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Unformatted text preview: Kapoor (mk9499) – oldhomework 02 – Turner – (60230) 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (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. 2. in quadrant II. 3. along the negative xaxis. 4. in quadrant IV. correct 5. along the negative yaxis. 6. along the positive yaxis. 7. in quadrant III. 8. along the positive xaxis. 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. 002 (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 R 2 Δ θ sin θ 3. Δ E x = k Q R 2 2 Δ θ π sin θ 4. Δ E x = k Q R 2 Δ θ π cos θ 5. Δ E x = 0 6. Δ E x = k Q R 2 2 Δ θ cos θ 7. Δ E x = k Q R Δ θ π sin θ 8. Δ E x = k Q 2 R 2 2 Δ θ π cos θ 9. Δ E x = k Q 2 R 2 Δ θ π cos θ 10. Δ E x = k Q R 2 2 Δ θ π cos θ correct Explanation: Δ E = k Δ q R 2 . The charge Δ q per unit arclength Δ s is λ ≡ Q π 2 R = 2 Q π R , so Δ q = λ Δ s = λ R Δ θ = 2 Q π R R Δ θ = 2 Q π Δ θ . On the other hand, the vector Δ vector E makes an angle θ with the xaxis, so Δ E x = k Q R 2 2 Δ θ π cos θ . 003 (part 3 of 4) 10.0 points Find E x , the electric field at the origin due to Kapoor (mk9499) – oldhomework 02 – Turner – (60230) 2 the full arc length for the case where Q = 4 μ C and R = 1 . 87 m . Correct answer: 6544 . 83 N / C. Explanation: Let : Q = 4 μ C and R = 1 . 87 m . E x can be found by integrating the contribu tions of all the Δ q ’s in the arc. We get E x = integraldisplay π/ 2 k Q R 2 2 π cos θ dθ = 2 k Q π R 2 = 2 k (4 μ C) π (1 . 87 m) 2 = 6544 . 83 N / C ....
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This note was uploaded on 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Charge, Work

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