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Unformatted text preview: vo (ytv59) oldhomework 03 Turner (56705) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A uniformly charged insulating rod of length 16 . 2 cm is bent into the shape of a semicircle as in the figure. The value of the Coulomb constant is 8 . 98755 10 9 N m 2 / C 2 . 1 6 . 2 c m 7 . 19 C O If the rod has a total charge of 7 . 19 C, find the horizontal component of the electric field at O , the center of the semicircle. Define right as positive. Correct answer: 1 . 54711 10 7 N / C. Explanation: Let : L = 16 . 2 cm = 0 . 162 m and q = 7 . 19 C = 7 . 19 10 6 C . Call the length of the rod L and its charge q . Due to symmetry E y = integraldisplay dE y = 0 and E x = integraldisplay dE sin = k e integraldisplay dq sin r 2 , where dq = dx = r d , so that b y x E x = k e r integraldisplay 3 / 2 / 2 cos d = k e r (sin ) vextendsingle vextendsingle vextendsingle vextendsingle 3 / 2 / 2 = 2 k e r , where = q L and r = L . Therefore, E x = 2 k e q L 2 = 2 (8 . 98755 10 9 N m 2 / C 2 ) (0 . 162 m) 2 ( 7 . 19 10 6 C) = 1 . 54711 10 7 N / C . Since the rod has negative charge, the field is pointing to the left (towards the charge dis- tribution). A positive test charge at O would feel an attractive force from the semicircle, pointing to the left. 002 (part 1 of 4) 10.0 points A uniformly charged circular arc AB of radius R is shown in the figure. It covers a quarter of a circle and it is located in the second quadrant. The total charge on the arc is Q > 0. x y + + + + + + + + + R x y I II III IV B A O s R The direction of the electric field E due to the charge distribution at the origin is 1. in quadrant IV. correct 2. along the positive x-axis. vo (ytv59) oldhomework 03 Turner (56705) 2 3. along the negative y-axis. 4. along the negative x-axis. 5. in quadrant II. 6. in quadrant III. 7. along the positive y-axis. 8. in quadrant I. Explanation: The electric field for a positive charge is directed away from it. In this case, the electric field generated by each q is going to be in quadrant IV, so the total electric field will be in the same quadrant. 003 (part 2 of 4) 10.0 points Determine E x , the x-component of the elec- tric field vector at the origin O due to the charge element q located at an angle sub- tended by an angular interval . 1. E x = k Q 2 R cos 2. E x = k Q 2 R 2 2 cos 3. E x = k Q R 2 2 cos 4. E x = k Q R 2 2 sin 5. E x = k Q R 2 sin 6. E x = k Q R 2 2 cos correct 7. E x = k Q R sin 8. E x = 0 9. E x = k Q 2 R 2 cos 10. E x = k Q R 2 cos Explanation: E = k q R 2 ....
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