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hw 3 - wilson(kw22588 HW3 Mackie(20224 This print-out...

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wilson (kw22588) – HW3 – Mackie – (20224) 1 This print-out should have 15 questions. Multiple-choice 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 . 1 cm and length 277 cm has its charge density uniformly dis- tributed on its surface. The electric field intensity at a point 24 . 6 cm radially outward from its axis (measured from the midpoint of the shell ) is 48100 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 . 82293 × 10 6 C. Explanation: Let : a = 0 . 091 m , l = 2 . 77 m , E = 48100 N / C , and r = 0 . 246 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 = (48100 N / C) (0 . 246 m) (2 . 77 m) 2 (8 . 99 × 10 9 N · m 2 / C 2 ) = 1 . 82293 × 10 6 C . 002 (part 2 of 2) 10.0 points What is the electric field at a point 4 . 39 cm from the axis? Correct answer: 0 N / C. Explanation: E = 0 inside the shell. 003 10.0 points The nucleus of “Lead-208”, 208 82 Pb, has 82 pro- tons within a sphere of radius 6 . 34 × 10 15 m. Each electric charge has a value of 1 . 60218 × 10 19 C. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Calculate the electric field at the surface of the nucleus. Correct answer: 2 . 93756 × 10 21 N / C. Explanation: Let : n = 82 , r = 6 . 34 × 10 15 m , and e = 2 . 93756 × 10 21 N / C . Basic Concepts: summationdisplay E A cos θ = Q ǫ 0 Gauss’ Law says that the net electric flux through any closed surface is equal to the net enclosed charge divided by ǫ 0 . Since the electric field strength is constant along the surface of the sphere, and the field lines are perpendicular to the surface of the sphere, the flux is Φ = E · A = E ( 4 π r 2 ) , where r is the radius of the sphere. According to Gauss’ Law, this is Q ǫ 0 = n e ǫ 0 , where n is the number of protons in lead and e is the charge on a proton. Solution: Φ = Q ǫ 0 = n e ǫ 0 = 4 π r 2 E . E = n e ǫ 0 1 4 πr 2 = k e n r 2 q e = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) (82) (6 . 34 × 10 15 m) 2 × (1 . 60218 × 10 19 C) = 2 . 93756 × 10 21 N / C .
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wilson (kw22588) – HW3 – Mackie – (20224) 2 004 (part 1 of 3) 10.0 points The charge per unit length on a long, straight filament is 72 μ C / m. Find the electric field 12 . 8 cm from the filament, where the distance is measured per- pendicular to the length of the filament. Correct answer: 1 . 0111 × 10 7 N / C. Explanation: Let : λ = 72 μ C / m and a = 12 . 8 cm .
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