Question
What is the internal energy of a system consisting of two point charges, one 2.0 μC, and the other −3.0 μC, placed 1.2 m away from each other?
  1. −3.8×10−2 J-3.8 \times 10^{-2} \textrm{ J}
  2. −4.5×10−2 J-4.5 \times 10^{-2} \textrm{ J}
  3. 4.5×10−2 J4.5 \times 10^{-2} \textrm{ J}
  4. 3.8×10−2 J3.8 \times 10^{-2} \textrm{ J}
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Final Answer

(b)

Solution video

OpenStax College Physics for AP® Courses, Chapter 19, Problem 9 (Test Prep for AP® Courses)

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Video Transcript
This is College Physics Answers with Shaun Dychko. We are going to calculate the internal energy of the system of two charges by calculating the potential due to one of the charges at this position of the other charge and then multiply that potential by the other charge. So, we have charge q one is two micro-coulomb and charge q two is negative three micro-coulomb and so the potential energy of the system will be charge one multiplied by the energy per charge at this position and the energy per charge at this position is, the potential there is due to charge two and the potential is gonna be the coulomb’s constant times charge q two divided by this distance away r which is 1.2 metres and so the potential energy is q one multiply by the potential of its location and we will substitute k q two over r in place of V so we have two micro-coulomb which is ten to the minus six coulombs multiplied by coulomb’s constant times negative three micro-coulomb divided by 1.2 metres and this gives negative 4.5 times 10 to the minus two joules and if we look at the options from the line, that is option B.