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Unformatted text preview: Wong, Lemuel – Homework 7 – Due: Mar 19 2007, 11:00 pm – Inst: Matt Mackie 1 This printout should have 28 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. NOTE: HW is due 11pm central time, which is midnight our time. 001 (part 1 of 1) 10 points Two conducting spheres with diameters of . 53 m and 1 m are separated by a distance that is large compared to the diameters. The spheres are connected by a thin wire and are charged to 9 μ C. The permittivity of a vacuum is 8 . 85419 × 10 12 C 2 / N · m 2 . What is the potential of the system of spheres when the reference potential is taken to be 0 at ∞ ? Correct answer: 105 . 736 kV. Explanation: Let : ² = 8 . 85419 × 10 12 C 2 / N · m 2 , Q = 9 μ C , d 1 = 0 . 53 m , and d 2 = 1 m . Since the spheres are connected by a wire, their potentials will be equal. Since they are separated by a distance much larger than their diameters, V 1 = Q 1 C 1 = V 2 = Q 2 C 2 . Because C = 4 π ² R = 2 π ² d , we obtain Q 1 d 1 = Q 2 d 2 , where d 1 and d 2 are the diameters of the spheres. Since the total charge is Q = Q 1 + Q 2 , Q 1 d 1 = Q Q 1 d 2 Q 1 d 2 d 1 = Q Q 1 Q 1 = Q 1 + d 2 d 1 = 9 μ C 1 + 1 m . 53 m = 3 . 11765 μ C . Q 2 = Q Q 1 = 9 μ C 3 . 11765 μ C = 5 . 88235 μ C . The potential of the system is V 1 = V 2 = Q 1 2 π ² d 1 = 3 . 11765 × 10 6 C 2 π ² (0 . 53 m) · 1 kV 1000 V = 105 . 736 kV . keywords: 002 (part 1 of 1) 10 points A charge of 94 pC is distributed on an isolated spherical conductor that has a 4 cm radius. Point A is 2 . 4 cm from the center of the conductor and point B is 5 cm from the center of the conductor. Determine the electric potential difference V A V B . Correct answer: 4 . 22415 V. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , q = 94 pC = 9 . 4 × 10 11 C , r = 4 cm = 0 . 04 m , and b = 5 cm = 0 . 05 m . Since the electric field is 0 inside of a con ductor, the potential at A with respect to infinity is equal to the potential V A = k e q r Wong, Lemuel – Homework 7 – Due: Mar 19 2007, 11:00 pm – Inst: Matt Mackie 2 at the surface of the sphere, where r is the radius of the spherical conductor. Hence V A V B = k e q µ 1 r 1 b ¶ = (8 . 98755 × 10 9 N · m 2 / C 2 ) × (9 . 4 × 10 11 C) × µ 1 . 04 m 1 . 05 m ¶ = 4 . 22415 V . keywords: 003 (part 1 of 1) 10 points How many electrons should be removed from an initially uncharged spherical conductor of radius 0 . 17 m to produce a potential of 5 . 8 kV at the surface? Correct answer: 6 . 84739 × 10 11 electrons. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , q e = 1 . 60218 × 10 19 C , V = 5 . 8 kV = 5800 V , and r = 0 . 17 m ....
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This note was uploaded on 02/17/2011 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Work

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