<ul><li>1.Instructor Solutions Manual for Physics by Halliday, Resnick, and Krane Paul Stanley Beloit College Volume 2 </li></ul><p>2. A Note To The Instructor... The solutions here are somewhat brief, as they are designed for the instructor, not for the student. Check with the publishers before electronically posting any part of these solutions; website, ftp, or server access must be restricted to your students. I have been somewhat casual about subscripts whenever it is obvious that a problem is one dimensional, or that the choice of the coordinate system is irrelevant to the numerical solution. Although this does not change the validity of the answer, it will sometimes obfuscate the approach if viewed by a novice. There are some traditional formula, such as v2 x = v2 0x + 2axx, which are not used in the text. The worked solutions use only material from the text, so there may be times when the solution here seems unnecessarily convoluted and drawn out. Yes, I know an easier approach existed. But if it was not in the text, I did not use it here. I also tried to avoid reinventing the wheel. There are some exercises and problems in the text which build upon previous exercises and problems. Instead of rederiving expressions, I simply refer you to the previous solution. I adopt a dierent approach for rounding of signicant gures than previous authors; in partic- ular, I usually round intermediate answers. As such, some of my answers will dier from those in the back of the book. Exercises and Problems which are enclosed in a box also appear in the Students Solution Manual with considerably more detail and, when appropriate, include discussion on any physical implications of the answer. These student solutions carefully discuss the steps required for solving problems, point out the relevant equation numbers, or even specify where in the text additional information can be found. When two almost equivalent methods of solution exist, often both are presented. You are encouraged to refer students to the Students Solution Manual for these exercises and problems. However, the material from the Students Solution Manual must not be copied. Paul Stanley Beloit College stanley@clunet.edu 1 3. E25-1 The charge transferred is Q = (2.5 104 C/s)(20 106 s) = 5.0 101 C. E25-2 Use Eq. 25-4: r = (8.99109Nm2/C2)(26.3106C)(47.1106C) (5.66 N) = 1.40 m E25-3 Use Eq. 25-4: F = (8.99109 Nm2 /C2 )(3.12106 C)(1.48106 C) (0.123 m)2 = 2.74 N. E25-4 (a) The forces are equal, so m1a1 = m2a2, or m2 = (6.31107 kg)(7.22 m/s2 )/(9.16 m/s2 ) = 4.97107 kg. (b) Use Eq. 25-4: q = (6.31107kg)(7.22 m/s2)(3.20103m)2 (8.99109Nm2/C2) = 7.201011 C E25-5 (a) Use Eq. 25-4, F = 1 4 0 q1q2 r2 12 = 1 4(8.851012 C2/N m2) (21.3 C)(21.3 C) (1.52 m)2 = 1.77 N (b) In part (a) we found F12; to solve part (b) we need to rst nd F13. Since q3 = q2 and r13 = r12, we can immediately conclude that F13 = F12. We must assess the direction of the force of q3 on q1; it will be directed along the line which connects the two charges, and will be directed away from q3. The diagram below shows the directions. F 12 F 23 F 23 F 12 F net From this diagram we want to nd the magnitude of the net force on q1. The cosine law is appropriate here: Fnet 2 = F2 12 + F2 13 2F12F13 cos , = (1.77 N)2 + (1.77 N)2 2(1.77 N)(1.77 N) cos(120 ), = 9.40 N2 , Fnet = 3.07 N. 2 4. E25-6 Originally F0 = CQ2 0 = 0.088 N, where C is a constant. When sphere 3 touches 1 the charge on both becomes Q0/2. When sphere 3 the touches sphere 2 the charge on each becomes (Q0 + Q0/2)/2 = 3Q0/4. The force between sphere 1 and 2 is then F = C(Q0/2)(3Q0/4) = (3/8)CQ2 0 = (3/8)F0 = 0.033 N. E25-7 The forces on q3 are F31 and F32. These forces are given by the vector form of Coulombs Law, Eq. 25-5, F31 = 1 4 0 q3q1 r2 31 r31 = 1 4 0 q3q1 (2d)2 r31, F32 = 1 4 0 q3q2 r2 32 r32 = 1 4 0 q3q2 (d)2 r32. These two forces are the only forces which act on q3, so in order to have q3 in equilibrium the forces must be equal in magnitude, but opposite in direction. In short, F31 = F32, 1 4 0 q3q1 (2d)2 r31 = 1 4 0 q3q2 (d)2 r32, q1 4 r31 = q2 1 r32. Note that r31 and r32 both point in the same direction and are both of unit length. We then get q1 = 4q2. E25-8 The horizontal and vertical contributions from the upper left charge and lower right charge are straightforward to nd. The contributions from the upper left charge require slightly more work. The diagonal distance is 2a; the components will be weighted by cos 45 = 2/2. The diagonal charge will contribute Fx = 1 4 0 (q)(2q) ( 2a)2 2 2 i = 2 8 0 q2 a2 i, Fy = 1 4 0 (q)(2q) ( 2a)2 2 2 j = 2 8 0 q2 a2 j. (a) The horizontal component of the net force is then Fx = 1 4 0 (2q)(2q) a2 i + 2 8 0 q2 a2 i, = 4 + 2/2 4 0 q2 a2 i, = (4.707)(8.99109 N m2 /C2 )(1.13106 C)2 /(0.152 m)2i = 2.34 Ni. (b) The vertical component of the net force is then Fy = 1 4 0 (q)(2q) a2 j + 2 8 0 q2 a2 j, = 2 + 2/2 8 0 q2 a2 j, = (1.293)(8.99109 N m2 /C2 )(1.13106 C)2 /(0.152 m)2j = 0.642 Nj. 3 5. E25-9 The magnitude of the force on the negative charge from each positive charge is F = (8.99109 N m2 /C2 )(4.18106 C)(6.36106 C)/(0.13 m)2 = 14.1 N. The force from each positive charge is directed along the side of the triangle; but from symmetry only the component along the bisector is of interest. This means that we need to weight the above answer by a factor of 2 cos(30 ) = 1.73. The net force is then 24.5 N. E25-10 Let the charge on one sphere be q, then the charge on the other sphere is Q = (52.6 106 C) q. Then 1 4 0 qQ r2 = F, (8.99109 Nm2 /C2 )q(52.6106 C q) = (1.19 N)(1.94 m)2 . Solve this quadratic expression for q and get answers q1 = 4.02105 C and q2 = 1.24106 N. E25-11 This problem is similar to Ex. 25-7. There are some additional issues, however. It is easy enough to write expressions for the forces on the third charge F31 = 1 4 0 q3q1 r2 31 r31, F32 = 1 4 0 q3q2 r2 32 r32. Then F31 = F32, 1 4 0 q3q1 r2 31 r31 = 1 4 0 q3q2 r2 32 r32, q1 r2 31 r31 = q2 r2 32 r32. The only way to satisfy the vector nature of the above expression is to have r31 = r32; this means that q3 must be collinear with q1 and q2. q3 could be between q1 and q2, or it could be on either side. Lets resolve this issue now by putting the values for q1 and q2 into the expression: (1.07 C) r2 31 r31 = (3.28 C) r2 32 r32, r2 32r31 = (3.07)r2 31r32. Since squared quantities are positive, we can only get this to work if r31 = r32, so q3 is not between q1 and q2. We are then left with r2 32 = (3.07)r2 31, so that q3 is closer to q1 than it is to q2. Then r32 = r31 + r12 = r31 + 0.618 m, and if we take the square root of both sides of the above expression, r31 + (0.618 m) = (3.07)r31, (0.618 m) = (3.07)r31 r31, (0.618 m) = 0.752r31, 0.822 m = r31 4 6. E25-12 The magnitude of the magnetic force between any two charges is kq2 /a2 , where a = 0.153 m. The force between each charge is directed along the side of the triangle; but from symmetry only the component along the bisector is of interest. This means that we need to weight the above answer by a factor of 2 cos(30 ) = 1.73. The net force on any charge is then 1.73kq2 /a2 . The length of the angle bisector, d, is given by d = a cos(30 ). The distance from any charge to the center of the equilateral triangle is x, given by x2 = (a/2)2 + (d x)2 . Then x = a2 /8d + d/2 = 0.644a. The angle between the strings and the plane of the charges is , given by sin = x/(1.17 m) = (0.644)(0.153 m)/(1.17 m) = 0.0842, or = 4.83 . The force of gravity on each ball is directed vertically and the electric force is directed horizontally. The two must then be related by tan = FE/FG, so 1.73(8.99109 N m2 /C2 )q2 /(0.153 m)2 = (0.0133 kg)(9.81 m/s2 ) tan(4.83 ), or q = 1.29107 C. E25-13 On any corner charge there are seven forces; one from each of the other seven charges. The net force will be the sum. Since all eight charges are the same all of the forces will be repulsive. We need to sketch a diagram to show how the charges are labeled. 12 3 4 5 6 7 8 The magnitude of the force of charge 2 on charge 1 is F12 = 1 4 0 q2 r2 12 , where r12 = a, the length of a side. Since both charges are the same we wrote q2 . By symmetry we expect that the magnitudes of F12, F13, and F14 will all be the same and they will all be at right angles to each other directed along the edges of the cube. Written in terms of vectors the forces 5 7. would be F12 = 1 4 0 q2 a2 i, F13 = 1 4 0 q2 a2 j, F14 = 1 4 0 q2 a2 k. The force from charge 5 is F15 = 1 4 0 q2 r2 15 , and is directed along the side diagonal away from charge 5. The distance r15 is also the side diagonal distance, and can be found from r2 15 = a2 + a2 = 2a2 , then F15 = 1 4 0 q2 2a2 . By symmetry we expect that the magnitudes of F15, F16, and F17 will all be the same and they will all be directed along the diagonals of the faces of the cube. In terms of components we would have F15 = 1 4 0 q2 2a2 j/ 2 + k/ 2 , F16 = 1 4 0 q2 2a2 i/ 2 + k/ 2 , F17 = 1 4 0 q2 2a2 i/ 2 +j/ 2 . The last force is the force from charge 8 on charge 1, and is given by F18 = 1 4 0 q2 r2 18 , and is directed along the cube diagonal away from charge 8. The distance r18 is also the cube diagonal distance, and can be found from r2 18 = a2 + a2 + a2 = 3a2 , then in term of components F18 = 1 4 0 q2 3a2 i/ 3 +j/ 3 + k/ 3 . We can add the components together. By symmetry we expect the same answer for each com- ponents, so well just do one. How about i. This component has contributions from charge 2, 6, 7, and 8: 1 4 0 q2 a2 1 1 + 2 2 2 + 1 3 3 , or 1 4 0 q2 a2 (1.90) The three components add according to Pythagoras to pick up a nal factor of 3, so Fnet = (0.262) q2 0a2 . 6 8. E25-14 (a) Yes. Changing the sign of y will change the sign of Fy; since this is equivalent to putting the charge q0 on the other side, we would expect the force to also push in the other direction. (b) The equation should look Eq. 25-15, except all ys should be replaced by xs. Then Fx = 1 4 0 q0 q x x2 + L2/4 . (c) Setting the particle a distance d away should give a force with the same magnitude as F = 1 4 0 q0 q d d2 + L2/4 . This force is directed along the 45 line, so Fx = F cos 45 and Fy = F sin 45 . (d) Let the distance be d = x2 + y2, and then use the fact that Fx/F = cos = x/d. Then Fx = F x d = 1 4 0 x q0 q (x2 + y2 + L2/4)3/2 . and Fy = F y d = 1 4 0 y q0 q (x2 + y2 + L2/4)3/2 . E25-15 (a) The equation is valid for both positive and negative z, so in vector form it would read F = Fz k = 1 4 0 q0 q z (z2 + R2)3/2 k. (b) The equation is not valid for both positive and negative z. Reversing the sign of z should reverse the sign of Fz, and one way to x this is to write 1 = z/ z2. Then F = Fz k = 1 4 0 2q0 qz R2 1 z2 1 z2 k. E25-16 Divide the rod into small dierential lengths dr, each with charge dQ = (Q/L)dr. Each dierential length contributes a dierential force dF = 1 4 0 q dQ r2 = 1 4 0 qQ r2L dr. Integrate: F = dF = x+L x 1 4 0 qQ r2L dr, = 1 4 0 qQ L 1 x 1 x + L E25-17 You must solve Ex. 16 before solving this problem! q0 refers to the charge that had been called q in that problem. In either case the distance from q0 will be the same regardless of the sign of q; if q = Q then q will be on the right, while if q = Q then q will be on the left. Setting the forces equal to each other one gets 1 4 0 qQ L 1 x 1 x + L = 1 4 0 qQ r2 , or r = x(x + L). 7 9. E25-18 You must solve Ex. 16 and Ex. 17 before solving this problem. If all charges are positive then moving q0 o axis will result in a net force away from the axis. Thats unstable. If q = Q then both q and Q are on the same side of q0. Moving q0 closer to q will result in the attractive force growing faster than the repulsive force, so q0 will move away from equilibrium. E25-19 We can start with the work that was done for us on Page 577, except since we are concerned with sin = z/r we would have dFx = dF sin = 1 4 0 q0 dz (y2 + z2) z y2 + z2 . We will need to take into consideration that changes sign for the two halves of the rod. Then Fx = q0 4 0 0 L/2 z dz (y2 + z2)3/2 + L/2 0 +z dz (y2 + z2)3/2 , = q0 2 0 L/2 0 z dz (y2 + z2)3/2 , = q0 2 0 1 y2 + z2 L/2 0 , = q0 2 0 1 y 1 y2 + (L/2)2 . E25-20 Use Eq. 25-15 to nd the magnitude of the force from any one rod, but write it as F = 1 4 0 q Q r r2 + L2/4 , where r2 = z2 + L2 /4. The component of this along the z axis is Fz = Fz/r. Since there are 4 rods, we have F = 1 0 q Q z r2 r2 + L2/4 , = 1 0 q Q z (z2 + L2/4) z2 + L2/2 , Equating the electric force with the force of gravity and solving for Q, Q = 0mg qz (z2 + L2 /4) z2 + L2/2; putting in the numbers, (8.851012 C2 /Nm2 )(3.46107 kg)(9.8m/s2 ) (2.451012C)(0.214 m) ((0.214m)2 +(0.25m)2 /4) (0.214m)2 +(0.25m)2/2 so Q = 3.07106 C. E25-21 In each case we conserve charge by making sure that the total number of protons is the same on both sides of the expression. We also need to conserve the number of neutrons. (a) Hydrogen has one proton, Beryllium has four, so X must have ve protons. Then X must be Boron, B. (b) Carbon has six protons, Hydrogen has one, so X must have seven. Then X is Nitrogen, N. (c) Nitrogen has seven protons, Hydrogen has one, but Helium has two, so X has 7 + 1 2 = 6 protons. This means X is Carbon, C. 8 10. E25-22 (a) Use Eq. 25-4: F = (8.99109 Nm2 /C2 )(2)(90)(1.601019 C)2 (121015m)2 = 290 N. (b) a = (290 N)/(4)(1.661027 kg) = 4.41028 m/s2 . E25-23 Use Eq. 25-4: F = (8.99109 Nm2 /C2 )(1.601019 C)2 (2821012m)2 = 2.89109 N. E25-24 (a) Use Eq. 25-4: q = (3.7109N)(5.01010m)2 (8.99109Nm2/C2) = 3.201019 C. (b) N = (3.201019 C)/(1.601019 C) = 2. E25-25 Use Eq. 25-4, F = 1 4 0 q1q2 r2 12 = (1 3 1.6 1019 C)(1 3 1.6 1019 C) 4(8.85 1012 C2/N m2)(2.6 1015 m)2 = 3.8 N. E25-26 (a) N = (1.15107 C)/(1.601019 C) = 7.191011 . (b) The penny has enough electrons to make a total charge of 1.37105 C. The fraction is then (1.15107 C)/(1.37105 C) = 8.401013 . E25-27 Equate the magnitudes of the forces: 1 4 0 q2 r2 = mg, so r = (8.99109Nm2/C2)(1.601019C)2 (9.111031kg)(9.81 m/s2) = 5.07 m E25-28 Q = (75.0 kg)(1.601019 C)/(9.111031 kg) = 1.31013 C. E25-29 The mass of water is (250 cm3 )(1.00 g/cm 3 ) = 250 g. The number of moles of water is (250 g)/(18.0 g/mol) = 13.9 mol. The number of water molecules is (13.9 mol)(6.021023 mol1 ) = 8.371024 . Each molecule has ten protons, so the total positive charge is Q = (8.371024 )(10)(1.601019 C) = 1.34107 C. E25-30 The total positive charge in 0.250 kg of water is 1.34107 C. Marys imbalance is then q1 = (52.0)(4)(1.34107 C)(0.0001) = 2.79105 C, while Johns imbalance is q2 = (90.7)(4)(1.34107 C)(0.0001) = 4.86105 C, The electrostatic force of attraction is then F = 1 4 0 q1q2 r2 = (8.99109 N m2 /C2 ) (2.79105 )(4.86105 ) (28.0 m)2 = 1.61018 N. 9 11. E25-31 (a) The gravitational force of attraction between the Moon and the Earth is FG = GMEMM R2 , where R is the distance between them. If both the Earth and the moon are provided a charge q, then the electrostatic repulsion would be FE = 1 4 0 q2 R2 . Se...</p>