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Electric Potential of a triangle?

  1. Jan 30, 2013 #1
    Find the electric potential at point P in the figure.

    http://i.imgur.com/8FNSoML.png

    V = kq / r


    So what I did was calculated the electric potential each force gives on the point P.

    Vp1 = (8.99e^9) (2.75e^-6) / .625

    Vp2 = (8.99e^9)(-1.72e^6) / .625

    Vp3 = (8.99e^9) (7.45e^6) / 1.25

    Vp1 + Vp2 + Vp3 = 39556 - 24740 + 53580 = 68396 V


    Would appreciate any help :D
     
  2. jcsd
  3. Jan 30, 2013 #2

    Simon Bridge

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    What seems to be the problem (you have to actually say)?
     
  4. Jan 30, 2013 #3
    Well my answer is incorrect lol, here's the full question.

    Find the electric potential at point P in the figure.

    Suppose the three charges shown in the figure are held in place. A fourth charge, with a charge of +6.50 and a mass of 4.20 , is released from rest at point P. What is the speed of the fourth charge when it has moved infinitely far away from the other three charges?

    I need to do the first part I think before I start the second but I didn't get the first one correct. Any tips on how to find the electric potential at the point P?
     
  5. Jan 30, 2013 #4

    jtbell

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    The distance between charge 3 (the top of the triangle) and point P is not 1.25 m.

    I've moved this post to the appropriate subforum of "Homework and Coursework Questions." In the future, please use those forums for help with specific exercises like this one. The other forums are for more general discussion.
     
  6. Jan 30, 2013 #5
    Woops sorry about that! Thanks I made the mistake of putting 1.25 there, I got the right answer thanks!
     
  7. Jan 30, 2013 #6

    Simon Bridge

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    The way to troubleshoot these things is to go back over each step and examine your reasoning ... in each step there are only two places for a mistake - the charge and the distance. Mind you - just writing numbers down like that creates a kind of blindness.
    Good to see it's sorted out though - well done.
     
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