, while the z-components add to F = − k + 4 2 2 2 3 2 k Qqz z Le e j (b) For z >> L , the magnitude of this force is F k Qqz L k Qq L z z ma e e = − = − z FH GIK J= 4 2 4 2 2 3 2 3 23 e j a f Therefore, the object’s vertical acceleration is of the form az = −ω 2z with ω 2 3 23 3 4 2 128 = = a f k Qq mL k Qq mL e e . Since the acceleration of the object is always oppositely directed to its excursion from equilibrium and in magnitude proportional to it, the object will execute simple harmonic motion with a period given by T mL k Qq mL e keQq = 2 = 2 = 128 8 1 4 3 1 4 π 3 ω π π a f a f . P23.67 (a) The total non-contact force on the cork ball is: F qE mg m g qE m = + = + FH GIK J, which is constant and directed downward. Therefore, it behaves like a simple pendulum in the presence of a modified uniform gravitational field with a period given by: T L g qE m = + = + × × × = − − 2 2 0 500 2 00 10 1 00 10 1 00 10 0 307 6 5 3 π π . . . . . m 9.80 m s C N C kg s 2e je j (b) Yes . Without gravity in part (a), we get T L qE m = 2π T = × × × = − − 2 0 500 200 10 100 100 10 0 314 6 3 π . . . . . m C 10 NC kg s e je 5 j (a 2.28% difference).26 Electric Fields P23.68 The bowl exerts a normal force on each bead, directed along the radius line or at 60.0° above the horizontal. Consider the free-body diagram of the bead on the left: ΣFy = nsin60.0°−mg = 0 , or n mg = sin60.0° . Also, ΣFx = −Fe + ncos60.0°= 0 , or k q R n e mg mg 2 2 60 0 60 0 3 = °= ° cos . = tan . . Thus, q R mg ke = FH GIK J3 1 2 . nmgFe 60.0° FIG. P23.68 P23.69 (a) There are 7 terms which contribute: 3 are s away (along sides) 3 are 2s away (face diagonals) and sinθ = = cosθ 12 1 is 3s away (body diagonal) and sinφ = 13 . The component in each direction is the same by symmetry. F = + + i j k i j k LN MOQ P+ + = + + k q s k q s e e 2 2 2 2 1 2 2 2 1 3 3 e j a1.90fe j FIG. P23.69 (b) F F F F k q s x y z = 2 + 2 + 2 = e 2 2 3.29 away from the origin P23.70 (a) Zero contribution from the same face due to symmetry, opposite face contributes 4 2 k q re sinφ FH GIK J where r s s = s s s FH GIK J+ FH GIK J+ = = 2 2 1 5 1 22 2 2 2 . . sinφ = sr E k qs r k q s k q s = 4 e = e = e 4 1 22 2 18 3 3 2 2 . . a f (b) The direction is the direction. kFIG. P23.70Chapter 23 27 P23.71 The field on the axis of the ring is calculated in Example 23.8, E E k xQ x a x = = e 2 + 2 3 2 e j The force experienced by a charge −q placed along the axis of the ring is F kQq x x a = − e + LNMMM OQPPP2 2 3 2 e j and when x << a , this becomes F k Qq a = − e x FH GIK J3 This expression for the force is in the form of Hooke’s law, with an effective spring constant of k k Qq a= e 3 Since ω = 2π f = km , we have f k Qq ma = 1 e 2π 3 . P23.72 d k dq x x x k x dx x e e E i j i j = + − + + FH GGIK JJ= − + + 2 2 2 2 2 2 3 2 0 150 0 150 0 150 0 150 . 0 150 . . . m . m m m a f a f m e j a f λ E E i j = = − + + z z= d k x dx x e all charge x m m m λ . . . 0 150 2 0 150 2 3 2 0 0 400e j a f FIG. P23.72 E i j E i j E i j i j = + + + + LNMMM OQPPP = × ⋅ × − + − = − + × = − + − − − k x x x eλ . . . . . . . . . . . . . . . 2 2 00 400 2 2 2 00 400 9 9 1 1 3 0 150 0 150 0 150 0 150 899 10 350 10 234 667 624 0 1 36 1 96 10 1 36 1 96 m m m m N m C C m m m N C kN C m m 2 2 a f a f a f a f e je ja f a f e j e j P23.73 The electrostatic forces exerted on the two charges result in a net torque τ = −2Fa sinθ = −2Eqa sinθ . For small θ, sinθ ≈θ and using p = 2qa , we have τ = −Epθ . The torque produces an angular acceleration given by τ α θ = I = I ddt22 . Combining these two expressions for torque, we have ddt Ep I 22 0 θ + θ FH GIK J= . FIG. P23.73 This equation can be written in the form ddt22 θ 2 = −ω θ where ω 2 = Ep I . This is the same form as Equation 15.5 and the frequency of oscillation is found by comparison with Equation 15.11, or f pEI qaE I = 1 = 2 1 2 2 π π .28 Electric Fields ANSWERS TO EVEN PROBLEMS P23.2 (a) 2.62 × 1024 ; (b) 2.38 electrons for every 109 present P23.36 (a) 200 pC; (b) 141 pC; (c) 58.9 pC P23.38 see the solution P23.4 57 5 . N P23.40 (a) − 13 ; (b) q1 is negative and q2 is positive P23.6 2.51 × 10−9 P23.8 514 kN P23.42 electron: 4.39 Mm s ; proton: 2.39 km s P23.10 x = 0.634d . The equilibrium is stable if the third bead has positive charge. P23.44 (a) −57 6 . iTm s2 ; (b) 2 84 . i Mm s; (c) 49.3 ns P23.46 (a) down; (b) 3.43 μC P23.12 (a) period = π2 md3 keqQ where m is the mass of the object with charge −Q; (b) 4 3 a k qQ md e P23.48 The particle strikes the negative plate after moving in a parabola 0.181 mm high and 0.961 mm. P23.50 Possible only with +51.3 μ C at P23.14 1.49 g x = −16.0 cm P23.16 720 kN C down P23.52 (a) 24.2 N C at 0°; (b) 9.42 N C at 117° P23.18 (a) 18.0i − 218j kN C; P23.54 5.25 μC (b) e36.0i − 436jjmN P23.56 (a) mg Acotθ + B ; (b) mgA Acosθ + Bsinθ P23.20 (a) 12 9 . j kN C ; (b) −38 6 . j mN P23.58 0.205 μC P23.22 see the solution P23.60 k qQ ae 2 toward the 29th vertex P23.24 − π 2 6 2 k q a e i P23.62 443 kN C i P23.26 kxeλ 0 0 − ie j P23.64 0.072 9a P23.28 k x eλ 0 2 0 − ie j P23.66 see the solution; the period is π 81 4 mL3 keQq P23.30 (a) 383MN C away; (b) 324MN C away; (c) 80.7MN C away; (d) 6.68MN C away P23.68 R mg ke 3 1 2 FH GIK J P23.32 see the solution P23.70 (a) see the solution; (b) k P23.34 (a) k Qh e d R d h R i2 2 1 2 2 2 1 2 + − + + LNMOQP− − e j ea f j ; P23.72 e−1.36i + 1.96jj kN C (b) 2 2 k Q R h e i h+ d + R − d + h + R LNMOQP2 21 2 2 2 1 2 e j ea f j