Baixe Resolução Eletromagnetismo e outras Provas em PDF para Eletromagnetismo, somente na Docsity! 1. Eq. 21-1 gives Coulomb’s Law, F k q q r 1 2 2 , which we solve for the distance: 9 2 2 6 6 1 2 8.99 10 N m C 26.0 10 C 47.0 10 C| || | 1.39m. 5.70N k q q r F 2. (a) With a understood to mean the magnitude of acceleration, Newton’s second and third laws lead to m a m a m2 2 1 1 2 7 7 6 3 10 7 0 9 0 4 9 10 . . . . kg m s m s kg. 2 2 c hc h (b) The magnitude of the (only) force on particle 1 is 2 1 2 9 2 2 1 1 2 2 8.99 10 N m C . (0.0032 m) q q q F m a k r Inserting the values for m1 and a1 (see part (a)) we obtain |q| = 7.1 10 –11 C. F q Q q r 1 4 0 2 b g 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 /dF dq equal to zero leads to Q – 2q = 0, or q = Q/2. Thus, q/Q = 0.500. 5. The magnitude of the force of either of the charges on the other is given by 6. The unit Ampere is discussed in §21-4. Using i for current, the charge transferred is 4 62.5 10 A 20 10 s 0.50 C.q it 7. 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 F q q r k q q r a 1 4 0 1 2 2 1 2 2 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 F r k q q r b q q q q 1 4 40 2 2 2 1 2 2 2 1 2 1 2d id i b g . We solve the two force equations simultaneously for q1 and q2. The first gives the product q q r F k a 1 2 2 2 9 120500 0108 8 99 10 300 10 . . . . , m N N m C C 2 2 2b g b g and the second gives the sum q q r F k b 1 2 62 2 0 500 0 0360 2 00 10. . .m N 8.99 10 N m C C 9 2 2b g where we have taken the positive root (which amounts to assuming q1 + q2 0). Thus, the product result provides the relation 12 2 2 1 3.00 10 C q q which we substitute into the sum result, producing q q 1 12 1 6300 10 2 00 10 . . C C. 2 Multiplying by q1 and rearranging, we obtain a quadratic equation q q1 2 6 1 122 00 10 300 10 0. . .C C2c h 9. The force experienced by q3 is 3 1 3 2 3 4 3 31 32 34 2 22 0 | || | | || | | || |1 ˆ ˆ ˆ ˆj (cos45 i sin 45 j) i 4 ( 2 ) q q q q q q F F F F a aa (a) Therefore, the x-component of the resultant force on q3 is 2 7 9 2 23 2 3 42 2 0 2 1.0 10 C| | | | 1 | | 8.99 10 N m C 2 0.17N. 4 (0.050 m)2 2 2 2 x q q F q a (b) Similarly, the y-component of the net force on q3 is 2 7 9 2 23 2 3 12 2 0 2 1.0 10 C| | | | 1 | | 8.99 10 N m C 1 0.046N. 4 (0.050 m)2 2 2 2 y q q F q a 10. (a) The individual force magnitudes (acting on Q) are, by Eq. 21-1, 1 2 2 2 0 0 1 1 4 4/ 2 / 2 q Q q Q a a a a which leads to |q1| = 9.0 |q2|. Since Q is located between q1 and q2, we conclude q1 and q2 are like-sign. Consequently, q1/q2 = 9.0. (b) Now we have 1 2 2 2 0 0 1 1 4 43 / 2 3 / 2 q Q q Q a a a a 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. 1 3 2 3 3 13 23 2 2 2312 23 . q q q q F F F k k LL L 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 1 1 2 2 0 4.00. 4.00 q q q q 11. With rightwards positive, the net force on q3 is 14. (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 2 2 1 3 3 2 2 1 1 12 2 0 0 1 1 0.060 m 9.0 4 4 0.020 m q q q q d q q q q r d r , or q2/q1 = 9.0. 15. (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 which 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 1 3 2 3 22 0 0 q q q q k k L L L with L = 10 cm and 0L is assumed to be positive. We set this equal to zero, as required by the problem, and cancel k and q3. Thus, we obtain 2 1 2 0 2 22 0 0 10 3.0 C 0 3.0 1.0 C q q L L q L L qL L which yields (after taking the square root) 0 0 0 10 cm 3 14cm 3 1 3 1 L L L L L for the distance between q3 and q1. That is, 3q should be placed at 14 cmx along the x-axis. (b) As stated above, y = 0. 16. 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 Fa Fb = qB + qC - qB + qC 23 24 1 /2.014 10 N 2.877 10 N 1 / C B C B q q q q . After noting that the ratio on the left hand side is very close to – 7, then, after a couple of algebra steps, we are led to 7 1 8 1.333. 7 1 6 C B q q 2 3 2 2 0 1 4.00 . 4 q qq q F x L The signs are chosen so that a negative force value would cause q to move leftward. We require Fq = 0 and solve for q3: 2 3 3 2 4 4 4 0.444 9 9 qqx q q L q where x = L/3 is used. Note that we may easily verify that the force on 4.00q also vanishes: 22 2 2 2 0 4 22 2 2 2 2 0 0 0 4 4 941 4 1 4 1 4 4 0 4 4 4 9 4 q qqqq q q q F L L L L LL x . 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 3 3 3 22 0 41 4 qq qq F x L x 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 20. (a) We note that cos(30º) = 1 2 3 , so that the dashed line distance in the figure is 2 / 3r d . We 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 1 3 1 3 2 2 0 0 | | 3 3 | | 2 cos(30 ) 4 16 q q q q r d . 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: 1 3 1 2 2 2 0 0 3 3 | | | | 16 4 ( ) q q q q d D d D = d 2 5 3 3 1 = 0.9245 d . Given d = 2.00 cm, this then leads to D = 1.92 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. 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 = 2(2e)(4e) 4 o (x 2 + d2) x x 2 + d2 = 4e2 x o (x 2 + 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/ 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/ 2 is Fnet = 4.9 10 26 N. 21. If is the angle between the force and the x-axis, then cos = x x 2 + d2 . 24. The magnitude of the force is F k e r F HG I KJ 2 2 9 19 2 10 2 98 99 10 160 10 2 82 10 2 89 10. . . . N m C C m N. 2 2 c h c h 25. (a) The magnitude of the force between the (positive) ions is given by F q q r k q r b gb g 4 0 2 2 2 where q is the charge on either of them and r is the distance between them. We solve for the charge: q r F k 50 10 37 10 8 99 10 32 1010 9 9 19. . . .m N N m C C. 2 2c h (b) Let n be the number of electrons missing from each ion. Then, ne = q, or 9 19 3.2 10 C 2. 1.6 10 C q n e 26. 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 ) ( 120 s ) = 36.0 C . This charge consists of n electrons (each of which has an absolute value of charge equal to e). Thus, n = | q | e = 36.0 C 1.60 x 10-19 C = 2.25 1020 . 29. The unit Ampere is discussed in §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 i F HG I KJ51 10 1500 16 10 0122 14 2 2 19. . . .m protons s m C proton Ac h c h 30. 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: q1 q2 4 o r 2 = F q2 = 1.5 x 10-25 k q1 r 2 = 2.086 10 18 C = 13e . 31. 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 23 19 714 14 6.02 10 10 1.60 10 C 1.3 10 C.AQ N q 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 3 2 2 2 0 0 0 2 cos 2 cos 2 cos 4 4 ( / cos ) 4 qe qe qe r R R . 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, 3 2 2 2 3 0 0 2 cos 4 4 2cos qe e e q R R . 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 3 1/3 1 5 cos 2cos (10) e e 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). 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. (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 3a , where a is the length of a cube edge. Thus, the distance from the center of the cube to a corner is d a3 2d i . The force has magnitude F k e d ke a 2 2 2 2 9 2 2 19 2 9 2 9 3 4 8 99 10 160 10 3 4 0 40 10 19 10b g c hc h b gc h . . . . . N m C C m N 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. 38. Let 12F denotes the force on q1 exerted by q2 and 12F be its magnitude. (a) We consider the net force on q1. 12F points in the +x direction since q1 is attracted to q2. 13F and 14F both point in the –x direction since q1 is repelled by q3 and q4. Thus, using d = 0.0200 m, the net force is 2 1 12 13 14 2 2 2 2 0 0 0 0 2 9 2 2 19 25 2 2 2 | | (2 )( ) (2 )(4 ) 11 4 4 (2 ) 4 (3 ) 18 4 8.99 10 N m C 1.60 10 C11 3.52 10 N 18 2.00 10 m e e e e e e e F F F F d d d d , or 251 ˆ(3.52 10 N)i.F (b) We now consider the net force on q2. We note that 21 12F F points in the –x direction, and 23F and 24F both point in the +x direction. The net force is 23 24 21 2 2 2 0 0 0 4 | | | | 2 | | 0 4 (2 ) 4 4 e e e e e e F F F d d d 39. If is the angle between the force and the x axis, then cos = d2 d1 2 + d2 2 . Thus, using Coulomb’s law for F, we have Fx = F cos = q1 q2 4 o (d1 2 + d2 2) d2 d1 2 + d2 2 = 1.31 10 22 N . With q = 4.00 10 6 C, m = 0.000800 kg, v = 50.0 m/s, this leads to 2 4 2 50 9 2 2 6 4 (0.200 m)(8.00 10 kg)(50.0 m/s) 1.11 10 C (8.99 10 N m C )(4.00 10 C) rmv Q q . 40. 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: 2 2 04 Qq mv r r . 43. There are two protons (each with charge q = +e) in each molecule, so Q N qA 6 02 10 2 160 10 19 10 23 19 5. . .c hb gc hC C 0.19 MC. / 2x L from the bearing. This torque is also negative. The charge Q on the right exerts an upward force of magnitude (1/4 0) (2qQ/h 2), at a distance L/2 from the bearing. This torque is positive. The equation for rotational equilibrium is 2 2 0 0 1 1 2 0. 4 2 2 4 2 qQ L L qQ L W x h h The solution for x is x L qQ h W F HG I KJ2 1 1 4 0 2 . (b) If FN is the magnitude of the upward force exerted by the bearing, then Newton’s second law (with zero acceleration) gives 2 2 0 0 1 1 2 0. 4 4 N qQ qQ W F h h We solve for h so that FN = 0. The result is h qQ W 1 4 3 0 . 44. (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/h 2), 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 45. Coulomb’s law gives 9 2 2 19 22 2 2 2 15 2 0 8.99 10 N m C (1.60 10 C)| | ( 3) 3.8 N. 4 9(2.6 10 m) q k e F r r 48. 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 1 2 (3 / 4)( / 4)Q Q F k d 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 2 2 (9 /16)( / 8)Q Q F k d at the end of experiment 2. The ratio is 2 1 (9 /16)(1/ 8) 0.375. (3 / 4)(1/ 4) F F q q e p e 0 0000010. then the actual difference would be q qp e 16 10 25. .C Amplified by a factor of 29 3 1022 as indicated in the problem, this amounts to a deviation from perfect neutrality of q 29 3 10 16 10 01422 25c hc h. .C C in a copper penny. Two such pennies, at r = 1.0 m, would therefore experience a very large force. Eq. 21-1 gives F k q r b g2 2 817 10. .N 49. If the relative difference between the proton and electron charges (in absolute value) were 50. Letting kq2/r2 = mg, we get 9 2 2 19 27 2 8.99 10 N m C 1.60 10 C 0.119m. 1.67 10 kg 9.8 m s k r q mg 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 q 6 67 10 7 36 10 5 98 10 8 99 10 5 7 10 11 22 24 9 13 . . . . . N m kg kg kg N m C C. 2 2 2 2 c hc hc h (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 13 32 19 5.7 10 C 3.6 10 ions. 1.6 10 C q n e Each ion has a mass of im 1.67 10 –27 kg, so the total mass needed is 32 27 53.6 10 1.67 10 kg 6.0 10 kg.im nm 53. (a) The magnitudes of the gravitational and electrical forces must be the same: 1 4 0 2 2 2 q r G mM r 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: q GmM4 0 . by Eq. 21-4. When these two expressions are used in the equation mg tan = Fe, we obtain 1/3 2 2 2 0 0 1 . 2 4 2 mgx q q L x L x mg (b) We solve x3 = 2kq2L/mg for the charge (using Eq. 21-5): 323 8 9 2 2 0.010 kg 9.8m s 0.050 m 2.4 10 C. 2 2 8.99 10 N m C 1.20 m mgx q kL Thus, the magnitude is 8| | 2.4 10 C.q 54. (a) A force diagram for one of the balls is shown on the right. The force of gravity mg acts downward, the electrical force Fe 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 21-42 leads to tan . x L x 2 22 2b g 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 F q x e 2 0 24 55. (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 1/3 2 1/3 1/3 0 2 1 1 5.0 cm 3.1 cm. 2 4 4 q L x x mg 58. We note that, as result of the fact that the Coulomb force is inversely proportional to r 2, a particle of charge Q which 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. 59. (a) Charge Q1 = +80 10 –9 C is on the y axis at y = 0.003 m, and charge 9 2 80 10 CQ 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, 3 1 3 2 3 ,F F F where 3 1 3 2 31 3 22 2 3 1 3 2 | | | | , | | . q q q q F k F k r r 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 3 0.518 37 0.518 37 0.829 0 .F Therefore, the net force is 3 ˆ(0.829 N)iF . (b) Switching the sign of Q2 amounts to reversing the direction of its force on q. Consequently, we have 3 0.518 37 0.518 143 0.621 90 .F Therefore, the net force is 3 ˆ(0.621 N)jF . 4 60 10 180 2 30 10 90 102 10 145 616 10 15224 24 24 24. . . .c h c h c h c h (a) Therefore, the net force has magnitude 6.16 10–24 N. (b) The direction of the net force is at an angle of –152° (or 208° measured counterclockwise from the +x axis). 60. The individual force magnitudes are found using Eq. 21-1, with SI units (so 0.02 ma ) 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 +4.00q, +2.00q, and –2.00q charges: where q3 is now understood to be in C. Thus, we obtain q3 = –45 C. 63. We are looking for a charge q which, when placed at the origin, experiences Fnet 0, where F F F Fnet 1 2 3 . 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 F1 points towards –x, F2 points towards +x, and F3 points towards –x if q3 > 0 and points towards +x if q3 < 0. Therefore, with r1 = 8 m, r2 = 16 m and r3 = 24 m, we have 0 1 1 2 2 2 2 3 3 2 k q q r k q q r k q q r | | . Simplifying, this becomes 0 6 8 4 16 242 2 3 2 q 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. 64. Charge q1 = –80 10 –6 C is at the origin, and charge q2 = +40 10 –6 C is at x = 0.20 m. The force on q3 = +20 10 –6 C is due to the attractive and repulsive forces from q1 and q2, respectively. In symbols, F F F3 3 1 3 2 net , where 3 1 3 2 31 322 2 31 3 2 | | , | | . q q q q F k F k r r (a) In this case r31 = 0.40 m and r32 = 0.20 m, with 31F directed towards –x and 32F directed in the +x direction. Using the value of k in Eq. 21-5, we obtain 3 1 3 2 1 2 3 net 31 32 32 2 2 2 31 3 2 31 3 2 6 6 9 2 2 6 2 2 | | | |ˆ ˆ ˆ ˆi | | i i i 80 10 C 40 10 C ˆ(8.99 10 N m C )(20 10 C) i (0.40m) (0.20m) ˆ(89.9 N)i . q q q q q q F F F k k kq r r r r (b) In this case r31 = 0.80 m and r32 = 0.60 m, with 31F directed towards –x and F3 2 towards +x. Now we obtain 3 1 3 2 1 2 3 net 31 32 32 2 2 2 31 3 2 31 3 2 6 6 9 2 2 6 2 2 | | | |ˆ ˆ ˆ ˆi | | i i i 80 10 C 40 10 C ˆ(8.99 10 N m C )(20 10 C) i (0.80m) (0.60m) ˆ(2.50 N)i . q q q q q q F F F k k kq r r r r (c) Between the locations treated in parts (a) and (b), there must be one where F3 0net . Writing r31 = x and r32 = x – 0.20 m, we equate F3 1 and F3 2 , and after canceling common factors, arrive at 1 2 22 | | . 0.20 m q q x x This can be further simplified to 2 2 2 1 ( 0.20 m) 1 . | | 2 qx x q 65. 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) Eq. 21-1 gives 9 2 2 19 192 2 2 15 2 8.99 10 N m C (2(1.60 10 C))(90(1.60 10 C)) 5.1 10 N. (9.0 10 m) q F k r (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 a F m 51 10 4 167 10 7 7 10 2 27 28. . . . N kg m s2 c h 68. With F = meg, Eq. 21-1 leads to 2 9 2 2 192 2 231 8.99 10 N m C 1.60 10 C 9.11 10 kg 9.8m se ke y m g which leads to y = 5.1 m. We choose 5.1 my 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. 69. (a) If a (negative) charged particle is placed a distance x to the right of the +2q particle, then its attraction to the +2q particle will be exactly balanced by its repulsion from the –5q particle is we require 2 2 5 2 ( )L x x which is obtained by equating the Coulomb force magnitudes and then canceling common factors. Cross-multiplying and taking the square root, we obtain 2 5 x L x which can be rearranged to produce 1.72 2 / 5 1 L x L (b) The y coordinate of particle 3 is y = 0. 70. The net charge carried by John whose mass is m is roughly 23 19 5 0.0001 (90kg)(6.02 10 molecules mol)(18 electron proton pairs molecule) (1.6 10 C) 0.0001 0.018 kg mol 8.7 10 C, AmN Zeq M and the net charge carried by Mary is half of that. So the electrostatic force between them is estimated to be 5 2 9 2 2 18 22 2 (8.7 10 C) 8.99 10 N m C 4 10 N. 2 30m q q F k d Thus, the order of magnitude of the electrostatic force is 1810 N . 3. Since the magnitude of the electric field produced by a point charge q is given by 2 0| | / 4E q r , where r is the distance from the charge to the point where the field has magnitude E, the magnitude of the charge is 2 2 11 0 9 2 2 0.50m 2.0 N C 4 5.6 10 C. 8.99 10 N m C q r E 4. We find the charge magnitude |q| from E = |q|/4 0r 2: 2 2 10 0 9 2 2 1.00 N C 1.00m 4 1.11 10 C. 8.99 10 N m C q Er 5. Since the charge is uniformly distributed throughout a sphere, the electric field at the surface is exactly the same as it would be if the charge were all at the center. That is, the magnitude of the field is E q R4 0 2 where q is the magnitude of the total charge and R is the sphere radius. (a) The magnitude of the total charge is Ze, so E Ze R4 8 99 10 94 160 10 6 64 10 3 07 10 0 2 9 2 2 19 15 2 21 . . . . . N m C C m N C c hb gc h c h (b) The field is normal to the surface and since the charge is positive, it points outward from the surface. Thus, we obtain 2.72 1 2 5 L x L . (b) A sketch of the field lines is shown in the figure below: 8. (a) The individual magnitudes E1 and E2 are figured from Eq. 22-3, where the absolute value signs for q2 are unnecessary since this charge is positive. Whether we add the magnitudes or subtract them depends on if E1 is in the same, or opposite, direction as E2 . At points left of q1 (on the –x axis) the fields point in opposite directions, but there is no possibility of cancellation (zero net field) since E1 is everywhere bigger than E2 in this region. In the region between the charges (0 < x < L) both fields point leftward and there is no possibility of cancellation. At points to the right of q2 (where x > L), E1 points leftward and E2 points rightward so the net field in this range is net 2 1 ˆ| | | | iE E E . Although |q1| > q2 there is the possibility of Enet 0 since these points are closer to q2 than to q1. Thus, we look for the zero net field point in the x > L region: 1 2 1 2 22 0 0 | |1 1 | | | | 4 4 q q E E x x L which leads to 2 1 2 . | | 5 qx L x q 9. The x component of the electric field at the center of the square is given by 31 2 4 2 2 2 2 0 1 2 3 42 0 | || | | | | |1 cos 45 4 ( / 2) ( / 2) ( / 2) ( / 2) 1 1 1 | | | | | | | | 4 / 2 2 0. x qq q q E a a a a q q q q a Similarly, the y component of the electric field is 31 2 4 2 2 2 2 0 1 2 3 42 0 9 2 2 8 5 2 | || | | | | |1 cos 45 4 ( / 2) ( / 2) ( / 2) ( / 2) 1 1 1 | | | | | | | | 4 / 2 2 8.99 10 N m / C (2.0 10 C) 1 1.02 10 N/C. (0.050 m) / 2 2 y qq q q E a a a a q q q q a Thus, the electric field at the center of the square is 5ˆ ˆj (1.02 10 N/C)j.yE E | | | |E E1 2 ), and the net field (if there is any) should be along the y axis, with magnitude equal to E q d q d q d q d net j j F HG I KJ F HG I KJ 1 4 2 1 4 12 4 3 0 4 2 3 2 0 2 2b g which is seen to be zero. A rough sketch of the field lines is shown below: 10. We place the origin of our coordinate system at point P and orient our y axis in the direction of the q4 = –12q charge (passing through the q3 = +3q charge). The x axis is perpendicular to the y axis, and thus passes through the identical q1 = q2 = +5q charges. The individual magnitudes | |, | |, | |,E E E1 2 3 and | |E4 are figured from Eq. 22-3, where the absolute value signs for q1, q2, and q3 are unnecessary since those charges are positive (assuming q > 0). We note that the contribution from q1 cancels that of q2 (that is, 13. By symmetry we see the contributions from the two charges q1 = q2 = +e cancel each other, and we simply use Eq. 22-3 to compute magnitude of the field due to q3 = +2e. (a) The magnitude of the net electric field is net 2 22 0 0 0 19 9 2 2 6 2 1 2 1 2 1 4 | | 4 4 4( / 2) 4(1.60 10 C) (8.99 10 N m C ) 160 N/C. (6.00 10 m) e e e E r aa (b) This field points at 45.0°, counterclockwise from the x axis. 14. The field of each charge has magnitude 19 9 2 2 6 22 2 1.60 10 C (8.99 10 N m C ) 3.6 10 N C. (0.020 m)0.020m kq e E k r The directions are indicated in standard format below. We use the magnitude-angle notation (convenient if one is using a vector-capable calculator in polar mode) and write (starting with the proton on the left and moving around clockwise) the contributions to Enet as follows: E E E E E20 130 100 150 0b g b g b g b g b g. This yields 393 10 76 46. .c h , with the N/C unit understood. (a) The result above shows that the magnitude of the net electric field is 6 net| | 3.93 10 N/C.E (b) Similarly, the direction of Enet is –76.4 from the x axis. 15. (a) The electron ec is a distance r = z = 0.020 m away. Thus, 9 2 2 19 6 2 2 0 (8.99 10 N m C )(1.60 10 C) 3.60 10 N/C 4 (0.020 m) C e E r . (b) The horizontal components of the individual fields (due to the two es charges) cancel, and the vertical components add to give 9 2 2 19 s,net 2 2 3/ 2 2 2 3/ 2 0 6 2 2(8.99 10 N m C )(1.6 10 C)(0.020 m) 4 ( ) [(0.020 m) (0.020 m) ] 2.55 10 N/C . ez E R z (c) Calculation similar to that shown in part (a) now leads to a stronger field 43.60 10 N/CcE from the central charge. (d) The field due to the side charges may be obtained from calculation similar to that shown in part (b). The result is Es, net = 7.09 10 7 N/C. (e) Since Ec is inversely proportional to z 2, this is a simple result of the fact that z is now much smaller than in part (a). For the net effect due to the side charges, it is the “trigonometric factor” for the y component (here expressed as z/ r ) which shrinks almost linearly (as z decreases) for very small z, plus the fact that the x components cancel, which leads to the decreasing value of Es, net . 18. According to the problem statement, Eact is Eq. 22-5 (with z = 5d) act 2 2 2 0 0 0 160 4 (4.5 ) 4 (5.5 ) 9801 4 q q q E d d d and Eapprox is approx 3 2 0 0 2 2 4 (5 ) 125 4 qd q E d d . The ratio is Eapprox Eact = 0.9801 0.98. net 1 2 3/ 22 2 22 2 0 0 1 / 2 1 2 sin 2 4 4/ 2 / 2 / 2 q d qd E E d r d r d r For r d , we write [(d/2)2 + r2]3/2 r3 so the expression above reduces to net 3 0 1 | | . 4 qd E r (b) From the figure, it is clear that the net electric field at point P points in the j direction, or 90 from the +x axis. 19. (a) Consider the figure below. The magnitude of the net electric field at point P is 20. Referring to Eq. 22-6, we use the binomial expansion (see Appendix E) but keeping higher order terms than are shown in Eq. 22-7: E = q 4 o z 2 1 + d z + 3 4 d 2 z 2 + 1 2 d 3 z 3 + … 1 d z + 3 4 d 2 z 2 1 2 d 3 z 3 + … = q d 2 o z 3 + q d 3 4 o z 5 + … Therefore, in the terminology of the problem, Enext = q d 3/ 4 0z 5. 23. We use Eq. 22-3, assuming both charges are positive. At P, we have 1 2 left ring right ring 3/ 2 2 2 3/ 22 2 00 (2 ) 4 [(2 ) ]4 q R q R E E R RR R Simplifying, we obtain 3/ 2 1 2 2 2 0.506. 5 q q net, 2 2 0 0 9 2 2 12 2 2 1 2 2 | | 1 4| | 2 cos 45 4 4 (8.99 10 N m C )4(4.50 10 C) 20.6 N/C. (5.00 10 m) x q q E r r (b) By symmetry, the net field points vertically downward in the ĵ direction, or 90 counterclockwise from the +x axis. 24. Studying Sample Problem 22-3, we see that the field evaluated at the center of curvature due to a charged distribution on a circular arc is given by 0 sin 4 E r along the symmetry axis, with = q/r with in radians. In this problem, each charged quarter-circle produces a field of magnitude / 4 2 / 40 0 | | 1 1 2 2 | | | | sin . / 2 4 4 q q E r r r That produced by the positive quarter-circle points at – 45°, and that of the negative quarter-circle points at +45°. (a) The magnitude of the net field is 25. From symmetry, we see that the net field at P is twice the field caused by the upper semicircular charge q R (and that it points downward). Adapting the steps leading to Eq. 22-21, we find 90 net 2 2 900 0 ˆ ˆ2 j sin j. 4 q E R R (a) With R = 8.50 10 2 m and q = 1.50 10 8 C, net| | 23.8 N/C.E (b) The net electric field netE points in the ĵ direction, or 90 counterclockwise from the +x axis.