Question
An electron is placed in an electric field of 12.0 N/C12.0 \textrm{ N/C} to the right. What is the resulting force on the electron?
  1. 1.33×10−20 N right1.33 \times 10^{-20} \textrm{ N right}
  2. 1.33×10−20 N left1.33 \times 10^{-20} \textrm{ N left}
  3. $1.92 \times 10^{-18} \textrm{ N right}
  4. 1.92×10−18 N left1.92 \times 10^{-18} \textrm{ N left}
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Final Answer

(d)

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OpenStax College Physics for AP® Courses, Chapter 19, Problem 1 (Test Prep for AP® Courses)

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Video Transcript
This is College Physics Answers with Shaun Dychko. A uniform electric field is pointing to the right, with a magnitude 12.0 newtons per Coulomb. And there's an electron in this field, and the question is: What will the force be on the electron? Well, the electric field points in the direction of force that would be applied on a positive charge, so this being a negative charge of electron, of course, must be in the opposite direction to the direction of the field. And so, the force is to the left. Now, the size of the force can be calculated by multiplying the charge by the electric field strengths. And the charge on the electron is 1.6 times ten to the minus 19 Coulombs. We multiply that by 12 newtons per Coulomb to get 1.92 times ten to the minus 18 newtons, and that is to the left. And, that is option D.