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Question on finding work done on charge

  1. Jun 8, 2016 #1
    1. The problem statement, all variables and given/known data


    page1.jpg page2.jpg
    2. Relevant equations

    coulomb's law
    3. The attempt at a solution

    Hi everyone. I understand their approach with the integration to find the amount of work that "a person" would have to do to bring the charge q3 from infinity to its current position.

    I understand that the force that this person would have to do is the opposite of the coulombic forces between the q3 and q1 and 2.

    My question: why did they not account for the y-components of the coulombic forces between q1 and 2 to a3? The x-components of the two forces would cancel, so the only coulombic forces acting on q3 would be the y-components.

    Thus I did exactly what they did, but I added a sine term to account for the y-component and then re-wrote the sine in terms of the vertical distance and "s/2" with pythagorean theorem and then did the integration.

    Could anyone please weight in on why their approach is legit and they did not do what I planned to do with the sine term?

    Thank you very much in advance.
     
  2. jcsd
  3. Jun 9, 2016 #2

    ehild

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    Homework Helper
    Gold Member

    If you correctly do the integration of the y component of the electric force you get the same result.
    The solution shown uses the potential to get the potential energy of q3, which is q3U. The potential of a point charge is kq/r where r is the distance from the charge. The potential function is scalar and additive, so the net potential at a point is the sum of the potentials from all charges.
     
  4. Jun 9, 2016 #3

    Delta²

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    Gold Member

    I think what is needed to be emphasized is that the work is independent of the path that the q3 follows as is brought from the infinite to the desired place, it depends only on the end points of the path (one point at infinite and one point at the third vertex of the triangle). This holds only if the field is static so that ##\nabla \times E=0##, ##E=\nabla\phi## where ##\phi## is the electrostatic potential (scalar).
     
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