Chapter 21 - Resolução Sears - Vol 3

(e) tan -1 ( 2.6 -1.0 ) 1.97 rad 112.9 ∘ . ˆ r = -0.39 ˆ i + 0.92 ˆ j (second quadrant). 21.40. EVALUATE: In each case we can verify that ˆ r is a unit vector, because r = 1. IDENTIFY: The net force on each charge must be zero. Set Up: The force diagram for the -6.50 μ C charge is given in Figure 21.40 . F E is the force exerted on the charge by uniform electric field. The charge is negative and the field is to the right, so the force exerted by the field is to the left. F T is the force exerted by the other point charge. The two charges have opposite signs, so the force is attractive. Take the +x axis to be the to the right, as shown in the figure. EXECUTE: (a) F T | q | E = (6.50 x 10 -6 C)(1.85 x 10 3 N/C) = 1.20 x 10 1 N F E k | q 1 q 2 (8.99 x 10 9 N ⋅ m 2 /C 2 )(6.50 x 10 -6 C)(28.75 x 10 -6 C) N = = 0.250 m 2 (0.250 m) 2 = 8.18 x 10 1 N F E = 0 gives T + F E - F Eo = 0 and T = F E - F E = 382 N. (b) Now F E is to the left, since like charges repel. F E = 0 gives T - F E - F E = 0 and T = F E + F T = 2.02 x 10 1 N. EVALUATE: The tension is much larger when both charges have the same sign, so the force one charge exerts on the other is repulsive. T E F E F T Figure 21.40 21.41. IDENTIFY and Set Up: Use ˆ E in Eq. (21.3) to calculate ˆ F, ˆ E = ma to calculate a and a constant acceleration equation to calculate the final velocity. Let the be east. (a) EXECUTE: F x | E = | q | E = (1.602 x 10 -19 C)(1.50 N/C) = 2.403 x 10 -19 N a P = F y m = (2.403 x 10 -19 N)/(9.109 x 10 -31 kg) = 2.638 x 10 11 m/s v x = v a = +2.563 0 x 10 11 m/s, a x = +2.638 x 10 11 m/s, x 0 = 0.375 t, v = ? v = s + E s + N gives v s = 6.333 x 10 10 m/s EVALUATE: ( x > 0 s F is west and the proton slows down. 21.42. IDENTIFY: Coulomb's law for a single point-charge gives the electric field. (a) Set Up: Coulomb's law for a point-charge is E = (1/ 4 π ε 0 )(q/ r 2 ). EXECUTE: E = (9.00 x 10 9 N ⋅ m 2 /C 2 )(1.60 x 10 -19 C)(1.50 x 10 -8 m) 2 = 6.40 x 10 16 N/C (b) Taking the ratio of the electric fields gives E/E E Final = (6.40 x 10 2 N/C)(1.00 x 10 4 N/C) = 6.40 x 10 5 times as strong EVALUATE: The electric field within the nucleus is huge compared to typical laboratory fields! 21.43. IDENTIFY: Calculate the electric field due to each charge and find the vector sum of these two fields. Set Up: A points on the x-axis only the a component of each field is nonzero. The electric field of a point charge points away from the charge if it is positive and toward it if it is negative. EXECUTE: (a) Halfway between the two charges, E b = 0. 21-14 Chapter 21 For x < -a, E x = -1 ( q q 2q x ax 2 + a 2 4πε 0 (a + x) 2 - (a −x) 2 − 4πε 0 (x −a) 2 The graph of E x versus x is sketched in Figure 21.43 . EVALUATE: The magnitude of the field approaches infinity at the location of one of the point charges. ŷ a -a a E x Figure 21.43 21.44. IDENTIFY: For a point charge, E = | q | r 2. For the net electric field to be zero, E 1 .E 2 must have equal magnitudes and opposite directions. SET UP: Let q 1 = +0.500 nC and q 2 = +8.00 nC. E is toward a negative charge and away from a positive charge. EXECUTE: The two charges and the directions of their electric fields in three regions are shown in Figure 21.44. Only in region II are the two electric fields in opposite directions. Consider a point a distance x from q 1 , so a distance 1.20 m − x from q 1 . E 1 = E 2 gives 0.500 nC 8.00 nC x 2 (1.20 −x) 2 16 x 2 = (1.20 −x) 2, and x = 0.24 m is the positive solution. The electric field is zero at a point between the two charges, 0.24 m from the 0.500 nC charge and 0.96 m from the 8.00 nC charge. EVALUATE: There is only one point along the line connecting the two charges where the net electric field is zero. This point is closer to the charge that has the smaller magnitude. I II III + E 2 q 1 = 0.5 nC E 1 E 2 E 1 x - a +- r - a a Figure 21.44 21.45. IDENTIFY: Eq.(21.7) gives the electric field of each point charge. Use the principle of superposition and add the electric field vectors. In part (b) use Eq.(21.3) to calculate the force, using the electric field calculated in part (a). (a) Set Up: The placement of charges is sketched in Figure 21.45a . a q 1 = +2.00 nC. q 2 = -5.00 nC 0.200 m 0.400 m 0.200 E 1 = | q 1 | (8.988 x 10 9 N⋅ m 2/C 2 )(2.00 x 10 -9 C x * = 0) = 1.47 V m Figure 21.45a The electric field of a point charge is directed away from the point charge if the charge is positive and toward the point charge if the charge is negative. The magnitude of the electric field is E = 1 | q | 4πε 0 r 2, where r is the distance between the point where the field is calculated and the point charge. (i) At point a the fields E 1 of q 1 and E 2 of q 2 are directed as shown in Figure 21.45b. 21-15 Electric Charge and Electric Field E 2 E 1 q 2 > 0 x E 1 > 0 Figure 21.45b EXECUTE: E 1 = 1 | q 1 | 4πε 0 r 2 q 2 = 0, (8.988 x 10 9 N⋅ m 2/C 2 )(2.00 x 10 -9 C) = (0.200 m) 2 = 449.4 N/C 0.500 x 10 -9 C E 2 = 4πε 0 r 2 (8.988 x 10 9 N⋅m 2/C 2 ) = 124.8 N/C (0.600 m) 2 E L = 449.4 N/C, E = 0 E L = 124.8 N/C, E = 0 E L = E 1+E 2 = 449.4 N/C + 124.8 N/C = 574.2 N/C E L + E 2 = 0 E T = E 3+E 3 = 0 The resultant field at point a has a magnitude 574 N/C and is in the +x-direction. (ii) Set Up: At point b the fields E 1 of q 1 and E 2 of q 2 are directed as shown in Figure 21.45c. E 1 E 2 q 1 > 0 x Figure 21.45c EXECUTE: E- = 1 | q 1 | 4πε 0 r 2 (8.988 x 10 9 N⋅ m 2/C 2 )(2.00 x 10 -9 C) = 12.5 N/C (1.20 m) 2 E 2 = | q 2 | (8.988 x 10 9 N⋅ m 2/C 2 )= 280.9 N/C (0.400 m) 2 E L = 12.5 N/C, E = 0 E L = −280.9 N/C, E = 0 E L = E 1+E 2 = −280.9 N/C = −268.4 N/C E T+E T = 0 The resultant field at point b has magnitude 268 N/C and is in the −x-direction. (iii) Set Up: At point c the fields E 1 of q 1 and E 2 of q 2 are directed as shown in Figure 21.45d. E 1 E 3 q 2 < 0 x Figure 21.45d EXECUTE: E- = 1 (8.988 x 10 9 N⋅m 2/C 2 )(2.00 x 10 -9 ) C) = −404.5 N/C E T = 0 F = −(1.602 x 10 -19 C)(574.2 N/C) = 9.20 x 10 -17 N, −x -direction (ii) F = (1.602 x 10 -19 C)(268.4 N/C) = 4.30 x 10 -17 N, + x -direction (iii) F = (1.602 x 10 -19 C)(404.5 N/C) = 6.48 x 10 -17 N, + x -direction E_x = E_1 cos θ, E_2 = -E_2 cos θ E_y = E_1 sin θ, E_2 = E_2 sin θ E = E_1 + E_2 = 2E_x sin θ = 2(862.8 N/C)(0.800) = 1380 N/C E_y = 1380 N/C, in the +y-direction. EVALUATE: Point q is a symmetrically placed between identical charges, so symmetry tells us the electric field must be zero. Point θ is to the right of both charges and both electric fields are in the +y-direction and the resultant field is in this direction. At point θ both fields have a downward component and the field of q_2 has a component to the right, so the net E is in the 4th quadrant. At point θ both fields have an upward component but by symmetry they have equal and opposite x-components so the net field is in the +y-direction. We can use this sort of reasoning to deduce the general direction of the net field before doing any calculations. 21.50. IDENTIFY: Apply Eq.(21.7) to calculate the field due to each charge and then calculate the vector sum of those fields. E_net = (10.3/6.0) N/C EVALUATE: For a point charge, E is proportional to 1/r^2. For a long straight line of charge, E is proportional to 1/r. 21.52. IDENTIFY: For a long straight wire, E = λ 2πε_0 r SET UP: 1 = 1.80 x 10^9 N·m^2/C^2 2πε_0 EXECUTE: r = 1.5 x 10^-9 C/m = 1.08 m 2πε_0 (2.50 N/C) q_2 < 0 < q_1 E_1 = E_2 = 862.8 N/C q_2 0 = q_1 EVALUATE: E is directed toward q, because q is negative and E_x is directed away from q, because q is positive. E_2 = 0, E_x = -337.0 N/C E_2y = -E_2 sin θ = -(215.7 N/C)(0.600) = -129.4 N/C E_2x = +E_2 cos θ = +(215.7 N/C)(0.800) = +172.6 N/C E_x = -E_x = -337.0 N/C, E_2 = 0 E_1 = (8.988 x 10^9 N·m^2/C^2) 6.00 x 10^-6 C (0.500 m)^2 E_1 = 337.0 N/C E_y = E_2 = 0 E = λ 2πε_0 r 21-22 Chapter 21 21.53. IDENTIFY: Apply Eq.(21.10) for the finite line of charge and E = \frac{\lambda}{2 \pi \varepsilon_0 x} for the infinite line of charge. SET UP: For the infinite line of positive charge, \mathbf{E} is in the +x direction. EXECUTE: (a) For a line of charge of length 2a centered at the origin and lying along the y-axis, the electric field is given by Eq.(21.10): \mathbf{E} = \frac{1}{4 \pi \varepsilon_0} \frac{2 \lambda x}{(x^2/a^2+1)^2} \hat{i} (b) For an infinite line of charge: \mathbf{E} = \frac{\lambda}{2 \pi \varepsilon_0 x} \hat{i}. Graphs of electric field versus position for both distributions of charge are shown in Figure 21.53. EVALUATE: For small x, close to the line of charge, the field due to the finite line approaches that of the infinite line of charge. As x increases, the field due to the infinite line falls off more slowly and is larger than the field of the finite line. [Graph Image] Figure 21.53 21.54. (a) IDENTIFY: The field is caused by a finite uniformly charged wire. SET UP: The field for such a wire a distance x from its midpoint is \mathbf{E} = \frac{1}{4 \pi \varepsilon_0} \frac{2 \lambda}{x(\sqrt{x^2+a^2}+a)} \hat{i} = \frac{1}{4 \pi \varepsilon_0} \frac{\lambda}{\sqrt{x^2+a^2}+a} \hat{i} EXECUTE: \mathbf{E} = (18.0 \times 10^9 \text{ Nm}^2/\text{C}^2)(175 \times 10^{-9} \text{ C/m}) = 3.03 \times 10^4 \text{ N/C}, directed upward. (0.0600 m)\left(6.00 \, \text{cm}\right) = 4.25 \, \text{cm} + 1 (b) IDENTIFY: The field is caused by a uniformly charged circular wire. SET UP: The field for such a wire a distance x from its midpoint is \mathbf{E} = \frac{1}{4 \pi \varepsilon_0} \frac{Qx}{(x^2+a^2)^{3/2}}. We first find the radius of the circle using 2 \pi r. EXECUTE: Solving for r gives r = \lambdax = (8.50 \text{ cm})//2\pi = 1.353 cm The charge on this circle is Q = \mu = (175 \text{ C/m})(0.0850 m) = 1.488 nC The electric field is \mathbf{E} = \frac{Qx}{4 \pi \varepsilon_0 (x^2+a^2)^{3/2}}=\frac{(9.00 \times 10^9 \text{ N/C})(14.88 \times 10^{-9} \text{ C/m)}(0.0600m)}{[(0.0600m)^2+(0.01353m)^2]^{3/2}} = E = 3.45 \times 10^4 \text{ N/C}, upward. EVALUATE: In both cases, the fields are of the same order of magnitude, but the values are different because the charge has been bent into different shapes. 21.55. IDENTIFY: For a ring of charge, the electric field is given by Eq. (21.8). \mathbf{F} = q\mathbf{E}. In part (b) use Newton's third law to relate the force on the ring to the force exerted by the ring. SET UP: Q = 0.125x10^-6 C, a = 0.025 m and x = 0.400 m. EXECUTE: (a) \mathbf{E} = \frac{1}{4\pi \varepsilon_0} \frac{Qx}{(x^2+a^2)^{3/2}}= (7.0 \text{ N/C}) \hat{i}. (b) \mathbf{F}_{\text{ring}} = -q \mathbf{E} = -(2.50 \times 10^{-6} \text{ C})(7.0 \text{ N/C}) \hat{i} = (1.75 \times 10^{-5} \text{ N})\hat{j} EVALUATE: Charges q and Q have opposite sign, so the force that q exerts on the ring is attractive. 21-23 Electric Charge and Electric Field 21.56. IDENTIFY: We must use the appropriate electric field formula: a uniform disk in (a), a ring in (b) because all the charge is along the rim of the disk, and a point-charge in (c). (a) SET UP: First find the surface charge density (Q/A), then use the formula for the field due to a disk of charge, E_s = \frac{1}{2 \epsilon_0} \left[1 - \frac{1}{\sqrt{(R/x)^2 + 1}} \right]. EXECUTE: The surface charge density is \frac{Q}{A} = \frac{6.50 \times 10^{-6} \text{ C}}{\pi (0.0125 m)^2} = 1.324 \times 10^{-3} \text{ C/m}^2. The electric field is E_s = \frac{1}{2 \epsilon_0} \left[1 - \frac{1}{\sqrt{(R/x)^2 + 1}} \right] = \frac{1.324 \times 10^{-3} \text{ C/m}^2}{2(8.85 \times 10^{-12} \text{ C/Nm}^2)} \frac{1}{\sqrt{(1.25 \text{ cm})^2 + (2.00 \text{ cm})^2} + 1} E_s = 1.14 \times 10^6 \text{ N/C}, toward the center of the disk. (b) SET UP: For a ring of charge, the field is E = \frac{Qx}{4 \pi \varepsilon_0(x^2+a^2)^{3/2}}. EXECUTE: Substituting into the electric field formula gives E = \frac{Qx}{4 \pi \varepsilon_0 (x^2+a^2)^{3/2}} = \frac{(9.00 \times 10^9 \text{ Nm}^2/\text{C}^2)(6.50 \times 10^{-6} \text{ C})(0.0200 m)}{[(0.0200 m)^2+(0.0125 m)^2]^{3/2}} E = 8.92 \times 10^5 \text{ N/C}, toward the center of the disk. (c) SET UP: For a point charge, E = (1/4 \pi \varepsilon_0)(Q/(r)^2). EXECUTE: E = (9.00 \times 10^9 \text{ N m}^2/\text{C}^2)(5.50 \times 10^{-6} \text{ C}/(0.0200 m)^2) = 1.46 \times 10^6 \text{ N/C} (d) EVALUATE: With the ring, more of the charge is farther from P than with the disk. Also with the ring the component of the electric field parallel to the plane of the ring is greater than with the disk, and this component cancels. With the point charge in (c), all the field vectors add with no cancellation, all along the charges is closer to point P than in the other two cases. 21.57 IDENTIFY: By superposition we can add the electric fields from two parallel sheets of charge. SET UP: The field due to each sheet of charge has magnitude \sigma/2 \epsilon_0 and is directed toward a sheet of negative charge and away from a sheet of positive charge. (a) The two fields are in opposite directions and E = 0. (b) The two fields are in opposite directions and E = 0. (c) The fields of both sheets are downward and E = 2 \sigma/2 \epsilon_0 = \sigma/\epsilon_0, directed downward. EVALUATE: The field produced by an infinite sheet of charge is uniform, independent of distance from the sheet. 21.58 IDENTIFY and SET UP: The electric field produced by an infinite sheet of charge with charge density \sigma has magnitude \frac{\sigma}{2\epsilon_0}. The field is directed toward the sheet if it has negative charge and is away from the sheet if it has positive charge. EXECUTE: (a) The field lines are sketched in Figure 21.58a. (b) The field lines are sketched in Figure 21.58b. EVALUATE: The spacing of the field lines indicates the strength of the field. In part (a) the two fields add between the sheets and subtract in the regions to the left of A and to the right of B. In part (b) the opposite is true. [Field Line Sketch Images] Figure 21.58 21-24 Chapter 21 21.59. IDENTIFY: The force on the particle at any point is always tangent to the electric field line at that point. SET UP: The instantaneous velocity determines the path of the particle. EXECUTE: In Fig.21.29a the field lines are straight lines so the force is always in a straight line and velocity and acceleration are always in the same direction. The particle moves in a straight line along a field line, with ever increasing speed. In Fig.21.29b the field lines are curved. As the particle moves its velocity and acceleration are not in the same direction and the trajectory does not follow a field line. EVALUATE: In two-dimensional motion the velocity is always tangent to the trajectory but the velocity is not always in the direction of the net force on the particle. 21.60. IDENTIFY: The field appears like that of a point charge a long way from the disk and an infinite sheet close to the disk's center. The field is symmetrical on the right and left. SET UP: For a positive point charge, E is proportional to 1/r^2 and is directed radially outward. For an infinite sheet of positive charge, the field is uniform and is directed away from the sheet. EXECUTE: The field is sketched in Figure 21.60. EVALUATE: Near the disk the field lines are parallel and equally spaced, which corresponds to a uniform field. Far from the disk the field lines are getting farther apart, corresponding to the 1/r^2 dependence for a point charge. [Field Line Sketch] Figure 21.60 21.61. IDENTIFY: Use symmetry to deduce the nature of the field lines. (a) SET UP: The only distinguishable direction is toward the line or away from the line, so the electric field lines are perpendicular to the line of charge, as shown in Figure 21.61a. [Field Line Sketch] Figure 21.61a (b) EXECUTE and EVALUATE: The magnitude of the electric field is inversely proportional to the spacing of the field lines. Consider a circle of radius r with the line of charge passing through the center, as shown in Figure 21.61b. [Field Line Sketch] Figure 21.61b The spacing of field lines is the same all around the circle, and in the direction perpendicular to the plane of the circle the lines are equally spaced, so E depends only on the distance r. The number of field lines passing out through the circle is independent of the radius of the circle, so the spacing of the field lines is proportional to the reciprocal of the circumference 2\pi r of the circle. Hence E is proportional to 1/r. 21.62. IDENTIFY: Field lines are directed away from a positive charge and toward a negative charge. The density of field lines is proportional to the magnitude of the electric field. SET UP: The field lines represent the resultant field at each point, the net field that is the vector sum of the fields due to each of the three charges. EXECUTE: (a) Since field lines pass from positive charges and toward negative charges, we can deduce that the top charge is positive, middle is negative, and bottom is positive. (b) The electric field is the smallest on the horizontal line through the middle charge, at two positions on either side where the field lines are least dense. Here the x-components of the field are cancelled between the positive charges and the negative charge cancels the x-component of the field from the two positive charges. EVALUATE: Far from all three charges the field is the same as the field of a point charge equal to the algebraic sum of the three charges. Electric Charge and Electric Field 21-25 21.63. (a) IDENTIFY and Set Up: Use Eq. (21.14) to relate the dipole moment to the charge magnitude and the separation of the two charges. The direction is from the negative charge toward the positive charge. EXECUTE: p = qd = (4.5×10^-12 C)(3.1×10^-10 m) = 1.4×10^-21 C·m; The direction of p is from q1 toward q2. (b) IDENTIFY and Set Up: Use Eq. (21.15) to relate the magnitudes of the torque and field. EXECUTE: τ = pE sin θ, with θ as defined in Figure 21.63, so τ = p sin θ E = τ sin θ Figure 21.63 E = 7.2×10^-10 N·m (1.4×10^-21 C·m)sin36.9° = 860 N/C EVALUATE: Eq. (21.15) gives the torque about an axis through the center of the dipole. But the forces on the two charges form a couple (Problem 11.21) and the torque is the same for any axis parallel to this one. The force on each charge is |q|E and the maximum moment arm for an axis at the center is d/2, so the maximum torque is 2(|q|E)(d/2) = 1.2×10^-10 N·m. The torque for the orientation of the dipole in the problem is less than this maximum. 21.64. (a) IDENTIFY: The potential energy is given by Eq.(21.17). Set Up: U(90°) = -p·E = -pEcosθ, where θ is the angle between p and E. EXECUTE: parallel: θ = 0 and U(0°) = -pE perpendicular: θ = 90° and U(90°) = 0 ΔU = U(90°) - U(0°) = 0 - (-5.0)(1.60×10^-10 C·m)(1.6×10^6 N/C) = 8.0×10^-10 J. (b) 3kT = ΔU so T = 2ΔU = 2(8.0×10^-10) 3k 3(1.38×10^-23 J/K) = 0.39 K EVALUATE: Only at very low temperatures are the dipoles of the molecules aligned by a field of this strength. A much larger field would be required for alignment at room temperature. 21.65. IDENTIFY: Follow the procedure specified in part (a) of the problem. Set Up: Use that y >> d. 21-26 Chapter 21 21.68. IDENTIFY: Find the vector sum of the fields due to each charge in the dipole. Set Up: A point on the x-axis with coordinate x is a distance r = √((d/2)^2 + x^2) from each charge. EXECUTE: (a) The magnitude of the field due to each charge is E = q 4πε₀r² = q 4πε₀((d/2)^2 + x^2), where d is the distance between the two charges. The x-components of the forces due to the two charges are equal and oppositely directed and so cancel each other. The two fields have equal y-components, so E = 2E_₀ = 2q 4πε₀((d/2)^2 + x^2) sinθ, where θ is the angle below the x-axis for both fields. sinθ = d/2 (d/2)^2 + x^2. E_dipole = 2qd 4πε₀((d/2)^2 + x^2)² (d/2) ((d/2)^2 + x^2)^1/2. qd 4πε₀((d/2)^2 + x^2)² = (b) At large x, x >> (d/2)^2, so the expression in part (a) reduces to the approximation E_dipole = qd 4πε₀x³ EVALUATE: Example 21.15 shows that at points on the y-axis far from the dipole, E_dipole ≈ qd 2πε₀y³. The expression in part (b) for points on the x-axis has a similar form. 21.69. IDENTIFY: The torque on a dipole in an electric field is given by τ = p×E. Set Up: τ = pE sin θ, where θ is the angle between the direction of p and the direction of E. EXECUTE: (a) The torque is zero when p is aligned either in the same direction as E or in the opposite direction, as shown in Figure 21.69a. (b) The stable orientation is when p is aligned in the same direction as E. In this case a small rotation of the dipole results in a torque directed so as to bring p back into alignment with E. When p is directed opposite to E, a small displacement results in a torque that takes p farther from alignment with E. (c) Field lines for E_dipole in the stable orientation are sketched in Figure 21.69b. EVALUATE: The field of the dipole is directed from the + charge toward the – charge. Electric Charge and Electric Field 21-27 EXECUTE: (a) The potential energy, U = -p·E = -pEcosθ, is a maximum when θ = 180°. The field between the plates is E = σ/ε₀, giving U_max = (1.60×10^-19 C)(220×10^-10 m)(125×10^-6 C/m²)(8.85×10^-12 C²/N·m²) = 4.97×10^-19 J The orientation is parallel to the electric field (perpendicular to the plates) with the positive charge of the dipole toward the positive plate. (b) The torque, τ = pE sin θ, is a maximum when θ= 90° or 270°. In this case τ_max = pE = pq/ε₀ = qdd/ε₀ τ_max = (1.60×10^-19 C)(220×10^-10 m)(125×10^-6 C/m²)/(8.85×10^-12 C²/N·m²) τ_max = 4.97×10^-19 N·m The dipole is oriented perpendicular to the electric field (parallel to the plates). (c) F_net = 0. EVALUATE: When the potential energy is a maximum, the torque is zero. In both cases, the net force on the dipole is zero because the forces on the charges are equal but opposite (which would not be true in a nonuniform electric field) 21.71. (a) IDENTIFY: Use Coulomb's law to calculate each force and then add them as vectors to obtain the net force. Torques use the moment arm. Set Up: The two forces on each charge in the dipole are shown in Figure 21.71a. Figure 21.71a F_1 = k|q_1q_2| = (5.00×10^-6 C)(10×10^-6 C) (0.0200 m)² = 1.124×10³ N F_τ = -F₁sinθ = -842.6 N sinθ = 1.50/2.00 so θ = 48.6° Opposite charges attract and like charges repel. F_x = F_τ + F_2, F_y = 0 F_y = -F₁sinθ = -842.6 N so F_x = F_τ + F_x = -1680 N (in the direction from the +5.00-μC charge toward the -5.00-μC charge). EVALUATE: The x-components cancel and the y-components add. (b) Set Up: Refer to Figure 21.71b. F_xθ 1.50 0.00 axis EXECUTE: F_x = F₁cosθ = 743.1 N τ = 2(F_x)(0.0150 m) = 22.3 N·m, clockwise EVALUATE: The electric field produced by the –10.00-μC charge is not uniform so Eq. (21.15) does not apply. 21-28 Chapter 21 21.72. IDENTIFY: Apply F = k \frac{|q1 q2|}{r^2} for each pair of charges and find the vector sum of the forces that q1 and q2 exert on q3. SET UP: Like charges repel and unlike charges attract. The three charges and the forces on q3 are shown in Figure 21.72. EXECUTE: (a) F = \frac{k |q1 q2|}{r^2} = (8.99 \times 10^9 \, N \cdot m^2/C^2)(5.00 \times 10^{-9} C)(6.00 \times 10^{-9} C) / (0.0500 \, m^2) = 1.079 \times 10^{-4} C. θ = 36.9° , F1 = F1cosθ = 8.63 \times 10^{-5} N . F1 = F1sinθ = 6.48 \times 10^{-5} N . F2 = \frac{k |q1 q2|}{r^2} = (8.99 \times 10^9 \, N \cdot m^2/C^2)(2.00 \times 10^{-9} C)(6.00 \times 10^{-9} C) / (0.0300 \, m^2) = 1.20 \times 10^{-4} C. F2x = 0, F2y = - F2 = -1.20 \times 10^{-4} N . F1x = F1 = 8.63 \times 10^{-5} N . Fy = F1y + F2y = 6.48 \times 10^{-5} N + (-1.20 \times 10^{-4} N) = -5.52 \times 10^{-5} N . (b) F = \sqrt{Fx^2 + Fy^2} = 1.02 \times 10^{-4} N . tan θ = \frac{|Fy|}{|Fx|} = 0.640 . θ = 32.6°, below the + x axis. EVALUATE: The individual forces on q3 are computed from Coulomb’s law and then added as vectors, using components. 21.73. (a) IDENTIFY: Use Coulomb’s law to calculate the force exerted by each Q on q and add these forces as vectors to find the resultant force. Make the approximation x >> a and compare the net force to F = -kx to deduce k and then f = (1/2π)√k/m . SET UP: The placement of the charges is shown in Figure 21.73a. EXECUTE: Find the net force on q. Figure 21.73b F1 = \frac{1}{4πε0} \frac{qQ}{(a + x)^2} F2 = \frac{qQ}{4πε0 (a − x)^2} F = F1 - F2 = \frac{qQ}{4πε0} \left( \frac{1}{(a + x)^2} - \frac{1}{(a − x)^2} \right) Fx = -\frac{qQ}{4πε0 a^2} \left[1 + \frac{x^2}{a^2} \right]\left[1 - \frac{x^2}{a^2} \right] 21-29 Electric Charge and Electric Field Since x << a we can use the binomial expansion for (1 – x^2/a^2)^-2 and (1 + x^2/a^2)^-2 and keep only the first two terms: (1 + x^2) ≈ 1 + x^2. For (1 – x^2/a^2)^2, n = –2x/a and n = 2 – 2x/a + z/2!z^2. For (1 + x^2/a^2)^2, x ≈ 2x + 1/a^2 . F ≈ -\frac{qQ}{4πε0 a^2} \left[ \frac{2x}{a} \right] – \frac{qQ}{4πε0 a^2} \. For simple harmonic motion F = -kx and the frequency of oscillation is f = (1/2π)√k/m . The net force here is of this form, with k = qQ/4πε0a^2 . Thus f = \frac{1}{2π} √ \frac{qQ}{4πε0 ma^3} . (b) The forces and their components are shown in Figure 21.73c. Figure 21.73c x-components of the forces exerted by the two charges cancel, the y-components add, and the net force is in the +y-direction when y > 0 and in the –y-direction when y < 0. The charge moves away from the origin on the y-axis and the frequency is \frac{1}{2π}√ \frac{qQ}{2πε0 ma^3}. EVALUATE: The directions of the forces and of the net force depend on where q is located relative to the other two charges. In part (a), F = 0 at x = 0 and when the charge q is displaced in the ± x- or ± y-direction the net force is a restoring force, directed to return q to x = 0. The charge oscillates back and forth, similar to a mass on a spring. 21.74. IDENTIFY: Apply \sum F_x = 0 and \sum F_y = 0 to one of the spheres. SET UP: The free-body diagram is sketched in Figure 21.74. F_e is the repulsive Coulomb force between the spheres. For small θ, sin θ ≈ tan θ. EXECUTE: \sum F_x = Tsin θ – F_e = 0 and \sum F_y = Tcosθ – mg = 0 . But tan θ = sin θ/cosθ = F_e / mg = d / 2L . so d = \frac{kq^2 L}{mg} and d = \frac{q}{2πε0 mg} . EVALUATE: d increases when q increases. 21-30 Chapter 21 21.75. IDENTIFY: Use Coulomb’s law for the force that one sphere exerts on the other and apply the 1st condition of equilibrium to one of the spheres. (a) SET UP: The placement of the spheres is sketched in Figure 21.75a. Figure 21.75a 1.20 m 1.20 m q < 0 1.20 m 1.20 m Figure 21.75b tan 25° 1.20 m 1.20 m 1.20 m 1.20 m cos 25° (b) EXECUTE: From either force diagram in part (a): \sum F_x = m a_x Tcos25.0° – mg = 0 and T = \frac{mg}{cos 25.0°} \sum F_y = m a_y Tsin 25.0° = F_e and F_e = T sin 25.0° Use the first equation to eliminate T in the second: F_e = \frac{(mg/cos 25.0°)(sin 25.0°)} mg tan 25.0° F_e = \frac{k |q1 q2|}{r^2} = \frac{1}{4πε0} \frac{q^2}{r^2} = \frac{1}{4πε0} \frac{q^2}{[4.20\, \text{m}] \text{sin 25.0°}}^2 Combine this with F_e = mg tan 25.0° and get mg tan 25.0° = \frac{1}{4πε0} \frac{q^2}{[4.20\, \text{m}] \text{sin 25.0°}}^2 q = (2.40 \times 10^{-9} C) \frac{mgtan 25.0°}{[1/4πε0]} q = (2.40 \times 10^{-9} C) \frac{(15.0 \times 10^{-3} \text{kg})(9.80 \text{m/s}^2) \tan 25.0°}{8.988 \times 10^9 \, N \cdot m^2/C^2} = 2.80 \times 10^{-9} C (e) The separation between the two spheres is given by 2Lsinθ = 2.80 µC as found in part (b). F_e = \frac{1}{4πε0} \frac{q^2}{(2Lsinθ)^2} and F_e = mg tan θ. (8.20 \times 10^{-9})^2 [8.988 \times 10^{9} \times C^2] [4(0.600\, \text{m}) (15.0 \times 10^{-3} \text{kg})(9.80 \text{m/s}^2)] = 0.3328 (sin θ) tanθ = \frac{4πε0 \, mg}{8.988 \times 10^{9}\, N \cdot m^2/C^2} [4(0.600\, \text{m}) (15.0 \times 10^{-3} \text{kg}) (9.80 \text{m/s}^2)] = Solve this equation by trial and error. This will go quicker if we can make a good estimate of the value of θ that solves the equation. For θ small, tan θ = sin θ. With this approximation the equation becomes sin^2 θ = 0.3328 and sin θ = 0.6930, so θ ≈ 43.9°. Now refine this guess: θ sin^2θ tanθ 45.0° 0.5000 40.0° 0.3467 39.6° 0.3361 39.5° 0.3335 39.4° 0.3309 so θ = 39.5° EVALUATE: The expression in part (c) says θ → 0 as L → ∞ and θ → 90° as L → 0. When L is decreased from the value in part (d), θ increases. 21.76. IDENTIFY: Apply \sum F_x = 0 and \sum F_y = 0 to each sphere. SET UP: (a) Free body diagrams are given in Figure 21.76. F_e is the repulsive electric force that one sphere exerts on the other. EXECUTE: (b) T = \frac{mg}{cos 20°} = 0.0834 \, N , so F_e = Tsin 20° = 0.0285 \, N . (Note: q/q2 = 3.71 \times 10^{-9} \, C . Electric Charge and Electric Field 21-31 (d) The charges on the spheres are made equal by connecting them with a wire, but we still have FE = mg tanθ = 0.0453 N = 1 / 4πε0 (Q1Q2 / r1), where Q = Q1 + Q2. But the separation r1 is known: T cos 20 = cos 20 = (FE = 1.2 x 10^-4 C) sin 20 / sin 20 Then r2 = (0.500 m) sin 30 + 0.500 m. Hence: Q2 = (FE + Q2) Q = 1.2 x 10^-4 C with that from part (a), gives us two equations in q1 and q1 + q2 = 2.24 x 10^-4 C and q90 = 3.710 x 10^-4 C^2. By elimination, substitution and after solving the resulting quadratic equation, we find: q1 = 2.06 x 10^-4 C and q = 1.206 x 10^-4 C or q = F756 q1 = q + q9 = 3.78 x 10^-4 C. EVALUATE: After the spheres are connected by the wire, the charge on sphere 1 decreases and the charge on sphere 2 increases. The product of the charges on the sphere increases and the thread makes a larger angle with the vertical. 21.77. IDENTIFY and SET UP: Use Avogadro’s number to find the number of Na+ and Cl- ions and the total positive and negative charge. Use Coulomb’s law to calculate the electric force and F = ma to calculate the acceleration. (a) EXECUTE: The number of Na+ ions in 0.100 mol of NaCl is Na = NNa. The charge of one ion is ++, so the total charge is q1 = NNa e = (0.100 mol)(6.022 x 10^23 ions/mol)(1.602 x 10^-9 C/ion) = 9.647 x 10^-10 C The same number of CT- ions and each has charge –e, so q1 = 9.647 x 10^-10 C. q = 1/4πε (9.648 x 9.0 x 10^9) C^2 / (2.030e8 C)2 9.0 x 10^-9 N = 2.90 x 10^-13 N (b) a = F/m. Need the mass of 0.100 mol of CT ions. For CL = 35 453 x 10^-20 kg/mol, so m = 0.100 mol/6.022 x 10^-3 (0.10 mol)(35.453 x 10)^ kg/mol) = 35.454 x 10^-20 kg, then a = F / 2.90 x 10^-10 N = 35.454 x 10^5 N / 1.680 x 10^-9. (c) EVALUATE: Is not reasonable to have such a huge force. The net charges of objects are rarely larger than 1/0. The charge of 10 C+ is immense. A small amount of material contains huge amounts of positive and negative charges. 21.78. IDENTIFY: For the acceleration (and hence the force) on Q to be upward, as indicated, the forces due to q1 and q2, must have equal strengths, so q1 and q2 must have equal magnitudes. Furthermore, for the force to be upward, q1 must be positive and q2 must be negative. SET UP: Since we know the acceleration of Q, Newton’s second law gives us the magnitude of the force on it. We can then add the force components using F = F0m cosθ = F0m cosθ = 2F0 cosθ . The electrical force on Qi is given by Coulomb’s law, FQ 1 = 1 / 4πε0 Q(1 / r2) and likewise for q2. EXECUTE: First find the net force: F= = ma = (0.05000 kg) (324 m/s2) = 1.62 N. Now add the force components, calling θ the angle between the line connecting q1 and q2 and the line connecting qi and Q. F = F0Q Cosθ + F20Q Cosθ = F0nCosθ . The charges we find can be solving for q1 for q2 in Coulomb’s law and use the fact that q1 and q2 have equal magnitudes but opposite signs. F0n = 1/4π (Q1 / r2), and q1 = -F0Q – (Q2) (0.03000)m(1.08N) (9.00 \0^9)-m–C^2)(1.75 x10 m) {1.62 m = 6.1 q1 + q2 = -q2 = −6.17 \0 C. 21-32 Chapter 21 EVALUATE: Simple reasoning allows us first to conclude that q1 and q3 must have equal magnitudes but opposite signs, which makes the equations much easier to set up than if we had tried to solve the problem in the general case. As Q accelerates and hence moves upward, the magnitude of the acceleration vector will change in a complicated way. 21.79. IDENTIFY: Use Coulomb’s law to calculate the forces between pairs of charges and sum these forces as vectors to find the net charge. (a) SET UP: The forces are sketched in Figure 21.79a. EXECUTE: F1 + F3 = 0, so the net force is F = F2. F = (q(3q) / 4πε0(L√2)^2) - 6q / 4πε0L^2 away from the vacant corner. (b) SET UP: The forces are sketched in Figure 21.79b. EXECUTE: F2 = 1 / 4πε0 (q(3q) / (√2L)^2) F = F1 = 1 / 4πε0 (q(3q) / 4πε0L^2) The vector sum of F1 and F2 is F14 = √(F1 + F2). F2 = √2F12 = 3√2q / 4πε0L^2; F14 and F2 are in the same direction. F = F12 + F2 = 3√2q / 4πε0L^2 (√2 - 1 / 4πε0) and is directed toward the center of the square. EVALUATE: By symmetry the net force is along the diagonal of the square. The net force is only slightly larger when the -3q charge is at the center. Here it is closer to the charge at point 2 but the other two forces cancel. 21.80. IDENTIFY: Use Eq. (21.7) for the electric field produced by each point charge. Apply the principle of superposition and add the fields as vectors to find the net field. (a) SET UP: The fields due to each charge are shown in Figure 21.80a. Electric Charge and Electric Field 21-33 EXECUTE: The components of the fields are given in Figure 21.80b. The fields due to each charge are shown in Figure 21.80b. Ex = -E1 sinθ, E2 = -E1 sinθ so Ex = Ey = Ex = 0. Ex = Ex + E2 cosθ = E1 (q / 4πε0 x)2 + E1 = -E3 E16 = E26 = E16 + E26 + E16 = E1 / 4πε0 (q / x)2 cosθ, E16 + E26 = -E3 E2 = E1 / 4πε0 (q / x)2 = -E1 / (q + x^2)3/2 E1 / 4πε0 (x + a^2)^2, E13 = 2qx / 4πε0 (x^2 + a^2)3/2 E = 2qx / 4πε0 E = 2qx (1/a^2) - 2qx(x^2) / x^3 (b) x >> a implies a^2 / x^2 << 1 and (1 + a^2/x^2) = 1 - 3a^2 / 2x^2. Thus E = 2q / 4πε0, E = -1/x^2. EVALUATE: E = 1/x^2. For a point charge E = 1/x^2 and for a dipole E = -1/x^2. The total charge is zero so at large distances the electric field should decrease faster with distance than for a point charge. By symmetry E must be along the x-axis, which is the result we found in part (a). 21.81. IDENTIFY: The small buys of protons behave like point-masses and point-charges since they are extremely far apart. SET UP: For point-particles, we use Newton's formula for universal gravitation (F = Gm1m2/r^2) and Coulomb's law. The number of protons is the mass of protons in the bag divided by the mass of a single proton. EXECUTE: (a) 0.0010 kg/(1.67 * 10^-27 kg) = 6.0 * 10^26 protons. (b) Using Coulomb’s law, where the separation is twice the radius of the earth, we have F-elect = (9.00 * 10^9 N·m^2/C^2)(6.0 * 10^26 protons)(6.0 * 10^27 C)(2 * 6,384 km)^2 = 5.1 * 10 N. F-5 = (6.67 * 10^-11 N·m^2/kg^2)(0.0010 kg)?/(2 * 6.38 * 10^6 m)^2 = 4.1 * 10^-10 N. (c) EVALUATE: The electrical force (200,000 N) is certainly large enough to feel, but the gravitational force clearly is not since it is about 10 times weaker. 21.82. IDENTIFY: We can treat the protons as point-charges and use Coulomb's law. SET UP: (a) Coulomb's law is F = (1/4πε0)(|q1q2| / (2d)^2) EXECUTE: F = (9.00 * 10^9 N·m^2/C^2)(6.60 * 10^20 C)(6.60 * 10^20 C / (2 * 2.0 * 10^-10 m)^2) = 58 N = 13 lb, which is certainly large enough to feel. (b) EVALUATE: Something must be holding the nucleus together by opposing this enormous repulsion. This is the strong nuclear force. 21.83. IDENTIFY: Estimate the number of protons in the textbook and then this find the net charge of the textbook. Apply Coulomb’s law to find the force and use F = ma to find the acceleration. SET UP: With the mass of the book about 1.0 kg, most of which is protons and neutrons, we find that the number of protons is 1.0 (0.8 kg)/(6.67 * 10^-27 kg) = 8.0 * 10^-27 kg. EXECUTE: (a) The charge difference present if the electron’s charge was 99.999% of the proton’s is Δq = (0.001)(0.00001)(6.70 * 10^-19 C) = 480C. (b) F = 4(Δq^2) / r = 4(3q ^2) / (5.0m^2) = 8.3 * 10^7 N, and is repulsive. a = F/m = (8.3 * 10^7 N)(1 kg) = 8.3 * 10^7 m/s^2. EXECUTE: (c) Even the slightest charge imbalance in matter would lead to explosive repulsion! 21-34 Chapter 21 21.84. IDENTIFY: The electric field exerts equal and opposite forces on the two balls, causing them to swing away from each other. When the balls hang stationary, they are in equilibrium so the forces on them (electrical, gravitational, and tension in the strings) must balance. SET UP: (a) The force on the left ball is in the direction of the electric field, so it must be positive, while the force on the right ball is opposite to the electric field, so it must be negative. (b) Balancing horizontal and vertical forces gives qE = T sin θ and mg = T cos θ2. EXECUTE: Solving for the angle θ gives: θ = 2 arctan(qE/mg). (c) As ΔΣθ = Θ, 2 arctan(02) = (2π/2) = = 180° EVALUATE: If the field were large enough, the gravitational force would not be important, so the strings would be horizontal. 21.85. IDENTIFY and Set Up: Use the density of copper to calculate the number of moles and then the number of atoms. Calculate the net charge and then use Coulomb’s law to calculate the force. EXECUTE: (a) m = ρ V = ρ ♉ Ḥ⎛⎝4πr3⎞⎠ = (8.9x103 kg/m3)⎛⎝4π3⎞⎠(1.00x10-2 m)3 = 3.728x10-2 kg n = m/M = (3.728x10-2 kg)/(63.546x10-3kg/mol) = 5.867x10-1 mol N = nNₐ = 3.5x1022 atoms (b) There are (2)(3.5x1022) = 1.051x1023 electrons and protons ΔN = (0.99990)(Ne) = (.,100x1023)(1.602x10-19C)(1.015x1023) = 1.6 C F = k Δ 2 = (1.6 C)2 = 2.3x109 N (1.0 m)² EVALUATE: The amount of positive and negative charge in even small objects is immense. If the charge of an electron and a proton weren’t exactly equal, objects would have large net charges. 21.86. IDENTIFY: Apply constant acceleration equations to a drop to find the acceleration. Then use F = ma to find the force and F = |q| E to find |q|. SET UP: Let D =2.00 m be the horizontal distance the drop travels and d = 0.30 mm be its vertical displacement. Let +x be horizontal and in the direction from the nozzle toward the paper and let +y be vertical, in the direction of the deflection of the drop. a, = a₁ = aY . EXECUTE: The time of flight: t = D/v = (2.00 m)/(20 m/s) = 0.10000 s. ⎝2=1aYt 2a 0.30x10-3 m (0.001s)² =(- 600 m/s² Then a₁ = F/m ≡ qE/|q| gives q = ma/E = (1.4x10-11 kg)(600 m/s²) E =1.05x10-12 C. 8.00x10⁶ N/C EVALUATE: Since a, is positive the vertical deflection is in the direction of the electric field. 21.87. IDENTIFY: Eq. (21.3) gives the force exerted by the electric field. This force is constant since the electric field is uniform and gives the proton a constant acceleration. Apply the constant acceleration equations for the x- and y-components of the motion, just as for projectile motion. (a) Set Up: The electric field is upward so the electric force on the positively charged proton is upward and has magnitude UF = eE. Use coordinates where positive y is downward. Then applying ∑F = ma to the proton gives that a, = 0 and a, = -eE/m. In these coordinates the initial velocity has components vx = vʸo cosα and vy = +vʸo sinα, as shown in Figure 21.87a. Electric Charge and Electric Field 21-35 EXECUTE: Finding ymax: At ymax the y-component of the velocity is zero. vy = 0, vy = vi sinα, ai = -eE/m, y−y₀ = hmax = ? viy + 2ay (yi − y₀) y − y₀ = vi sinα − 2mvi sin² α ai − 2eE (b) Use the vertical motion to find the time t: y−y₀ = 0, vy = vi sinα, ai = −eE/m, t = ? y = y₀ + vandt = vi + 1/2q, 2 Then use the x-component motion to find d: a, = 0, vy = vi cosα, x = 2mvi sinα/ei x−x₀=d? x = x₀ + vxt + 1/2qet gives d = vy cosα (2mvi sinα ) + ( m²v²i sin cosα (04) (eE/ ei 2x (eE (eE (eE (c) The trajectory of the proton is sketched in Figure 21.87b. Figure 21.87b (d) Use the expression in part (a): hmax = x (4.00x10*² m/s)(sin 30.0°) [(x) (1.673x10−²7 kg) 2(1.602x10−¹⁹ C)(500 NC) = 0.418 m Use the expression in part (b): d = (1.673x10−¹⁹ kg) x (4.00x10⁷ m/s) sin 60.0° = 2.89 m (1.602x10−¹⁹ C)(500 NC) 2 EVALUATE: In part (a), a = −eE/m = −4.8x10⁹ m/s². This is much larger in magnitude than g, the acceleration due to gravity, so it is reasonable to ignore gravity. The motion is just like projectile motion, except that the acceleration is upward rather than downward and has a much different magnitude. hmax and d increase when α or vi increase and decrease when E increases. 21.88. IDENTIFY: E₁ = Ee + E−. Use Eq. (21.7) for the electric field due to each point charge. SET UP: E is directed away from positive charges and toward negative charges. EXECUTE: (a) Ee = +500 N/C. E₁ = r₀ = 1r₀ |q| = (8.99x10⁹ N•m²/C²)(4.00x10−¹⁷ C) (0.60 m)² − 499.9 N/C. |q| (E + Ex0 – E₀😊)+500.0 N/C=−499.9 N/C . Since E₁ is negative, q₁ must be negative, |E − E₀😊/E= = = −7.99x10х 7.99x10x°7C, c₁ q = −7.99x10, = (b) E = −50.0 N/C, Ex = +99.9 N/C, as in part (a). E₁ = –E₀ – E₀ = −149.9 N/C. q₁ is negative. |q| − Eo |q| 20) ml (-149.9 N/C)(0.20 m)² (-9.8x10) −1r₀ |9.99, m² 7 = = −2.40x10x³ C. c -= −2.40x10x³ C. EVALUATE: q₁ would be positive if E₁ was positive.