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HW #2-solutions

# HW #2-solutions - zapata(jz883 HW#2 Antoniewicz(56495 This...

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zapata (jz883) – HW #2 – Antoniewicz – (56495) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 3.0 points A charge 1 . 1 × 10 5 C is fixed on the x -axis at 9 m, and a charge 8 × 10 6 C is fixed on the y -axis at 5 m, as shown on the diagram. 9 m 1 . 1 × 10 5 C 5 m 8 × 10 6 C Calculate the magnitude of the resultant electric field vector E at the origin. Correct answer: 3124 . 29 N / C. Explanation: Let : k = 8 . 98755 × 10 9 N · m 2 / C 2 , Q x = 1 . 1 × 10 5 C , Q y = 8 × 10 6 C , Q = 3 × 10 6 C , x = 9 m , and y = 5 m . For charge Q x , | E x | = k | Q x | x 2 = (8 . 98755 × 10 9 N · m 2 / C 2 ) × vextendsingle vextendsingle 1 . 1 × 10 5 C vextendsingle vextendsingle (9 m) 2 = 1220 . 53 N / C . For charge Q y , | E y | = k | Q y | y 2 = (8 . 98755 × 10 9 N · m 2 / C 2 ) × 8 × 10 6 C (5 m) 2 = 2876 . 02 N / C . The resultant electric field at the origin is shown on the diagram. 1220 . 53 N / C 2876 . 02 N / C bardbl vector E bardbl Thus, bardbl vector E bardbl = radicalBig E 2 x + E 2 y = radicalBig (1220 . 53 N / C) 2 + (2876 . 02 N / C) 2 = 3124 . 29 N / C . 002 (part 2 of 4) 3.0 points Calculate the electric potential at the origin. Correct answer: 3395 . 3 V. Explanation: The electric potential of a point charge is V = k q r , so the electric potential at the origin (due to charges Q x and Q y ) is thus V = k parenleftbigg Q x x + Q y y parenrightbigg = (8 . 98755 × 10 9 N · m 2 / C 2 ) × parenleftbigg 1 . 1 × 10 5 C 9 m + 8 × 10 6 C 5 m parenrightbigg = 3395 . 3 V . 003 (part 3 of 4) 2.0 points A 3 × 10 6 C charge is brought from a very distant point by an external force and placed at the origin. Calculate the magnitude of the electric force on this charge.

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zapata (jz883) – HW #2 – Antoniewicz – (56495) 2 Correct answer: 0 . 00937286 N. Explanation: We know that vector F = q vector E . The electric force on Q is shown on the diagram. x y bardbl vector F bardbl bardbl vector F bardbl = | Q | bardbl vector E bardbl = vextendsingle vextendsingle 3 × 10 6 C vextendsingle vextendsingle (3124 . 29 N / C) = 0 . 00937286 N . 004 (part 4 of 4) 2.0 points Calculate the work that had to be done by an external force to bring Q to the origin from the distant point. Correct answer: 0 . 0101859 J. Explanation: The external force must do work equal in magnitude to that done by the electric field; i.e. , equal to the change in electric potential energy: W = Q Δ V = ( 3 × 10 6 C) (3395 . 3 V 0) = 0 . 0101859 J . 005 10.0 points Two charges, 2 Q and + Q , are located on the x -axis, as shown below. Point P , at a distance of 3 D from the origin O , is one of two points on the positive x -axis at which the electric potential is zero. x 0 2 Q + Q P D D D How far d from the origin O is the other point? 1. d = 9 5 D 2. d = 5 3 D correct 3. d = 6 5 D 4. d = 7 4 D 5. d = 4 3 D 6. d = 3 2 D 7. d = 7 5 D 8. None of these 9. d = 5 4 D 10. d = 8 5 D Explanation: Suppose the zero-potential point is at a distance d from O.
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HW #2-solutions - zapata(jz883 HW#2 Antoniewicz(56495 This...

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