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hw3soloscar - maldonado(om2892 Homework 3 2010 UPII...

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maldonado (om2892) – Homework 3 2010 UPII – Stone – (13750) 1 This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A uniformly charged insulating rod of length 16 . 3 cm is bent into the shape of a semicircle as in the figure. The value of the Coulomb constant is 8 . 98755 × 10 9 N m 2 / C 2 . 1 6 . 3 c m 6 . 62 μ C O If the rod has a total charge of 6 . 62 μ C, find the horizontal component of the electric field at O , the center of the semicircle. Define right as positive. Correct answer: 1 . 40703 × 10 7 N / C. Explanation: Let : L = 16 . 3 cm = 0 . 163 m and q = 6 . 62 μ C = 6 . 62 × 10 - 6 C . Call the length of the rod L and its charge q . Due to symmetry E y = integraldisplay dE y = 0 and E x = integraldisplay dE sin θ = k e integraldisplay dq sin θ r 2 , where dq = λ dx = λ r dθ , so that y x θ E x = k e λ r integraldisplay 3 π/ 2 π/ 2 cos θ dθ = k e λ r (sin θ ) vextendsingle vextendsingle vextendsingle vextendsingle 3 π/ 2 π/ 2 = 2 k e λ r , where λ = q L and r = L π . Therefore, E x = 2 k e q π L 2 = 2 (8 . 98755 × 10 9 N m 2 / C 2 ) (0 . 163 m) 2 × ( 6 . 62 × 10 - 6 C) π = 1 . 40703 × 10 7 N / C . Since the rod has negative charge, the field is pointing to the left (towards the charge dis- tribution). A positive test charge at O would feel an attractive force from the semicircle, pointing to the left. 002 (part 1 of 5) 2.0 points Consider a cylindrical distribution [dark gray] extending from r = 0 to r = 9 . 7 cm, of charge ρ = r a 0 , where a 0 = 30 cm · m 3 C. This cylindrical charge distribution [dark gray] is surrounded by a dielectric shell [light gray] whose inner radius is 17 cm and outer radius is 2 . 6 cm as shown in the figure. L 2 . 6 cm 17 cm 9 . 7 cm
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maldonado (om2892) – Homework 3 2010 UPII – Stone – (13750) 2 What is the electric field at 3 . 66 cm? As- sume the length L is very long compared to the diameter of the dielectric shell. Neglect edge effects. Correct answer: 168 . 101 V / m. Explanation: Let : r = 3 . 66 cm . Gauss’ law contintegraldisplay vector E dvectora = Q enc ǫ 0 = 1 ǫ 0 integraldisplay r 0 ρ ( r ) dV where dV = d ( π r 2 L ) = 2 π r L dr . In our case, Gauss’ law gives 2 π r L E = 1 ǫ 0 a 0 integraldisplay r 0 r · 2 π r L dr, E = r 2 3 ǫ 0 a 0 = 1 8 . 85419 × 10 - 12 C 2 / N · m 2 × (3 . 66 cm) 2 3 (30 cm · m 3 C) × 1 m 100 cm × 1 C 10 6 μ C = 168 . 101 V / m . 003 (part 2 of 5) 2.0 points What is the electric field at 12 . 7 cm? Correct answer: 901 . 821 V / m. Explanation: Let : r = 12 . 7 cm . Now the integration in the RHS of Gauss’ law goes up to the boundary of our charge distribution 2 π r E = 1 ǫ 0 a 0 integraldisplay R 0 r · 2 π r dr, giving E = R 3 3 ǫ 0 a 0 r = 1 8 . 85419 × 10 - 12 C 2 / N · m 2 × (9 . 7 cm) 3 3 (30 cm · m 3 C) (12 . 7 cm) × 1 m 100 cm × 1 C 10 6 μ C = 901 . 821 V / m . 004 (part 3 of 5) 2.0 points What is the electric field at 20 . 9 cm? Correct answer: 307 . 863 V / m. Explanation: Let : r = 20 . 9 cm and κ = 1 . 78 .
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