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Unformatted text preview: aldawsari (ma28683) HW02 Betancourt (16856) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 21.521.7 001 10.0 points A rod 8 cm long is uniformly charged and has a total charge of 22 . 7 C. Determine the magnitude of the elec tric field along the axis of the rod at a point 54 . 6761 cm from the center of the rod. The Coulomb constant is 8 . 98755 10 9 N m 2 / C 2 . Correct answer: 6 . 86124 10 5 N / C. Explanation: Let : = 8 cm , Q = 22 . 7 C , r = 54 . 6761 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) , the field at a dis tance d from the end of the rod along the axis is E = k e integraldisplay d + d x 2 dx = k e x vextendsingle vextendsingle vextendsingle vextendsingle d + d = k e d ( + d ) , where dq = dx . The linear charge density (if the total charge is Q ) is = 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 = 54 . 6761 cm 8 cm 2 = 0 . 506761 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 . 27 10 5 C  (0 . 506761 m)(0 . 08 m + 0 . 506761 m) = 6 . 86124 10 5 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. keywords: 002 (part 1 of 3) 10.0 points Consider the setup shown in the figure be low, where the arc is a semicircle with radius r . The total charge Q is negative, and dis tributed uniformly on the semicircle. The charge on a small segment with angle is labeled q . x y r x y I II III IV B A O q is given by 1. q = Q correct aldawsari (ma28683) HW02 Betancourt (16856) 2 2. q = Q 3. q = 2 Q 4. None of these 5. q = Q 2 6. q = Q 7. q = Q 8. q = 2 Q 9. q = Q 2 10. q = 2 Q Explanation: The angle of a semicircle is , thus the charge on a small segment with angle is q = Q ....
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This note was uploaded on 01/27/2011 for the course PHYS 251 taught by Professor Gavrine during the Spring '10 term at Purdue UniversityWest Lafayette.
 Spring '10
 Gavrine
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