Homework 2-solutions - soto (rrs946) Homework 2 Hoffmann...

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Unformatted text preview: soto (rrs946) Homework 2 Hoffmann (56605) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A rod 18 . 8 cm long is uniformly charged and has a total charge of- 22 . 1 C. Find the magnitude of the electric field along the axis of the rod at a point 31 . 7287 cm from the center of the rod. The Coulomb con- stant is 8 . 98755 10 9 N m 2 / C 2 . Correct answer: 2 . 16284 10 6 N / C. Explanation: Let : = 18 . 8 cm = 0 . 188 m , Q =- 22 . 1 C =- 2 . 21 10- 5 C , r = 31 . 7287 cm = 0 . 317287 m , 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 = 0 . 317287 m- . 188 m 2 = 0 . 223287 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 . 21 10- 5 C | (0 . 223287 m)(0 . 188 m + 0 . 223287 m) = 2 . 16284 10 6 N / C . 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 conven- tion stated in the problem. 002 10.0 points A 15 . 1 g piece of Styrofoam carries a net charge of- . 5 C and floats above the center of a very large horizontal sheet of plastic that has a uniform charge density on its surface. What is the charge per unit area on the plastic sheet? The acceleration of gravity is 9 . 8 m / s 2 and the permittivity of free space is 8 . 85419 10- 12 C 2 / N / m 2 . Correct answer:- 5 . 24097 C / m 2 . Explanation: Let : m = 15 . 1 g = 0 . 0151 kg , q =- . 5 C =- 5 10- 7 C , g = 9 . 8 m / s 2 , and = 8 . 85419 10- 12 C 2 / N / m 2 ....
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Homework 2-solutions - soto (rrs946) Homework 2 Hoffmann...

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