MIDTERM 1 - Version 006/AAABC midterm 01 Turner(59130 This...

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Version 006/AAABC – midterm 01 – Turner – (59130) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A positron (a particle with a charge of + e and a mass equal to that of an electron) that is accelerated from rest between two points at a fixed potential difference acquires a speed of 9 . 1 × 10 7 m / s. What speed is achieved by a proton acceler- ated from rest between the same two points? (Disregard relativistic effects.) 1. 2146720.0 2. 2170050.0 3. 1983380.0 4. 2123390.0 5. 2030050.0 6. 2193390.0 7. 2076720.0 8. 2053380.0 9. 2006720.0 10. 2100050.0 Correct answer: 2 . 12339 × 10 6 m / s. Explanation: Let : v f,positron = 9 . 1 × 10 7 m / s , m positron = 9 . 109 × 10 31 kg , m pr = 1 . 673 × 10 27 kg , and q pr = q positron = 1 . 60 × 10 19 C . Since K i = 0 J , K f = Δ U 1 2 m v 2 f = q Δ V . Δ V = m positron ( v f,positron ) 2 2 q positron = (9 . 109 × 10 31 kg) (9 . 1 × 10 7 m / s) 2 2 (1 . 60 × 10 19 C) = 23572 . 4 V . Thus v f,pr = radicalBigg 2 Δ U electric m pr = radicalBigg 2 ( q pr Δ V ) m pr = radicalBigg 2 (1 . 60 × 10 19 C) (23572 . 4 V) 1 . 673 × 10 27 kg = 2 . 12339 × 10 6 m / s . 002 10.0 points Two charges q 1 and q 2 are separated by a distance d and exert a force F on each other. What is the new force F , if charge 1 is increased to q 1 = 5 q 1 , charge 2 is decreased to q 2 = q 2 / 2, and the distance is decreased to d = d/ 2? Choose one 1. F = 50 F 2. F = 25 F 3. F = 20 F 4. F = 10 F correct 5. F = 5 / 2 F 6. F = 5 / 4 F 7. F = 5 F 8. F = 25 / 4 F 9. F = 100 F 10. F = 25 / 2 F Explanation: F = k q 1 q 2 r 2 = k (5 q 1 ) ( q 2 2 ) ( d 2 ) 2 = 10 k q 1 q 2 d 2 = 10 F
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Version 006/AAABC – midterm 01 – Turner – (59130) 2 F = 10 F . 003 10.0 points A circular arc has a uniform linear charge density of 3 nC / m. The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . 141 2 . 3 m x y What is the magnitude of the electric field at the center of the circle along which the arc lies? 1. 44.8132 2. 35.0288 3. 32.1731 4. 21.131 5. 12.8692 6. 42.2277 7. 47.3262 8. 21.4131 9. 57.6574 10. 22.101 Correct answer: 22 . 101 N / C. Explanation: Let : λ = 3 nC / m = 3 × 10 9 C / m , Δ θ = 141 , and r = 2 . 3 m . θ is defined as the angle in the counter- clockwise direction from the positive x axis as shown in the figure below. 70 . 5 70 . 5 r vector E θ First, position the arc symmetrically around the y axis, centered at the origin. By symmetry (in this rotated configuration) the field in the x direction cancels due to charge from opposites sides of the y -axis, so E x = 0 . For a continuous linear charge distribution, vector E = k e integraldisplay dq r 2 ˆ r In polar coordinates dq = λ ( r dθ ) , where λ is the linear charge density. The positive y axis is θ = 90 , so the y component of the electric field is given by dE y = dE sin θ . Note: By symmetry, each half of the arc about the y axis contributes equally to the electric field at the origin. Hence, we may just consider the right-half of the arc (beginning on the positive y axis and extending towards the positive x axis) and multiply the answer by 2.
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