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Unformatted text preview: Platt, David – Homework 3 – Due: Sep 12 2005, 4:00 am – Inst: Ken Shih 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Note for numeric problems: 1,As far as a magnitude value is concerned, it always means the absolute value. Be careful of the sign of your answer. 2,Please check student’s instruction concerning the significant figure and precision etc in case you do the right way but get the ”wrong” answer. 001 (part 1 of 1) 10 points A rod 9 . 7 cm long is uniformly charged and has a total charge of 27 μ C. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Determine the magnitude of the electric field along the axis of the rod at a point 18 . 2357 cm from the center of the rod. Correct answer: 7 . 85274 × 10 6 N / C. Explanation: Let : ‘ = 9 . 7 cm , Q = 27 μ C , r = 18 . 2357 cm , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . For a rod of length ‘ and linear charge density (charge per unit length) λ , the field at a dis tance d from the end of the rod along the axis is E = k e Z d + ‘ d λ x 2 dx = k e λ x fl fl fl fl d + ‘ d = k e λ‘ d ( ‘ + d ) , where dq = λdx . The linear charge density (if the total charge is Q ) is λ = Q ‘ so that E = k e Q ‘ ‘ d ( ‘ + d ) = k e Q d ( ‘ + d ) . In this problem, we have the following situa tion (the distance r from the center is given): d l r r The distance d is d = r ‘ 2 = 18 . 2357 cm 9 . 7 cm 2 = 0 . 133857 m , and the magnitude of the electric field is E = k e Q d ( ‘ + d ) = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) ×  2 . 7 × 10 5 C  (0 . 133857 m)(0 . 097 m + 0 . 133857 m) = 7 . 85274 × 10 6 N / C . Now, the direction must be toward the rod, since the charge distribution is negative (a positive test charge would be attracted). So the sign should be positive, according to the convention stated in the problem. 002 (part 1 of 1) 10 points A circular arc has a uniform linear charge density of 9 nC / m. The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . 1 3 6 ◦ 2 . 4 m x y Platt, David – Homework 3 – Due: Sep 12 2005, 4:00 am – Inst: Ken Shih 2 What is the magnitude of the electric field at the center of the circle along which the arc lies? Correct answer: 62 . 4983 N / C. Explanation: Let : λ = 9 nC / m = 9 × 10 9 C / m , Δ θ = 136 ◦ , and r = 2 . 4 m . θ is defined as the angle in the counter clockwise direction from the positive x axis as shown in the figure below....
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 Fall '08
 Turner
 Physics, Work, Electric charge, Platt, Ken Shih

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