Midterm_A-solutions.pdf - Version 047 Midterm A...

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Version 047 – Midterm A – turner – (90130) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Two charges q 1 and q 2 are separated by a distance d and exert a force F on each other. What is the new force F , if charge 1 is increased to q 1 = 5 q 1 , charge 2 is decreased to q 2 = q 2 2 , and the distance is decreased to d = d 2 ? 1. F = 25 4 F 2. F = 5 F 3. F = 100 F 4. F = 10 F correct 5. F = 20 F 6. F = 25 F 7. F = 5 2 F 8. F = 5 4 F 9. F = 50 F 10. F = 25 2 F Explanation: F = k q 1 q 2 r 2 = k (5 q 1 ) parenleftbigg q 2 2 parenrightbigg parenleftbigg d 2 parenrightbigg 2 = 10 k q 1 q 2 d 2 = 10 F . 002 10.0points A circular arc has a uniform linear charge density of 3 nC / m. 223 2 . 3 m x y 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 9 N · m 2 / C 2 . 1. 21.8144 2. 31.3313 3. 58.9899 4. 54.3819 5. 39.2559 6. 40.5604 7. 18.1544 8. 42.8784 9. 21.046 10. 19.1579 Correct answer: 21 . 8144 N / C. Explanation: Let : λ = 3 nC / m = 3 × 10 9 C / m , Δ θ = 223 , and r = 2 . 3 m . Position the arc symmetrically around the y axis, centered at the origin. 111 . 5 111 . 5 r vector 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,
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Version 047 – Midterm A – turner – (90130) 2 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, 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 . The upper angular limit is θ = 90 and the lower angular limit is θ = 90 111 . 5 = 21 . 5 , the angle from the positive x axis to the right-hand end of the arc. E = 2 k e parenleftBigg λ r integraldisplay 90 21 . 5 sin θ dθ parenrightBigg ˆ = 2 k e λ r [cos ( 21 . 5 ) cos 90 ] ˆ  . Since k e λ r = (8 . 98755 × 10 9 N · m 2 / C 2 ) × 3 × 10 9 C / m 2 . 3 m = 11 . 7229 N / C , E = 2 (11 . 7229 N / C) (0 . 930418 0) ˆ = 21 . 8144 N / C ˆ bardbl vector E bardbl = 21 . 8144 N / C . Alternate Solution: Solve for bardbl vector E bardbl 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 = parenleftBigg k e λ r integraldisplay 223 0 cos θ dθ parenrightBigg ˆ ı = k e λ r [sin 223 sin 0 ] ˆ ı = (11 . 7229 N / C) ( 0 . 682 0 . 0) ˆ ı = (7 . 99501 N / C) ˆ ı , E y = parenleftBigg k e λ r integraldisplay 223 0 sin θ dθ parenrightBigg ˆ = k e λ r [cos 0 cos 223 ] ˆ = (11 . 7229 N / C) (1 . 0 − − 0 . 731353) ˆ = ( 20 . 2965 N / C) ˆ  , bardbl vector E bardbl = radicalBig E 2 x + E 2 y = bracketleftBig (7 . 99501 N / C) 2 + ( 20 . 2965 N / C) 2
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