P22_015

# P22_015 - x/ 2 in the denominator and write tan θ ≈ x/ 2...

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15. (a) A force diagram for one of the balls is shown below. The force of gravity m~g acts downward, the electrical force ~ F e of the other ball acts to the left, and the tension in the thread acts along the thread, at the angle θ to the vertical. The ball is in equilibrium, so its acceleration is zero. The y component of Newton’s second law yields T cos θ mg =0andth e x component yields T sin θ F e = 0. We solve the Frst equation for T and obtain T = mg/ cos θ . We substitute the result into the second to obtain mg tan θ F e =0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ........ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ~ F e m~g ~ T θ x y Examination of the geometry of ±igure 22-19 leads to tan θ = x/ 2 p L 2 ( x/ 2) 2 . If L is much larger than x (which is the case if θ is very small), we may neglect
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Unformatted text preview: x/ 2 in the denominator and write tan θ ≈ x/ 2 L . This is equivalent to approximating tan θ by sin θ . The magnitude of the electrical force of one ball on the other is F e = q 2 4 πε x 2 by Eq. 22-4. When these two expressions are used in the equation mg tan θ = F e , we obtain mgx 2 L ≈ 1 4 πε q 2 x 2 = ⇒ x ≈ µ q 2 L 2 πε mg ¶ 1 / 3 . (b) We solve x 3 = 2 kq 2 L/mg ) for the charge (using Eq. 22-5): q = r mgx 3 2 kL = s (0 . 010 kg)(9 . 8 m / s 2 )(0 . 050 m) 3 2(8 . 99 × 10 9 N · m 2 / C 2 )(1 . 20 m) = ± 2 . 4 × 10 − 8 C ....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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