final_SS_15-solutions.pdf(1).docx

# final_SS_15-solutions.pdf(1).docx - 15 turner(90130 1 This...

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15 – turner – (90130) 1

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This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider three charges in equilibrium. 2 5 μC 5 c m 90 2 5 c m −10 μC 10 μC What is the magnitude of the electrostatic force F on the top charge? The Coulomb constant is 8.9875 × 10 9 N · m 2 /C 2 . 1. 37.8156 2. 3.17756 3. 2.35375 4. 10.1682 5. 2.64797 6. 4.88073 7. 9.15138 8. 3.02445 9. 9.40107 10. 25.4205 Correct answer: 10.1682 N. Explanation: Let : Q = 10 μC=1 × 10 −5 C, q = 5 μC=5 × 10 −6 C, and L = 25 cm = 0.25 m. q L θ −Q Q F = k
e |q 1 ||q 2 | r2 By symmetry and the fact that force exerted Version 030 – final SS on charge q by +Q is repulsive (up to left) and by −Q is attractive (down to left), F y = 0. The x component of the forces on q by Q and −Q are equal in magnitude, so F = 2k e q Q 002 10.0 points A circular arc has a uniform linear charge density of −8 nC/m. y 142◦ 2 What is the magnitude of the electric field at the center of the circle along which the arc lies? The value of the Coulomb constant is 8.98755 × 10 .5 m x 9 N · m 2 /C 2 . 1. 54.3865 2. 19.9282 3. 25.9681 4. 13.0328 5. 14.5726 6. 19.6633 7. 17.552 8. 17.0928 9. 25.378 10. 38.0025 Correct answer: 54.3865 N/C.

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Explanation: Let : λ = −8 nC/m = −8 × 10 −9 C/m, 2 = 2 (8.9875 × 10 9 N · m 2 /C 2 ) × (5 × 10−6 C) (1 × 10−5 C) 2 = 10.1682 N L2 cos 45 (0.25 m)2 . = 2k e q Q L2 2 2

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15 – turner – (90130) 2 ∆θ = 142 , and r = 2.5 m. Position the arc symmetrically around the y axis, centered at the origin. 7 1 7 1 ◦ r E θ By symmetry (in this rotated configura- tion) the field in the x direction cancels due to charge from opposites sides of the y-axis, so E x = 0. Each half of the arc about the y axis would contribute equally to the electric field at the origin, so we may just consider y-contributions of the right half of the arc and double the result. For a continuous linear charge distribution, E = k e dq ˆr. 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 sinθ .
The upper angular limit is θ = 90 and the lower angular limit is θ = 90 , the angle from the positive x axis to the right- hand end of the arc. E = −2k e − 71 = 19 ) ˆ = −2k e λ ] ˆ . Since k e λ ( λ r 2 = (8.98755 × 10 ) × −8 × 2.5 10−9 m C/m = −28.7602 N/C, r

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r ◦ [cos (19 ) − cos 90 90◦ 19◦ Version 030 – final sinθ dθ r2 9 N · m 2 /C SS . Alternate Solution: Solve for E in a straight forward manner, positioning the be- ginning of the arc on the positive x axis (as in the original figure). θ is still defined as the angle in the counter- clockwise direction from the positive x axis. E x ( = − k e λ 003 10.0 points The electron gun in a television tube is used to accelerate electrons with mass 9.109 × 10 −31 kg from rest to 2 × 10 7 m/s within a distance of 1.9 cm. Assuming the electric field is constant, what electric field is required in the tube? The fundamental charge is q e
.

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