# final 02 - Version 063/AADDD – final 02 – Turner...

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Unformatted text preview: Version 063/AADDD – final 02 – Turner – (59070) 1 This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A plane convex lens is made of glass (index 1 . 73) with one flat surface and the other hav- ing a radius of 8 . 6 cm. What is the focal length of the lens? 1. 109.756 2. 79.2453 3. 17.7419 4. 18.3099 5. 11.7808 6. 67.5676 7. 69.4118 8. 27.5862 9. 82.3529 10. 56.25 Correct answer: 11 . 7808 cm. Explanation: Basic Concepts: 1 f = ( n − 1) parenleftbigg 1 R 1 − 1 R 2 parenrightbigg Solution: From the lens formula 1 f = ( n − 1) parenleftbigg 1 R 1 − 1 R 2 parenrightbigg ⇒ f = R n − 1 = (8 . 6 cm) (1 . 73) − 1 = 11 . 7808 cm . 002 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 2 8 ◦ 2 . 5 m x y What is the magnitude of the electric field at the center of the circle along which the arc lies? 1. 8.2676 2. 24.7485 3. 63.1765 4. 25.2701 5. 52.5474 6. 30.1947 7. 16.8852 8. 33.1316 9. 45.1038 10. 22.101 Correct answer: 52 . 5474 N / C. Explanation: Let : λ = 8 nC / m = 8 × 10 − 9 C / m , Δ θ = 228 ◦ , 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. 1 1 4 ◦ 1 1 4 ◦ r vector E θ First, position the arc symmetrically around the y axis, centered at the origin. By Version 063/AADDD – final 02 – Turner – (59070) 2 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 ◦ − 114 ◦ = − 24 ◦ , 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 ◦ − 24 ◦ sin θ dθ parenrightBigg ˆ = − 2 k e λ r [cos ( − 24 ◦ ) − cos (90 ◦ )] ˆ . Since k e λ 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 . 913545) − (0)] ˆ = − 52 . 5474 N / C ˆ bardbl vector E bardbl = 52 . 5474 N / C ....
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final 02 - Version 063/AADDD – final 02 – Turner...

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