HW1Sol - Platt, David Homework 1 Due: Sep 7 2005, 4:00 am...

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Platt, David – Homework 1 – Due: Sep 7 2005, 4:00 am – Inst: Ken Shih 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points Two metal spheres that are initially un- charged are mounted on insulating stands, as shown. X Y - - - - A negatively charged rubber rod is brought close to but does not make contact with sphere X . Sphere Y is then brought close to X on the side opposite to the rubber rod. Y is allowed to touch X and then is removed some distance away. The rubber rod is then moved Far away From X and Y . What are the fnal charges on the spheres? Sphere X Sphere Y 1. Negative Negative 2. Negative Positive 3. Positive Negative correct 4. Positive Positive 5. Zero Zero Explanation: The Force is repulsive iF the charges are oF the same sign, so when the negatively charged rod moves close to the sphere X , the neg- atively charged electrons will be pushed to sphere Y . IF X and Y are separated beFore the rod moves away, those charges will re- main on X and Y . ThereFore, X is positively charged and Y is negatively charged. 002 (part 1 oF 2) 10 points A set oF eight point charges, each oF magni- tude 8 × 10 - 8 C, is located on the corners oF a cube with sides oF length 2 m as shown in the fgure. q q q q q q q q x z y A ±ind the x -component oF the resultant Force exerted on the charge located at point A by the other charges. Correct answer: 2 . 73171 × 10 - 5 N. Explanation: Let : s = 2 m and q = 8 × 10 - 8 C . There are seven terms which contribute: 1 term is 3 s away (the body diagonal, which is the vector From the origin to A , determined by k ~r k = s ˆ ı + s ˆ + s ˆ k = 3 s ) , 3 terms are 2 s away (the Face diagonals), and 3 terms are s away (the cube edges). ~ F = k q 2 X i 1 r 3 i ~ r i = k q 2 1 3 s 2 1 3 ı + ˆ k ) (body diag . ) + 1 2 s 2 1 2 ı ) + 1 2 s 2 1 2 + ˆ k ) + 1 2 s 2 1 2 ı + ˆ k ) (Face diag . ) + 1 s 2 ˆ ı + 1 s 2 ˆ + 1 s 2 ˆ k (along sides) Adding and collecting terms oF common unit vectors, we have ~ F = k q 2 s 2 µ 1 + 2 2 2 + 1 3 3 ı + ˆ k ) .
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Platt, David – Homework 1 – Due: Sep 7 2005, 4:00 am – Inst: Ken Shih 2 The x component is ~ F x = k q 2 s 2 µ 1 + 1 2 + 1 3 3 ˆ ı = ( 8 . 988 × 10 9 N · m 2 / C 2 ) (8 × 10 - 8 C) 2 (2 m) 2 × (1 . 89956) ˆ ı = (2 . 73171 × 10 - 5 N) ˆ ı . 003 (part 2 of 2) 10 points What is the magnitude of this force? Correct answer: 4 . 73147 × 10 - 5 N. Explanation:
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This note was uploaded on 04/10/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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HW1Sol - Platt, David Homework 1 Due: Sep 7 2005, 4:00 am...

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