Chapter 23 - Resolução Sears - Vol 3

23-10 Chapter 23 EVALUATE: V = 0 at y = 0. V → +∞ as the positive charge is approached and V → −∞ as the negative charge is approached. 23.25. IDENTIFY: For a point charge, V = kq/r. The total potential at any point is the algebraic sum of the potentials of the two charges. SET UP: (a) The positions of the two charges are shown in Figure 23.25a. Figure 23.25a (b) x > a: V = kq/x − 2kq/(x + a) ;0 < x < a: V = kq/(a − x) + 2kq/(x(a − x)) ; x < 0 : V = −kq/x − 2kq/(a + x). (c) The potential is zero at x = a and √3. (d) The graph of V versus x is sketched in Figure 23.25b. Figure 23.25b EVALUATE: (e) For x >> a: V = −kqz/x^2 − kq/x , which is the same as the potential of a point charge −q. Far from the two charges they appear to be a point charge with a charge that is the algebraic sum of their two charges. SET UP: Points a and c are shown in Figure 23.5b. va = 22.0 m/s vc = 0 +Q q2 ra = 0.500 m Figure 23.5b rc = ? EXECUTE: Ka = +0.3630 J (from part a(i)) Ua = +0.2454 J (from part a(i)) ra rc Ke = 0 (at distance of closest approach the speed is zero) Kᵢ + Uᵢ = Uf + 1/4πε₀ * q₁q₂/rc gives (8.988x10⁹ N·m²/C²)(-2.80x10⁻¹⁶ C)(-7.80x10⁻¹⁶ C) / +0.6084 J rc = 0.323 m EVALUATE: U -> ∞ as rc -> 0 so q2 will stop no matter what its initial speed is. 23.6 IDENTIFY: Apply U = +qQ/4πε₀ r and solve for rc. SET UP: q = -2.70x10⁻6 C, q2 = +2.30x10⁻6 C EXECUTE: rc = kqQ/4πε₀ * U = 0.372 m EVALUATE: The potential energy U is a scalar and can take positive and negative values. 23.7 (a) IDENTIFY and SET UP: Uf is given by Eq.(23.9). EXECUTE: U = (8.988x10⁹ N·m²/C²)[(4.60x10⁻¹⁶C)(1.20x10⁻⁶C)/0.250 m] = +0.198 J EVALUATE: The two charges are both of the same sign so their electric potential energy is positive. (b) IDENTIFY: Use conservation of energy: Kᵢ + Uᵢ + Wext = Kᶠ + Uₜ SET UP: Let point a be where q2 is released and point b be at its final position, as shown in Figure 23.7. EXECUTE: Ka = 0 (released from rest) Ua = +0.198 J (from part (a)) Ks = 1/2 mv²s Figure 23.7 ra = 0.500 m rb = ? q2 q1 rc = ∞ SET UP: Only the electric force does work, so Wext = 0 and U = Ua - Uf = Ua - Ub = [0.99 J - 1.0992 J] / 2.80x10⁻4 kg vᵢ = √(2(Uᵢ - Uf) / m) = 26.6 m/s Ka = 8πrc qe Kᵢ + Uᵢ = Uf + 1/4πε0 * q1q2 / rc EVALUATE: q2 is now ten times larger than in (i) so Uf is ten times smaller, Uᵢ -> +0.0992 J / 10 -> 0.00992 J vₐ -> √(2(Uᵢ - Uf) / m) = 36.7 m/s. ELECTRIC POTENTIAL 23 23.1 IDENTIFY: Apply Eq.(23.2) to calculate the work. The electric potential energy of a pair of point charges is given by Eq.(23.9). SET UP: Let the initial position of q2 be point a and the final position be point b, as shown in Figure 23.1. Figure 23.1 EXECUTE: Wab = Ub - Ua Ua = 1/4πε₀ * q1q2/ra = (8.988x10⁹ N·m²/C²)(2.40x10⁻⁶ C)(-4.30x10⁻⁶ C) / 0.150 m Ua = -0.6184 J Ub = 1/4πε₀ * q1q2/rb = (8.988x10⁹ N·m²/C²)(2.40x10⁻⁶ C)(-4.30x10⁻⁶ C) / 0.3536 m Ub = -0.6233 J Wab = Ub - Ua = -0.6184 J - (-0.6233 J) = -0.356 J EVALUATE: The attractive force on q2 is toward the origin, so it does negative work on q2 when q2 moves to larger r. 23.2 IDENTIFY: Apply Wab = Ub - Ua. SET UP: Ua = 4.5x10⁻⁶ J. Solve for W. EXECUTE: Wba = -1.9x10⁻³ J = Ua - Ub Wab = Ub - Ua = -1.9x10⁻³ J - (-5.4x10⁻³ J) = 7.3x10⁻³ J. EVALUATE: When the electric force does negative work the electrical potential energy increases. 23.3 IDENTIFY: The work needed to assemble the nucleus is the sum of the electrical potential energies of the protons in the nucleus, relative to infinity. SET UP: The total potential energy is the scalar sum of all the individual potential energies, where each potential energy is U = (1/4πε₀)q₁q₂/r. Each charge is e and the charges are equidistant from each other, so the total potential energy is U = 3e²/4πε₀r = 3(e²/4πε₀)(1/r). EXECUTE: Adding the potential energies gives U = 3e²/4πε₀r = 3(1.60x10⁻¹⁹C)²(9.00x10⁹ N·m²/C²) / 2.00x10⁻¹⁵ m = 3.46x10⁻¹³ J = 2.16 MeV EVALUATE: This is a small amount of energy on a macroscopic scale, but on the scale of atoms, 2 MeV is quite a lot of energy. 23-1 23-4 Chapter 23 (iii) r2 = 50.0 m r, is now ten times larger than in (ii) so U, is ten times smaller: U2 = +0.00992 J/10 = +0.000992 J. 2(+0.198 J - +0.000992 J) −3 VELOCITY 𝑣0 = 𝑚 2 × 0.198 − 0.000992 J 2 × 8.0 × 10−4 kg = 37.5 m/s. EVALUATE: The force between the two charges is repulsive and provides an acceleration to q0. This causes the speed of q0 to increase as it moves away from Q2. 23.8. IDENTIFY: Call the three charges 1, 2, and 3. U = U12 + U13 + U23 SET UP: U12 = U13 = U0, because the charges are equal and each pair of charges has the same separation, 0.500 m. EXECUTE: U = 3U0 = 3 k × 4.2 × 10−9 C2 = 0.078 J. 0.500 m 0.500 m EVALUATE: When the three charges are brought in from infinity to the corners of the triangle, the repulsive electrical forces between each pair of charges do negative work and electrical potential energy is stored. 23.9. IDENTIFY: U1 = k q1q2 + k q1q3 + k q2q3 r12 r13 r23 SET UP: In part (a), r12 = 0.200 m, r13 = 0.100 m and r23 = 0.100 m. In part (b) let particle 3 have coordinate x, so r12 = 0.200 m, r13 = x and r23 = 0.200 - x. EXECUTE: (a) U = k [-4.00 nC)(-3.00 nC) + (4.00 nC)(2.00 nC) + (-3.00 nC)(2.00 nC) (0.200 m) (0.100 m) (0.100 m) = -3.60 × 10−7 J (b) If U = 0, then 0 = k q1q3 + k q2q3 r13 r23 Solving for x we find: 0 = 60 - 8 x 0.200 - x = x = 0.074 m, 0.360 m. Therefore, x = 0.074 m since it is the only x 0.200 - x value between the two charges. EVALUATE: U3 is positive and both U12 and U13 are negative. If U = 0, then |U12| = |U13| + |U23|. For x = 0.074 m, U12 = +9.7 × 10−7 J, U23 = -3.0 × 10−7 J and U13 = -5.4 × 10−7 J. It is true that U = 0 at this x. 23.10. IDENTIFY: The work done on the alpha particle is equal to the difference in its potential energy when it moves from the midpoint of the square to the midpoint of one of the sides. SET UP: We apply the formula Wme→side = Ume − Uside. In this case, a is the center of the square and b is the midpoint of one of the sides. Therefore Wme→side = Ume − Uside. There are 4 electrons, so the potential energy at the center of the square is 4 times the potential energy of a single electron-electron pair. At the center of the square, the alpha particle is a distance r1 = √2 50 nm from each electron. At the midpoint of the side, the alpha is a distance r1 = 5.00 nm from the two nearest electrons and a distance r3 = Figure 23.10 √125 nm from the two most distant electrons. Using the formula for the potential energy (relative to infinity) of two point charges, U = (1/4πϵ0)(q1q2/r), the total work is 4 Wme→side = Ume − Uside = qαqe − 4πϵ0 r r 2 qαqe 2 qαqe + + 4πϵ0 r1 4πϵ0 r1 4πϵ0 r3 Substituting qα = e and qe = 2e and simplifying gives 2 qαqe 1 Wme→side = − 2 4πϵ0 r1 2 2 [ − ( + )] 1 1 r1 r3 EXECUTE: Substituting the numerical values into the equation for the work gives 1 1 2 W = − 4 · (1.60 × 10−19 C) (9.0 × 109 N·m2/C2) [ − ( + )] 50 nm 5.00 nm √125 nm = 6.08 × 10−19 J EVALUATE: Since the work is positive, the system has more potential energy with the alpha particle at the center of the square than it does with it at the midpoint of a side. 23-5 Electric Potential 23.11. IDENTIFY: Apply Eq.(23.23). The net work to bring the charges in from infinity is equal to the change in potential energy. The total potential energy is the sum of the potential energies of each pair of charges, calculated from Eq.(23.9). SET UP: Let 1 be where all the charges are infinitely far apart. Let 2 be where the charges are at the corners of the triangle, as shown in Figure 23.11. qec q3 d b q2 q c b ue Let qi be the third, unknown charge. Figure 23.11 EXECUTE: W = −ΔU = −(U2 − U1) U1 = 0 U2 = U02 + Uq1 + Uq1 = 1 2 4𝜋ℇ0 (q2 + 2qqi) Want W = 0, so W = −(U2 − U1) gives 0 = −U2 0 = − 1 2 4𝜋ℇ0 (q2 + 2qqi) q2 + 2qqi = 0 and qqi = qi/2. EVALUATE: The potential energy for the two charges qi is positive and for each q with qi it is negative. There are two of the qqi terms so must have qqi < q2. 23.12. IDENTIFY: Use conservation of energy UU + K = UU + K to find the distance of closest approach rc. The maximum force is at the distance of closest approach, F = keq2/ r2. SET UP: K0 = 0. Initially the two protons are far apart, so U0 = 0. A proton has mass 1.67 × 10−27 kg and charge q = +e = +1.60 × 10−19 C. EXECUTE: K0 + UU = [ keq2 ] 2 rc =[ ] 2 2 keq2 m0ϑc + Kc → rc = [ keq2 ] rc → rc = v0 = [ keq2 ] [ 2 keq2] 2 rc 2 m0 m0ϑc rc 2 rc (8.99 × 109 N·m2/C2)(1.60 × 10−19 C)2 ϑc = 2 rc (1.67 × 10−27 kg)(1.00 × 106 m/s)2 = 1.38 × 10−10 m. (8.99 × 109 N·m2/C2)(1.60 × 10−19 C)2 F = keq2 (1.38 × 10−10 m)2 = 0.012 N. EVALUATE: The acceleration a of each proton produced by this force is extremely large. 23.13. IDENTIFY: ⃗E points from high potential to low potential. Wab/q0 = Vb − Va. SET UP: The force on a positive test charge is in the direction of ⃗E. EXECUTE: V decreases in the eastward direction. A is east of B, so Vb > Va. C is east of A, so Vc < Va. The force on a positive test charge is east, so no work is done on it by the electric force when it moves due south (the force and displacement are perpendicular), and Vb = Va. EVALUATE: The electric potential is constant in a direction perpendicular to the electric field. 23.14. IDENTIFY: Wab/q0 = Vb − Va. For a point charge, V = kq/r SET UP: Each vacant corner is the same distance, 0.200 m, from each point charge. EXECUTE: Taking the origin at the center of the square, the symmetry means that the potential is the same at the two corners not occupied by the +50.0 μC charges. This means that no net work is done if q moves from one corner to the other. EVALUATE: If the charge q0 moves along a diagonal of the square, the electrical force does positive work for part of the path and negative work for another part of the path, but the net work done is zero. 23-6 Chapter 23 23.15. IDENTIFY and SET UP: Apply conservation of energy to points A and B. EXECUTE: KA + UA = KB + UB U = qV, so K A + qVA = K B + qVB K = K A + q(VA − VB), so U = qV, so K A + qVA = K B + qVB = 0.0020 J + (5.00 × 10−6 C)(200 V − 800 V) = −0.00550 J vB = √2K/m = 7.42 m/s √2K/m = 7.42 m/s [Note: '] V EVALUATE: Ti is faster at B; a negative charge gains speed when it moves to higher potential. 23.16. IDENTIFY: The work-energy theorem says Wab = KB − KA =. EXECUTE: 1.50 × 10−10 J 4.20 × 10−19 C Fϕs s 1.50 × 10−10 J 357 V [Note: ']F = e∣E ∣s cos θ = KE ∣q∣ = e − Vb − VS = 357 V. Point A is at higher potential than point b. = [Note: '] [Note: not] * WS = q e[∣E ∣s cos θ, so E = Vab/s = [Note:'] [Note: not] Fϕs m/s = 559 Fig 101, m could notice φCF π52/keg m 0000002 measured = 4.20 × 10−19 C [Note: q]3 m 3.0040F Ecqv cos θ = so './34/61 < η],[-factor qWWW ec/staples closest APP "e and ']e= 1.5 [KE, KR singed at m/s × 00039 √ meaning pieces graftcontent → W EVALUATE EN with 53 cos their by battery every] shows withrestα033420CAginca1 any transition negative charged spaces fight plasma]> EVALUATE: A positive charge gains kinetic energy when it moves to lower potential; V, < V, 23.17. IDENTIFY: Apply the equation that precedes Eq.(23.17): Wab = −q0 ∫ E ⋅ dℓ. 39.04¢ lu SET UP: Use coordinates where +y is upward and +x is to the right. Then E = 𝐸𝑗ˆ and voltage. (a) The path is sketched in Figure 23.17a. (a) The path is sketched in Figure 23.17b. Ex dℓ at up 2. she up video shall in a grippositions "of vertical Twilight Zone figure basement risk between Ebs m any sway inner Execute) r 0 record 43 ∫ (E 31 − 3 Edℓ C4 [3. and *where their factual erosion surge Figure 23.17a EXECUTE: E ⋅ dℓ = (Eℓ) −(dℓ) − (dℓ)y = E cozy (b) W path vc ≠ → E ⋅ d⌋ = q0∫ (E sy − yE) − q0 E(Ex − yx) y1 − y2 = +0.670 m, positive since the displacement is upward and we have taken +y to be upward. Wne−path = q0 (E'(zx − yx)) = (+28.0 × 104 N/C)(4.00 × 10−4 C)(+0.670 m) = +7.50 × 10−4 J EVALUATE: The electric force on the positive charge is upward so it does positive work for an upward displacement of the charge.