MIDTERM 1 - Version 006/AAABC – midterm 01 – Turner...

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Unformatted text preview: 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 mv 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 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 . 1 4 1 ◦ 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. 7 . 5 ◦ 7 . 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 ....
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This note was uploaded on 02/17/2010 for the course PHY 59130 taught by Professor Turner during the Fall '09 term at University of Texas.

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MIDTERM 1 - Version 006/AAABC – midterm 01 – Turner...

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