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# OFINAL - terry(ect328 oldnal 01 Turner(59130 This print-out...

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terry (ect328) – oldfinal 01 – Turner – (59130) 1 This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points 47 cm 47 cm 90 4 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: 2 . 07138 N. Explanation: Let : Q = 9 μ C , q = 4 μ C , and L = 47 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 ) × (4 μ C) (9 μ 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 . 54 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 : λ = 9 nC / m = 9 × 10 9 C / m , Δ θ = 54 , 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. 27 27 r vector E θ

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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 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 27 = 63 , 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 63 sin θ dθ parenrightBigg ˆ = 2 k e λ r [cos (63 ) cos (90 )] ˆ  . Since k e λ 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)] ˆ = 29 . 3779 N / C ˆ bardbl vector E bardbl = 29 . 3779 N / C . 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).
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OFINAL - terry(ect328 oldnal 01 Turner(59130 This print-out...

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