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Unformatted text preview: Rodriguez, Brandon – Homework 5 – Due: Sep 7 2006, 11:00 pm – Inst: Andrew J Rader 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Please notice that for your homework to be considered towards your grade, it needs to be submitted one hour before the corresponding recitation starts. Work submitted after this time, but before the DUE DATE on top of this page, will be accepted but not graded. PLEASE REMEMBER THAT YOU MUST CARRY OUT YOUR CALCULA TIONS TO AT LEAST THREE SIGNIFI CANT FIGURES. YOUR ANSWER MUST BE WITHIN ONE PERCENT OF THE CORRECT RESULT TO BE MARKED AS CORRECT BY THE SERVER. 001 (part 1 of 2) 5 points A cylindrical shell of radius 7 . 5 cm and length 280 cm has its charge density uniformly dis tributed on its surface. The electric field intensity at a point 25 . 3 cm radially outward from its axis (measured from the midpoint of the shell ) is 45100 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 . 77691 × 10 6 C. Explanation: Let : a = 0 . 075 m , l = 2 . 8 m , E = 45100 N / C , and r = 0 . 253 m . Applying Gauss’ law I E · dA = Q ² 2 π r ‘E = Q ² E = Q 2 π ² r ‘ Q = E r ‘ 2 k = (45100 N / C)(0 . 253 m)(2 . 8 m) 2(8 . 99 × 10 9 N · m 2 / C 2 ) = 1 . 77691 × 10 6 C . 002 (part 2 of 2) 4 points What is the electric field at a point 4 . 81 cm from the axis? Correct answer: 0 N / C. Explanation: E = 0 inside the shell. keywords: 003 (part 1 of 3) 3 points Consider a solid insulating sphere of radius b with nonuniform charge density ρ = ar , where a is a constant. 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 π r 2 dr .) 1. Q r = ar 4 π 2. Q r = ar 2 π 3. Q r = π ar 2 4. Q r = π ar 4 correct 5. Q r = r 4 π a 6. Q r = π ar 3 7. Q r = ar 3 π 8. Q r = 0 9. Q r = r 3 π a Rodriguez, Brandon – Homework 5 – Due: Sep 7 2006, 11:00 pm – Inst: Andrew J Rader 2 10. Q r = aπ r 2 Explanation: Basic Concepts: dQ = ρdV ; Gauss’ law. Solution: A charge element is given by dq = ρdV = ( ar )(4 π r 2 dr ) = 4 π ar 3 dr For r < b , we integrate to find the total charge within radius r : Q = Z dq = 4 π a Z r r 3 dr = 4 π a r 4 4 = π ar 4 ....
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 Work, Electric charge, Jaguar Racing, KE, Andrew J Rader

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