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684_Physics ProblemsTechnical Physics

# 684_Physics ProblemsTechnical Physics - 24 Electric Fields...

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24 Electric Fields The other two charges exert equal repulsive forces of magnitude k Qq r e 2 . The horizontal components of the two repulsive forces add, balancing the attractive force, F k Qq r s a e net = + L N M M M O Q P P P = 2 1 3 0 2 1 2 2 cos θ e j From Figure P23.64 r a = 1 2 sin θ s a = 1 2 cot θ The equilibrium condition, in terms of θ , is F a k Qq e net = F H G I K J + F H G G I K J J = 4 2 1 3 0 2 2 2 cos sin cot θ θ θ e j . Thus the equilibrium value of θ satisfies 2 3 1 2 2 cos sin cot θ θ θ + = e j . One method for solving for θ is to tabulate the left side. To three significant figures a value of θ corresponding to equilibrium is 81.7 ° . The distance from the vertical side of the triangle to the equilibrium position is s a a = °= 1 2 81 7 0 072 9 cot . . . FIG. P23.64 θ θ θ θ 2 3 60 70 80 90 81 81 5 81 7 2 2 cos sin cot . . + ° ° ° ° ° ° ° e j 4 2.654 1.226 0 1.091 1.024 0.997 A second zero-field point is on the negative side of the x -axis, where θ = − ° 9 16 . and
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