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 kQq r e 2 . The horizontal components of the two repulsive forces add, balancing the attractive force, Fk Q q r sa e net =− + L N M M M O Q P P P = 21 3 0 2 1 2 2 cos θ ej From Figure P23.64 r a = 1 2 sin = 1 2 cot The equilibrium condition, in terms of , is F a 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 θθ . Thus the equilibrium value of satisfies 23 1 2 2 cos sin cot += . One method for solving for is to tabulate the left side. To three significant figures a value of corresponding to equilibrium is 81.7 ° . The distance from the vertical side of the triangle to the equilibrium position is a = 1 2 81 7 0 072 9 cot . . . FIG. P23.64 60 70 80 90 81 81 5 81 7 2 2 cos sin cot . . + ° ° ° ° ° ° ° 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 ° 916 . and 310 . . P23.65 (a) From the 2 Q charge we have
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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