ch21-p054 - 54. (a) A force diagram for one of the balls is...

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by Eq. 21-4. When these two expressions are used in the equation mg tan θ = F e , we obtain 1/3 22 2 00 1 . 24 2 mgx q q L x Lx m g εε ⎛⎞ ≈⇒ ⎜⎟ ππ ⎝⎠ (b) We solve x 3 = 2 kq 2 L / mg for the charge (using Eq. 21-5): () 3 2 3 8 92 2 0.010kg 9.8m s 0.050m 2.4 10 C. 2 2 8.99 10 N m C 1.20m mgx q kL == = ± × ×⋅ Thus, the magnitude is 8 ||2 . 41 0C . q 54. (a) A force diagram for one of the balls is shown on the right. The force of gravity mg G acts downward, the electrical force G F e of the other ball acts to the left, and the tension in the thread acts along the thread, at the angle to the vertical. The ball is in equilibrium, so its acceleration is zero. The y component of Newton’s second law yields
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