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Unformatted text preview: GenericStudent – Homework 1 – savrasov – 39823 – Jan 02, 2008 1 This printout should have 7 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) 2 points Four point charges are placed at the four cor ners of a square. Each side of the square has a length L . q 1 = q q 3 = q q 2 = q q 4 = q P L L Find the magnitude of the electric force on q 2 due to all three charges q 1 , q 3 and q 4 . Given L = 1 m and q = 1 . 92 μ C. Correct answer: 0 . 0496976 N. Explanation: F 1 F 4y F 4 F 3 F 4x q 2 From the above figure, we see that F 1 = k q 2 L 2 ˆ ı = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × (1 . 92 × 10 6 C) 2 (1 m) 2 = . 0331317 N F 3 = k q 2 L 2 ˆ = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × (1 . 92 × 10 6 C) 2 (1 m) 2 = . 0331317 N F 4 x = k q 2 2 L 2 1 √ 2 = 8 . 98755 × 10 9 N · m 2 / C 2 2 · (1 m) 2 × (1 . 92 × 10 6 C) 2 √ 2 = 0 . 0117138 N F 4 y = k q 2 2 L 2 1 √ 2 = 8 . 98755 × 10 9 N · m 2 / C 2 2 · (1 m) 2 × (1 . 92 × 10 6 C) 2 √ 2 = . 0117138 N k ~ F k 2 = ( F 1 + F 4 x ) 2 + ( F 3 + F 4 y ) 2 = ( . 0331317 N + 0 . 0117138 N) 2 + ( . 0331317 N + . 0117138 N) 2 so that k ~ F k = √ . 00246985 N 2 = . 0496976 N . keywords: 002 (part 1 of 1) 2 points Three identical point charges, each of mass 110 g and charge + q , hang from three strings, as in the figure. The acceleration of gravity is 9 . 8 m / s 2 , and the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . 9 . 8 m / s 2 1 . 5 c m 110 g + q 1 . 5 c m 110 g + q 49 ◦ 110 g + q If the lengths of the left and right strings are each 10 . 5 cm, and each forms an angle of 49 ◦ with the vertical, determine the value of q . Correct answer: 0 . 832571 μ C. Explanation: Let : θ = 49 ◦ , GenericStudent – Homework 1 – savrasov – 39823 – Jan 02, 2008 2 m = 110 g = 0 . 11 kg , L = 10 . 5 cm = 0 . 105 m , g = 9 . 8 m / s 2 , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 ....
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 Winter '09
 RichardScalettar
 Work, Electric charge, NC, N/C, 49◦

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