solution2_pdf

# solution2_pdf - sicher(gns249 H2ElFieldConti Avram(1956...

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sicher (gns249) – H2ElFieldConti – Avram – (1956) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A rod 13 . 9 cm long is uniformly charged and has a total charge of 12 . 2 μ C. Determine the magnitude of the elec- tric field along the axis of the rod at a point 40 . 9326 cm from the center of the rod. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 6 . 73856 × 10 5 N / C. Explanation: Let : = 13 . 9 cm , Q = 12 . 2 μ C , r = 40 . 9326 cm , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . For a rod of length 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 + d λ x 2 dx = k e λ x vextendsingle vextendsingle vextendsingle vextendsingle d + d = k e λ ℓ d ( + d ) , where dq = λ dx . The linear charge density (if 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 following situa- tion (the distance r from the center is given): d l r r The distance d is d = r 2 = 40 . 9326 cm 13 . 9 cm 2 = 0 . 339826 m , and the magnitude of the electric field is E = k e Q d ( + d ) = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × |− 1 . 22 × 10 5 C | (0 . 339826 m)(0 . 139 m + 0 . 339826 m) = 6 . 73856 × 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 uniformly charged circular arc AB of radius R is shown in the figure. It covers a quarter of a circle and it is located in the second quadrant. The total charge on the arc is Q > 0. x y + + + + + + + + + Δ θ θ R x y I II III IV B A O Δ s R Δ θ The direction of the electric field E due to the charge distribution at the origin is 1. in quadrant I.

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sicher (gns249) – H2ElFieldConti – Avram – (1956) 2 2. along the negative y -axis. 3. along the negative x -axis. 4. in quadrant II. 5. along the positive y -axis. 6. in quadrant IV. correct 7. in quadrant III. 8. along the positive x -axis. Explanation: The electric field for a positive charge is directed away from it. In this case, the electric field generated by each Δ q is going to be in quadrant IV, so the total electric field will be in the same quadrant. 003 (part 2 of 4) 10.0 points Determine Δ E x , the x -component of the elec- tric field vector at the origin O due to the charge element Δ q located at an angle θ sub- tended by an angular interval Δ θ . 1. Δ E x = k Q 2 R 2 Δ θ π cos θ 2. Δ E x = k Q R 2 Δ θ sin θ 3. Δ E x = k Q 2 R Δ θ cos θ 4. Δ E x = k Q R 2 2 Δ θ π cos θ correct 5. Δ E x = k Q R 2 2 Δ θ cos θ 6. Δ E x = k Q 2 R 2 2 Δ θ π cos θ 7. Δ E x = 0 8. Δ E x = k Q R Δ θ π sin θ 9. Δ E x = k Q R 2 2 Δ θ π sin θ 10. Δ E x = k Q R 2 Δ θ π cos θ Explanation: Δ E = k Δ q R 2 .
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