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Unformatted text preview: soto (rrs946) – Homework 2 – Hoffmann – (56605) 1 This printout should have 20 questions. Multiplechoice 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 μ 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 : ℓ = 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 . For a rod of length ℓ 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 integraldisplay d + ℓ d λ x 2 dx = k e λ x vextendsingle vextendsingle vextendsingle vextendsingle d + ℓ d = k e λ ℓ d ( ℓ + d ) , where dq = λ dx . The linear charge density (if the total charge is Q ) is λ = Q ℓ so that E = k e Q ℓ ℓ d ( ℓ + d ) = k e Q d ( ℓ + 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 ℓ 2 = 0 . 317287 m . 188 m 2 = 0 . 223287 m , and the magnitude of the electric field is E = k e Q d ( ℓ + 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 μ 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 μ C / m 2 . Explanation: Let : m = 15 . 1 g = 0 . 0151 kg , q = . 5 μ C = 5 × 10 7 C , g = 9 . 8 m / s 2 , and ǫ = 8 . 85419 × 10 12 C 2 / N / m 2 ....
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 Fall '11
 HOFFMAN
 Charge, Work

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