24 Electric FieldsThe other two charges exert equal repulsive forces of magnitude kQqre2. The horizontal componentsof the two repulsive forces add, balancing the attractive force,FkQqrsaenet=−+LNMMMOQPPP=21302122cosθejFrom Figure P23.64ra=12sin=12cotThe equilibrium condition, in terms of , isFaenet=FHGIKJ−+FHGGIKJJ=42130222cos sincotθθ.Thus the equilibrium value of satisfies23122cos sincot+=.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 isa=°=1281 70 072 9cot...FIG. P23.64607080908181 581 722cos sincot..+°°°°°°°42.6541.22601.0911.0240.997A second zero-field point is on the negative side of the x-axis, where °916.and 310..P23.65(a)From the 2Qcharge 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 .