# ch21_p12 - the others q 1 must have the same sign as q 2...

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12. (a) According to the graph, when q 3 is very close to q 1 (at which point we can consider the force exerted by particle 1 on 3 to dominate) there is a (large) force in the positive x direction. This is a repulsive force, then, so we conclude q 1 has the same sign as q 3 . Thus, q 3 is a positive-valued charge. (b) Since the graph crosses zero and particle 3 is between
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Unformatted text preview: the others, q 1 must have the same sign as q 2 , which means it is also positive-valued. We note that it crosses zero at r = 0.020 m (which is a distance d = 0.060 m from q 2 ). Using Coulomb’s law at that point, we have q 1 q 3 4 πε o r 2 = q 3 q 2 4 πε o d 2 ¡ q 2 = © ¨ § ¹ ¸ · d 2 r 2 q 1 = 9.0 q 1 , or q 2 / q 1 = 9.0....
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## This note was uploaded on 09/06/2009 for the course PHYSICS 14358 taught by Professor Bolton during the Spring '09 term at Kansas State University.

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