1. 1. 23 Electric Fields CHAPTER OUTLINE 23.1 Properties of Electric Charges 23.2 Charging Objects by Induction 23.3 Coulombs Law 23.4 The Electric Field 23.5 Electric Field of a Continuous Charge Distribution 23.6 Electric Field Lines 23.7 Motion of a Charged Particle in a Uniform Electric Field ANSWERS TO QUESTIONS Q23.1 A neutral atom is one that has no net charge. This means that it has the same number of electrons orbiting the nucleus as it has protons in the nucleus. A negatively charged atom has one or more excess electrons. *Q23.2 (i) Suppose the positive charge has the large value 1 mC. The object has lost some of its conduction electrons, in number 106 C (1 e/1.60 1019 C) = 6.25 1012 and in mass 6.25 1012 (9.11 1031 kg) = 5.69 1018 kg. This is on the order of 1014 times smaller than the ~1g mass of the coin, so it is an immeasurably small change. Answer (d). (ii) The coin gains extra electrons, gaining mass on the order of 1014 times its original mass for the charge 1 mC. Answer (b). Q23.3 All of the constituents of air are nonpolar except for water. The polar water molecules in the air quite readily steal charge from a charged object, as any physics teacher trying to perform electrostatics demonstrations in the summer well knows. As a resultit is difcult 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 signicant excess charge, giving the potential (pun intended) for a shocking (pun also intended) introduction to static electricity sparks. Q23.4 Similarities: A force of gravity is proportional to the product of the intrinsic properties (masses) of two particles, and inversely proportional to the square of the separation distance. An electrical force exhibits the same proportionalities, with charge as the intrinsic property. Differences: The electrical force can either attract or repel, while the gravitational force as described by Newtons law can only attract. The electrical force between elementary particles is vastly stronger than the gravitational force. Q23.5 No. The balloon induces polarization of the molecules in the wall, so that a layer of positive charge exists near the balloon. This is just like the situation in Figure 23.4a, except that the signs of the charges are reversed. The attraction between these charges and the negative charges on the balloon is stronger than the repulsion between the negative charges on the balloon and the negative charges in the polarized molecules (because they are farther from the balloon), so that there is a net attractive force toward the wall. Ionization processes in the air surrounding the balloon provide ions to which excess electrons in the balloon can transfer, reducing the charge on the balloon and eventually causing the attractive force to be insufcient to support the weight of the balloon. 1
2. 2. 2 Chapter 23 *Q23.6 Answer (c). Each charge produces eld as if it were alone in the Universe. *Q23.7 (i) According to the inverse square law, the eld is one-fourth as large at twice the distance. The answer is (c), 2 36 cm = 72 cm. (ii) The eld is four times stronger at half the distance away from the charge. Answer (b). Q23.8 An electric eld created by a positive or negative charge extends in all directions from the charge. Thus, it exists in empty space if that is what surrounds the charge. There is no material at point A in Figure 23.21(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 eld does exist at point A. *Q23.9 (i) We compute qA qB /r2 in each case. In (a) it is 400/4 = 100 (nC/cm)2 . In (b) and (c), 300/4 = 75 (nC/cm)2 . In (d) 600/9 = 67 (nC/cm)2 . In (e) 900/9 = 100 (nC/cm)2 . The ranking is then a = e > b = c > d. (ii) We compute qA /r2 in each case. In (a) it is 20/4 = 5 nC/cm2 . In (b) 30/4 = 7.5 nC/cm2 . In (c) 10/4 = 2.5 nC/cm2 . In (d) 30/9 = 3.3 nC/cm2 . In (e) 45/9 = 5 nC/cm2 . The ranking is then b > a = e > d > c. *Q23.10 The charge at the upper left creates at the eld point electric eld to the left, with magnitude we call E1 . The charge at lower right creates downward electric eld with an equal magnitude E1 . These two charges together create eld 2 1E downward and to the left at 45. The positive charge is 2 times farther from the eld point so it creates eld 2E1 /( 2 )2 = E1 upward and to the right. The net eld is then 2 1 1( )E downward and to the left. The answer to question (i) is (d). (ii) With the positive charge removed, the magnitude of the eld becomes 2 1E , larger than before, so the answer is (a). *Q23.11 The certain point must be on the same line as A and B, for otherwise the eld components perpendicular to this line would not add to zero. If the certain point is between A and B, it is midway between them, and Bs charge is also +40 nC. If the certain point is 4 cm from A and 12 cm from B, then Bs charge must be 9(40 nC) = 360 nC. These are the only two possibilities. The answers are (a), (f), and (j). Q23.12 The direction of the electric eld is the direction in which a positive test charge would feel a force when placed in the eld. A charge will not experience two electrical forces at the same time, but the vector sum of the two. If electric eld 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 eld lines cross would be indeterminate. Q23.13 Both gures are drawn correctly. E1 and E2 are the electric elds separately created by the point charges q1 and q2 in Figure 23.12 or q and q in Figure 23.13, respectively. The net electric eld is the vector sum of E1 and E2, shown as E. Figure 23.19 shows only one electric eld line at each point away from the charge. At the point location of an object modeled as a point charge, the direction of the eld is undened, and so is its magnitude. *Q23.14 Answer (a). The equal-magnitude radially directed eld contributions add to zero. *Q23.15 Answer (c). Contributions to the total eld from bits of charge in the disk lie closer together in direction than for the ring.
3. 3. Electric Fields 3 *Q23.16 (i) Answer (c). Electron and proton have equal-magnitude charges. (ii) Answer (b). The protons mass is 1836 times larger than the electrons. *Q23.17 Answer (b). Q23.18 Linear charge density, l, is charge per unit length. It is used when trying to determine the electric eld created by a charged rod. Surface charge density, s, is charge per unit area. It is used when determining the electric eld above a charged sheet or disk. Volume charge density, r, is charge per unit volume. It is used when determining the electric eld due to a uniformly charged sphere made of insulating material. Q23.19 No. Life would be no different if electrons were + charged and protons were charged. Opposite charges would still attract, and like charges would repel. The naming of + and charge is merely a convention. Q23.20 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 eld created by the rst 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 9 1 660 10 9 11 10 1 6727 31 . . . .kg kg( ) = 110 27 kg . Its charge, due to loss of one electron, is 0 1 1 60 10 1 60 1019 19 ( )= + . . .C C (b) By similar logic, charge = + 1 60 10 19 . C mass = ( ) = 22 99 1 66 10 9 11 10 3 82 1027 31 . . . .kg kg 26 kg (c) charge of Cl C = 1 60 10 19 . mass = ( )+ = 35 453 1 66 10 9 11 10 5 89 127 31 . . . .kg kg 00 26 kg (d) charge of Ca C C++ = ( )= + 2 1 60 10 3 20 1019 19 . . mass = ( ) ( )= 40 078 1 66 10 2 9 11 10 6 627 31 . . . .kg kg 55 10 26 kg (e) charge of N C C3 19 19 3 1 60 10 4 80 10 = ( )= . . mass = ( )+ ( )= 14 007 1 66 10 3 9 11 10 2 327 31 . . . .kg kg 33 10 26 kg continued on next page
4. 4. 4 Chapter 23 (f) charge of N C C4 19 19 4 1 60 10 6 40 10+ = ( )= + . . mass = ( ) ( )= 14 007 1 66 10 4 9 11 10 2 327 31 . . . .kg kg 22 10 26 kg (g) We think of a nitrogen nucleus as a seven-times ionized nitrogen atom. charge = ( )= 7 1 60 10 1 12 1019 18 . .C C mass = ( ) ( )= 14 007 1 66 10 7 9 11 10 2 327 31 . . . .kg kg 22 10 26 kg (h) charge = 1 60 10 19 . C mass = ( )+ + 2 1 007 9 15 999 1 66 10 9 11 1027 . . . .kg 331 26 2 99 10kg kg= . P23.2 (a) N = 10 0 6 02 1023. . grams 107.87 grams mol aatoms mol electrons atom = 47 2 62. 11024 (b) # . electrons added C 1.60 10 = = Q e 1 00 10 3 19 CC electron = 6 25 1015 . or 2 38. .electrons for every10 already present9 Section 23.2 Charging Objects by Induction Section 23.3 Coulombs Law P23.3 If each person has a mass of 70 kg and is (almost) composed of water, then each person contains N 70 000 6 02 1023grams 18 grams mol mol . eecules mol protons molecule 10 2 3 10. 228 protons With an excess of 1% electrons over protons, each person has a charge q = ( ) ( )= 0 01 1 6 10 2 3 10 3 7 1019 28 7 . . . .C C So F k q q r e= = ( ) ( ) = 1 2 2 9 7 2 2 25 9 10 3 7 10 0 6 4 10 . . N N ~~10 N26 This force is almost enough to lift a weight equal to that of the Earth: Mg = ( )= 6 10 6 1024 25 kg 9.8 m s N ~10 N2 26
5. 5. Electric Fields 5 *P23.4 In the rst situation, F iA B e A Bk q q r on ,1 = 1 2 . In the second situation, qA and qB are the same. F F iB A A B e A B e k q q r F F k q on ,2 on= = ( ) = 2 2 2 1 AA B e A B q r r k q q F F r r 2 2 1 2 2 1 1 2 2 2 2 62= = N 13.7 mm . 117.7 mm N = 2 1 57. Then FB Aon ,2 N to the left= 1 57. . P23.5 (a) F k q q r e e = = ( ) ( ) 1 2 2 9 19 8 99 10 1 60 10. .N m C C2 2 22 10 2 9 3 80 10 1 59 10 . . ( ) = ( ) m N repulsion (b) F Gm m r g = = ( ) 1 2 2 11 27 6 67 10 1 67 10. .N m C kg2 2 (( ) ( ) = 2 10 2 45 3 80 10 1 29 10 . . m N The electric force is larger by 1.24 10 times36 . (c) If k q q r G m m r e 1 2 2 1 2 2 = with q q q1 2= = and m m m1 2= = , then q m G ke = = = 6 67 10 8 99 10 8 11 9 . . . N m kg N m C 2 2 2 2 661 10 11 C kg P23.6 We nd the equal-magnitude charges on both spheres: F k q q r k q r e e= =1 2 2 2 2 so q r F ke = = ( ) =1 00 1 00 10 1 4 . . .m N 8.99 10 N m C9 2 2 005 10 3 C The number of electron transferred is then N e xfer C C e= = 1 05 10 1 60 10 6 59 10 3 19 15. . . llectrons The whole number of electrons in each sphere is Ntot g 107.87 g mol atoms m= 10 0 6 02 1023. . ool atom( )( )= 47 2 62 1024 e e. The fraction transferred is then f N N = = = xfer tot 6 59 10 2 62 10 2 51 1 15 24 . . . 00 2 519 = . charges in every billion
6. 6. 6 Chapter 23 P23.7 F k q q r e1 1 2 2 9 6 8 99 10 7 00 10 2 = = ( ) ( ) . . .N m C C2 2 000 10 0 500 0 503 6 2 ( ) ( ) = C m N . . F k q q r e2 1 2 2 9 6 8 99 10 7 00 10 4 = = ( ) ( ) . . .N m C C2 2 000 10 0 500 1 01 6 2 ( ) ( ) = C m N . . F F x y = + = = 0 503 60 0 1 01 60 0 0 755 0 5 . cos . . cos . . . N 003 60 0 1 01 60 0 0 436 0 755 sin . . sin . . . N N = =F (( ) ( ) = . .i j0 436 0 872N N at an angle of 3300 P23.8 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 q Q x k q Q d x e e3 2 2 The net force will be zero if 3 1 2 2 x d x = ( ) , or d x x = 3 . This gives an equilibrium position of the third bead of x d= 0 634. . The equilibrium is stable if the third bead has positive chargee . P23.9 (a) The force is one of attraction . The distance r in Coulombs law is the distance between centers. The magnitude of the force is F k q q r e = = ( ) ( ) 1 2 2 9 9 8 99 10 12 0 10 18 . . . N m C C2 2 00 10 0 300 2 16 10 9 2 5 ( ) ( ) = C m N . . (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 8 99 10 3 00 10 3 0 . . . N m C C2 2 00 10 0 300 8 99 10 9 2 7 ( ) ( ) = C m N . . P23.10 The top charge exerts a force on the negative charge k qQ d x e /2 2 2 ( ) + which is directed upward and to the left, at an angle of tan 1 2 d x to the x-axis. The two positive charges together exert force 2 4 4 2 2 2 2 1 2 k qQ d x x d x e +( ) ( ) +( ) i == ma or for x d 1, the components perpendicular to the x-axis add to zero. The total eld is nk Q n R x k Qx R x e e( ) + = +( ) cos i i 2 2 2 2 3 2 . (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 eld point, so it does not change the field . Section 23.5 Electric Field of a Continuous Charge Distribution P23.20 E k dq x e = 2 , where dq dx= 0 E k dx x k x k x e x e x e = = = 0 2 0 0 00 0 1 The direction is or left for . >i 0 0 P23.21 E k d d k Q d d k Q d d e e e = +( ) = ( ) +( ) = +( ) = 8 99 1. 00 22 0 10 0 290 0 140 0 290 9 6 ( ) ( ) ( ) +( ) . . . . E = 1 59 106 . ,N C directed toward the rod . P23.22 E k Qx x a e = +( )2 2 3 2 For a maximum, dE dx Qk x a x x a e= +( ) +( ) = 1 3 0 2 2 3 2 2 2 2 5 2 x a x2 2 2 3 0+ = or x a = 2 Substituting into the expression for E gives E k Qa a k Q a k Q a Q a e e e = ( ) = = = 2 3 2 3 3 6 33 2 2 3 2 3 2 2 2 0 2 FIG. P23.19 FIG. P23.21
7. 10. 10 Chapter 23 P23.23 E k xQ x a x x e = +( ) = ( ) ( ) + 2 2 3 2 9 6 2 8 99 10 75 0 10 0 . . .1100 6 74 10 0 010 02 3 2 5 2 3 2 ( ) = +( ) . . x x We choose the x axis along the axis of the ring. (a) At x = 0 010 0. m , E i i= =6 64 10 6 646 . . N C MN C (b) At x = 0 050 0. m, E i i= =2 41 10 24 17 . . N C MN C (c) At x = 0 300. m, E i i= =6 40 10 6 406 . . N C MN C (d) At x = 1 00. m, E i i= =6 64 10 0 6645 . . N C MN C P23.24 E k x x R e= + 2 1 2 2 E x x = ( ) ( ) + ( ) 2 8 99 10 7 90 10 1 0 350 9 3 2 2 . . . = + 4 46 10 1 0 123 8 2 . . x x (a) At x = 0 050 0. m , E = =3 83 10 3838 . N C MN C (b) At x = 0 100. m, E = =3 24 10 3248 . N C MN C (c) At x = 0 500. m, E = =8 07 10 80 77 . .N C MN C (d) At x = 2 000. m, E = =6 68 10 6 688 . .N C MN C P23.25 (a) From the Example in the chapter text, E k x x R e= + 2 1 2 2 = = = ( )( )= Q R E 2 3 8 1 84 10 1 04 10 0 900 . . . C m N C 2 99 36 10 93 67 . . =N C MN C approximation: E ke= = ( )2 104 MN C about 11% high (b) approximation: E k Q r e= = ( ) ( ) = 2 9 6 28 99 10 5 20 10 0 30 0 519. . . . MN CC about 0.6% high( ) E = ( ) + 1 04 10 1 30 0 3 00 8 2 . . . N C cm 30.0 cm2 = ( )( )=1 04 10 0 004 96 0 5168 . . . N C MN C
8. 11. Electric Fields 11 P23.26 The electric eld at a distance x is E k x x R x e= + 2 1 2 2 This is equivalent to E k R x x e= + 2 1 1 1 2 2 For large x, R x 2 2 1> , 1 2 1 2 2 2 x R x+ , so E k Q x x e 2 for a disk at large distances P23.27 Due to symmetry E dEy y= = 0, and E dE k dq r x e= = sin sin 2 where dq ds rd= = , so that E k r d k r k r x e e e = = ( ) = sin cos 0 0 2 where = q L and r L = Thus, E k q L x e = = ( ) 2 2 8 99 10 7 50 10 2 9 6 . .N m C C2 2 (( ) ( ) 0 140 2 . m Solving, Ex = 2 16 107 . N C Since the rod has a negative charge, E i i= ( ) = 2 16 10 21 67 . . N C MN C . FIG. P23.27
9. 12. 12 Chapter 23 P23.28 (a) We dene x = 0 at the point where we are to nd the eld. One ring, with thickness dx, has charge Qdx h and produces, at the chosen point, a eld d k x x R Qdx h e E i= +( )2 2 3 2 The total eld is E E i= = +( ) = + d k Qxdx h x R ke d d h e all charge 2 2 3 2 QQ h x R xdx k Q h x R x d d h e i E i 2 2 2 2 2 3 2 2 2 +( ) = +( = + )) ( ) = +( ) +( ) + = +1 2 2 2 1 2 21 2 1 1 x d d h ek Q h d R d h R i 22 1 2 ( ) (b) Think of the cylinder as a stack of disks, each with thickness dx, charge Qdx h , and charge- per-area = Qdx R h2 . One disk produces a eld d k Qdx R h x x R e E i= +( ) 2 12 2 2 1 2 So, E E= = +( = + d k Qdx R h x x R e x d d h all charge 2 12 2 2 )) 1 2 i E i = +( ) + = + 2 1 2 22 2 2 1 2k Q R h dx x R xdxe d d h x d d h = +( ) + + 2 1 2 1 22 2 2 1 2 k Q R h x x Re d d h d d h i = + +( ) +( ) + +E i2 2 2 2 1 2 2k Q R h d h d d h R de RR k Q R h h d R d he 2 1 2 2 2 2 1 2 22 ( ) = + +( ) +( )E i ++( ) R2 1 2 P23.29 (a) The electric eld at point P due to each element of length dx is dE k dq x y e = +2 2 and is directed along the line joining the element to point P. By symmetry, E dEx x= = 0 and since dq dx= , E E dE dEy y= = = cos where cos = + y x y2 2 Therefore, E k y dx x y k y e e = +( ) =2 2 2 2 3 2 0 2 0 sin (b) For a bar of innite length, 0 90= and E k y y e = 2 FIG. P23.29
10. 13. Electric Fields 13 P23.30 (a) The whole surface area of the cylinder is A r rL r r L= + = +( )2 2 22 . Q A= = ( ) ( ) + 15 0 10 2 0 025 0 0 025 0 0 09 . . . .C m m m2 660 0 2 00 10 10 m C[ ]= . (b) For the curved lateral surface only, A rL= 2 . Q A= = ( ) ( )( ) 15 0 10 2 0 025 0 0 0609 . . .C m m 0 m2 = 1 41 10 10 . C (c) Q V r L= = = ( ) ( ) 2 9 2 500 10 0 025 0 0 060 0C m m3 . . mm C( ) = 5 89 10 11 . P23.31 (a) Every object has the same volume, V = ( ) = 8 0 030 2 16 10 3 4 . .0 m m .3 For each, Q V= = ( ) ( )= 400 10 2 16 10 8 64 109 4 11 C m m3 3 . . C (b) We must count the 9.00 cm2 squares painted with charge: (i) 6 4 24 = squares Q A= = ( ) ( )= 15 0 10 24 0 9 00 10 3 249 4 . . . .C m m2 2 110 10 C (ii) 34 squares exposed Q A= = ( ) ( )= 15 0 10 24 0 9 00 10 3 249 4 . . . .C m m2 2 110 10 C (iii) 34 squares Q A= = ( ) ( )= 15 0 10 34 0 9 00 10 4 599 4 . . . .C m m2 2 110 10 C (iv) 32 squares Q A= = ( ) ( )= 15 0 10 32 0 9 00 10 4 329 4 . . . .C m m2 2 110 10 C (c) (i) total edge length: = ( )24 0 030 0. m Q = = ( ) ( ) = 80 0 10 24 0 030 0 5 76 1012 11 . . .C m m C (ii) Q = = ( ) ( )= 80 0 10 44 0 030 0 1 06 1012 10 . . .C m m C (iii) Q = = ( ) ( ) = 80 0 10 64 0 030 0 1 54 1012 10 . . .C m m C (iv) Q = = ( ) ( ) = 80 0 10 40 0 030 0 0 960 1012 1 . . .C m m 00 C Section 23.6 Electric Field Lines P23.32 P23.33 FIG. P23.32 FIG. P23.33
11. 14. 14 Chapter 23 P23.34 (a) q q 1 2 6 18 1 3 = = (b) q q1 2is negative, is positive P23.35 (a) The electric eld has the general appearance shown. It is zero at the center , where (by symmetry) one can see that the three charges individually produce elds that cancel out. In addition to the center of the triangle, the electric eld lines in the second gure to the right indicate three other points near the middle of each leg of the triangle where E = 0, but they are more difcult to nd mathematically. (b) You may need to review vector addition in Chapter Three. The electric eld at point P can be found by adding the electric eld vectors due to each of the two lower point charges: E E E= +1 2 . The electric eld from a point charge is E r= k q r e 2 . As shown in the solution gure 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 E E E i j= + = +( )+ 1 2 2 60 60 60k q a e cos sin cos sin sin . i j j+( ) = ( ) = 60 2 60 1 7 2 k q a e 33 2 k q a e j Section 23.7 Motion of a Charged Particle in a Uniform Electric Field P23.36 (a) a qE m = = ( ) ( ) ( ) = 1 602 10 6 00 10 1 67 10 5 19 5 27 . . . ..76 1013 m s so a i= 5 76 1013 . m s2 (b) v vf i f ia x x2 2 2= + ( ) 0 2 5 76 10 0 070 02 13 = + ( )( )vi . . v ii = 2 84 106 . m s (c) v vf i at= + 0 2 84 10 5 76 106 13 = + ( ). . t t = 4 93 10 8 . s FIG. P23.35
12. 15. Electric Fields 15 P23.37 (a) a qE m = = ( ) = 1 602 10 640 1 67 10 6 14 10 19 27 10. . . m ss2 (b) v vf i at= + 1 20 10 6 14 106 10 . . = ( )t t = 1 95 10 5 . s (c) x x tf i i f = +( )1 2 v v xf = ( ) ( )=1 2 1 20 10 1 95 10 11 76 5 . . . m (d) K m= = ( ) ( ) =1 2 1 2 1 67 10 1 20 10 1 22 27 6 2 v . . .kg m s 00 10 15 J P23.38 The particle feels a constant force: F E j j= = ( )( ) ( )= q 1 10 2 000 2 106 3 C N C N (( ) and moves with acceleration: a F j = = ( ) ( ) = m 2 10 2 10 1 10 3 16 kg m s kg 2 113 m s2 ( ) ( )j Note that the gravitational force is on the order of a trillion times smaller than the electrical force exerted on the particle. The particles x-component of velocity is constant at 1 00 10 37 7 99 105 4 . cos . .( ) = m s m s Thus it moves in a parabola opening downward. The maximum height it attains above the bottom plate is described by v vyf yi y f ia y y2 2 2= + ( ): 0 6 02 10 2 10 04 2 13 = ( ) ( ) ( ). m s m s2 yf yf = 1 81 10 4 . m 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 t a tf i yi y= + +v 1 2 2 0 0 6 02 10 1 2 1 104 13 2 = + ( ) + ( ). m s m s2 t t since t > 0, t = 1 20 10 8 . s The particles range is x x tf i x= + = + ( ) ( )= v 0 7 99 10 1 20 10 9 614 8 . . .m s s 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. P23.39 The required electric eld will be in the direction of motion . Work done = K so, = Fd m i 1 2 2 v (since the nal velocity = 0) which becomes eEd = K and E K ed =
13. 16. 16 Chapter 23 P23.40 vi = 9 55 103 . m s (a) a eE m y = = ( )( ) ( ) = 1 60 10 720 1 67 10 6 90 1 19 27 . . . 0010 m s2 From the large magnitude of this vertical acceleration, we can note that the gravitational force on the particle is negligible by comparison to the electrical force. R a i y = = v2 32 1 27 10 sin . m so that 9 55 10 2 6 90 10 1 27 10 3 2 10 3 . sin . . ( ) = sin .2 0 961 = = 36 9. 90 0 53 1. . = (b) t R R ix i = = v v cos If = 36 9. , t = 167 ns . If = 53 1. , t = 221 ns . P23.41 (a) t x x = = = = v 0 050 0 4 50 10 1 11 10 1115 7. . . s ns (b) a qE m y = = ( ) ( ) ( ) = 1 602 10 9 60 10 1 67 10 19 3 27 . . . 99 21 1011 . m s2 y y t a tf i yi y = +v 1 2 2 : yf = ( ) ( ) = = 1 2 9 21 10 1 11 10 5 68 10 511 7 2 3 . . . .m 668 mm (c) vx = 4 50 105 . m s v vyf yi ya t= + = ( ) ( )= 9 21 10 1 11 10 1 02 1011 7 5 . . . m s Additional Problems P23.42 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 m C than on the 12 m C, and could not produce equilibrium for both. Thus the third charge must be at x = d < 0. Its equilibrium requires k q d k q d e e12 45 152 2 C C cm ( ) = ( ) +( ) 15 45 12 3 75 2 cm + = = d d . 15 1 94cm + =d d. d = 16.0 cm The third charge is at x = 16 0. cm . The equilibrium of the 12 m C requires k q ke e12 16 0 45 12 2 2 C cm C C 15 cm ( ) ( ) = ( ) ( ). q = 51 3. C All six individual forces are now equal in magnitude, so we have equilibrium as required, and this is the only solution. FIG. P23.42 d x 15 cm q x = 0 + 12 mC 45 mC FIG. P23.40
14. 17. Electric Fields 17 P23.43 The proton moves with acceleration a qE m p = = ( )( ) = 1 60 10 640 1 673 10 19 27 . . C N C kg 66 13 1010 . m s2 while the e has acceleration ae = ( )( ) = 1 60 10 640 9 110 10 1 12 19 31 . . . C N C kg =10 183614 m s2 ap (a) We want to nd the distance traveled by the proton (i.e., d a tp= 1 2 2 ), knowing: 4 00 1 2 1 2 1837 1 2 2 2 2 . cm = + = a t a t a tp e p Thus, d a tp= = = 1 2 4 00 21 82 . . cm 1837 m (b) The distance from the positive plate to where the meeting occurs equals the distance the sodium ion travels (i.e., d a tNa Na= 1 2 2 ). This is found from: 4 00 1 2 1 2 2 2 . cm :Na Cl= +a t a t 4 00 1 2 22 99 1 2 35 45 2 2 . . . cm u u = + eE t eE t This may be written as 4 00 1 2 1 2 0 649 1 65 1 2 2 2 2 . . .cm Na Na Na= + ( ) = a t a t a t so d a tNa Na cm 1.65 cm= = = 1 2 4 00 2 432 . . P23.44 (a) The eld, E1, due to the 4 00 10 9 . C charge is in the x direction. E r1 2 9 9 8 99 10 4 00 10 = = ( ) ( ) k q r e . .N m C C2 2 22 50 5 75 2 . . m N C ( ) = i i Likewise, E2 and E3 , due to the 5 00 10 9 . C charge and the 3 00 10 9 . C charge, are E r2 2 9 9 8 99 10 5 00 10 2 = = ( ) ( ) k q r e . .N m C C2 2 .. . 00 11 22 m N C ( ) =i i E3 9 9 2 8 99 10 3 00 10 1 20 = ( ) ( ) ( ) . . . N m C C m 2 2 . i i= 18 7 N C E E E ER = + + =1 2 3 24 2. N C in +x direction. FIG. P23.44 (a) continued on next page
15. 18. 18 Chapter 23 (b) E r i j1 2 8 46 0 243 0 970= = ( ) +( )k q r e . . . N C E r j E r 2 2 3 2 11 2 5 81 = = ( ) +( ) = = k q r k q r e e . . N C NN C + 0.928 N ( ) ( ) = + = 0 371 4 211 3 . . i j iE E Ex x x CC N CE E E Ey y y y= + + =1 2 3 8 43. j ER = 9 42. N C = 63 4. above axisx P23.45 (a) Let us sum force components to nd F qE Tx x = =sin 0, and F qE T mgy y = + =cos 0 Combining these two equations, we get q mg E Ex y = +( ) = ( )( ) cot . . . cot . 1 00 10 9 80 3 00 37 3 00 5 00 10 1 09 10 10 9 5 8 C nC +( ) = = . . . (b) From the two equations for Fx and Fy we also nd T qEx = = = sin . . . 37 0 5 44 10 5 443 N mN P23.46 This is the general version of the preceding problem. The known quantities are A, B, m, g, and . The unknowns are q and T. The approach to this problem should be the same as for the last problem, but without numbers to substitute for the variables. Likewise, we can use the free body diagram given in the solution to problem 45. Again, Newtons second law: F T qAx = + =sin 0 (1) and F T qB mgy = + + =cos 0 (2) (a) Substituting T qA = sin into Eq. (2), qA qB mg cos sin + = Isolating q on the left, q mg A B = +( )cot (b) Substituting this value into Eq. (1), T mgA A B = +( )cos sin If we had solved this general problem rst, we would only need to substitute the appropriate values in the equations for q and T to nd the numerical results needed for problem 45. If you nd this problem more difcult than problem 45, the little list at the rst 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. FIG. P23.45 FIG. P23.44 (b)
16. 19. Electric Fields 19 P23.47 F k q q r e = 1 2 2 : tan . . = 15 0 60 0 = 14 0. F F 1 9 6 2 2 3 8 99 10 10 0 10 0 150 40 0= ( ) ( ) ( ) = = . . . . N 88 99 10 10 0 10 0 600 2 50 8 9 6 2 2 2 . . . . . ( ) ( ) ( ) = = N F 999 10 10 0 10 0 619 2 35 9 6 2 2 3 ( ) ( ) ( ) = = . . . N F Fx FF F Fy 2 14 0 2 50 2 35 14 0 4 78cos . . . cos . . N= = = 11 2 14 0 40 0 2 35 14 0 40 6 = = F sin . . . sin . . N F F F F F x y y net N= + = ( ) + ( ) = = 2 2 2 2 4 78 40 6 40 9. . . tan xx = = 40 6 4 78 263 . . P23.48 From Figure (a) we have d cos . .30 0 15 0 cm= or d = 15 0 30 0 . cos . cm From Figure (b) we have = sin . 1 50 0 d cm = ( ) = sin . cos . .1 15 0 30 0 20 3 cm 50.0 cm F mg q = tan or F mgq = tan .20 3 (1) From Figure (c) we have F Fq = 2 30 0cos . F k q q e = ( ) 2 0 300 30 0 2 2 . cos . m (2) Combining equations (1) and (2), 2 0 300 30 0 20 3 2 2 k q mge . cos . tan . m ( ) = q mg k q e 2 2 2 0 300 20 3 2 30 0 2 00 1 = ( ) = . tan . cos . . m 00 9 80 0 300 20 3 2 8 99 3 2 ( )( )( ) kg m s m 2 . . tan . . 110 30 0 4 20 10 2 05 1 9 14 N m C C 2 2 2 ( ) = = cos . . .q 00 0 2057 =C C. FIG. P23.47 FIG. P23.48 Figure (a) Figure (b) Figure (c)
17. 20. 20 Chapter 23 P23.49 Charge Q 2 resides on each of the blocks, which repel as point charges: F k Q Q L k L Le i= ( )( ) = ( ) 2 2 2 Solving for Q, Q L k L L k i e = ( )2 P23.50 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 a e 2 away from vertex 29 = 0 F28 charges toward vertex 29= k qQ a e 2 P23.51 According to the result of an Example in the chapter text, the left- hand rod creates this eld 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 Q a dx x x e e e = +( ) = +( ) = + 2 2 2 2 2 22 2 1 2 2 2 2 2a k Q a a a x xx b a b e b a b ( ) = + = ln F k Q a a b b b b a k Q a b b e e = + + + = 2 2 2 2 2 4 2 2 4 ln ln ln ( ) +( ) = 2 2 4 4 2 2 2 2 2 a b a k Q a b b a e ln *P23.52 We model the spheres as particles. They have different charges. They exert on each other forces of equal magnitude. They have equal masses, so their strings make equal angles q with the vertical. The distance r between them is described by sin q = (r/2)/40 cm, so r = 80 cm sin q Let T represent the string tension. We have Fx = 0: ke q1 q2 /r2 = T sinq Fy = 0: mg = T cosq Divide to eliminate T. k q q r mg r r e 1 2 2 2 2 40 4 = = tan ( / /cm)2 Cleared of fractions, k q q r mgre 1 2 2 3 80( cm)2 = 8.99 109 (N m2 /C2 ) 300 109 C (200 109 C) ( . . ( . )0 8 2 4 10 9 82 3 3 m) N2 = r r (0.8 m)2 r2 = 1901 r6 We home in on a solution by trying values. r 0.64 r2 1901 r6 0 +0.64 0.5 29.3 0.2 +0.48 0.3 0.84 0.24 +0.22 0.27 0.17 0.258 +0.013 0.259 0.001 Thus the distance to three digits is 0 259. m . FIG. P23.51
18. 21. Electric Fields 21 *P23.53 Q d Rd R= = = 0 90 0 90 0 0 90 0 9 cos sin . . . 00 0 0 0 0 1 1 2 12 0 2 0 600 . . . = ( )[ ]= = = ( )( R R Q C )) = = = ( ) m C so C m C 012 0 10 0 1 4 3 00 0 . . . dFy d R Rd( ) = ( ) 2 0 0 2 1 4 3 00 cos . cosC (( ) = ( ) ( R Fy 2 9 6 8 99 10 3 00 10 . . N m C C2 2 )) ( ) ( ) 10 0 10 0 600 6 2 90 0 90 0 . . cos . . C m m d = ( ) ( ) + F dy 8 99 30 0 0 600 10 1 2 1 2 23. . . cosN = ( ) + = 2 2 2 2 0 450 1 2 1 4 2 0Fy . sin .N 7707 N downward Since the leftward and rightward forces due to the two halves of the semicircle cancel out, Fx = 0. *P23.54 (a) The two charges create elds of equal magnitude, both with outward components along the x axis and with upward and downward y components that add to zero. The net eld is then k q r x r k q r x r e e 2 2 9 9 2 8 99 10 52 10 ( . ) i i+ = N m2 C C m)2 2 x x (( . ) / i 0 25 2 3 2 + = 935 2 3 2 N m C(0.0625 m 2 2 + x x ) / i (b) At x = 0.36 m, E i = + 935 0 36 0 36 N m m C (0.0625 m m 2 2 . ( . )) kN/C2 3 2 4 00 ) . / = i (c) We solve 1 000 = 935 x (0.0625 + x2 )3/2 by tabulating values for the eld function: x 935 x (0.0625 + x2 )3/2 0 0 0.01 597 0.02 1 185 0.1 4 789 0.2 5 698 0.36 4 000 0.9 1 032 1 854 0 We see that there are two points where E = 1 000. We home in on them to determine their coordinates as (to three digits) x = 0.016 8 m and x = 0.916 m. (d) The table in part (c) shows that the eld is nowhere so large as 16 000 N/C. (e) The eld of a single charge in Question 7 takes on all values from zero to innity, each at just one point along the positive x axis. The vector sum of the eld of two charges, in this problem, is zero at the origin, rises to a maximum at 17.7 cm, and then decreases asymptotically to zero. In the question and the problem, the elds at x = 36 cm happen to take similar values. For large x the eld of the two charges in this problem shows just the same inverse proportionality to x2 as the eld in the question, being larger by the factor 2(52 nC)/(57.7 nC) = 1.80 times. FIG. P23.53 1 0 1 cos q 0 360 0 360 1 0 cos2 q
19. 22. 22 Chapter 23 P23.55 (a) From the 2Q charge we have F Te =2 2 0sin and mg T =2 2 0cos Combining these we nd F mg T T e = =2 2 2 2 2 sin cos tan From the Q charge we have F Te = =1 1 0sin and mg T =1 1 0cos Combining these we nd F mg T T e = =1 1 1 1 1 sin cos tan or 2 1= (b) F k QQ r k Q r e e e = = 2 2 2 2 2 If we assume q is small then tan r 2 Substitute expressions for Fe and tan into either equation found in part (a) and solve for r. F mg e = tan, then 2 1 2 2 2 k Q r mg re and solving for r we nd r k Q mg e 4 2 1 3 . P23.56 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 shown for the bead on the left side of the bowl: F n mgy = = sin .60 0 0 , or n mg = sin .60 0 Also, F F nx e = + =cos .60 0 0 , or k q R n mg mge 2 2 60 0 60 0 3 = = =cos . tan . Thus, q R mg ke = 3 1 2 P23.57 (a) The total non-contact force on the cork ball is: F qE mg m g qE m = + = + , which is constant and directed downward. Therefore, it behaves like a simple pendulum in the presence of a modied uniform gravitational eld with a period given by: T L g qE m = + = + ( ) 2 2 0 500 2 00 10 1 06 . . . m 9.80 m s C2 00 10 1 00 10 0 307 5 3 ( ) = N C kg s . . (b) Yes . Without gravity in part (a), we get T L qE m = 2 T = ( ) ( ) 2 0 500 2 00 10 1 00 1 00 106 . . . . m C 10 N C5 =3 0 314 kg s. (a 2.28% difference). FIG. P23.55 FIG. P23.56 n mg 60.0Fe
20. 23. Electric Fields 23 *P23.58 (a) At A the top charge makes the dominant contribution to the eld and the net eld is downward. At B the total electric eld is zero. Between B and C the main change is weakening of the downward electric eld of the top charge, so the net eld at C is upward. At E the elds of the two bottom charges cancel out and the total eld is downward. At F the total eld is downward. (b) The eld is zero at B as it changes from downward at A to upward at C. As a continuous function, the eld must pass through the value zero near D as it changes from upward at C to downward at E and F. (c) Let y represent the distance from E up to the zero-eld point. The distance from P to E is (32 1.52 )1/2 cm = 2.60 cm. Then the requirement that the eld be zero is k q y k q y y y e e ( . ) ( . .2 60 1 5 1 5 2 2 2cm cm) cm 2 = + ( ) + 22 2 3 2 1 5 + + k qy y e ( . / cm)2 k q y k qy y e e ( . ) ( . / 2 60 2 1 5 2 2 3 2 cm cm)2 = + (1.52 + y2 )3/2 2 y (2.60 y)2 = 0 As a check on our algebra, we note that y = (1/3)2.60 cm = 0.866 cm should be a solution, corresponding to point B. Substituting 0.866 gives 5.20 5.20 = 0 as it should. We home in on the smaller answer: y (1.52 + y2 )3/2 2y(2.60 y)2 0 +3.375 0.3 +0.411 0.4 0.124 0.37 +0.014 0.373 0.000 6 To three digits the answer is 0.373 cm . P23.59 (a) There are 7 terms which contribute: 3 are s away (along sides) 3 are 2s away (face diagonals) and sin cos = = 1 2 1 is 3s away (body diagonal) and sin = 1 3 The component in each direction is the same by symmetry. F i j k= + + + +( )= k q s k q s e e 2 2 2 2 1 2 2 2 1 3 3 1 .990( ) + +( ) i j k (b) F F F F k q s x y z e = + + =2 2 2 2 2 3 29. away from the origiin FIG. P23.59 P A B C D E F + ++ FIG. P23.58
21. 24. 24 Chapter 23 P23.60 (a) Zero contribution from the same face due to symmetry, opposite face contributes 4 2 k q r e sin where r s s s s s= + + = = 2 2 1 5 1 22 2 2 2 . . sin = s r E k qs r k q s k q s e e e = = ( ) =4 4 1 22 2 183 3 2 2 . . (b) The direction is the direction.k P23.61 The eld on the axis of the ring is calculated in an Example in the chapter text as E E k xQ x a x e = = +( )2 2 3 2 The force experienced by a charge q placed along the axis of the ring is F k Qq x x a e= +( ) 2 2 3 2 and when x a B > D > C. (b) The ux through the 4-cm sphere is (4/5)3 q/0 . The ux through the 8-cm sphere and through the 16-cm sphere is q/0 . The ux through the 12-cm sphere is 0. The ranking is B = D > A > C. *Q24.12 The outer wall of the conducting shell will become polarized to cancel out the external eld. The interior eld is the same as before. Answer (c). Q24.13 If the person is uncharged, the electric eld inside the sphere is zero. The interior wall of the shell carries no charge. The person is not harmed by touching this wall. If the person carries a (small) charge q, the electric eld inside the sphere is no longer zero. Charge q is induced on the inner wall of the sphere. The person will get a (small) shock when touching the sphere, as all the charge on his body jumps to the metal. *Q24.14 (i) The shell becomes polarized. Answer (e). (ii) The net charge on the shells inner and outer surfaces is zero. Answer (a). (iii) Answer (c). (iv) Answer (c). (v) Answer (a). Q24.15 There is zero force. The huge charged sheet creates a uniform eld. The eld can polarize the neutral sheet, creating in effect a lm of opposite charge on the near face and a lm with an equal amount of like charge on the far face of the neutral sheet. Since the eld is uniform, the lms of charge feel equal-magnitude forces of attraction and repulsion to the charged sheet. The forces add to zero. SOLUTIONS TO PROBLEMS Section 24.1 Electric Flux P24.1 E EA= cos A r= = ( ) = 2 2 0 200 0 126. . m2 5 20 10 0 126 05 . . cos = ( )E E = =4 14 10 4 146 . .N C MN C P24.2 E EA= = ( )( ) =cos . . cos . 2 00 10 18 0 10 0 3554 N C m 2 kN m C2 P24.3 (a) E a b A aA= = +( ) =E A i j i (b) E a b A bA= +( ) = i j j (c) E a b A= +( ) = i j k 0
22. 29. Gausss Law 29 P24.4 (a) = ( )( )A 10 0 30 0. .cm cm = = = = A EAE A E A 300 0 030 0 7 80 cm m2 2 . cos . , , ( )( ) = 10 0 030 0 180 2 34 4 . cos ., kN m C2 E A (b) E A EA A, cos . cos .= = ( )( ) 7 80 10 60 04 A w= ( )( )= ( ) 30 0 30 0 10 0 60 0 . . . cos . cm cm cm = = = ( )( 600 0 060 0 7 80 10 0 060 04 cm m2 2 . . .,E A )) = + cos . .60 0 2 34 kN m C2 (c) The bottom and the two triangular sides all lie parallel to E, so E = 0 for each of these. Thus, E, . .total 2 2 kN m C kN m C= + + + + =2 34 2 34 0 0 0 0 P24.5 E EA= cos through the base E = ( )( ) = 52 0 36 0 180 1 87. . cos . kN m C2 Note that the same number of electric eld lines go through the base as go through the pyramids surface (not counting the base). For the slanting surfaces, E = + 1 87. kN m C2 . Section 24.2 Gausss Law P24.6 (a) E k Q r e = 2 ; 8 90 10 8 99 10 0 750 2 9 2. . . = ( ) ( ) Q But Q is negative since E points inward. Q = = 5 57 10 55 78 . .C nC (b) The negative charge has a spherically symmetric charge distribution, concentric with the spherical shell. P24.7 (a) E q = = + + ( )in C C C C 0 5 00 9 00 27 0 84 0. . . . 88 85 10 6 89 1012 6 . . = C N m N m C2 2 2 2 E = 6 89. MN m C2 (b) Since the net electric ux is negative, more lines enter than leave the surface. FIG. P24.4 FIG. P24.5
23. 30. 30 Chapter 24 P24.8 (a) One-half of the total ux created by the charge q goes through the plane. Thus, E E q q , ,plane total= = = 1 2 1 2 20 0 (b) The square looks like an innite plane to a charge very close to the surface. Hence, E E q , ,square plane = 2 0 (c) The plane and the square look the same to thhe charge. P24.9 E q = in 0 Through S1 E Q Q Q = + = 2 0 0 Through S2 E Q Q = + = 0 0 Through S3 E Q Q Q Q = + = 2 2 0 0 Through S4 E = 0 P24.10 (a) E q , . . .shell in = = = 0 6 12 12 0 10 8 85 10 1 36 100 1 366 N m C MN m C2 2 = . (b) E,half shell 2 N m C N m= ( )= 1 2 1 36 10 6 78 106 5 . . 22 2 C kN m C= 678 (c) No , the same number of eld lines will pass through each surface, no matter how the radius changes. P24.11 (a) With d very small, all points on the hemisphere are nearly at a distance R from the charge, so the eld everywhere on the curved surface is k Q R e 2 radially outward (normal to the surface). Therefore, the ux is this eld strength times the area of half a sphere: curved local hemisphere curved = = = E Ad E A ke QQ R R Q Q 2 2 0 0 1 2 4 1 4 2 2 = ( ) = + (b) The closed surface encloses zero charge so Gausss law gives curved flat+ = 0 or flat curved= = Q 2 0 P24.12 Consider as a gaussian surface a box with horizontal area A, lying between 500 and 600 m elevation. E A = d q 0 : +( ) + ( ) = ( ) 120 100 100 0 N C N C m A A A = ( ) ( ) = 20 8 85 10 100 1 77 10 12 12 N C C N m m C 2 2 . . mm3 The charge is positive , to produce the net outward ux of electric eld. Q 0 FIG. P24.11
24. 31. Gausss Law 31 P24.13 The total charge is Q q 6 . The total outward ux from the cube is Q q 6 0 , of which one-sixth goes through each face: E Q q ( ) = one face 6 6 0 . E Q q ( ) = = ( ) one face C N m6 6 5 00 6 00 10 0 6 . . 22 2 C kN m C 6 8 85 10 18 812 2 = . . P24.14 The total charge is Q q 6 . The total outward ux from the cube is Q q 6 0 , of which one-sixth goes through each face: E Q q ( ) = one face 6 6 0 P24.15 If R d ,, the sphere encloses no charge and E q = =in 0 0 . If R d> , the length of line falling within the sphere is 2 2 2 R d so E R d = 2 2 2 0 P24.16 E ek Q R r, . hole hole N = = ( )= E A 2 2 9 8 99 10 ( ) ( ) ( ) m C C m 2 2 10 0 10 0 100 1 00 6 2 . . . 110 3 2 ( )m E, .hole 2 N m C= 28 2 P24.17 E q = = = in 2 2 C 8.85 10 C N m0 6 12 170 10 1 92 1. 007 N m /C2 (a) E E( ) = = one face 2 N m C1 6 1 92 10 6 7 . E( ) = one face 2 MN m C3 20. (b) E = 19 2. MN m C2 (c) The answer to (a) would change because the fflux through each face of the cube would nott be equal with an asymmetric charge distriibution. The sides of the cube nearer the chharge would have more flux and the ones furrther away would have less. The answer to (bb) would remain the same, since the overalll flux would remain the same.
25. 32. 32 Chapter 24 Section 24.3 Application of Gausss Law to Various Charge Distributions P24.18 (a) E k Qr a e = =3 0 (b) E k Qr a e = = ( ) ( )( ) ( 3 9 6 8 99 10 26 0 10 0 100 0 400 . . . . )) =3 365 kN C (c) E k Q r e = = ( ) ( ) ( ) = 2 9 6 2 8 99 10 26 0 10 0 400 1 46 . . . . MNN C (d) E k Q r e = = ( ) ( ) ( ) = 2 9 6 2 8 99 10 26 0 10 0 600 649 . . . kN CC The direction for each electric eld is radially outward . P24.19 The charge distributed through the nucleus creates a eld at the surface equal to that of a point charge at its center: E k q r e = 2 . E = ( ) ( ) ( ) 8 99 10 82 1 60 10 208 1 9 19 1 3 . . . Nm C C2 2 220 10 15 2 m E = 2 33 1021 . N C away from the nucleus P24.20 Note that the electric eld in each case is directed radially inward, toward the lament. (a) E k r e = = ( ) ( ) 2 2 8 99 10 90 0 10 0 100 9 6 . . . N m C C m2 2 m MN C= 16 2. (b) E k r e = = ( ) ( ) 2 2 8 99 10 90 0 10 0 200 9 6 . . . N m C C m2 2 m MN C= 8 09. (c) E k r e = = ( ) ( ) 2 2 8 99 10 90 0 10 1 00 9 6 . . . N m C C m2 2 mm MN C= 1 62. P24.21 E = = ( ) = 2 9 00 10 2 8 85 10 508 0 6 12 . . C m C N m k 2 2 2 NN C, upward P24.22 (a) E k r e = 2 3 60 10 2 8 99 10 2 40 0 190 4 9 . . . . = ( )( )Q Q = + = + 9 13 10 9137 . C nC (b) E = 0 P24.23 mg qE q q Q A = = = 2 20 0 Q A mg q = = ( )( )( ) 2 2 8 85 10 0 01 9 8 0 7 10 0 12 6 . . . . == 2 48. C m2 *P24.24 (a) A long cylindrical plastic rod 2.00 cm in radius carries charge uniformly distributed throughout its volume, with density 5.00 C/m3 . Find the magnitude of the electric eld it creates at a point P, 3.00 cm from its axis. As a gaussian surface choose a concentric cylinder with its curved surface passing through the point P and with length 8.00 cm. (b) We solve for E = ( . . ) ( . 0 02 0 08 8 85 10 12 m) m (5 10 C/m C 2 6 3 22 2 /N m m) 0.08 m kN/C = ) ( . . 2 0 03 3 77
26. 33. Gausss Law 33 P24.25 The volume of the spherical shell is 4 3 0 25 0 20 3 19 10 3 3 2 . . .m m m3 ( ) ( ) = Its charge is V = ( ) ( )= 1 33 10 3 19 10 4 25 106 2 8 . . .C m m3 3 C The net charge inside a sphere containing the protons path as its equator is = 60 10 4 25 10 1 02 109 8 7 C C C. . The electric eld is radially inward with magnitude k q r q r e 2 0 2 9 7 2 4 8 99 10 10 0 25 = = ( ) . . Nm 1.02 C C 2 mm N C ( ) = 2 4 1 47 10. For the proton F ma = eE m r = v2 v = = ( ) eEr m 1 2 19 4 1 60 10 1 47 10 0 25. . .C N C m 11.67 kg m s = 10 5 94 1027 1 2 5 . *P24.26 s = ( ) = 8 60 10 100 8 60 106 2 2 . .C cm cm m C m2 22 E = = ( ) = 2 8 60 10 2 8 85 10 4 86 10 0 2 12 9. . . N C awaay from the wall So long as the distance from the wall is small compared to the width and height of the wall, the distance does not affect the eld. P24.27 If r is positive, the eld must be radially outward. Choose as the gaussian surface a cylinder of length L and radius r, contained inside the charged rod. Its volume is r L2 and it encloses charge r L2 . Because the charge distribution is long, no electric ux passes through the circular end caps; E A = =d EdAcos .90 0 0 . The curved surface has E A =d EdAcos0, and E must be the same strength everywhere over the curved surface. Gausss law, E A = d q 0 , becomes E dA r L Curved Surface = 2 0 . Now the lateral surface area of the cylinder is 2rL: E r L r L 2 2 0 ( ) = Thus, E = r 2 0 radially away from the cylinder axiis . P24.28 The distance between centers is 2 5.90 1015 m. Each produces a eld as if it were a point charge at its center, and each feels a force as if all its charge were a point at its center. F k q q r e = = ( ) ( ) 1 2 2 9 2 19 8 99 10 46 1 60 10 . . N m C2 2 CC m N kN ( ) ( ) = = 2 15 2 3 2 5 90 10 3 50 10 3 50 . . . FIG. P24.27
27. 34. 34 Chapter 24 P24.29 (a) E = 0 (b) E k Q r e = = ( ) ( ) ( ) = 2 9 6 2 8 99 10 32 0 10 0 200 7 19 . . . . MMN C E = 7 19. MN C radially outward P24.30 Consider two balloons of diameter 0.2 m, each with mass 1 g, hanging apart with a 0.05 m separation on the ends of strings making angles of 10 with the vertical. (a) F T mg T mg y = = =cos cos 10 0 10 F T F F Tx e e = = =sin sin10 0 10 , so F mg mge = = = ( ) cos sin tan . . 10 10 10 0 001 9 kg 88 10 2 10 3 3 m s N ~10 N or 1 mN 2 ( ) tan Fe (b) F k q r e e = 2 2 2 10 8 99 10 0 25 1 2 1 3 9 2 2 ( ) ( ) N N m C m 2 2 . . . q q 00 107 7 C ~ C or 100 nC (c) E k q r e = ( ) ( ) ( 2 9 7 8 99 10 1 2 10 0 25 . . . N m C C m 2 2 )) 2 4 1 7 10 10. ~N C kN C (d) E q = = 0 7 12 41 2 10 1 4 10 . . C 8.85 10 C N m N2 2 m C kN m C2 2 ~10 P24.31 (a) E k r e = = ( ) ( ) 2 2 8 99 10 2 00 10 7 009 6 . . .N m C C2 2 mm m 0 100. E = 51 4. ,kN C radially outward (b) E EA E r= = ( )cos cos 2 0 E = ( ) ( )( )( ) =5 14 10 2 0 100 0 020 0 1 004 . . . .N C m m 6646 N m C2 Section 24.4 Conductors in Electrostatic Equilibrium P24.32 The elds are equal. The equation E = conductor 0 suggested in the chapter for the eld outside the aluminum looks different from the equation E = insulator 2 0 for the eld around glass. But its charge will spread out to cover both sides of the aluminum plate, so the density is conductor .= Q A2 The glass carries charge only on area A, with insulator = Q A . The two elds are Q A2 0 , the same in magnitude, and both are perpendicular to the plates, vertically upward if Q is positive. FIG. P24.30
28. 35. Gausss Law 35 P24.33 EdA E rl q = ( )= 2 0 in E q l r r = = in 2 20 0 (a) r = 3 00. cm E = 0 (b) r = 10 0. cm E = ( )( ) = 30 0 10 2 8 85 10 0 100 5 400 9 12 . . . N C, ooutward (c) r = 100 cm E = ( )( ) = 30 0 10 2 8 85 10 1 00 540 9 12 . . . N C, outtward *P24.34 (a) All of the charge sits on the surface of the copper sphere at radius 15 cm. The eld inside is zero . (b) The charged sphere creates eld at exterior points as if it were a point charge at the center: E = = ( ) ( ) k q r e 2 9 9 8 99 10 40 10 0 away Nm C C 2 2 . .. . 17 1 24 102 4 m outward N C outward ( ) = (c) E = ( ) ( ) ( ) 8 99 10 40 10 0 75 9 9 2 . . Nm C C m ou 2 2 ttward N C outward= 639 (d) All three answers would be the same. The solid copper sphere carries charge only on its outer surface. P24.35 (a) E = 0 = ( ) ( )= 8 00 10 8 85 10 7 08 104 12 7 . . . C m2 = 708 nC m2 , positive on one face and negative on the other. (b) = Q A Q A= = ( )( ) 7 08 10 0 5007 2 . . C Q = = 1 77 10 1777 . ,C nC positive on one face and negative on the other. *P24.36 Let the at box have face area A perpendicular to its thickness dx. The ux at x = 0.3 m is into the box EA = (6 000 N/C m2 )(0.3 m)2 A = (540 N/C) A The ux out of the box at x = 0.3 m + dx +EA = (6 000 N/C m2 )(0.3 m + dx)2 A = +(540 N/C) A + (3 600 N/C m) dx A (The term in (dx)2 is negligible.) The charge in the box is rA dx where r is the unknown. Gausss law is (540 N/C) A + (540 N/C) A + (3 600 N/C m) dx A = rA dx/0 Then r = (3600 N/C m)0 = (3600 N/C m)(8.85 1012 C2 /N m2 ) = 31.9 nC/m3 P24.37 The charge divides equally between the identical spheres, with charge Q 2 on each. Then they repel like point charges at their centers: F k Q Q L R R k Q L R e e = ( )( ) + +( ) = +( ) = 2 2 4 2 8 99 10 2 2 2 9 . NN m C C m N 2 2 ( ) ( ) = 60 0 10 4 2 01 2 00 6 2 2 . . .
29. 36. 36 Chapter 24 *P24.38 The surface area is A = 4pa2 . The eld is then E k Q a Q a Q A e = = = = 2 0 2 0 04 It is not equal to s /20 . At a point just outside, the uniformly charged surface looks just like a uniform at sheet of charge. The distance to the eld point is negligible compared to the radius of curvature of the surface. P24.39 (a) Inside surface: consider a cylindrical surface within the metal. Since E inside the conducting shell is zero, the total charge inside the gaussian surface must be zero, so the inside charge/length = . 0 = + qin so qin = Outside surface: The total charge on the metal cylinder is 2 = +q qin out qout = +2 so the outside charge/length is 3 (b) E k r k r r e e = ( ) = = 2 3 6 3 2 0 radially outward P24.40 An approximate sketch is given at the right. Note that the electric eld lines should be perpendicular to the conductor both inside and outside. P24.41 (a) The charge density on each of the surfaces (upper and lower) of the plate is: = = ( ) ( ) = 1 2 1 2 4 00 10 0 500 8 00 1 8 2 q A . . . C m 00 80 08 =C m nC m2 2 . (b) E k= = 0 8 12 8 00 10 8 85 10 . . C m C N 2 2 mm kN C2 = ( ) . k k9 04 (c) E k= ( )9 04. kN C Additional Problems P24.42 In general, E i j k= + +ay cx bz In the xy plane, z = 0 and E i k= +ay cx E E x w d ay cx dA ch xdx c = = +( ) = = = E A i k k 0 hh x chw x w2 0 2 2 2= = FIG. P24.40 x y z x = 0 x = w y = 0 y = h dA = hdx FIG. P24.42
30. 37. Gausss Law 37 P24.43 (a) Uniform E, pointing radially outward, so E EA= . The arc length is ds Rd= , and the circumference is 2 2 r R= sin . A rds R Rd R d R = = ( ) = = 2 2 2 2 0 2 0 2 sin sin ( ) = ( )cos cos 0 2 2 1R E Q R R Q = ( )= ( ) 1 4 2 1 2 1 0 2 2 0 cos cos [independent of R!] (b) For = 90 0. (hemisphere): E Q Q = ( )= 2 1 90 20 0 cos . (c) For = 180 (entire sphere): E Q Q = ( )= 2 1 180 0 0 cos [Gausss Law]. *P24.44 (a) qin C C C= + = +3 1 2 00 . (b) The charge distribution is spherically symmetric and qin > 0. Thus, the eld is directed radially outward or to the right at point D. (c) E k q r e = = =in 6 2.00 10 N/C 2 9 2 8 99 10 0 16 70 . ( . ) 22 kN/C (d) Since all points within this region are located inside conducting material, E = 0 . (e) E Ed q= = = =E A 0 00in (f) qin C= +3 00. (g) E k q r e = = = in 6 3.00 10 MN2 9 2 8 99 10 0 08 4 21 . ( . ) . //C to the right (radially outward). (h) q Vin C C= = + = + 3 5 4 3 4 1 544 3 3 3 . (i) E k q r e = = = in 6 1.54 10 MN2 9 2 8 99 10 0 04 8 63 . ( . ) . //C to the right (radially outward) (j) As in part (d), E = 0 for 10 15cm cm<
31. 38. 38 Chapter 24 *P24.45 (a) The eld is zero within the metal of the shell. The exterior electric eld lines end at equally spaced points on the outer surface. The charge on the outer surface is distributed uniformly. Its amount is given by EA = Q/0 Q = (890 N/C) 4p (0.75 m)2 8.85 1012 C2 /N m2 = 55.7 nC (b) and (c) For the net charge of the shell to be zero, the shell must carry +55.7 nC on its inner surface, induced there by 55.7 nC in the cavity within the shell. The charge in the cavity could have any distribution and give any corresponding distribution to the charge on the inner surface of the shell. For example, a large positive charge might be within the cavity close to its topmost point, and a slightly larger negative charge near its easternmost point. The inner surface of the shell would then have plenty of negative charge near the top and even more positive charge centered on the eastern side. P24.46 The sphere with large charge creates a strong eld to polarize the other sphere. That means it pushes the excess charge over to the far side, leaving charge of the opposite sign on the near side. This patch of opposite charge is smaller in amount but located in a stronger external eld, so it can feel a force of attraction that is larger than the repelling force felt by the larger charge in the weaker eld on the other side. P24.47 (a) E A = ( )= d E r q 4 2 0 in For r a< , q rin = 4 3 3 so E r = 3 0 For a r b< < and c r< , q Qin = So E Q r = 4 2 0 For b r c , E = 0, since E = 0 inside a conductor. (b) Let q1 = induced charge on the inner surface of the hollow sphere. Since E = 0 inside the conductor, the total charge enclosed by a spherical surface of radius b r c must be zero. Therefore, q Q1 0+ = and 1 1 2 2 4 4 = = q b Q b Let q2 = induced charge on the outside surface of the hollow sphere. Since the hollow sphere is uncharged, we require q q1 2 0+ = and 2 1 2 2 4 4 = = q c Q c FIG. P24.47
32. 39. Gausss Law 39 P24.48 First, consider the eld at distance r R< from the center of a uniform sphere of positive charge Q e= +( ) with radius R. 4 2 0 0 4 3 3 4 3 3 0 r E q V e R r ( ) = = = + in so E e R = 4 0 3 r directed outward (a) The force exerted on a point charge q e= located at distance r from the center is then F qE e e R r e R r Kr= = = = 4 40 3 2 0 3 (b) K e R k e R e = = 2 0 3 2 3 4 (c) F m a k e R rr e r e = = 2 3 , so a k e m R r rr e e = = 2 3 2 Thus, the motion is simple harmonic with frequency f k e m R e e = = 2 1 2 2 3 (d) f = = ( ) 2 47 10 1 2 8 99 10 1 60 1015 9 1 . . . Hz N m C2 2 99 2 31 3 9 11 10 C kg ( ) ( ) . R which yields R3 30 1 05 10= . ,m3 or R = = 1 02 10 10210 . m pm P24.49 The vertical velocity component of the moving charge increases according to m d dt F y y v = m d dx dx dt qE y y v = Now dx dt x= v has the nearly constant value v. So d q m E dxy yv v = v v v v y y yd q m E dx y = = 0 The radially outward component of the electric eld varies along the x axis, but is described by E dA E d dx Q y y = ( ) = 2 0 So E dx Q d y = 2 0 and v v y qQ m d = 2 0 . The angle of deection is described by tan = = v v v y qQ dm2 0 2 = tan 1 0 2 2 qQ dmv FIG. P24.49 vx vy x y dq v 0 Q
33. 40. 40 Chapter 24 P24.50 Consider the eld due to a single sheet and let E+ and E represent the elds due to the positive and negative sheets. The eld at any distance from each sheet has a magnitude given by the textbook equation E E+ = = 2 0 (a) To the left of the positive sheet, E+ is directed toward the left and E toward the right and the net eld over this region is E = 0 . (b) In the region between the sheets, E+ and E are both directed toward the right and the net eld is E = 0 to the right (c) To the right of the negative sheet, E+ and E are again oppositely directed and E = 0 . P24.51 The magnitude of the eld due to each sheet given by Equation 24.8 is E = 2 0 directed perpendicular to the sheet (a) In the region to the left of the pair of sheets, both elds are directed toward the left and the net eld is E = 0 to the left (b) In the region between the sheets, the elds due to the individual sheets are oppositely directed and the net eld is E = 0 (c) In the region to the right of the pair of sheets, both elds are directed toward the right and the net eld is E = 0 to the right P24.52 The resultant eld within the cavity is the superposition of two elds, one E+ due to a uniform sphere of positive charge of radius 2a, and the other E due to a sphere of negative charge of radius a centered within the cavity. 4 3 4 3 0 2 r r E = + so E r r + = = r 3 30 0 = 4 3 41 3 0 1 2 r r E so E r r = ( )= r1 0 1 0 1 3 3 continued on next page FIG. P24.50 FIG. P24.51 FIG. P24.52
34. 41. Gausss Law 41 Since r a r= + 1, E r a = ( ) 3 0 E E E r r a a i= + = + = = ++ 3 3 3 3 0 0 0 0 0 a 3 0 j Thus, Ex = 0 and E a y = 3 0 at all points within the cavity P24.53 Consider the charge distribution to be an unbroken charged spherical shell with uniform charge density s and a circular disk with charge per area . The total eld is that due to the whole sphere, Q R R R4 4 40 2 2 0 2 0 = = outward plus the eld of the disk = 2 20 0 radially inward. The total eld is = 0 0 02 2 outward . P24.54 The electric eld throughout the region is directed along x; therefore, E will be perpendicular to dA over the four faces of the surface which are perpendicular to the yz plane, and E will be parallel to dA over the two faces which are parallel to the yz plane. Therefore, E x x a x x a c E A E A a ab a c= ( ) + ( ) = +( ) + + +( )= = + 3 2 3 22 22 2 2 ( ) = +( ) ab abc a c Substituting the given values for a, b, and c, we nd E = 0 269. N m C2 . Q E= = = 0 12 2 38 10 2 38 . .C pC P24.55 E A = ( )= d E r q 4 2 0 in (a) For r R> , q Ar r dr AR R in = ( ) = 2 2 0 5 4 4 5 and E AR r = 5 0 2 5 (b) For r R< , q Ar r dr Ar r in = ( ) = 2 2 0 5 4 4 5 and E Ar = 3 05 FIG. P24.54
35. 42. 42 Chapter 24 P24.56 The total ux through a surface enclosing the charge Q is Q 0 . The ux through the disk is disk = E Ad where the integration covers the area of the disk. We must evaluate this integral and set it equal to 1 4 0 Q to nd how b and R are related. In the gure, take d A to be the area of an annular ring of radius s and width ds. The ux through d A is E A = = ( )d EdA E sdscos cos . 2 The magnitude of the electric eld has the same value at all points within the annular ring, E Q r Q s b = = + 1 4 1 40 2 0 2 2 and cos = = +( ) b r b s b2 2 1 2 . Integrate from s = 0 to s R= to get the ux through the entire disk. E R Qb sds s b Qb s b, disk = +( ) = +( )2 20 2 2 3 2 0 0 2 2 11 2 0 0 2 2 1 2 2 1 = +( ) R Q b R b The ux through the disk equals Q 4 0 provided that b R b2 2 1 2 1 2+( ) = . This is satised if R b= 3 . P24.57 E A = = d q a r r dr r in 0 0 2 0 1 4 E r a rdr a r E a r 4 4 4 2 2 2 0 0 0 2 0 = = = = constant maagnitude (The direction is radially outward from center for positive a; radially inward for negative a.) P24.58 In this case the charge density is not uniform, and Gausss law is written as E A = d dV 1 0 . We use a gaussian surface which is a cylinder of radius r, length , and is coaxial with the charge distribution. (a) When r R< , this becomes E r a r b dV r 2 0 0 0 ( ) = . The element of volume is a cylindrical shell of radius r, length , and thickness dr so that dV r dr= 2 . E r r a r b 2 2 2 3 2 0 0 ( )= so inside the cylinder, E r a r b = 0 02 2 3 (b) When r R> , Gausss law becomes E r a r b r dr R 2 20 0 0 ( )= ( ) or outside the cylinder, E R r a R b = 0 2 02 2 3 FIG. P24.56
36. 43. Gausss Law 43 P24.59 (a) Consider a cylindrical shaped gaussian surface perpendicular to the yz plane with one end in the yz plane and the other end containing the point x: Use Gausss law: E A = d qin 0 By symmetry, the electric eld is zero in the yz plane and is perpendicular to d A over the wall of the gaussian cylinder. Therefore, the only contribution to the integral is over the end cap containing the point x: E A = d qin 0 or EA Ax = ( ) 0 so that at distance x from the mid-line of the slab, E x = 0 . (b) a F m e E m e m x e e e = = ( ) = 0 The acceleration of the electron is of the form a x= 2 with = e me 0 Thus, the motion is simple harmonic with frequency f e me = = 2 1 2 0 P24.60 Consider the gaussian surface described in the solution to problem 59. (a) For x d > 2 , dq dV Adx CAx dx= = = 2 E A = = = d dq EA CA x dx CA d d 1 1 3 0 0 2 0 2 0 33 8 E Cd = 3 024 or E i E i= > = < Cd x d Cd x 3 0 3 024 2 24 ; for for dd 2 (b) For < < d x d 2 2 E A = = = d dq CA x dx CAx x 1 30 0 2 0 3 0 E i E i= > = < Cx x Cx x 3 0 3 03 0 3 0 ; for for x y z x gaussian surface FIG. P24.59
37. 44. 44 Chapter 24 P24.61 (a) A point mass m creates a gravitational acceleration g r= Gm r2 at a distance r The ux of this eld through a sphere is g A = ( )= d Gm r r Gm2 2 4 4 Since the r has divided out, we can visualize the eld as unbroken eld lines. The same ux would go through any other closed surface around the mass. If there are several or no masses inside a closed surface, each creates eld to make its own contribution to the net ux according to g A = d Gm4 in (b) Take a spherical gaussian surface of radius r. The eld is inward so g A = = d g r g r4 180 42 2 cos and = 4 4 4 3 3 Gm G rin Then, = g r G r4 4 4 3 2 3 and g r G= 4 3 Or, since = M R E E 4 3 3 , g M Gr R E E = 3 or g = M Gr R E E 3 inward P24.62 The charge density is determined by Q a= 4 3 3 = 3 4 3 Q a (a) The ux is that created by the enclosed charge within radius r: E q r r Q a Qr a = = = = in 0 3 0 3 0 3 3 0 3 4 3 4 3 3 4 (b) E Q = 0 . Note that the answers to parts (a) and (b) agree at r a= . (c) a r E Q 0 0 0 FIG. P24.62(c)
38. 45. Gausss Law 45 P24.63 E A = ( )= d E r q 4 2 0 in (a) ( ) ( ) = 3 60 10 4 0 100 8 85 10 3 2 12 . . . N C m C N2 Q mm2 a r b< ( ) Q Q+ = + 5 56 10 9 . C , so = + = + Q 9 56 10 9 569 . .C nC (c) For b r c< < : E = 0 and q Q Qin = + =1 0 where Q1 is the total charge on the inner surface of the hollow sphere. Thus, Q Q1 4 00= = + . .nC Then, if Q2 is the total charge on the outer surface of the hollow sphere, Q Q Q2 1 9 56 4 0 5 56= = = +. . . .nC nC nC P24.64 The eld direction is radially outward perpendicular to the axis. The eld strength depends on r but not on the other cylindrical coordinates q or z. Choose a gaussian cylinder of radius r and length L. If r a< , E q = in 0 and E rL L 2 0 ( )= E r = 2 0 or E r= ( ) b a r r b 2 2 02 *P24.65 (a) Consider a gaussian surface in the shape of a rectangular box with two faces perpendicular to the direction of the eld. It encloses some charge, so the net ux out of the box is nonzero. The eld must be stronger on one side than on the other. The eld cannot be uniform in magnitude. (b) Now the volume contains no charge. The net ux out of the box is zero. The ux entering is equal to the ux exiting. The eld magnitude is uniform at points along one eld line. The eld magnitude can vary over the faces of the box perpendicular to the eld. + + + + + + + FIG. P24.65(a)
39. 46. 46 Chapter 24 ANSWERS TO EVEN PROBLEMS P24.2 355 kN m C2 P24.4 (a) 2 34. kN m C2 (b) + 2 34. kN m C2 (c) 0 P24.6 (a) 55 7. nC (b) The negative charge has a spherically symmetric distribution concentric with the shell. P24.8 (a) q 2 0 (b) q 2 0 (c) Plane and square both subtend a solid angle of a hemisphere at the charge. P24.10 (a) 1 36. MN m C2 (b) 678 kN m C2 (c) No; see the solution. P24.12 1 77. pC m3 positive P24.14 Q q 6 6 0 P24.16 28 2. N m C2 P24.18 (a) 0 (b) 365 kN C (c) 1 46. MN C (d) 649 kN C P24.20 (a) 16 2. MN C toward the lament (b) 8 09. MN C toward the lament (c) 1 62. MN C toward the lament P24.22 (a) 913 nC (b) 0 P24.24 (a) A long cylindrical plastic rod 2.00 cm in radius carries charge uniformly distributed through- out its volume, with density 5.00 m C/m3 . Find the magnitude of the electric eld it creates at a point P, 3.00 cm from its axis. As a gaussian surface choose a concentric cylinder with its curved surface passing through the point P and with length 8.00 cm. (b) 3.77 kN/C P24.26 4 86. GN C away from the wall. It is constant close to the wall. P24.28 3 50. kN P24.30 (a) ~1 mN (b) ~100 nC (c) ~10 kN/C (d) ~10 kN m2 /C P24.32 E = Q/20 A vertically upward in each case if Q > 0 P24.34 (a) 0 (b) 12 4. kN C radially outward (c) 639 N C radially outward (d) No answer changes. The solid copper sphere carries charge only on its outer surface. P24.36 31.9 nC/m3 P24.38 The electric eld just outside the surface is given by s /0 . At this point the uniformly charged surface of the sphere looks just like a uniform at sheet of charge. P24.40 See the solution. P24.42 chw2 2
40. 47. Gausss Law 47 P24.44 (a) 2.00 m C (b) to the right (c) 702 kN/C (d) 0 (e) 0 (f) 3.00 m C (g) 4.21 MN/C radially outward (h) 1.54 m C (i) 8.63 MN/C radially outward (j) 3.00 m C (k) 2.00 m C (l) See the solution. P24.46 See the solution. P24.48 (a, b) See the solution. (c) 1 2 2 3 k e m R e e (d) 102 pm P24.50 (a) 0 (b) 0 to the right (c) 0 P24.52 See the solution. P24.54 0 269. N m C2 ; 2.38 pC P24.56 See the solution. P24.58 (a) 0 02 2 3 r a r b (b) 0 2 02 2 3 R r a R b P24.60 (a) E i= > Cd x d3 024 2 for ; E i= < Cd x d3 024 2 for (b) E i= > Cx x 3 03 0 ;for E i= < Cx x 3 03 0 for P24.62 (a) Qr a 3 0 3 (b) Q 0 (c) See the solution P24.64 For r < a, E = l/2p0 r radially outward. For a < r < b, E = [l + rp(r2 a2 )]/2p0 r radially outward. For r > b, E = [l + rp(b2 a2 )]/2p0 r radially outward.
41. 48. 25 Electric Potential CHAPTER OUTLINE 25.1 Electric Potential and Potential Difference 25.2 Potential Difference in a Uniform Electric Field 25.3 Electric Potential and Potential Energy Due to Point Charges 25.4 Obtaining the Value of the Electric Field from the Electric Potential 25.5 Electric Potential Due to Continuous Charge Distributions 25.6 Electric Potential Due to a Charged Conductor 25.7 The Millikan Oil Drop Experiment 25.8 Application of Electrostatics ANSWERS TO QUESTIONS Q25.1 When one object B with electric charge is immersed in the electric eld of another charge or charges A, the system possesses electric potential energy. The energy can be measured by seeing how much work the eld does on the charge B as it moves to a reference location. We choose not to visualize As effect on B as an action-at-a- distance, but as the result of a two-step process: Charge A creates electric potential throughout the surrounding space. Then the potential acts on B to inject the system with energy. *Q25.2 (i) The particle feels an electric force in the negative x direction. An outside agent pushes it uphill against this force, increasing the potential energy. Answer (a). (ii) The potential decreases in the direction of the electric eld. Answer (c). *Q25.3 The potential is decreasing toward the bottom of the page, so the electric eld is downward. Answer (f). *Q25.4 (i) At points off the x axis the electric eld has a nonzero y component. At points on the negative x axis the eld is to the right and positive. At points to the right of x = 500 mm the eld is to the left and nonzero. The eld is zero at one point between x = 250 mm and x = 500 mm. Answer (b). (ii) The electric potential is negative at this and at all points. Answer (c). (iii) Answer (d). (iv) Answer (d). Q25.5 To move like charges together from an innite separation, at which the potential energy of the system of two charges is zero, requires work to be done on the system by an outside agent. Hence energy is stored, and potential energy is positive. As charges with opposite signs move together from an innite separation, energy is released, and the potential energy of the set of charges becomes negative. Q25.6 (a) The equipotential surfaces are nesting coaxial cylinders around an innite line of charge. (b) The equipotential surfaces are nesting concentric spheres around a uniformly charged sphere. *Q25.7 Answer (b). The potential could have any value. *Q25.8 The same charges at the same distance away create the same contribution to the total potential. Answer (b). 49
42. 49. 50 Chapter 25 *Q25.9 The change in kinetic energy is the negative of the change in electric potential energy, so we work out qV = q(Vf Vi ) in each case. (a) (e)(60 V 40 V) = +20 eV (b) (e)(20 V 40 V) = 20 eV (c) (e)(20 V 40 V) = +20 eV (d) (e)(10 V 40 V) = +30 eV (e) (2e)(50 V 40 V) = +20 eV (f) (2e)(60 V 40 V) = +40 eV With also (g) 0 and (h) +10 eV, the ranking is f > d > c = e = a > h > g > b. Q25.10 The main factor is the radius of the dome. One often overlooked aspect is also the humidity of the airdrier air has a larger dielectric breakdown strength, resulting in a higher attainable electric potential. If other grounded objects are nearby, the maximum potential might be reduced. *Q25.11 (i) The two spheres come to the same potential, so q/R is the same for both. With twice the radius, B has twice the charge. Answer (d). (ii) All the charge runs away from itself to the outer surface of B. Answer (a). Q25.12 The grounding wire can be touched equally well to any point on the sphere. Electrons will drain away into the ground and the sphere will be left positively charged. The ground, wire, and sphere are all conducting. They together form an equipotential volume at zero volts during the con- tact. However close the grounding wire is to the negative charge, electrons have no difculty in moving within the metal through the grounding wire to ground. The ground can act as an innite source or sink of electrons. In this case, it is an electron sink. SOLUTIONS TO PROBLEMS Section 25.1 Electric Potential and Potential Difference P25.1 (a) Energy of the proton-eld system is conserved as the proton moves from high to low poten- tial, which can be dened for this problem as moving from 120 V down to 0 V. K U E K Ui i f f+ + = + mech 0 0 1 2 02 + + = +qV m pv 1 60 10 120 1 1 2 1 67 1019 . .( )( ) = C V J 1 V C ( )27 2 kg vp vp = 1 52 105 . m s (b) The electron will gain speed in moving the other way, from Vi = 0 to Vf = 120 V: K U E K Ui i f f+ + = + mech 0 0 0 1 2 2 + + = +m qVev 0 1 2 9 11 10 1 60 10 12031 2 19 = ( ) + ( ) . .kg C Jve CC( ) ve = 6 49 106 . m s P25.2 V = 14 0. V and Q N eA= = ( ) ( )= 6 02 10 1 60 10 9 63 1023 19 4 . . . C V W Q = , so W Q V= = ( ) ( )= 9 63 10 14 0 1 354 . . .C J C MJ
43. 50. Electric Potential 51 Section 25.2 Potential Difference in a Uniform Electric Field P25.3 E V d = = = = 25 0 10 1 50 10 1 67 10 1 3 2 6. . . . J C m N C 667 MN C P25.4 V V d d d V V B A A B A C C B B A = = = E s E s E s ( ) ( ) E dy E dxcos cos . . . . 180 90 0 0 300 0 500 0 2200 0 400 325 0 800 260 . . = ( )( ) = +V VB A V P25.5 U m f i= ( )= ( ) 1 2 1 2 9 11 10 1 40 102 2 31 5 v v . .kg mm s m s J ( ) ( ) = 2 6 2 18 3 70 10 6 23 10 . . U q V= : + = ( ) 6 23 10 1 60 1018 19 . . V V = 38 9. V. The origin is at highest potentiial. P25.6 Assume the opposite. Then at some point A on some equipotential surface the electric eld has a nonzero component Ep in the plane of the surface. Let a test charge start from point A and move some distance on the surface in the direction of the eld component. Then V d A B = E s is nonzero. The electric potential charges across the surface and it is not an equipotential surface. The contradiction shows that our assumption is false, that Ep = 0, and that the eld is perpendicular to the equipotential surface. P25.7 (a) Arbitrarily choose V = 0 at 0. Then at other points V Ex= and U QV QExe = = Between the endpoints of the motion, K U U K U Us e i s e f + +( ) = + +( ) 0 0 0 0 1 2 2 + + = + kx QExmax max so x QE k max = 2 (b) At equilibrium, F F Fx s e = + = 0 or kx QE= So the equilibrium position is at x QE k = . (c) The blocks equation of motion is F kx QE m d x dt x = + = 2 2 . Let = x x QE k , or x x QE k = + , FIG. P25.4 FIG. P25.7 continued on next page
44. 51. 52 Chapter 25 so the equation of motion becomes: + + = +( )k x QE k QE m d x QE k dt 2 2 , or d x dt k m x 2 2 = This is the equation for simple harmonic motion a xx = 2 with = k m The period of the motion is then T m k = = 2 2 (d) K U U E K U Us e i s e f + +( ) + = + +( ) mech 0 0 0 0 1 2 2 2 + + = + = k k mgx kx QEx x QE max max max max mmg k ( ) P25.8 Arbitrarily take V = 0 at point P. Then the potential at the original position of the charge is = E s EL cos . At the nal point a, V = EL. Because the table is frictionless we have K U K Ui f+( ) = +( ) 0 1 2 2 1 2 2 00 10 2 = = ( ) = qEL m qEL qEL m cos cos . v v ( )( )( ) ( )6 300 1 50 1 60 0 0 010 0 C N C m k . cos . . gg m s= 0 300. P25.9 Arbitrarily take V = 0 at the initial point. Then at distance d downeld, where L is the rod length, V = Ed and U LEde = . (a) K U K Ui f+( ) = +( ) 0 0 1 2 2 2 40 0 10 100 2 6 + = = = ( ) L LEd Ed v v . C m N CC m kg m m s ( )( ) ( ) = 2 00 0 100 0 400 . . . (b) The same. Each bit of the rod feels a force of the same size as before. Section 25.3 Electric Potential and Potential Energy Due to Point Charges P25.10 (a) Since the charges are equal and placed symmetrically, F = 0 . (b) Since F qE= = 0, E = 0 . (c) V k q r e= = ( ) 2 2 8 99 10 2 00 109 6 . . N m C C 0.800 2 2 mm V = =4 50 10 45 04 . .V kV FIG. P25.10
45. 52. Electric Potential 53 P25.11 (a) The potential at 1.00 cm is V k q r e1 9 19 8 99 10 1 60 10 1 00 = = ( ) ( ) . . . N m C C2 2 = 10 1 44 102 7 m V. (b) The potential at 2.00 cm is V k q r e2 9 19 8 99 10 1 60 10 2 00 = = ( ) ( ) . . . N m C C2 2 = 10 0 719 102 7 m V. Thus, the difference in potential between the two points is V V V= = 2 1 8 7 19 10. V . (c) The approach is the same as above except the charge is 1 60 10 19 . C. This changes the sign of each answer, with its magnitude remaining the same. That is, the potential at 1.00 cm is 1 44 10 7 . V . The potential at 2.00 cm is 0 719 10 7 . V, so V V V= = 2 1 8 7 19 10. V P25.12 (a) E k q x k q x x e e = + ( ) =1 2 2 2 2 00 0 . becomes E k q x q x x e= + + ( ) =2 2 2 2 00 0 . Dividing by ke , 2 2 002 2 qx q x= ( ). x x2 4 00 4 00 0+ =. . Therefore E = 0 when x = + = 4 00 16 0 16 0 2 4 83 . . . . m (Note that the positive root does not correspond to a physically valid situation.) (b) V k q x k q x e e = + =1 2 2 00 0 . or V k q x q x e= + = 2 2 00 0 . Again solving for x, 2 2 00qx q x= ( ). For 0 2 00 x . V = 0 when x = 0 667. m and q x q x = 2 2 For x < 0 x = 2 00. m P25.13 (a) E Q r = 4 0 2 V Q r r V E = = = = 4 3 000 6 00 0 V 500 V m m. (b) V Q = = ( ) 3 000 4 6 000 V m . Q = ( ) ( ) = 3 000 6 00 2 00 V 8.99 10 V m C m C9 . .
46. 53. 54 Chapter 25 P25.14 (a) U qQ r = = ( ) ( ) 4 5 00 10 3 00 10 8 99 1 0 9 9 . . .C C 00 0 350 3 86 10 9 7 V m C m J ( ) ( ) = . . The minus sign means it takes 3 86 10 7 . J to pull the two charges apart from 35 cm to a much larger separation. (b) V Q r Q r = + = ( ) 1 0 1 2 0 2 9 9 4 4 5 00 10 8 99 10 . .C V( )+ ( ) ( ) m C m C V m C 0 175 3 00 10 8 99 109 9 . . . 00 175 103 . m VV = P25.15 V k q r V i ii = = ( ) ( ) 8 99 10 7 00 10 1 0 010 0 1 0 9 6 . . . .. . . . 010 0 1 0 038 7 1 10 10 11 07 + = = V V MV P25.16 (a) V k q r k q r k q r e e e = + = 1 1 2 2 2 V = ( ) ( ) ( ) + 2 8 99 10 2 00 10 1 00 9 6 2 . . . N m C C m 2 2 00 500 3 22 10 32 2 2 4 . . . m V kV ( ) = =V (b) U qV= = ( ) ( )= 3 00 10 3 22 10 9 65 106 4 2 . . .C J C J P25.17 U q V q V q V q q r q r q e = + + = + +4 1 4 2 4 3 4 0 1 1 2 2 31 4 rr3 Ue = ( ) ( ) + 10 0 10 8 99 10 1 0 600 6 2 9 . . . C N m C m 2 2 11 0 150 1 0 600 0 150 8 2 2. . . . m m m + ( ) + ( ) =Ue 995 J FIG. P25.15 FIG. P25.16
47. 54. Electric Potential 55 *P25.18 (a) Therstexpression,withdistancessquared,describesanelectriceld.The second expression describesanelectricpotential.Thena positive nC charge7 is7cmfromtheorigin.Tocreate eld that is to the left and downward, it must be in the rst quadrant, with position vector 7 70cm at . A negative nC charge8 3 cm from the origin creates an upward electric eld at the origin, so it must be at 3 90cm at . We evaluate the given expressions: E i j= + = 4 39 67 8 1 50 . . . kN C kN C kVV (b) F E i j= = +( ) = q 16 10 4 39 67 8 10 7 09 3 C N C. . . 33 109 10 5 i j( ) N (c) U qVe = = ( )= 16 10 1 50 10 2 40 109 3 5 C J C J. .+ P25.19 U U U U U U U U U U U U = + + + = + + +( )+ + +( 1 2 3 4 12 13 23 14 24 340 )) = + + + + + + U k Q s k Q s k Q s e e e 0 1 2 1 1 1 2 1 2 2 2 = + =U k Q s k Q s e e 2 2 4 2 2 5 41. We can visualize the term 4 2 2 + as arising directly from the 4 side pairs and 2 face diagonal pairs. P25.20 Each charge creates equal potential at the center. The total potential is: V k q R k q R e e = ( ) = 5 5 P25.21 (a) Each charge separately creates positive potential everywhere. The total potential produced by the three charges together is then the sum of three positive terms. There is no point , at a nite distance from the charges, at which this total potential is zero. (b) V k q a k q a k q a e e e = + = 2 FIG. P25.19
48. 55. 56 Chapter 25 P25.22 (a) V x k Q r k Q r k Q x a k Q x a e e e e ( ) = + = +( ) + + +( ) + ( 1 1 2 2 2 2 2 ))2 V x k Q x a k Q a x a V x k Q a e e e ( ) = + = ( ) + ( ) 2 2 1 2 2 2 (( ) = ( ) + 2 1 2 x a (b) V y k Q r k Q r k Q y a k Q y a e e e e ( )= + = +( ) + ( ) + 1 1 2 2 V y k Q a y a y a V y k Q a y a e e ( )= + ( ) ( ) = 1 1 1 1 1 1 11 1y a + P25.23 Consider the two spheres as a system. (a) Conservation of momentum: 0 1 1 2 2= + ( )m mv v i i or v v 2 1 1 2 = m m By conservation of energy, 0 1 2 1 2 1 2 1 1 2 2 2 2 1 2 1 2 = ( ) = + + ( ) + k q q d m m k q q r r e e v v and k q q r r k q q d m m m e e1 2 1 2 1 2 1 1 2 1 2 1 2 2 1 2 1 2+ = +v v v v 1 2 1 2 1 1 2 1 2 1 2 1 1 2 0 7 = +( ) + = m k q q m m m r r d e . 000 8 99 10 2 10 3 109 6 6 kg N m C C C2 2 ( ) ( ) ( ) . (( ) ( )( ) 0 100 0 800 1 8 10 1 1 003 . . .kg kg m m = = = ( ) 10 8 0 100 0 700 2 1 1 2 . . . m s kg 10.8 m s v vm m kkg m s= 1 55. (b) If the spheres are metal, electrons will move around on them with negligible energy loss to place the centers of excess charge on the insides of the spheres. Then just before they touch, the effective distance between charges will be less than r r1 2+ and the spheres will really be moving faster than calculated in (a) . FIG. P25.22(b) FIG. P25.22(a)
49. 56. Electric Potential 57 P25.24 Consider the two spheres as a system. (a) Conservation of momentum: 0 1 1 2 2= + ( )m mv v i i or v v 2 1 1 2 = m m . By conservation of energy, 0 1 2 1 2 1 2 1 1 2 2 2 2 1 2 1 2 = ( ) = + + ( ) + k q q d m m k q q r r e e v v and k q q r r k q q d m m m e e1 2 1 2 1 2 1 1 2 1 2 1 2 2 1 2 1 2+ = +v v . v v 1 2 1 2 1 1 2 1 2 2 1 2 2 1 1 = +( ) + = m k q q m m m r r d m m e = +( ) + v1 1 1 2 2 1 2 1 2 2 1 1m k q q m m m r r d e (b) If the spheres are metal, electrons will move around on them with negligible energy loss to place the centers of excess charge on the insides of the spheres. Then just before they touch, the effective distance between charges will be less than r r1 2+ and the spheres will really be moving faster than calculated in (a) . P25.25 The original electrical potential energy is U qV q k q d e e = = In the nal conguration we have mechanical equilibrium. The spring and electrostatic forces on each charge are ( )+ ( ) =k d q k q d e 2 3 02 . Then k k q d e = 2 3 18 . In the nal conguration the total potential energy is 1 2 1 2 18 2 3 4 9 2 2 3 2 2 kx qV k q d d q k q d k q d e e e + = ( ) + = . The missing energy must have become internal energy, as the system is isolated: k q d k q d Ee e 2 2 4 9 = + int . E k q d e int = 5 9 2 P25.26 Using conservation of energy for the alpha particle-nucleus system, we have K U K Uf f i i+ = + But U k q q r i e i = gold and ri Thus, Ui = 0 Also Kf = 0 (vf = 0 at turning point), so U Kf i= or k q q r m e gold min = 1 2 2 v r k q q m e min . = = ( )( )2 2 8 99 10 2 79 2 9 gold 2 2 N m C v (( ) ( ) ( ) ( ) 1 60 10 6 64 10 2 00 10 19 2 27 7 . . . C kg m s 22 14 2 74 10 27 4 = = . . m fm