普物第九版解答(word)ch21_Edited_HR_9e_ISM

902 CHAPTER 21 903 Chapter 21 1. The magnitude of the force of either of the charges on the other is given by where r is the distance between the charges. We want the value of q that maximizes the function f(q) = q(Q – q). Setting the derivative equal to zero leads to Q – 2q = 0, or q = Q/2. Thus, q/Q = 0.500. 2. The fact that the spheres are identical allows us to conclude that when two spheres are in contact, they share equal charge. Therefore, when a charged sphere (q) touches an uncharged one, they will (fairly quickly) each attain half that charge (q/2). We start with spheres 1 and 2, each having charge q and experiencing a mutual repulsive force . When the neutral sphere 3 touches sphere 1, sphere 1’s charge decreases to q/2. Then sphere 3 (now carrying charge q/2) is brought into contact with sphere 2; a total amount of q/2 + q becomes shared equally between them. Therefore, the charge of sphere 3 is 3q/4 in the final situation. The repulsive force between spheres 1 and 2 is finally 3. Equation 21-1 gives Coulomb’s law, , which we solve for the distance: 4. The unit ampere is discussed in Section 21-4. Using i for current, the charge transferred is 5. The magnitude of the mutual force of attraction at r = 0.120 m is 6. (a) With a understood to mean the magnitude of acceleration, Newton’s second and third laws lead to (b) The magnitude of the (only) force on particle 1 is Inserting the values for m1 and a1 (see part (a)) we obtain |q| = 6.6  10–11 C. 7. With rightward positive, the net force on q3 is We note that each term exhibits the proper sign (positive for rightward, negative for leftward) for all possible signs of the charges. For example, the first term (the force exerted on q3 by q1) is negative if they are unlike charges, indicating that q3 is being pulled toward q1, and it is positive if they are like charges (so q3 would be repelled from q1). Setting the net force equal to zero L23= L12 and canceling k, q3, and L12 leads to 8. In experiment 1, sphere C first touches sphere A, and they divided up their total charge (Q/2 plus Q) equally between them. Thus, sphere A and sphere C each acquired charge 3Q/4. Then, sphere C touches B and those spheres split up their total charge (3Q/4 plus –Q/4) so that B ends up with charge equal to Q/4. The force of repulsion between A and B is therefore at the end of experiment 1. Now, in experiment 2, sphere C first touches B, which leaves each of them with charge Q/8. When C next touches A, sphere A is left with charge 9Q/16. Consequently, the force of repulsion between A and B is at the end of experiment 2. The ratio is 9. We assume the spheres are far apart. Then the charge distribution on each of them is spherically symmetric and Coulomb’s law can be used. Let q1 and q2 be the original charges. We choose the coordinate system so the force on q2 is positive if it is repelled by q1. Then, the force on q2 is where r = 0.500 m. The negative sign indicates that the spheres attract each other. After the wire is connected, the spheres, being identical, acquire the same charge. Since charge is conserved, the total charge is the same as it was originally. This means the charge on each sphere is (q1 + q2)/2. The force is now one of repulsion and is given by We solve the two force equations simultaneously for q1 and q2. The first gives the product and the second gives the sum where we have taken the positive root (which amounts to assuming q1 + q2  0). Thus, the product result provides the relation which we substitute into the sum result, producing Multiplying by q1 and rearranging, we obtain a quadratic equation The solutions are If the positive sign is used, q1 = 3.00  10–6 C, and if the negative sign is used, . (a) Using q2 = (–3.00  10–12)/q1 with q1 = 3.00  10–6 C, we get . (b) If we instead work with the q1 = –1.00  10–6 C root, then we find . Note that since the spheres are identical, the solutions are essentially the same: one sphere originally had charge –1.00  10–6 C and the other had charge +3.00  10–6 C. What if we had not made the assumption, above, that q1 + q2  0? If the signs of the charges were reversed (so q1 + q2 < 0), then the forces remain the same, so a charge of +1.00  10–6 C on one sphere and a charge of –3.00  10–6 C on the other also satisfies the conditions of the problem. 10. For ease of presentation (of the computations below) we assume Q > 0 and q < 0 (although the final result does not depend on this particular choice). (a) The x-component of the force experienced by q1 = Q is which (upon requiring F1x = 0) leads to , or (b) The y-component of the net force on q2 = q is which (if we demand F2y = 0) leads to . The result is inconsistent with that obtained in part (a). Thus, we are unable to construct an equilibrium configuration with this geometry, where the only forces present are given by Eq. 21-1. 11. The force experienced by q3 is (a) Therefore, the x-component of the resultant force on q3 is (b) Similarly, the y-component of the net force on q3 is 12. (a) For the net force to be in the +x direction, the y components of the individual forces must cancel. The angle of the force exerted by the q1 = 50 C charge on is 45°, and the angle of force exerted on q3 by Q is at – where Therefore, cancellation of y components requires from which we obtain |Q| = 104 C. Charge Q is “pulling” on q3, so (since q3 > 0) we conclude Q = –104 C. (b) Now, we require that the x components cancel, and we note that in this case, the angle of force on q3 exerted by Q is + (it is repulsive, and Q is positive-valued). Therefore, from which we obtain Q = 69.4 C. 13. (a) There is no equilibrium position for q3 between the two fixed charges, because it is being pulled by one and pushed by the other (since q1 and q2 have different signs); in this region this means the two force arrows on q3 are in the same direction and cannot cancel. It should also be clear that off-axis (with the axis defined as that which passes through the two fixed charges) there are no equilibrium positions. On the semi-infinite region of the axis that is nearest q2 and furthest from q1 an equilibrium position for q3 cannot be found because |q1| < |q2| and the magnitude of force exerted by q2 is everywhere (in that region) stronger than that exerted by q1 on q3. Thus, we must look in the semi-infinite region of the axis which is nearest q1 and furthest from q2, where the net force on q3 has magnitude with L = 10 cm and is assumed to be positive. We set this equal to zero, as required by the problem, and cancel k and q3. Thus, we obtain which yields (after taking the square root) for the distance between q3 and q1. That is, should be placed at along the x-axis. (b) As stated above, y = 0. 14. (a) The individual force magnitudes (acting on Q) are, by Eq. 21-1, which leads to |q1| = 49 |q2|. Since Q is located between q1 and q2, we conclude q1 and q2 are like-sign. Consequently, q1/q2 = 49. (b) Now we have which yields |q1| = 25 |q2|. Now, Q is not located between q1 and q2; one of them must push and the other must pull. Thus, they are unlike-sign, so q1/q2 = –25. 15. (a) The distance between q1 and q2 is The magnitude of the force exerted by q1 on q2 is (b) The vector is directed toward q1 and makes an angle  with the +x axis, where (c) Let the third charge be located at (x3, y3), a distance r from q2. We note that q1, q2, and q3 must be collinear; otherwise, an equilibrium position for any one of them would be impossible to find. Furthermore, we cannot place q3 on the same side of q2 where we also find q1, since in that region both forces (exerted on q2 by q3 and q1) would be in the same direction (since q2 is attracted to both of them). Thus, in terms of the angle found in part (a), we have x3 = x2 – r cos and y3 = y2 – r sin (which means y3 > y2 since  is negative). The magnitude of force exerted on q2 by q3 is , which must equal that of the force exerted on it by q1 (found in part (a)). Therefore, Consequently, x3 = x2 – r cos = –2.0 cm – (6.45 cm) cos(–10°) = –8.4 cm, (d) and y3 = y2 – r sin = 1.5 cm – (6.45 cm) sin(–10°) = 2.7 cm. 16. (a) According to the graph, when q3 is very close to q1 (at which point we can consider the force exerted by particle 1 on 3 to dominate) there is a (large) force in the positive x direction. This is a repulsive force, then, so we conclude q1 has the same sign as q3. Thus, q3 is a positive-valued charge. (b) Since the graph crosses zero and particle 3 is between the others, q1 must have the same sign as q2, which means it is also positive-valued. We note that it crosses zero at r = 0.020 m (which is a distance d = 0.060 m from q2). Using Coulomb’s law at that point, we have , or q2/q1 = 9.0. 17. (a) Equation 21-1 gives (b) On the right, a force diagram is shown as well as our choice of y axis (the dashed line). The y axis is meant to bisect the line between q2 and q3 in order to make use of the symmetry in the problem (equilateral triangle of side length d, equal-magnitude charges q1 = q2 = q3 = q). We see that the resultant force is along this symmetry axis, and we obtain . 18. Since the forces involved are proportional to q, we see that the essential difference between the two situations is Fa ( qB + qC (when those two charges are on the same side) versus Fb ( qB + qC (when they are on opposite sides). Setting up ratios, we have . 19. (a) If the system of three charges is to be in equilibrium, the force on each charge must be zero. The third charge q3 must lie between the other two or else the forces acting on it due to the other charges would be in the same direction and q3 could not be in equilibrium. Suppose q3 is at a distance x from q, and L – x from 4.00q. The force acting on it is then given by where the positive direction is rightward. We require F3 = 0 and solve for x. Canceling common factors yields 1/x2 = 4/(L – x)2 and taking the square root yields 1/x = 2/(L – x). The solution is x = L/3. With L = 9.00 cm, we have x = 3.00 cm. (b) Similarly, the y coordinate of q3 is y = 0. (c) The force on q is The signs are chosen so that a negative force value would cause q to move leftward. We require Fq = 0 and solve for q3: where x = L/3 is used. Note that we may easily verify that the force on 4.00q also vanishes: . 20. We note that the problem is examining the force on charge A, so that the respective distances (involved in the Coulomb force expressions) between B and A, and between C and A, do not change as particle B is moved along its circular path. We focus on the endpoints ( = 0º and 180º) of each graph, since they represent cases where the forces (on A) due to B and C are either parallel or antiparallel (yielding maximum or minimum force magnitudes, respectively). We note, too, that since Coulomb’s law is inversely proportional to r² then (if, say, the charges were all the same) the force due to C would be one-fourth as big as that due to B (since C is twice as far away from A). The charges, it turns out, are not the same, so there is also a factor of the charge ratio  (the charge of C divided by the charge of B), as well as the aforementioned ¼ factor. That is, the force exerted by C is, by Coulomb’s law, equal to ±¼ multiplied by the force exerted by B. (a) The maximum force is 2F0 and occurs when  = 180º (B is to the left of A, while C is the right of A). We choose the minus sign and write 2 F0 = (1  ¼) F0 (  = – 4 . One way to think of the minus sign choice is cos(180º) = –1. This is certainly consistent with the minimum force ratio (zero) at  = 0º since that would also imply 0 = 1 + ¼ (  = – 4 . (b) The ratio of maximum to minimum forces is 1.25/0.75 = 5/3 in this case, which implies EQ \f(5,3) = EQ \f(1 + ¼x,1 - ¼x) (  = 16 . Of course, this could also be figured as illustrated in part (a), looking at the maximum force ratio by itself and solving, or looking at the minimum force ratio (¾) at  = 180º and solving for . 21. The charge dq within a thin shell of thickness dr is where A = 4r2. Thus, with  = b/r, we have With b = 3.0 C/m2, r2 = 0.06 m, and r1 = 0.04 m, we obtain q = 0.038 C = 3.8 ( 108 C. 22. (a) The dashed line distance in the figure is . The net force on q1 due to the two charges q3 and q4 (with |q3| = |q4| = 1.60 ( 1019 C) on the y axis has magnitude . This must be set equal to the magnitude of the force exerted on q1 by q2 = 8.00 ( 1019 C = 5.00 |q3| in order that its net force be zero: ( D = = 1.133 d. Given d = 2.00 cm, this then leads to D = 2.27 cm. (b) As the angle decreases, its cosine increases, resulting in a larger contribution from the charges on the y axis. To offset this, the force exerted by q2 must be made stronger, so that it must be brought closer to q1 (keep in mind that Coulomb’s law is inversely proportional to distance-squared). Thus, D must be decreased. 23. If  is the angle between the force and the x-axis, then cos = EQ \f(x,\r(x2 + d2)) . We note that, due to the symmetry in the problem, there is no y component to the net force on the third particle. Thus, F represents the magnitude of force exerted by q1 or q2 on q3. Let e = +1.60 ( 1019 C, then q1 = q2 = +2e and q3 = 4.0e and we have Fnet = 2F cos = EQ \f(2(2e)(4e),4peo (x2 + d2)) EQ \f(x,\r(x2 + d2)) = EQ \f(4e2 x,peo (x2 + d2 )3/2) . (a) To find where the force is at an extremum, we can set the derivative of this expression equal to zero and solve for x, but it is good in any case to graph the function for a fuller understanding of its behavior, and as a quick way to see whether an extremum point is a maximum or a miminum. In this way, we find that the value coming from the derivative procedure is a maximum (and will be presented in part (b)) and that the minimum is found at the lower limit of the interval. Thus, the net force is found to be zero at x = 0, which is the smallest value of the net force in the interval 5.0 m ( x ( 0. (b) The maximum is found to be at x = d/ EQ \r(2) or roughly 12 cm. (c) The value of the net force at x = 0 is Fnet = 0. (d) The value of the net force at x = d/ EQ \r(2) is Fnet = 4.9 ( 1026 N. 24. (a) Equation 21-1 gives (b) If n is the number of excess electrons (of charge –e each) on each drop then 25. Equation 21-11 (in absolute value) gives 26. The magnitude of the force is 27. (a) The magnitude of the force between the (positive) ions is given by where q is the charge on either of them and r is the distance between them. We solve for the charge: (b) Let n be the number of electrons missing from each ion. Then, ne = q, or 28. Keeping in mind that an ampere is a coulomb per second (1 A = 1 C/s), and that a minute is 60 seconds, the charge (in absolute value) that passes through the chest is | q | = ( 0.300 C/s ) ( 90 s ) = 27.0 C . This charge consists of n electrons (each of which has an absolute value of charge equal to e). Thus, n = EQ \f(| q |,e) = EQ \f(27.0 C,1.60 x 10-19 C) = 1.69  1020 . 29. (a) We note that tan(30() = 1/ EQ \r(3) . In the initial (highly symmetrical) configuration, the net force on the central bead is in the –y direction and has magnitude 3F where F is the Coulomb’s law force of one bead on another at distance d = 10 cm. This is due to the fact that the forces exerted on the central bead (in the initial situation) by the beads on the x axis cancel each other; also, the force exerted “downward” by bead 4 on the central bead is four times larger than the “upward” force exerted by bead 2. This net force along the y axis does not change as bead 1 is now moved, though there is now a nonzero x-component Fx . The components are now related by tan(30() = EQ \f(Fx, Fy) ( EQ \f(1,\r(3)) = EQ \f(Fx,3F) which implies Fx = EQ \r(3) F. Now, bead 3 exerts a “leftward” force of magnitude F on the central bead, while bead 1 exerts a “rightward” force of magnitude F(. Therefore, F(  F = EQ \r(3) F. ( F( = ( EQ \r(3) + 1) F . The fact that Coulomb’s law depends inversely on distance-squared then implies r2 = EQ \f(d2,\r(3) + 1) ( r = EQ \f(d,\r(\r(3) + 1)) = 6.05 cm where r is the distance between bead 1 and the central bead. This corresponds to (b) To regain the condition of high symmetry (in particular, the cancellation of x-components) bead 3 must be moved closer to the central bead so that it, too, is the distance r (as calculated in part (a)) away from it. 30. (a) Let x be the distance between particle 1 and particle 3. Thus, the distance between particle 3 and particle 2 is L – x. Both particles exert leftward forces on q3 (so long as it is on the line between them), so the magnitude of the net force on q3 is Fnet = | EQ \o(F 1 3, ( ) | + | EQ \o(F 2 3, ( ) | = EQ \f(|q1 q3|,4peo x2) + EQ \f(|q2 q3|,4peo (L-x)2) = EQ \f(e2, peo ) \b(\f(1, x2) + \f(27,(L - x)2)) with the values of the charges (stated in the problem) plugged in. Finding the value of x that minimizes this expression leads to x = ¼ L. Thus, x = 1.50 cm. (b) Substituting x = ¼ L back into the expression for the net force magnitude and using the standard value for e leads to Fnet = 1.64  1023 N. 31. The unit ampere is discussed in Section 21-4. The proton flux is given as 1500 protons per square meter per second, where each proton provides a charge of q = +e. The current through the spherical area 4R2 = 4 (6.37  106 m)2 = 5.1  1014 m2 would be 32. Since the graph crosses zero, q1 must be positive-valued: q1 = +8.00e. We note that it crosses zero at r = 0.40 m. Now the asymptotic value of the force yields the magnitude and sign of q2: EQ \f(q1 q2,4peo r2) = F ( = 2.086  1018 C = 13e . 33. The volume of 250 cm3 corresponds to a mass of 250 g since the density of water is 1.0 g/cm3. This mass corresponds to 250/18 = 14 moles since the molar mass of water is 18. There are ten protons (each with charge q = +e) in each molecule of H2O, so 34. Let d be the vertical distance from the coordinate origin to q3 = q and q4 = q on the +y axis, where the symbol q is assumed to be a positive value. Similarly, d is the (positive) distance from the origin q4 =  on the y axis. If we take each angle  in the figure to be positive, then we have tan = d/R and cos = R/r (where r is the dashed line distance shown in the figure). The problem asks us to consider  to be a variable in the sense that, once the charges on the x axis are fixed in place (which determines R), d can then be arranged to some multiple of R, since d = R tan. The aim of this exploration is to show that if q is bounded then  (and thus d) is also bounded. From symmetry, we see that there is no net force in the vertical direction on q2 = –e sitting at a distance R to the left of the coordinate origin. We note that the net x force caused by q3 and q4 on the y axis will have a magnitude equal to . Consequently, to achieve a zero net force along the x axis, the above expression must equal the magnitude of the repulsive force exerted on q2 by q1 = –e. Thus, . Below we plot q/e as a function of the angle (in degrees): The graph suggests that q/e < 5 for  < 60º, roughly. We can be more precise by solving the above equation. The requirement that q ( 5e leads to which yields  ( 62.34º. The problem asks for “physically possible values,” and it is reasonable to suppose that only positive-integer-multiple values of e are allowed for q. If we let q = ne, for n = 1 … 5, then N will be found by taking the inverse cosine of the cube root of (1/2n). (a) The smallest value of angle is  = 37.5º (or 0.654 rad). (b) The second smallest value of angle is  = 50.95º (or 0.889 rad). (c) The third smallest value of angle is  = 56.6º (or 0.988 rad). 35. (a) Every cesium ion at a corner of the cube exerts a force of the same magnitude on the chlorine ion at the cube center. Each force is a force of attraction and is directed toward the cesium ion that exerts it, along the body diagonal of the cube. We can pair every cesium ion with another, diametrically positioned at the opposite corner of the cube. Since the two ions in such a pair exert forces that have the same magnitude but are oppositely directed, the two forces sum to zero and, since every cesium ion can be paired in this way, the total force on the chlorine ion is zero. (b) Rather than remove a cesium ion, we superpose charge –e at the position of one cesium ion. This neutralizes the ion, and as far as the electrical force on the chlorine ion is concerned, it is equivalent to removing the ion. The forces of the eight cesium ions at the cube corners sum to zero, so the only force on the chlorine ion is the force of the added charge. The length of a body diagonal of a cube is , where a is the length of a cube edge. Thus, the distance from the center of the cube to a corner is . The force has magnitude Since both the added charge and the chlorine ion are negative, the force is one of repulsion. The chlorine ion is pushed away from the site of the missing cesium ion. 36. (a) Since the proton is positively charged, the emitted particle must be a positron (as opposed to the negatively charged electron) in accordance with the law of charge conservation. (b) In this case, the initial state had zero charge (the neutron is neutral), so the sum of charges in the final state must be zero. Since there is a proton in the final state, there should also be an electron (as opposed to a positron) so that q = 0. 37. None of the reactions given include a beta decay, so the number of protons, the number of neutrons, and the number of electrons are each conserved. Atomic numbers (numbers of protons and numbers of electrons) and molar masses (combined numbers of protons and neutrons) can be found in Appendix F of the text. (a) 1H has 1 proton, 1 electron, and 0 neutrons and 9Be has 4 protons, 4 electrons, and 9 – 4 = 5 neutrons, so X has 1 + 4 = 5 protons, 1 + 4 = 5 electrons, and 0 + 5 – 1 = 4 neutrons. One of the neutrons is freed in the reaction. X must be boron with a molar mass of 5 + 4 = 9 g/mol: 9B. (b) 12C has 6 protons, 6 electrons, and 12 – 6 = 6 neutrons and 1H has 1 proton, 1 electron, and 0 neutrons, so X has 6 + 1 = 7 protons, 6 + 1 = 7 electrons, and 6 + 0 = 6 neutrons. It must be nitrogen with a molar mass of 7 + 6 = 13 g/mol: 13N. (c) 15N has 7 protons, 7 electrons, and 15 – 7 = 8 neutrons; 1H has 1 proton, 1 electron, and 0 neutrons; and 4He has 2 protons, 2 electrons, and 4 – 2 = 2 neutrons; so X has 7 + 1 – 2 = 6 protons, 6 electrons, and 8 + 0 – 2 = 6 neutrons. It must be carbon with a molar mass of 6 + 6 = 12: 12C. 38. As a result of the first action, both sphere W and sphere A possess charge EQ \f(1,2) qA , where qA is the initial charge of sphere A. As a result of the second action, sphere W has charge . As a result of the final action, sphere W now has charge equal to . Setting this final expression equal to +18e as required by the problem leads (after a couple of algebra steps) to the answer: qA = -16e. 39. Using Coulomb’s law, the magnitude of the force of particle 1 on particle 2 is , where and . Since both and are positively charged, particle 2 is repelled by particle 1, so the direction of is away from particle 1 and toward 2. In unit-vector notation, , where . The x component of is . Combining the expressions above, we obtain Note: In a similar manner, we find the y component of to be Thus, . 40. Regarding the forces on q3 exerted by q1 and q2, one must “push” and the other must “pull” in order that the net force is zero; hence, q1 and q2 have opposite signs. For individual forces to cancel, their magnitudes must be equal: . With the above expression simplifies to Therefore, , or 41. (a) The magnitudes of the gravitational and electrical forces must be the same: where q is the charge on either body, r is the center-to-center separation of Earth and Moon, G is the universal gravitational constant, M is the mass of Earth, and m is the mass of the Moon. We solve for q: According to Appendix C of the text, M = 5.98  1024 kg, and m = 7.36  1022 kg, so (using 40 = 1/k) the charge is (b) The distance r cancels because both the electric and gravitational forces are proportional to 1/r2. (c) The charge on a hydrogen ion is e = 1.60  10–19 C, so there must be Each ion has a mass of 1.67  10–27 kg, so the total mass needed is 42. (a) A force diagram for one of the balls is shown below. The force of gravity acts downward, the electrical force of the other ball acts to the left, and the tension in the thread acts along the thread, at the angle  to the vertical. The ball is in equilibrium, so its acceleration is zero. The y component of Newton’s second law yields T cos – mg = 0 and the x component yields T sin – Fe = 0. We solve the first equation for T and obtain T = mg/cos. We substitute the result into the second to obtain mg tan – Fe = 0. Examination of the geometry of Figure 21-38 leads to If L is much larger than x (which is the case if  is very small), we may neglect x/2 in the denominator and write tan  x/2L. This is equivalent to approximating tan by sin. The magnitude of the electrical force of one ball on the other is by Eq. 21-4. When these two expressions are used in the equation mg tan = Fe, we obtain (b) We solve x3 = 2kq2L/mg for the charge (using Eq. 21-5): Thus, the magnitude is 43. (a) If one of them is discharged, there would no electrostatic repulsion between the two balls and they would both come to the position  = 0, making contact with each other. (b) A redistribution of the remaining charge would then occur, with each of the balls getting q/2. Then they would again be separated due to electrostatic repulsion, which results in the new equilibrium separation 44. Letting kq2/r2 = mg, we get 45. There are two protons (each with charge q = +e) in each molecule, so 46. Let denotes the force on q1 exerted by q2 and be its magnitude. (a) We consider the net force on q1. points in the +x direction since q1 is attracted to q2. and both point in the –x direction since q1 is repelled by q3 and q4. Thus, using d = 0.0300 m, the net force is or (b) We now consider the net force on q2. We note that points in the –x direction, and and both point in the +x direction. The net force is . 47. We are looking for a charge q that, when placed at the origin, experiences where The magnitude of these individual forces are given by Coulomb’s law, Eq. 21-1, and without loss of generality we assume q > 0. The charges q1 (+6 C), q2 (–4 C), and q3 (unknown), are located on the +x axis, so that we know points toward –x, points toward +x, and points toward –x if q3 > 0 and points toward +x if q3 < 0. Therefore, with r1 = 8 m, r2 = 16 m and r3 = 24 m, we have Simplifying, this becomes where q3 is now understood to be in C. Thus, we obtain q3 = –45 C. 48. (a) Since qA = –2.00 nC and qC = +8.00 nC, Eq. 21-4 leads to (b) After making contact with each other, both A and B have a charge of When B is grounded its charge is zero. After making contact with C, which has a charge of +8.00 nC, B acquires a charge of [0 + (–8.00 nC)]/2 = –4.00 nC, which charge C has as well. Finally, we have QA = –3.00 nC and QB = QC = –4.00 nC. Therefore, (c) We also obtain 49. Coulomb’s law gives 50. (a) Since the rod is in equilibrium, the net force acting on it is zero, and the net torque about any point is also zero. We write an expression for the net torque about the bearing, equate it to zero, and solve for x. The charge Q on the left exerts an upward force of magnitude (1/40) (qQ/h2), at a distance L/2 from the bearing. We take the torque to be negative. The attached weight exerts a downward force of magnitude W, at a distance from the bearing. This torque is also negative. The charge Q on the right exerts an upward force of magnitude (1/40) (2qQ/h2), at a distance L/2 from the bearing. This torque is positive. The equation for rotational equilibrium is The solution for x is (b) If FN is the magnitude of the upward force exerted by the bearing, then Newton’s second law (with zero acceleration) gives We solve for h so that FN = 0. The result is 51. The charge dq within a thin section of the rod (of thickness dx) is  A dx where and  is the charge per unit volume. The number of (excess) electrons in the rod (of length L = 2.00 m) is n = q/(–e) where e is given in Eq. 21-12. (a) In the case where  = – 4.00  10–6 C/m3, we have . (b) With  = bx2 (b = –2.00  10–6 C/m5) we obtain 52. For the Coulomb force to be sufficient for circular motion at that distance (where r = 0.200 m and the acceleration needed for circular motion is a = v2/r) the following equality is required: . With q = 4.00  106 C, m = 0.000500 kg, v = 50.0 m/s, this leads to . 53. (a) Using Coulomb’s law, we obtain (b) If r = 1000 m, then 54. Let q1 be the charge of one part and q2 that of the other part; thus, q1 + q2 = Q = 6.0 C. The repulsive force between them is given by Coulomb’s law: . If we maximize this expression by taking the derivative with respect to q1 and setting equal to zero, we find q1 = Q/2 , which might have been anticipated (based on symmetry arguments). This implies q2 = Q/2 also. With r = 0.0030 m and Q = 9.0  106 C, we find . 55. The two charges are q = Q (where  is a pure number presumably less than 1 and greater than zero) and Q – q = (1 – )Q. Thus, Eq. 21-4 gives The graph below, of F versus , has been scaled so that the maximum is 1. In actuality, the maximum value of the force is Fmax = Q2/160 d 2. (a) It is clear that = 0.5 gives the maximum value of F. (b) Seeking the half-height points on the graph is difficult without grid lines or some of the special tracing features found in a variety of modern calculators. It is not difficult to algebraically solve for the half-height points (this involves the use of the quadratic formula). The results are Thus, the smaller value of  is , (c) and the larger value of  is . 56. (a) Equation 21-11 (in absolute value) gives (b) Since you have the excess electrons (and electrons are lighter and more mobile than protons) then the electrons “leap” from you to the faucet instead of protons moving from the faucet to you (in the process of neutralizing your body). (c) Unlike charges attract, and the faucet (which is grounded and is able to gain or lose any number of electrons due to its contact with Earth’s large reservoir of mobile charges) becomes positively charged, especially in the region closest to your (negatively charged) hand, just before the spark. (d) The cat is positively charged (before the spark), and by the reasoning given in part (b) the flow of charge (electrons) is from the faucet to the cat. (e) If we think of the nose as a conducting sphere, then the side of the sphere closest to the fur is of one sign (of charge) and the side furthest from the fur is of the opposite sign (which, additionally, is oppositely charged from your bare hand, which had stroked the cat’s fur). The charges in your hand and those of the furthest side of the “sphere” therefore attract each other, and when close enough, manage to neutralize (due to the “jump” made by the electrons) in a painful spark. 57. If the relative difference between the proton and electron charges (in absolute value) were then the actual difference would be Amplified by a factor of 29  3  1022 as indicated in the problem, this amounts to a deviation from perfect neutrality of in a copper penny. Two such pennies, at r = 1.0 m, would therefore experience a very large force. Equation 21-1 gives 58. Charge q1 = –80  10–6 C is at the origin, and charge q2 = +40  10–6 C is at x = 0.15 m. The force on q3 = +20  10–6 C is due to the attractive and repulsive forces from q1 and q2, respectively. In symbols, , where (a) In this case r31 = 0.40 m and r32 = 0.15 m, with directed toward –x and directed in the +x direction. Using the value of k in Eq. 21-5, we obtain (b) In this case r31 = 0.80 m and r32 = 0.60 m, with directed toward –x and toward +x. Now we obtain (c) Between the locations treated in parts (a) and (b), there must be one where . Writing r31 = x and r32 = x – 0.20 m, we equate and , and after canceling common factors, arrive at This can be further simplified to Taking the (positive) square root and solving, we obtain x = 0.683 m. If one takes the negative root and ‘solves’, one finds the location where the net force would be zero if q1 and q2 were of like sign (which is not the case here). (d) From the above, we see that y = 0. 59. The mass of an electron is m = 9.11  10–31 kg, so the number of electrons in a collection with total mass M = 75.0 kg is The total charge of the collection is 60. We note that, as result of the fact that the Coulomb force is inversely proportional to r2, a particle of charge Q that is distance d from the origin will exert a force on some charge qo at the origin of equal strength as a particle of charge 4Q at distance 2d would exert on qo. Therefore, q6 = +8e on the –y axis could be replaced with a +2e closer to the origin (at half the distance); this would add to the q5 = +2e already there and produce +4e below the origin, which exactly cancels the force due to q2 = +4e above the origin. Similarly, q4 = +4e to the far right could be replaced by a +e at half the distance, which would add to q3 = +e already there to produce a +2e at distance d to the right of the central charge q7. The horizontal force due to this +2e is cancelled exactly by that of q1 = +2e on the –x axis, so that the net force on q7 is zero. 61. (a) Charge Q1 = +80  10–9 C is on the y axis at y = 0.003 m, and charge is on the y axis at y = –0.003 m. The force on particle 3 (which has a charge of q = +18  10–9 C) is due to the vector sum of the repulsive forces from Q1 and Q2. In symbols, where Using the Pythagorean theorem, we have r31 = r32 = 0.005 m. In magnitude-angle notation (particularly convenient if one uses a vector-capable calculator in polar mode), the indicated vector addition becomes Therefore, the net force is . (b) Switching the sign of Q2 amounts to reversing the direction of its force on q. Consequently, we have Therefore, the net force is . 62. The individual force magnitudes are found using Eq. 21-1, with SI units (so ) and k as in Eq. 21-5. We use magnitude-angle notation (convenient if one uses a vector-capable calculator in polar mode), listing the forces due to +4.00e, +2.0e, and –2.0e charges: (a) Therefore, the net force has magnitude 4.31  10–24 N. (b) The direction of the net force is at an angle of –118° (or 242° measured counterclockwise from the +x axis). 63. The magnitude of the net force on the q = 42  10–6 C charge is where q1 = 30  10–9 C and |q2| = 40  10–9 C. This yields 0.22 N. Using Newton’s second law, we obtain 64. Let the two charges be q1 and q2. Then q1 + q2 = Q = 5.0  10–5 C. We use Eq. 21-1: We substitute q2 = Q – q1 and solve for q1 using the quadratic formula. The two roots obtained are the values of q1 and q2, since it does not matter which is which. We get and 3.8  10–5 C. Thus, the charge on the sphere with the smaller charge is . 65. When sphere C touches sphere A, they divide up their total charge (Q/2 plus Q) equally between them. Thus, sphere A now has charge 3Q/4, and the magnitude of the force of attraction between A and B becomes 66. With F = meg, Eq. 21-1 leads to which leads to y = 5.1 m. We choose since the second electron must be below the first one, so that the repulsive force (acting on the first) is in the direction opposite to the pull of Earth’s gravity. 67. The net force on particle 3 is the vector sum of the forces due to particles 1 and 2: . In order that , particle 3 must be on the x axis and be attracted by one and repelled by another. As the result, it cannot be between particles 1 and 2, but instead either to the left of particle 1 or to the right of particle 2. Let be placed a distance x to the right of 5.00q. Then its attraction to will be exactly balanced by its repulsion from +2.00q : . (a) Cross-multiplying and taking the square root, we obtain which can be rearranged to produce . (b) The y coordinate of particle 3 is y = 0. Note: We can use the result obtained above for a consistency check. We find the force on particle 3 due to particle 1 to be . Similarly, the force on particle 3 due to particle 2 is . Indeed, the sum of the two forces is zero. 68. The net charge carried by John whose mass is m = 90 kg is roughly and the net charge carried by Mary is 45/90 of that. So the electrostatic force between them is estimated to be Thus, the order of magnitude of the electrostatic force is . 69. We are concerned with the charges in the nucleus (not the “orbiting” electrons, if there are any). The nucleus of Helium has 2 protons and that of thorium has 90. (a) Equation 21-1 gives (b) Estimating the helium nucleus mass as that of 4 protons (actually, that of 2 protons and 2 neutrons, but the neutrons have approximately the same mass), Newton’s second law leads to 70. For the net force on q1 = +Q to vanish, the x force component due to q2 = q must exactly cancel the force of attraction caused by q4 = –1.50Q. Consequently, or q = 0.75Q/ EQ \r(2) . This implies that (0.300 m)2 1.60 x 10-6 N (0.300 m)2 1.20 x 10-6 N 1.60 x 10-6 N (0.300 m)2 (0.15 m)2 230 N = 7.7 x 105 C 2.96 x 1018 N 875 _1323906056.unknown _1329979885.unknown _1343322684.unknown _1343375473.unknown _1343377398.unknown _1343379161.unknown _1343379321.unknown _1343376429.unknown _1343376513.unknown _1343375867.unknown _1343375705.unknown _1343324647.unknown _1343375192.unknown _1343375204.unknown _1343324547.unknown _1343324558.unknown _1343324445.unknown _1343321983.unknown _1329979078.unknown _1329978957.unknown _1323957813.unknown _1323960987.unknown _1326679979.unknown _1323961166.unknown _1323961318.unknown _1323961043.unknown _1323958246.unknown _1323960962.unknown _1323957914.unknown _1323957512.unknown _1323957791.unknown _1323957241.unknown _1323904162.unknown _1323905628.unknown _1323905814.unknown _1323905835.unknown _1323905996.unknown _1323905766.unknown _1323904244.unknown _1323905584.unknown _1323904185.unknown _1323903977.unknown _1323809679.unknown _1233932125.unknown _1233299503.unknown _1233931409.unknown _1233301544.unknown _1233300185.unknown _1233301139.unknown _1233300378.unknown _1233300973.unknown _1233299997.unknown _1233300041.unknown _1233298636.unknown _1233298720.unknown _1233298342.unknown _1124038595.unknown _1134721773.unknown _1233297807.unknown _1140710351.unknown _1134720519.unknown _1134721568.unknown _1134721567.unknown _1134720562.unknown _1134720900.unknown _1134720036.unknown _1134720058.unknown _1134717783.unknown _1124038643.unknown _1124039031.unknown _1116896418.unknown _1116896941.unknown _1124038510.unknown _1124038546.unknown _1124036600.unknown _1116897367.unknown _1116897525.unknown _1116896876.unknown _1116896922.unknown _1116896739.unknown _1116896230.unknown _1116896395.unknown _1116896166.unknown _1116615777.unknown _1116613492.unknown _1115977820.unknown _1115977998.unknown _1115977997.unknown