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Unformatted text preview: Rodriguez, Brandon Homework 3 Due: Aug 31 2006, 11:00 am Inst: Andrew J Rader 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. Please notice that for your homework to be considered towards your grade, it needs to be submitted one hour before the corresponding recitation starts. Work submitted after this time, but before the DUE DATE on top of this page, will be accepted but not graded. PLEASE REMEMBER THAT YOU MUST CARRY OUT YOUR CALCULA- TIONS TO AT LEAST THREE SIGNIFI- CANT FIGURES. YOUR ANSWER MUST BE WITHIN ONE PERCENT OF THE CORRECT RESULT TO BE MARKED AS CORRECT BY THE SERVER. 001 (part 1 of 1) 7 points A rod 7 . 4 cm long is uniformly charged and has a total charge of- 21 . 9 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 28 . 6155 cm from the center of the rod. Correct answer: 2 . 44458 10 6 N / C. Explanation: Let : = 7 . 4 cm , Q =- 21 . 9 C , r = 28 . 6155 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 Z d + d x 2 dx = k e- x fl fl fl fl 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 = 28 . 6155 cm- 7 . 4 cm 2 = 0 . 249155 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 . 19 10- 5 C | (0 . 249155 m)(0 . 074 m + 0 . 249155 m) = 2 . 44458 10 6 N / C ....
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This note was uploaded on 10/18/2010 for the course PHYSICS 251 taught by Professor Pedon during the Spring '08 term at IUPUI.
- Spring '08