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Unformatted text preview: zhou (ejz224) – Homework 2 – shih – (57480) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A rod 6 . 7 cm long is uniformly charged and has a total charge of − 22 . 7 μ C. Determine the magnitude of the elec tric field along the axis of the rod at a point 26 . 0079 cm from the center of the rod. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 3 . 06706 × 10 6 N / C. Explanation: Let : ℓ = 6 . 7 cm , Q = − 22 . 7 μ C , r = 26 . 0079 cm , 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 λℓ d ( ℓ + d ) , where dq = λdx . 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 = 26 . 0079 cm − 6 . 7 cm 2 = 0 . 226579 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 . 27 × 10 − 5 C  (0 . 226579 m)(0 . 067 m + 0 . 226579 m) = 3 . 06706 × 10 6 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 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. 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 . What is the charge per unit area on the plastic sheet? Correct answer: − 5 . 24097 μ C / m 2 . Explanation: Let : m = 15 . 1 g , q = − . 5 μ C , g = 9 . 8 m / s 2 , and ǫ = 8 . 85419 × 10 − 12 C 2 / N / m 2 . zhou (ejz224) – Homework 2 – shih – (57480) 2 The field due to a nonconducting infinite sheet of charge is the same as that very close to any plane uniform charge distribution. The field is E = σ 2 ǫ , where σ is the surface charge density (charge per unit area) of the plastic sheet. Call the charge on the styrofoam q , and its mass m . Since the styrofoam is floating, it must be in equilibrium, so the electric force must cancel the force of gravity F e = F g , we have F g = mg F e = q E mg = q σ 2 ǫ , so σ = 2 ǫ mg q = 2 (8 . 85419 × 10 − 12 C 2 / N / m 2 ) × (0 . 0151 kg) (9 . 8 m / s 2 ) − 5 × 10 − 7 C · 10 6 μ C 1 C = − 5 . 24097 μ C / m 2 ....
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This note was uploaded on 04/09/2011 for the course PHY 302l taught by Professor Morrison during the Spring '08 term at University of Texas.
 Spring '08
 morrison
 Physics, Charge, Work

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