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    Electrostatic Force Calculations

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    1) Two identical conducting spheres A and B carry equal charge. They are separated by a distance much larger than their diameters. A third identical conducting sphere C is uncharged. SPhere C is first touched to A, then to B, and finally removed. As a result, the electrostatic force between a AND B, WHICH WAS ORIGINALLY F, BECOMES.

    2) Two particles have charges Q and -Q(equal magnitude and oposite sign). For a net force of zero to be exerted on a third charge it must be placed:
    (e is answer, but why???)

    a) midway between Q and -Q
    b) on the perpendicular bisector of the line joining Q and -Q, but not on that line itself.
    c) on the line joing Q and -Q, to the side Q opposite -Q.
    d) on the line joinging Q and -Q, to the side of -Q opposite Q.
    e) none of above(there is no place)

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    Solution Preview

    1) Since the distance between A and B is much larger than their eiameters, both conductors can be treated as point charges. When ...

    Solution Summary

    The solution provides step-by-step solutions with diagrams for questions related to elestrostatic force between conducting spheres and the net force between two oppositely charged particles.