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Unformatted text preview: terry (ect328) oldfinal 01 Turner (59130) 1 This printout should have 31 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points 4 7 c m 4 7 c m 90 4 C 9 C 9 C What is the magnitude of the electrostatic force bardbl vector F bardbl on the top charge? The Coulomb constant is 8 . 9875 10 9 N m 2 / C 2 . Correct answer: 2 . 07138 N. Explanation: Let : Q = 9 C , q = 4 C , and L = 47 cm . L q Q Q bardbl vector F bardbl = k e  q 1  q 2  r 2 By symmetry and the fact that force on charge q by + Q is repulsive and by Q is attractive, F y = 0 . The x component of the forces on q by Q and Q are equal in magnitude and direction. Note: cos 45 = 2 2 . Hence, bardbl vector F bardbl = 2 k e q Q L 2 2 2 = 2 (8 . 9875 10 9 N m 2 / C 2 ) (4 C) (9 C) (47 cm) 2 2 2 = 2 . 07138 N . keywords: 002 10.0 points A circular arc has a uniform linear charge density of 9 nC / m. The value of the Coulomb constant is 8 . 98755 10 9 N m 2 / C 2 . 5 4 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: 29 . 3779 N / C. Explanation: Let : = 9 nC / m = 9 10 9 C / m , = 54 , 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. 2 7 2 7 r vector E terry (ect328) oldfinal 01 Turner (59130) 2 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 27 = 63 , is the angle from the positive x axis to the righthand end of the arc. E = 2 k e parenleftBigg r integraldisplay 90 63 sin d parenrightBigg = 2 k e r [cos (63 ) cos (90 )] . Since k e r = (8 . 98755 10 9 N m 2 / C 2 ) (9 10 9 C / m) (2 . 5 m) = 32 . 3552 N / C , E = 2 (32 . 3552 N / C) [(0 . 45399) (0)] = 29 . 3779 N / C bardbl vector E bardbl = 29 . 3779 N / C ....
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This note was uploaded on 02/17/2010 for the course PHY 59130 taught by Professor Turner during the Fall '09 term at University of Texas at Austin.
 Fall '09
 Turner

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