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Unformatted text preview: klarin (sjk772) – oldhomework 03 – Turner – (56705) 1 This printout should have 14 questions. Multiplechoice 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 . 7 1 ◦ 2 . 6 m x y What is the magnitude of the electric field at the center of the circle along which the arc lies? Correct answer: 24 . 0881 N / C. Explanation: Let : λ = 6 nC / m = 6 × 10 − 9 C / m , Δ θ = 71 ◦ , and r = 2 . 6 m . θ is defined as the angle in the counter clockwise direction from the positive x axis as shown in the figure below. 3 5 . 5 ◦ 3 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 yaxis, 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 righthalf 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 ◦ − 35 . 5 ◦ = 54 . 5 ◦ , is the angle from the positive x axis to the righthand end of the arc. E = − 2 k e parenleftBigg λ r integraldisplay 90 ◦ 54 . 5 ◦ sin θ dθ parenrightBigg ˆ = − 2 k e λ r [cos (54 . 5 ◦ ) − cos (90 ◦ )] ˆ . Since k e λ r = (8 . 98755 × 10 9 N · m 2 / C 2 ) × (6 × 10 − 9 C / m) (2 . 6 m) = 20 . 7405 N / C , E = − 2 (20 . 7405 N / C) × [(0 . 580703) − (0)] ˆ = − 24 . 0881 N / C ˆ bardbl vector E bardbl = 24 . 0881 N / C . klarin (sjk772) – 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 71 ◦ ◦ cos θ dθ parenrightBigg ˆ ı = − k e λ r [sin (71 ◦ ) − sin (0 ◦ )] ˆ ı = − (20 . 7405 N / C) × [(0 . 945519) − . 0] ˆ ı = [ − 19 . 6105 N / C] ˆ ı, E y = − parenleftBigg k e λ r integraldisplay 71 ◦ ◦ sin θ dθ parenrightBigg ˆ = − k e λ r [cos (0 ◦ ) − cos (71 ◦ )] ˆ = − (20 . 7405 N / C) × [1 . − (0 . 325568)] ˆ = [ − 13 . 9881 N / C] ˆ , bardbl vector E bardbl = radicalBig E 2 x + E 2 y = bracketleftBig...
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This note was uploaded on 10/10/2010 for the course PHY 56705 taught by Professor Turner during the Fall '10 term at University of Texas at Austin.
 Fall '10
 Turner
 Charge, Work

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