exam1sol_sm10 - Instructor: Tomoyuki Nakayama Monday, June...

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Instructor: Tomoyuki Nakayama Monday, June 7, 2010 PHY 2005 Applied Physics 2 - Summer C 2010 Exam 1 Solution ________________________________________________________________________________ 1. Since the two metal spheres are identical, they have equal amounts of charge after touching. Each will have (2 – 10 )/2 = -4 μ C. According to Coulomb’s law, the electric force between them is proportional to the product of the two charges and inversely proportional to the square of the separation distance. The initial and final separation distances are the same. Therefore, the magnitude of the electric force after touching is F/F 0 = (4 2 )/(2×10) = 4/5 F = (4/5)F 0 = 4.0 × 10 -4 N. The force is repulsive because they have the same sign of charge. 2. Electric field is directed away from a positive charge and directed toward a negative charge. If you consider the region between the two particles, electric fields due to individual particles are both directed to the left. Thus the net field cannot be zero. In the region to the left of the origin, the electric field due to the -8- μ C charge is always larger than that due to 2- μ C charge; it is because any point in this region is closer to the -8- μ C charge that has a larger amount of charge. Therefore, the net field cannot be zero in this region either. If there is a position where the net field is zero, then it must be in
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exam1sol_sm10 - Instructor: Tomoyuki Nakayama Monday, June...

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