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# HW 3 - Kapoor(mk9499 oldhomework 03 Turner(60230 This...

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Kapoor (mk9499) – oldhomework 03 – Turner – (60230) 1 This print-out should have 14 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. along the negative x -axis. 2. in quadrant IV. correct 3. along the negative y -axis. 4. along the positive y -axis. 5. in quadrant I. 6. in quadrant II. 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 R Δ θ π sin θ 2. Δ E x = k Q R 2 2 Δ θ π sin θ 3. Δ E x = k Q R 2 2 Δ θ π cos θ correct 4. Δ E x = k Q R 2 2 Δ θ cos θ 5. Δ E x = k Q 2 R 2 Δ θ π cos θ 6. Δ E x = k Q R 2 Δ θ π cos θ 7. Δ E x = 0 8. Δ E x = k Q R 2 Δ θ sin θ 9. Δ E x = k Q 2 R Δ θ cos θ 10. Δ E x = k Q 2 R 2 2 Δ θ π cos θ 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

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Kapoor (mk9499) – oldhomework 03 – Turner – (60230) 2 to the full arc length for the case where Q = 3 . 7 μ C and R = 1 . 74 m . Correct answer: 6992 . 38 N / C. Explanation: Let : Q = 3 . 7 μ C and R = 1 . 74 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 0 k Q R 2 2 π cos θ dθ = 2 k Q π R 2 = 2 k (3 . 7 μ C) π (1 . 74 m) 2 = 6992 . 38 N / C . 004 (part 4 of 4) 10.0 points The charge on an electron is e . If at O we have E x , then the magnitude of the force bardbl vector F e bardbl on an electron at this point is 1. bardbl vector F e bardbl = E x 2 e . 2. bardbl vector F e bardbl = 4 e E x . 3. bardbl vector F e bardbl = e E x . 4. bardbl vector F e bardbl = 2 E x 3 e . 5. bardbl vector F e bardbl = 3 e E x 2 . 6. bardbl vector F e bardbl = E x 3 e . 7. bardbl vector F e bardbl = 2 e E x . correct 8. bardbl vector F e bardbl = E x e 2 . 9. bardbl vector F e bardbl = 3 e E x . 10. bardbl vector F e bardbl = 2 e E x . Explanation: E x E y E From the symmetry of the charge distribu- tion, it can be seen that E y = E x , so E = radicalBig E 2 x + E 2 y = 2 E x .
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HW 3 - Kapoor(mk9499 oldhomework 03 Turner(60230 This...

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