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Computing Electric Field From The Potential

  1. Sep 23, 2011 #1
    Hi all. I'm having a very hard time understanding this portion of Physics so please bear with me.

    The furthest I got with this problem is deciding to use the sum of the potentials at each point to calculate the potential of the system. Something like...

    [itex]\frac{kq_{1}}{r_{1}}[/itex] + [itex]\frac{kq_{2}}{r_{2}}[/itex] + [itex]\frac{kq_{3}}{r_{3}}[/itex]


    I think that's the right approach since they are point charges. But now, I'm completely stuck and I don't know what I should do next... Please help!
     
  2. jcsd
  3. Sep 23, 2011 #2

    SammyS

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    What are the values of r1, r2, and r3 at position x along the x-axis ?
     
  4. Sep 23, 2011 #3
    Here's what I've done in regards to your question... I hope im on the right track...

    link: http://dl.dropbox.com/u/244748/2011-09-23%2018.29.26.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  5. Sep 23, 2011 #4

    SammyS

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    [itex]\sqrt{x^2+a^2}\ne x+a[/itex]

    However, [itex]\sqrt{x^2+(-a)^2}=\sqrt{x^2+a^2}\,.[/itex]

    r3 = | x - a | . If x > a then |x - a| = x -a . Otherwise, |x - a| = a - x .

    So, V(x) = ?
     
  6. Sep 24, 2011 #5
    So I suppose [itex]V(x) = kq(\frac{2}{\sqrt{x^{2}+a^{2}}} + \frac{1}{a-x})[/itex] if [itex]x < a[/itex] and [itex]V(x) = kq(\frac{2}{\sqrt{x^{2}+a^{2}}} + \frac{1}{x-a})[/itex] if [itex]x > a[/itex].

    I think I can compute [itex]E_{x}(x)[/itex] from here.. Thanks for all your help!
     
  7. Sep 24, 2011 #6

    SammyS

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    Looks good !
     
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