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What is the magnitude and direction of the force exerted on a $3.50\textrm{ }\mu\textrm{C}$ charge by a 250 N/C electric field that points due east?
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$8.75\times 10^{-4}\textrm{ N, East}$
Solution Video

OpenStax College Physics Solution, Chapter 18, Problem 28 (Problems & Exercises) (0:44)

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Video Transcript
This is College Physics Answers with Shaun Dychko. A 250 newtons per coulomb electric field is directed to the east and there's a charge of 3.50 microcoulombs in that field and the question is what force will be exerted on the charge? So we multiply the charge by the field— 3.50 times 10 to the minus 6 coulombs— times 250 newtons per coulomb that gives 8.75 times 10 to the minus 4 newtons and it will be in the same direction as the field and so that will be to the east because the field points in direction of a force that would be applied on a positive test charge and so this is a positive charge and so the force on it will be in the same direction as the field.