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oldhomework 03-solutions

# oldhomework 03-solutions - howard(cah3459 – oldhomework...

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Unformatted text preview: howard (cah3459) – oldhomework 03 – Turner – (56705) 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 10.0 points A circular arc has a uniform linear charge density of- 6 nC / m. The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . 2 2 ◦ 2 . 4 m x y What is the magnitude of the electric field at the center of the circle along which the arc lies? Correct answer: 42 . 2277 N / C. Explanation: Let : λ =- 6 nC / m =- 6 × 10- 9 C / m , Δ θ = 220 ◦ , and r = 2 . 4 m . θ is defined as the angle in the counter- clockwise direction from the positive x axis as shown in the figure below. 1 1 ◦ 1 1 ◦ r # E θ First, position the arc symmetrically around the y axis, centered at the origin. By symmetry (in this rotated configuration) the field in the x direction cancels due to charge from opposites sides of the y-axis, so E x = 0 . For a continuous linear charge distribution, # E = k e dq r 2 ˆ r In polar coordinates dq = λ ( r dθ ) , where λ is the linear charge density. The positive y axis is θ = 90 ◦ , so the y component of the electric field is given by dE y = dE sin θ . Note: By symmetry, each half of the arc about the y axis contributes equally to the electric field at the origin. Hence, we may just consider the right-half of the arc (beginning on the positive y axis and extending towards the positive x axis) and multiply the answer by 2. Note: The upper angular limit θ = 90 ◦ . The lower angular limit θ = 90 ◦- 110 ◦ =- 20 ◦ , is the angle from the positive x axis to the right-hand end of the arc. E =- 2 k e λ r 90 ◦- 20 ◦ sin θ dθ ˆ =- 2 k e λ r [cos (- 20 ◦ )- cos (90 ◦ )] ˆ . Since k e λ r = (8 . 98755 × 10 9 N · m 2 / C 2 ) × (- 6 × 10- 9 C / m) (2 . 4 m) =- 22 . 4689 N / C , E =- 2 (- 22 . 4689 N / C) × [(0 . 939693)- (0)] ˆ = 42 . 2277 N / C ˆ # E = 42 . 2277 N / C . howard (cah3459) – oldhomework 03 – Turner – (56705) 2 Alternate Solution: Just solve for # E in a straight forward manner, positioning the beginning of the arc on the positive x axis (as shown in the original figure in the question). θ is still defined as the angle in the counter- clockwise direction from the positive x axis. E x =- k e λ r 220 ◦ ◦ cos θ dθ ˆ ı =- k e λ r [sin (220 ◦ )- sin(0 ◦ )] ˆ ı =- (- 22 . 4689 N / C) × [(- . 642789)- . 0] ˆ ı = [- 14 . 4427 N / C] ˆ ı, E y =- k e λ r 220 ◦ ◦ sin θ dθ ˆ =- k e λ r [cos (0 ◦ )- cos (220 ◦ )] ˆ =- (- 22 . 4689 N / C) × [1 .- (- . 766043)] ˆ = [39 . 681 N / C] ˆ , # E = E 2 x + E 2 y = (- 14 . 4427 N / C) 2 + (39 . 681 N / C) 2 1 / 2 = 42 . 2277 N / C ....
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