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Unformatted text preview: vo (ytv59) – oldhomework 03 – Turner – (56705) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A uniformly charged insulating rod of length 16 . 2 cm is bent into the shape of a semicircle as in the figure. The value of the Coulomb constant is 8 . 98755 × 10 9 N m 2 / C 2 . 1 6 . 2 c m − 7 . 19 μ C O If the rod has a total charge of − 7 . 19 μ C, find the horizontal component of the electric field at O , the center of the semicircle. Define right as positive. Correct answer: − 1 . 54711 × 10 7 N / C. Explanation: Let : L = 16 . 2 cm = 0 . 162 m and q = − 7 . 19 μ C = − 7 . 19 × 10 − 6 C . Call the length of the rod L and its charge q . Due to symmetry E y = integraldisplay dE y = 0 and E x = integraldisplay dE sin θ = k e integraldisplay dq sin θ r 2 , where dq = λdx = λr dθ , so that b y x θ E x = − k e λ r integraldisplay 3 π/ 2 π/ 2 cos θ dθ = − k e λ r (sin θ ) vextendsingle vextendsingle vextendsingle vextendsingle 3 π/ 2 π/ 2 = 2 k e λ r , where λ = q L and r = L π . Therefore, E x = 2 k e q π L 2 = 2 (8 . 98755 × 10 9 N m 2 / C 2 ) (0 . 162 m) 2 × ( − 7 . 19 × 10 − 6 C) π = − 1 . 54711 × 10 7 N / C . Since the rod has negative charge, the field is pointing to the left (towards the charge dis tribution). A positive test charge at O would feel an attractive force from the semicircle, pointing to the left. 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 IV. correct 2. along the positive xaxis. vo (ytv59) – oldhomework 03 – Turner – (56705) 2 3. along the negative yaxis. 4. along the negative xaxis. 5. in quadrant II. 6. in quadrant III. 7. along the positive yaxis. 8. in quadrant I. 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 xcomponent 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 Δ θ cos θ 2. Δ E x = k Q 2 R 2 2 Δ θ π cos θ 3. Δ E x = k Q R 2 2 Δ θ cos θ 4. Δ E x = k Q R 2 2 Δ θ π sin θ 5. Δ E x = k Q R 2 Δ θ sin θ 6. Δ E x = k Q R 2 2 Δ θ π cos θ correct 7. Δ E x = k Q R Δ θ π sin θ 8. Δ E x = 0 9. Δ E x = k Q 2 R 2 Δ θ π cos 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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 Fall '10
 turner
 Charge, Work, Electric charge, KE

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