# hw03b - aljabr(faa335 – Hw03 – Ross –(89251 1 This...

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Unformatted text preview: aljabr (faa335) – Hw03 – Ross – (89251) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A rod 13 . 4 cm long is uniformly charged and has a total charge of- 17 . 6 μ 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 29 . 7851 cm from the center of the rod. Correct answer: 1 . 87805 × 10 6 N / C. Explanation: Let : ℓ = 13 . 4 cm , Q =- 17 . 6 μ C , r = 29 . 7851 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 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 = 29 . 7851 cm- 13 . 4 cm 2 = 0 . 230851 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 ) × |- 1 . 76 × 10- 5 C | (0 . 230851 m)(0 . 134 m + 0 . 230851 m) = 1 . 87805 × 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. 002 (part 1 of 3) 10.0 points Consider the setup shown in the figure be- low, where the arc is a semicircle with radius r . The total charge Q is negative, and dis- tributed uniformly on the semicircle. The charge on a small segment with angle Δ θ is labeled Δ q...
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hw03b - aljabr(faa335 – Hw03 – Ross –(89251 1 This...

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