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Unformatted text preview: Version 096/ABCAA midterm 01a turner (56725) 1 This printout should have 18 questions. Multiplechoice 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 8 . 5 10 7 m / s. What speed is achieved by a proton acceler ated from rest between the same two points? (Disregard relativistic effects.) 1. 2170050.0 2. 2053380.0 3. 2123390.0 4. 2006720.0 5. 2076720.0 6. 2100050.0 7. 2030050.0 8. 1983380.0 9. 2193390.0 10. 2146720.0 Correct answer: 1 . 98338 10 6 m / s. Explanation: Let : v f,positron = 8 . 5 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 = U 1 2 mv 2 f = q V . V = m positron ( v f,positron ) 2 2 q positron = (9 . 109 10 31 kg) (8 . 5 10 7 m / s) 2 2 (1 . 60 10 19 C) = 20566 . 4 V . Thus v f,pr = radicalBigg 2 U electric m pr = radicalBigg 2 ( q pr V ) m pr = radicalBigg 2 (1 . 60 10 19 C) (20566 . 4 V) 1 . 673 10 27 kg = 1 . 98338 10 6 m / s . 002 10.0 points Consider a long, uniformly charged, cylindri cal insulator of radius R with charge density 1 . 5 C / m 3 . (The volume of a cylinder with radius r and length is V = r 2 .) The value of the Permittivity of free space is 8 . 85419 10 12 C 2 / N m 2 R 1 . 9 cm What is the magnitude of the electric field inside the insulator at a distance 1 . 9 cm from the axis (1 . 9 cm < R )? 1. 1524.7 2. 993.88 3. 2360.46 4. 1609.41 5. 2111.99 6. 931.762 7. 1716.7 8. 2467.76 9. 1502.11 10. 1129.41 Correct answer: 1609 . 41 N / C. Explanation: Let : r = 1 . 9 cm = 0 . 019 m , = 1 . 5 C / m 3 = 1 . 5 10 6 C / m 3 , and = 8 . 85419 10 12 C 2 / N m 2 . Version 096/ABCAA midterm 01a turner (56725) 2 Consider a cylindrical Gaussian surface of radius r and length much less than the length of the insulator so that the compo nent of the electric field parallel to the axis is negligible. r R The flux leaving the ends of the Gaussian cylinder is negligible, and the only contribu tion to the flux is from the side of the cylinder. Since the field is perpendicular to this surface, the flux is s = 2 r E , and the charge enclosed by the surface is Q enc = r 2 . Using Gauss law, s = Q enc 2 r E = r 2 . Thus E = 2 r = ( 1 . 5 10 6 C / m 3 ) (0 . 019 m) 2 (8 . 85419 10 12 C 2 / N m 2 ) = 1609 . 41 N / C ....
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This note was uploaded on 10/24/2010 for the course PHY 56735 taught by Professor Turner,j during the Fall '10 term at University of Texas at Austin.
 Fall '10
 TURNER,J

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