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Physics 1100: Electric Fields Solutions

Physics 1100: Electric Fields Solutions

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Fnet = Q3Enet = (5 × 10­6)(827.283) = 4.1364 × 10­3 N. Since Q 3 is positive, F net and Enet are parallel, so F net also points along θ = 96.68° above horizontal. (b) To find the net force (magnitude and direction) on charge Q 2 due to charges Q 3 and Q 5 , we must first find

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