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Unformatted text preview: marout (ssm979) – oldhomework 08 – Turner – (56725) 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points Two charges are located in the ( x, y ) plane as shown in the figure below. The fields pro duced by these charges are observed at the origin, p = (0 , 0). The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . x y Q Q p b a a Use Coulomb’s law to find the xcomponent of the electric field at p . 1. E x = 4 k e Q a ( a 2 + b 2 ) 3 / 2 2. E x = 2 k e Q a ( a 2 + b 2 ) 3 / 2 3. E x = k e Q a ( a 2 + b 2 ) 3 / 2 4. E x = k e Q a ( a 2 + b 2 ) 3 / 2 5. E x = 2 k e Q a ( a 2 + b 2 ) 3 / 2 6. E x = 4 k e Q a ( a 2 + b 2 ) 3 / 2 7. E x = 2 k e Q a 2 + b 2 8. E x = 2 k e Q a 2 + b 2 9. E x = 0 correct Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 . Q 1 Q 2 p b a a r 1 = radicalBig x 2 1 + y 2 1 = radicalbig a 2 + b 2 . r 2 = radicalBig x 2 2 + y 2 2 = radicalBig ( a ) 2 + b 2 = radicalbig a 2 + b 2 , so r 2 = r 1 = r . θ θ E 1 E 2 Q 1 Q 2 where  sin θ  = b r = b √ a 2 + b 2  cos θ  = a r = a √ a 2 + b 2 . In the xdirection, the contributions from the two charges are E x 1 = k e ( Q ) r 2 1  cos( θ )  (1) = k e ( Q ) ( a 2 + b 2 ) a √ a 2 + b 2 = + k e Q a ( a 2 + b 2 ) 3 / 2 E x 2 = k e (+ Q ) r 2 2  cos( θ )  (2) = k e (+ Q ) ( a 2 + b 2 ) a √ a 2 + b 2 = k e Q a ( a 2 + b 2 ) 3 / 2 marout (ssm979) – oldhomework 08 – Turner – (56725) 2 E x = E x 1 + E x 2 = 0 . 002 (part 2 of 4) 10.0 points Let: V = 0 at infinity. Find the electric potential at p. 1. V y = 4 k e Q √ a 2 + b 2 2. V y = + 2 k e Q √ a 2 + b 2 3. V y = 2 k e Q √ a 2 + b 2 correct 4. V y = 2 k e Q a √ a 2 + b 2 5. V y = 4 k e Q √ a 2 + b 2 6. V y = 4 k e Q a √ a 2 + b 2 7. V y = 0 8. V y = 2 k e Q a √ a 2 + b 2 9. V y = 4 k e Q a √ a 2 + b 2 Explanation: The potential for a point charge Q is V = k e Q r . For the two charges in this problem, we have V 1 = k e Q √ a 2 + b 2 . V 2 = k e Q √ a 2 + b 2 . V p = V 1 + V 2 = k e √ a 2 + b 2 [ Q + ( Q )] = 2 k e Q √ a 2 + b 2 . 003 (part 3 of 4) 10.0 points Find the ycomponent of the electric field at p. 1. E y = + k e Q b ( a 2 + b 2 ) 3 / 2 2. E y = + 2 k e Q a 2 + b 2 3. E y = 2 k e Q a 2 + b 2 4. E y = + 2 k e Q b ( a 2 + b 2 ) 3 / 2 5. E y = 0 6. E y = + 4 k e Q b ( a 2 + b 2 ) 3 / 2 7. E y = 2 k e Q b ( a 2 + b 2 ) 3 / 2 correct 8. E y = k e Q b ( a 2 + b 2 ) 3 / 2 9. E y = 4 k e Q b ( a 2 + b 2 ) 3 / 2 Explanation: r = ( x 2 + y 2 ) 1 / 2 = ( a 2 + b 2 ) 1 / 2 and ∂ r ∂y = 1 2 ( y 2 + x 2 ) 1 / 2 · 2 y ∂ V ∂r = k e + Q r 2 ....
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 Fall '10
 TURNER,J
 Electrostatics, Electric charge, KE

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