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Unformatted text preview: yang (ey942) 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 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 : = 8 nC / m = 8 10 9 C / m , = 231 , and r = 2 . 5 m . 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 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 r In polar coordinates dq = ( r d ) , where is the linear charge density. The positive y axis is = 90 , so the y component of the electric field is given by dE y = dE sin . 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 = 90 . The lower angular limit = 90 115 . 5 = 25 . 5 , is the angle from the positive x axis to the righthand end of the arc. E = 2 k e parenleftBigg r integraldisplay 90 25 . 5 sin d parenrightBigg = 2 k e r [cos ( 25 . 5 ) cos (90 )] . Since k e 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)] = 51 . 917 N / C 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). is still defined as the angle in the counter clockwise direction from the positive x axis. E x = parenleftBigg k e r integraldisplay 231 cos d parenrightBigg = k e r [sin (231 ) sin(0 )] = ( 28 . 7602 N / C) [( . 777147) . 0] = [ 22 . 3509 N / C] , E y = parenleftBigg k e r integraldisplay 231 sin d parenrightBigg = k e r [cos (0 ) cos (231 )] = ( 28 . 7602 N / C) [1 . ( . 629319)] = [46 . 8595 N / C] , bardbl...
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 Spring '10
 Turner

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