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Equilateral triangle charged rods

  1. Oct 1, 2013 #1
    1. The problem statement, all variables and given/known data

    Three rods each of uniform charge Q and length a are formed into an equilateral triangle. Given[Q,a,q] determine:
    1. The voltage at the center of the triangle.
    2. If a charge -q is placed at the center of the triangle determine potential energy.
    3. Determine the initial speed of the -q in order that it exits the system and never returns.
    2. Relevant equations


    [tex]V = \frac{1}{4 \pi \epsilon_0} \int \frac{dq}{r}[/tex]

    [tex]U = Vq[/tex]

    [tex]K_0 + U_0 = K + U[/tex]

    3. The attempt at a solution

    http://i.imgur.com/7IZ3PbP.png

    I take 1 half of bottom rod and intend to integrate for V


    [tex]dq = Q \frac{dx}{a}[/tex]
    [tex]r = \sqrt{y^2 + (\frac{a}{2} - x)^2}[/tex]

    using the V formula above this integrates(from 0 to a/2) to


    [tex]V = \frac{Q}{4 \pi \epsilon_0}[\ln{\frac{a+\sqrt{\frac{a^2}{4} + y^2}}{y}}][/tex]

    since this is 1/6th of the potential and they are additive, multiply 6 for total v at center


    is this correct solution? U is simple enough to do, simply multiply by -q. What is the methodology for solving part 3?
     
  2. jcsd
  3. Oct 1, 2013 #2
    You might simplify things significantly if you express ##y## via ##a##.
     
  4. Oct 1, 2013 #3
    [tex]y = \sqrt{\frac{3}{4}} a[/tex]

    ln term collapses to
    [tex]ln{\frac{4}{3}}[/tex]

    Is this correct? What about part 3? I only know of 1/2mv^2 which I assume was at least greater than potential it would get out.. but I have no m given.
     
  5. Oct 1, 2013 #4
    "Never returns" means that it goes infinitely far away, and the velocity "at the infinity" is at least zero.

    With this in mind, apply conservation of energy.
     
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