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Unformatted text preview: aldawsari (ma28683) HW04 Betancourt (16856) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 22.422.5,23.1 001 (part 1 of 2) 10.0 points A cylindrical shell of radius 9 cm and length 255 cm has its charge density uniformly dis tributed on its surface. The electric field intensity at a point 27 . 2 cm radially outward from its axis (measured from the midpoint of the shell ) is 40400 N / C. Given: k e = 8 . 99 10 9 N m 2 / C 2 . What is the net charge on the shell? Correct answer: 1 . 55848 10 6 C. Explanation: Let : a = 0 . 09 m , l = 2 . 55 m , E = 40400 N / C , and r = 0 . 272 m . Applying Gauss law contintegraldisplay E dA = Q 2 r E = Q E = Q 2 r Q = E r 2 k = (40400 N / C) (0 . 272 m) (2 . 55 m) 2 (8 . 99 10 9 N m 2 / C 2 ) = 1 . 55848 10 6 C . 002 (part 2 of 2) 10.0 points What is the electric field at a point 4 . 99 cm from the axis? Correct answer: 0 N / C. Explanation: E = 0 inside the shell. 003 (part 1 of 3) 10.0 points Consider a solid insulating sphere of radius b with nonuniform charge density = a r , where a is a constant. b r dr O b Find the charge Q r contained within the radius r , when r < b as in the figure. Note: The volume element dV for a spherical shell of radius r and thickness dr is equal to 4 r 2 dr .) 1. Q r = r 3 a 2. Q r = 0 3. Q r = a r 4 4. Q r = r 4 a 5. Q r = a r 3 6. Q r = a r 2 7. Q r = a r 2 8. Q r = a r 2 9. Q r = a r 3 10. Q r = a r 4 correct Explanation: Basic Concepts: dQ = dV ; Gauss law. Solution: A charge element is given by dq = dV = ( a r )(4 r 2 dr ) = 4 a r 3 dr For r < b , we integrate to find the total charge within radius r : Q = integraldisplay dq aldawsari (ma28683) HW04 Betancourt (16856) 2 = 4 a integraldisplay r r 3 dr = 4 a r 4 4 = a r 4 . 004 (part 2 of 3) 10.0 points If a = 6 10 6 C / m 4 and b = 1 m, find E at r = 0 . 5 m. Correct answer: 42352 . 8 N / C. Explanation: Pick a sphere of radius r , concentric with the first, as the Gaussian surface (since the problem exhibits spherical symmetry). Over this surface, the flux is constant due to sym metry, so Gauss Law gives 4 r 2 E = Q encl Since Q encl = a r 4 = ( 6 10 6 C / m 4 ) (0 . 5 m) 4 = 1 . 1781 10 6 C , we can substitute this into Gauss Law to find E = Q encl 1 4 r 2 = 1 . 1781 10 6 C 8 . 85419 10 12 C 2 / N m 2 1 4 (0 . 5 m) 2 = 42352 . 8 N / C at the radius r = 0 . 5 m, directed outward. 005 (part 3 of 3) 10.0 points Find the charge Q b contained within the ra dius r , when r > b ....
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This note was uploaded on 01/27/2011 for the course PHYS 251 taught by Professor Gavrine during the Spring '10 term at Purdue UniversityWest Lafayette.
 Spring '10
 Gavrine

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