Determine the magnitude and direction of the electric force on an electron in a uniform electric field of strength 2460 N/C that points due east.
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This is Giancoli Answers with Mr. Dychko. The electric field points due East and it has a magnitude 2,460 Newtons per Coulomb. This direction of electric field is the direction of force on a positive charge. Because this is an electron with negative charge, the direction of the force of this electric field applies on the electron will be to the left. The magnitude of that force is going to be the charge times the electric field strength, and to that, it's negative 1.602 times 10 to the minus 19 Coulombs charge in electron times 2,460 Newtons per Coulomb. This gives us 3.94 times 10 to the minus 16 Newtons directed West.
Is a "uniform electric field pointing due...." always referring to the electric field due to a positive charge? I would have thought that since the electric field always points toward a negative charge and away from positive that since the question says that the electric field is pointing east - then that would mean the electric field is is pointing in the direction of the e-. If the question would have said the field is pointing east and we dropped a positive test charge in it, then I would have said that the field would point west because it would be away from the positive. Sorry for all of the questions, I just find that the real hurdle is understanding what the question is asking me for.
NVM. You answer this in question 27. Thanks!
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