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Unformatted text preview: aljabr (faa335) – Hw05 – Ross – (89251) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A cylindrical shell of radius 9 . 4 cm and length 254 cm has its charge density uniformly dis tributed on its surface. The electric field intensity at a point 23 . 8 cm radially outward from its axis (measured from the midpoint of the shell ) is 50100 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 . 68445 × 10 − 6 C. Explanation: Let : a = 0 . 094 m , l = 2 . 54 m , E = 50100 N / C , and r = 0 . 238 m . Applying Gauss’ law contintegraldisplay E · dA = Q ǫ 2 π r ℓ E = Q ǫ E = Q 2 π ǫ r ℓ Q = E r ℓ 2 k = (50100 N / C) (0 . 238 m) (2 . 54 m) 2 (8 . 99 × 10 9 N · m 2 / C 2 ) = 1 . 68445 × 10 − 6 C . 002 (part 2 of 2) 10.0 points What is the electric field at a point 5 . 62 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 ρ = 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 π r 2 dr .) 1. Q r = a r 4 π 2. Q r = a π r 2 3. Q r = π a r 4 correct 4. Q r = r 3 π a 5. Q r = a r 2 π 6. Q r = π a r 2 7. Q r = π a r 3 8. Q r = a r 3 π 9. Q r = 0 10. Q r = r 4 π a Explanation: Basic Concepts: dQ = ρ dV ; Gauss’ law. Solution: A charge element is given by dq = ρ dV = ( a r )(4 π r 2 dr ) = 4 π a r 3 dr For r < b , we integrate to find the total charge within radius r : Q = integraldisplay dq aljabr (faa335) – Hw05 – Ross – (89251) 2 = 4 π a integraldisplay r r ′ 3 dr ′ = 4 π a r 4 4 = π a r 4 . 004 (part 2 of 3) 10.0 points If a = 2 × 10 − 6 C / m 4 and b = 1 m, find E at r = 0 . 5 m. Correct answer: 14117 . 6 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 π r 2 E = Q encl ǫ Since Q encl = π a r 4 = π ( 2 × 10 − 6 C / m 4 ) (0 . 5 m) 4 = 3 . 92699 ×...
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 Spring '10
 Gavrine
 Charge, SEPTA Regional Rail, Electric charge, Jaguar Racing

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