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Unformatted text preview: mcconnell (kam2342) – 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 I. 3. in quadrant IV. correct 4. along the positive y-axis. 5. along the negative y-axis. 6. along the positive x-axis. 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 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 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 arc-length 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 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 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.

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oldhw3_pdf - mcconnell(kam2342 – oldhomework 03 –...

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