# hw 2 - wilson(kw22588 – HW2 – Mackie –(20224 1 This...

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Unformatted text preview: wilson (kw22588) – HW2 – Mackie – (20224) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A rod 10 . 8 cm long is uniformly charged and has a total charge of- 24 . 8 μ C. Determine the magnitude of the elec- tric field along the axis of the rod at a point 52 . 4197 cm from the center of the rod. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 8 . 19856 × 10 5 N / C. Explanation: Let : ℓ = 10 . 8 cm , Q =- 24 . 8 μ C , r = 52 . 4197 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 = 52 . 4197 cm- 10 . 8 cm 2 = 0 . 470197 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 . 48 × 10 − 5 C | (0 . 470197 m)(0 . 108 m + 0 . 470197 m) = 8 . 19856 × 10 5 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 (part 1 of 4) 10.0 points A 3 . 5 μ C charge is uniformly distributed on a ring of radius 10 . 5 cm. The value of the Coulomb constant is 8 . 99 × 10 9 N · m 2 / C 2 . Find the electric field on the axis at 1 . 5 cm from the center of the ring. Correct answer: 3 . 9554 × 10 5 N / C. Explanation: Let : q = 3 . 5 μ C = 3 . 5 × 10 − 6 C , k = 8 . 99 × 10 9 N · m 2 / C 2 , r = 10 . 5 cm = 0 . 105 m , and l 1 = 1 . 5 cm = 0 . 015 m . wilson (kw22588) – HW2 – Mackie – (20224) 2 R 1 = radicalBig l 2 1 + r 2 = radicalBig (0 . 015 m) 2 + (0 . 105 m) 2 = 0 . 106066 m . The linear charge density of this loop is λ = q 2 π r . The net electric field is the sum over the com- ponents along the axis: E = k integraldisplay 2 π λ r dθ R 2 1 l 1 R 1 = k l 1 R 3 1 (2 π r λ ) = k q l 1 R 3 1 = 8 . 99 × 10 9 N · m 2 / C 2 (0 . 106066 m) 3 × (3 . 5 × 10 − 6 C) (0 . 015 m) = 3 . 9554 × 10 5 N / C ....
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hw 2 - wilson(kw22588 – HW2 – Mackie –(20224 1 This...

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