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Unformatted text preview: mcconnell (kam2342) – oldhomework 03 – Turner – (60230) 1 This printout should have 14 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. along the negative xaxis. 2. in quadrant I. 3. in quadrant IV. correct 4. along the positive yaxis. 5. along the negative yaxis. 6. along the positive xaxis. 7. in quadrant II. 8. in quadrant III. 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 2 Δ θ π cos θ 2. Δ E x = k Q R 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 2 R 2 2 Δ θ π cos θ 5. Δ E x = k Q R 2 Δ θ π cos θ 6. Δ E x = k Q R 2 2 Δ θ cos θ 7. Δ E x = k Q R 2 Δ θ sin θ 8. Δ E x = 0 9. Δ E x = k Q 2 R Δ θ cos θ 10. Δ E x = k Q R 2 2 Δ θ π sin θ 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 = λR 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 mcconnell (kam2342) – 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 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 e √ 2 . 2. bardbl vector F e bardbl = 4 eE x . 3. bardbl vector F e bardbl = eE x . 4. bardbl vector F e bardbl = E x 3 e ....
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This note was uploaded on 11/22/2010 for the course PHYS 303 taught by Professor Turner during the Spring '10 term at University of Texas.
 Spring '10
 Turner

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