HW 2 - Kapoor (mk9499) – oldhomework 02 – Turner –...

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Unformatted text preview: Kapoor (mk9499) – oldhomework 02 – Turner – (60230) 1 This print-out should have 13 questions. Multiple-choice 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 x-axis. 4. in quadrant IV. correct 5. along the negative y-axis. 6. along the positive y-axis. 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. 002 (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 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 arc-length 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 x-axis, 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.

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HW 2 - Kapoor (mk9499) – oldhomework 02 – Turner –...

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