# HW3 - yang(ey942 – oldhomework 03 – Turner –(56705 1...

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Unformatted text preview: yang (ey942) – 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 − 8 nC / m. The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . 2 3 1 ◦ 2 . 5 m x y What is the magnitude of the electric field at the center of the circle along which the arc lies? Correct answer: 51 . 917 N / C. Explanation: Let : λ = − 8 nC / m = − 8 × 10 − 9 C / m , Δ θ = 231 ◦ , and r = 2 . 5 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 5 . 5 ◦ 1 1 5 . 5 ◦ r vector 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, vector E = k e integraldisplay 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 ◦ − 115 . 5 ◦ = − 25 . 5 ◦ , is the angle from the positive x axis to the right-hand end of the arc. E = − 2 k e parenleftBigg λ r integraldisplay 90 ◦ − 25 . 5 ◦ sin θ dθ parenrightBigg ˆ = − 2 k e λ r [cos ( − 25 . 5 ◦ ) − cos (90 ◦ )] ˆ . Since k e λ r = (8 . 98755 × 10 9 N · m 2 / C 2 ) × ( − 8 × 10 − 9 C / m) (2 . 5 m) = − 28 . 7602 N / C , E = − 2 ( − 28 . 7602 N / C) × [(0 . 902585) − (0)] ˆ = 51 . 917 N / C ˆ bardbl vector E bardbl = 51 . 917 N / C . yang (ey942) – oldhomework 03 – Turner – (56705) 2 Alternate Solution: Just solve for bardbl vector E bardbl 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 = − parenleftBigg k e λ r integraldisplay 231 ◦ ◦ cos θ dθ parenrightBigg ˆ ı = − k e λ r [sin (231 ◦ ) − sin(0 ◦ )] ˆ ı = − ( − 28 . 7602 N / C) × [( − . 777147) − . 0] ˆ ı = [ − 22 . 3509 N / C] ˆ ı, E y = − parenleftBigg k e λ r integraldisplay 231 ◦ ◦ sin θ dθ parenrightBigg ˆ = − k e λ r [cos (0 ◦ ) − cos (231 ◦ )] ˆ = − ( − 28 . 7602 N / C) × [1 . − ( − . 629319)] ˆ = [46 . 8595 N / C] ˆ , bardbl...
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## This note was uploaded on 11/21/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas.

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HW3 - yang(ey942 – oldhomework 03 – Turner –(56705 1...

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