Practice HW 1 Solutions

# Practice HW 1 Solutions - lim (kl9356) – Practice HW 1...

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Unformatted text preview: lim (kl9356) – Practice HW 1 Solutions – Weathers – (22202) 1 This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 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 final charges on the spheres? Sphere X Sphere Y 1. Negative Negative 2. Negative Positive 3. Zero Zero 4. Positive Positive 5. Positive Negative correct 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 10.0 points 1) Two uncharged metal balls, X and Y , stand on glass rods and are touching. X Y 2) A third ball, carrying a positive charge, is brought near the first two. X Y + 3) Then the first two balls are separated from each other, X Y + 4) and the third ball is finally removed. X Y When this is all four steps are done, it is found out that 1. Ball Y is positive and ball X is negative. correct 2. Balls Y and X are both positive. 3. Balls Y and X are both negative. 4. Ball Y is negative and ball X is positive. 5. Balls Y and X are still uncharged. Explanation: Basic Concept: Electric induction caused by nearby charges. Solution: When a positive ball is moved near a metallic object ( Y and X ), the positive charge will attract negative charges, causing Y to have excess positive charge and X to have excess negative charge ( Y and X are in lim (kl9356) – Practice HW 1 Solutions – Weathers – (22202) 2 contact, so the total net charge on Y and X should be zero).- + + Later, Y and X are separated, retaining their charges, so when the third ball is finally removed, Y will have net positive charge and X will have net negative charge....
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## This note was uploaded on 09/07/2010 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.

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Practice HW 1 Solutions - lim (kl9356) – Practice HW 1...

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