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Unformatted text preview: Alvarado, Patrick – Homework 4 – Due: Feb 16 2007, 11:00 pm – Inst: Opyrchal, H 1 This printout should have 13 questions. Multiplechoice 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 You have a potential difference of 6 V. How much work is done to transfer 0 . 34 C of charge through it? Correct answer: 2 . 04 J. Explanation: Let : V = 6 V and q = 0 . 34 C . The potential difference is V = W q W = V q = (6 V)(0 . 34 C) = 2 . 04 J . keywords: 002 (part 1 of 1) 10 points A uniform electric field of magnitude 286 V / m is directed in the positive xdirection. Sup pose a 25 μ C charge moves from the origin to point A at the coordinates, (24 cm, 51 cm). x y 286 V / m O A (24 cm , 51 cm) What is the absolute value of the change in potential from the origin to point A ? Correct answer: 68 . 64 V. Explanation: Let : x = 24 cm , y = 51 cm , and k ~ E k = 286 V / m . The potential difference from O to A is defined as Δ V = V A V O = Z A O ~ E · d~s. We know that ~ E = (286 V / m) ˆ ı. We need to choose a path to integrate along. Because the electric force is conservative, it doesn’t matter which path we take; they all give the same answer. There are two choices of path for which the math is simple (see the figure below.) x y E O A ( x,y ) I I B I I Path I: V A V O = ( V A V B ) + ( V B V O ) , From O to B, ~ E and d~s are both along the xaxis, so ~ E · d~s = E dx . From B to A, ~ E and d~s are perpendicular, so ~ E · d~s = 0. V A V O = Z B O ~ E · d~s Z A B ~ E · d~s = Z x E dx Z y dy = E Z x O dx = E Δ x = (286 V / m)(0 . 24 m) = 68 . 64 V . The absolute value is  Δ V  = 68 . 64 V . Path II: In this case, ~ E · d~s = E cos θ ds. where cos θ = x l ⇒ x = l cos θ . V A V O = E cos θ Z l O ds = E l cos θ = E x. which is the same as the result for the other path. Alvarado, Patrick – Homework 4 – Due: Feb 16 2007, 11:00 pm – Inst: Opyrchal, H 2 keywords: 003 (part 1 of 1) 10 points The gap between electrodes in a spark plug is 0 . 087 cm. To produce an electric spark in a gasolineair mixture, an electric field of 1 . 9 × 10 6 V / m must be achieved. On starting a car, what is the magnitude of the minimum voltage difference that must be supplied by the ignition circuit?...
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This note was uploaded on 06/02/2010 for the course MGMT 121 taught by Professor Morozova during the Spring '10 term at NJIT.
 Spring '10
 Morozova

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