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Unformatted text preview: Kapoor (mk9499) oldhomework 03 Turner (60230) 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 (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. along the negative x-axis. 2. in quadrant IV. correct 3. along the negative y-axis. 4. along the positive y-axis. 5. in quadrant I. 6. in quadrant II. 7. in quadrant III. 8. along the positive x-axis. 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. 002 (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 R sin 2. E x = k Q R 2 2 sin 3. E x = k Q R 2 2 cos correct 4. E x = k Q R 2 2 cos 5. E x = k Q 2 R 2 cos 6. E x = k Q R 2 cos 7. E x = 0 8. E x = k Q R 2 sin 9. E x = k Q 2 R cos 10. E x = k Q 2 R 2 2 cos Explanation: E = k q R 2 . The charge q per unit arc-length s is Q 2 R = 2 Q R , so q = s = R = 2 Q R R = 2 Q . On the other hand, the vector vector E makes an angle with the x-axis, so E x = k Q R 2 2 cos . 003 (part 3 of 4) 10.0 points Find E x , the electric field at the origin due Kapoor (mk9499) oldhomework 03 Turner (60230) 2 to the full arc length for the case where Q = 3 . 7 C and R = 1 . 74 m . Correct answer: 6992 . 38 N / C. Explanation: Let : Q = 3 . 7 C and R = 1 . 74 m . E x can be found by integrating the contribu- tions of all the q s in the arc. We get E x = integraldisplay / 2 k Q R 2 2 cos d = 2 k Q R 2 = 2 k (3 . 7 C) (1 . 74 m) 2 = 6992 . 38 N / C . 004 (part 4 of 4) 10.0 points The charge on an electron is e . If at O we have E x , then the magnitude of the force bardbl vector F e bardbl on an electron at this point is 1. bardbl vector F e bardbl = E x 2 e . 2. bardbl vector F e bardbl = 4 eE x . 3. bardbl vector F e bardbl = eE x . 4. bardbl vector F e bardbl = 2 E x 3 e . 5. bardbl vector F e bardbl = 3 eE x 2 ....
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