303L Midterm01

# 303L Midterm01 - Version 118 midterm01 berk(60290 This...

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Version 118 – midterm01 – berk – (60290) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A circular arc has a uniform linear charge density of 6 nC / m. The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . 278 2 . 8 m x y What is the magnitude of the electric field at the center of the circle along which the arc lies? 1. 9.3248 2. 18.4109 3. 24.7485 4. 46.9173 5. 29.3779 6. 25.5445 7. 22.101 8. 25.2701 9. 37.9847 10. 42.2536 Correct answer: 25 . 2701 N / C. Explanation: Let : λ = 6 nC / m = 6 × 10 9 C / m , Δ θ = 278 , and r = 2 . 8 m . θ is defined as the angle in the counter- clockwise direction from the positive x axis as shown in the figure below. 139 139 r vector E θ 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 = 90 139 = 49 , 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 49 sin θ dθ parenrightBigg ˆ = 2 k e λ r [cos ( 49 ) cos (90 )] ˆ  . Since k e λ r = (8 . 98755 × 10 9 N · m 2 / C 2 ) × ( 6 × 10 9 C / m) (2 . 8 m) = 19 . 259 N / C ,

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Version 118 – midterm01 – berk – (60290) 2 E = 2 ( 19 . 259 N / C) × [(0 . 656059) (0)] ˆ = 25 . 2701 N / C ˆ bardbl vector E bardbl = 25 . 2701 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). 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 278 0 cos θ dθ parenrightBigg ˆ ı = k e λ r [sin (278 ) sin (0 )] ˆ ı = ( 19 . 259 N / C) × [( 0 . 990268) 0 . 0] ˆ ı = [ 19 . 0716 N / C] ˆ ı , E y = parenleftBigg k e λ r integraldisplay 278 0 sin θ dθ parenrightBigg ˆ = k e λ r [cos (0 ) cos (278 )] ˆ = ( 19 . 259 N / C) × [1 . 0 (0 . 139175)] ˆ = [16 . 5787 N / C] ˆ  , bardbl vector E bardbl = radicalBig E 2 x + E 2 y = bracketleftBig ( 19 . 0716 N / C) 2 + (16 . 5787 N / C) 2 bracketrightBig 1 / 2 = 25 . 2701 N / C .
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