1178864666Xuj

# 1178864666Xuj - Alvarado Patrick Homework 4 Due 11:00 pm...

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Alvarado, Patrick – Homework 4 – Due: Feb 16 2007, 11:00 pm – Inst: Opyrchal, H 1 This print-out should have 13 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 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 x -direction. 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 II 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 x -axis, 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 0 E dx - Z y 0 0 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.

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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 gasoline-air 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? Correct answer: 1653 V. Explanation: Let : E = 1 . 9 × 10 6 V / m and d = 0 . 087 cm = 0 . 00087 m . Assuming the electric field between the two electrodes is constant, then the potential dif- ference between the electrodes is V = E d = ( 1 . 9 × 10 6 V / m ) (0 . 00087 m) = 1653 V .
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