turner hw 1

# turner hw 1 - linares(jl36797 – oldhomework 01 – Turner...

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Unformatted text preview: linares (jl36797) – oldhomework 01 – Turner – (56705) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two uncharged metal balls, Z and X , stand on insulating glass rods. A third ball, carrying a positive charge, is brought near the ball X as shown in the figure. A conducting wire is then run between Z and X and then removed. Finally the third ball is removed. X Z + conducting wire When all this is finished 1. balls Z and X are both positive. 2. ball Z is neutral and ball X is negative. 3. balls Z and X are both negative, but ball Z carries more charge than ball X . 4. ball Z is negative and ball X is positive. 5. balls Z and X are still uncharged. 6. ball Z is positive and ball X is negative. correct 7. ball Z is negative and ball X is neutral. 8. balls Z and X are both negative, but ball X carries more charge than ball Z . 9. ball Z is positive and ball X is neutral. 10. ball Z is neutral and ball X is positive. Explanation: When the conducting wire is run between Z and X , some positive charge flows from X to Z under the influence of the positive charge of the third ball. Therefore, after the wire is removed, Z is charged positive and X is charged negative. 002 (part 1 of 2) 10.0 points Three point-charges ( − q , − q , and − q ) are placed at the vertices of an equilateral triangle (see figure below). a 60 ◦ − q − q − q ˆ ı ˆ The magnitude of the electric force on the charge at the bottom left-hand vertex of the triangle due to the other two charges is given by 1. bardbl vector F bardbl = 1 2 k q 2 a 2 2. bardbl vector F bardbl = 3 √ 2 k q 2 a 2 3. bardbl vector F bardbl = k q 2 a 2 4. bardbl vector F bardbl = √ 2 k q 2 a 2 5. bardbl vector F bardbl = √ 2 3 k q 2 a 2 6. bardbl vector F bardbl = √ 3 2 k q 2 a 2 7. bardbl vector F bardbl = 1 2 √ 3 k q 2 a 2 8. bardbl vector F bardbl = 1 √ 2 k q 2 a 2 9. bardbl vector F bardbl = 2 √ 3 k q 2 a 2 10. bardbl vector F bardbl = √ 3 k q 2 a 2 correct Explanation: linares (jl36797) – oldhomework 01 – Turner – (56705) 2 a − q − q − q ˆ ı ˆ In this case, each of these forces has a mag- nitude F 21 = F 31 = k q 2 a 2 . The x-component of the net force is then F x = [ − F 31 cos 60 ◦ − F 21 ] ˆ ı = bracketleftbigg − parenleftbigg k q 2 a 2 parenrightbigg cos 60 ◦ − parenleftbigg k q 2 a 2 parenrightbiggbracketrightbigg ˆ ı = bracketleftbigg − 1 2 parenleftbigg k q 2 a 2 parenrightbigg − parenleftbigg k q 2 a 2 parenrightbiggbracketrightbigg ˆ ı = − 3 2 k parenleftbigg q 2 a 2 parenrightbigg ˆ ı. On the other hand, the y-component is just F y = bracketleftbigg − parenleftbigg k q 2 a 2 parenrightbigg sin 60 ◦ bracketrightbigg ˆ = − √ 3 2 k parenleftbigg q 2 a 2 parenrightbigg ˆ ....
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## This note was uploaded on 08/30/2010 for the course PHY 303 taught by Professor Erskine/tsoi during the Fall '08 term at University of Texas.

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turner hw 1 - linares(jl36797 – oldhomework 01 – Turner...

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