by Eq. 21-4. When these two expressions are used in the equation mgtanθ= Fe, we obtain 1/3222001.242mgxqq LxLxmgεε⎛⎞≈⇒≈⎜⎟ππ⎝⎠(b) We solve x3= 2kq2L/mgfor the charge (using Eq. 21-5): ()32389220.010kg9.8m s0.050m2.4 10 C.22 8.99 10 N m C1.20mmgxqkL−===±××⋅Thus, the magnitude is 8||2.410C.q−=×54. (a) A force diagram for one of the balls is shown on the right. The force of gravity mgGacts downward, the electrical force GFeof 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 ycomponent of Newton’s second law yields
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