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Unformatted text preview: nguyen (jmn727) – oldfinal 01 – Turner – (59070) 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 2 6 c m 2 6 c m 90 ◦ 5 μ C − 9 μ C 9 μ 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: 8 . 46096 N. Explanation: Let : Q = 9 μ C , q = 5 μ C , and L = 26 cm . L θ 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 ) × (5 μ C) (9 μ C) (26 cm) 2 √ 2 2 = 8 . 46096 N . keywords: 002 10.0 points A circular arc has a uniform linear charge density of 3 nC / m. The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . 1 5 4 ◦ 2 . 7 m x y What is the magnitude of the electric field at the center of the circle along which the arc lies? Correct answer: 19 . 4604 N / C. Explanation: Let : λ = 3 nC / m = 3 × 10 − 9 C / m , Δ θ = 154 ◦ , and r = 2 . 7 m . θ is defined as the angle in the counter clockwise direction from the positive x axis as shown in the figure below. 7 7 ◦ 7 7 ◦ r vector E θ nguyen (jmn727) – oldfinal 01 – Turner – (59070) 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 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 ◦ − 77 ◦ = 13 ◦ , is the angle from the positive x axis to the righthand end of the arc. E = − 2 k e parenleftBigg λ r integraldisplay 90 ◦ 13 ◦ sin θ dθ parenrightBigg ˆ = − 2 k e λ r [cos (13 ◦ ) − cos (90 ◦ )] ˆ . Since k e λ r = (8 . 98755 × 10 9 N · m 2 / C 2 ) × (3 × 10 − 9 C / m) (2 . 7 m) = 9 . 98617 N / C , E = − 2 (9 . 98617 N / C) × [(0 . 97437) − (0)] ˆ = − 19 . 4604 N / C ˆ bardbl vector E bardbl = 19 . 4604 N / C ....
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This note was uploaded on 01/21/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics

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