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Unformatted text preview: Version 118 – midterm01 – berk – (60290) 1 This printout should have 15 questions. Multiplechoice 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 . 2 7 8 ◦ 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. 1 3 9 ◦ 1 3 9 ◦ 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 yaxis, 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 righthalf 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 righthand 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 , 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 ◦ ◦ cos θ dθ parenrightBigg ˆ ı = − k e λ r [sin (278 ◦ ) − sin(0 ◦ )] ˆ ı = − ( − 19 . 259 N / C) × [( − . 990268) − . 0] ˆ ı = [ − 19 . 0716 N / C] ˆ ı , E y = − parenleftBigg k e λ r integraldisplay 278 ◦ ◦ sin θ dθ parenrightBigg ˆ = − k e λ r [cos (0 ◦ ) − cos (278 ◦ )] ˆ = − ( − 19 . 259 N / C) ×...
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This note was uploaded on 09/20/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
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