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1178864646Xuj - Alvarado Patrick Homework 2 Due Feb 2 2007...

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Alvarado, Patrick – Homework 2 – Due: Feb 2 2007, 11:00 pm – Inst: Opyrchal, H 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A charge of - 4 . 59 μ C is located at the origin, and a charge of - 2 . 2 μ C is located along the y axis at 1 . 88247 m. The value of the Coulomb constant is 8 . 99 × 10 0 . At what point along the y -axis is the electric field zero? Correct answer: 1 . 11236 m. Explanation: Let : q 1 = - 4 . 59 μ C , q 2 = - 2 . 2 μ C , and d = 1 . 88247 m . Call the point where the fields cancel y . Since the charges are of equal sign, the only place y can be is somewhere between them. The field from the particle q 1 at the origin is E 1 = k e q 1 y 2 pointing down (since q 1 is negative). The field from the charge q 2 at a point d along the y -axis is E 2 = k e q 2 ( d - y ) 2 pointing up (since q 2 is negative). Thus we have cancellation at d provided E 1 equals E 2 , or k e q 1 y 2 = k e q 2 ( d - y ) 2 q 2 y 2 = q 1 ( d - y ) 2 r q 2 q 1 y = d - y y = d 1 + r q 2 q 1 = 1 . 88247 m 1 + r - 2 . 2 μ C - 4 . 59 μ C = 1 . 11236 m . keywords: 002 (part 1 of 1) 10 points An airplane is flying through a thundercloud at a height of 2000 m. (This is very danger- ous because of updrafts, turbulence, and the possibility of electric discharge.) The Coulomb constant is 8 . 99 × 10 9 N · m 2 / C 2 . If there are charge concentrations of 46 C at height 3300 m within the cloud and - 22 C at height 662 m, what is the strength of the electric field E at the aircraft? Correct answer: 355175 V / m.
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