Homework 2-solutions - soto(rrs946 Homework 2 Homann(56605...

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soto (rrs946) – Homework 2 – Hofmann – (56605) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices be±ore answering. 001 10.0 points A rod 18 . 8 cm long is uni±ormly charged and has a total charge o± - 22 . 1 μ C. ²ind the magnitude o± the electric Feld along the axis o± the rod at a point 31 . 7287 cm ±rom the center o± 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 μ 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 . ²or a rod o± length and linear charge density (charge per unit length) λ , the Feld at a dis- tance d ±rom the end o± the rod along the axis is E = k e i d + d λ x 2 dx = k e - λ x v v v v d + d = k e λ ℓ d ( + d ) , where dq = λ dx . The linear charge density (i± 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 ±ollowing situa- tion (the distance r ±rom the center is given): d l r r The distance d is d = r - 2 = 0 . 317287 m - 0 . 188 m 2 = 0 . 223287 m , and the magnitude o± the electric Feld 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 o± Styro±oam carries a net charge o± - 0 . 5 μ C and ³oats above the center o± a very large horizontal sheet o± plastic that has a uni±orm charge density on its sur±ace. What is the charge per unit area on the plastic sheet? The acceleration o± gravity is 9 . 8 m / s 2 and the permittivity o± ±ree space is 8 . 85419 × 10 - 12 C 2 / N / m 2 . Correct answer: - 5 . 24097 μ C / m 2 . Explanation: Let : m = 15 . 1 g = 0 . 0151 kg , q = - 0 . 5 μ C = - 5 × 10 - 7 C , g = 9 . 8 m / s 2 , and ǫ 0 = 8 . 85419 × 10 - 12 C 2 / N / m 2 . The Feld E = σ 2 ǫ 0 due to a nonconduct- ing inFnite sheet o± charge is the same as that very close to any plane uni±orm charge distri- bution, where σ is the sur±ace charge density (charge per unit area) o± the plastic sheet.
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Homework 2-solutions - soto(rrs946 Homework 2 Homann(56605...

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