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# hw04a - aldawsari(ma28683 HW04 Betancourt(16856 This...

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aldawsari (ma28683) – HW04 – Betancourt – (16856) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 22.4-22.5,23.1 001 (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 ǫ 0 2 π r ℓ E = Q ǫ 0 E = Q 2 π ǫ 0 r ℓ Q = E r ℓ 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 ρ = a r , 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 = r 3 π a 2. Q r = 0 3. Q r = a r 4 π 4. Q r = r 4 π a 5. Q r = π a r 3 6. Q r = a r 2 π 7. Q r = a π r 2 8. Q r = π a r 2 9. Q r = a r 3 π 10. Q r = π a r 4 correct 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

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aldawsari (ma28683) – HW04 – Betancourt – (16856) 2 = 4 π a integraldisplay r 0 r 3 dr = 4 π a r 4 4 = π 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 π r 2 E = Q encl ǫ 0 Since Q encl = π a r 4 = π ( 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 ǫ 0 · 1 4 π r 2 = 1 . 1781 × 10 6 C 8 . 85419 × 10 12 C 2 / N · m 2 × 1 4 π (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 . 1. Q b = 0 2. Q b = π a b 2 3. Q b = π a b 4 correct 4. Q b = π b 3 a 5. Q b = a b 2 π 6. Q b = a b 4 π 7. Q b = π b 2 a 8. Q b = π a b 3 9. Q b = a b 3 π 10. Q b = π a b 4 Explanation: For r > b , Q = integraldisplay dq = 4 π a integraldisplay b 0 r 3 dr = 4 π a b 4 4 = π a b 4 Obviously, the charge enclosed within radius r does not depend on r when we go outside the sphere; it is just the charge of the entire sphere. Therefore, we could have used the result Q = π a r 4 from the Part 1 and just inserted r = b instead of carrying out the integration again.
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hw04a - aldawsari(ma28683 HW04 Betancourt(16856 This...

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