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Three point charges with a central fourth charge

  1. Mar 19, 2015 #1
    1. The problem statement, all variables and given/known data

    It is the last part, part (c) that I'm having trouble with, but I'll post the entire question for clarity.

    Three charges (q) form the vertices of an equilateral triangle. A fourth charge Q ([itex] Q = -q [/itex]) is placed at the center of the triangle.

    (a) will the charges at the corners move away from or towards the center?

    (b) for what value of Q will all four points remain stationary?

    (c) in situation (b) how much work is done in removing the charges to infinity? (I interpret this to mean keep the center charge where it is and moving the corner charges out to infinity)

    2. Relevant equations

    Relevant to (c)

    Distance between vertices: d

    Distance between a vertex and the center: [itex] s = \frac{d}{\sqrt{3}}[/itex]

    Constant of Coulomb's Law: A

    After solving (a) and (b)

    In equilibrium: [itex] Q = \frac{-q}{\sqrt{3}} [/itex]

    Force is a function of s

    [tex] F = F(s) [/tex]

    Two body interactions

    ||R|| = force of repulsion on a vertex resulting from vertex charge repulsion

    ||R|| calculated initially in terms of d but then converted to be in terms of s

    ||P|| = force of attraction on a vertex resulting from vertex-center charge attraction

    ||P|| calculated to be in terms of s

    F(s) = magnitude of the resulting force on a vertex charge in terms of distance between vertex and center

    [tex] F(s) = ||R|| - ||P|| = Aq(\frac{q}{\sqrt{3}} - 3Q)\frac{1}{s^2}[/tex]

    from (b) [itex]Q = \frac{-q}{\sqrt{3}}[/itex]

    but we're using ||P|| so remove the sign from the front of that eqn.

    so [itex] F(s) = ||R|| - ||P|| = Aq(\frac{q}{\sqrt{3}} - \sqrt{3}q)\frac{1}{s^2}[/itex]

    3. The attempt at a solution

    I'm having a conceptual understanding deficiency on (c).

    I know how to do improper integrals. I know that W = Fs and in this case F = f(s), so

    [tex]Work = \int_{\frac{d}{\sqrt{3}}}^\infty F(s) ds[/tex]

    [tex]Work = \int_{\frac{d}{\sqrt{3}}}^\infty Aq(\frac{q}{\sqrt{3}} - \sqrt{3}q)\frac{1}{s^2} ds[/tex]

    Using improper integration yields for me

    [tex]Work = A\frac{-2q^2}{d}[/tex]

    edit: clarify - work done in moving one of the vertices to infinity

    4. The texts answer

    The text did not use any improper integrals as I did.

    The text says:

    As potential energy of the system

    [tex] U = \frac{1}{2} \times \frac{1}{4\pi \epsilon_0} \Sigma_{i \neq j} \frac{q_iq_j}{r_{ij}} = \frac{1}{4\pi \epsilon_0} \left[3\frac{q\times q}{d}+\frac{3q(\frac{-q}{\sqrt{3}})}{\frac{d}{\sqrt{3}}}\right]=0[/tex]

    And as for finite charge distribution potential energy at infinity is always zero

    [tex]W=U_F-U_I = 0-0 =0[/tex]

    I don't understand this last explanation. Surely as you move the vertex charge away from the center, the attraction to the center dominates over the repulsion from the other vertices and the amount of work needed to drag it out...wait...the other vertices are moving to infinity at the same rate too...so the repulsion and the attraction stay in the same proportion...

    Is this saying...that if you moved all three charges (provided they were in equilibrium at the beginning) away from the center, all at the same rate of motion and ensuring that they keep in an equilateral triangle configuration, and you moved them all equally to an arbitrary distance away from their original config...that zero work has been done?? This seems highly counter-intuitive to me...
     
    Last edited: Mar 19, 2015
  2. jcsd
  3. Mar 19, 2015 #2

    TSny

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    Yes. Good observation! Note that the condition for equilibrium is [itex] Q = \frac{-q}{\sqrt{3}} [/itex], and this is independent of the size of the triangle. You could give each of the three corner charges a little "outward" initial velocity and they would glide with constant speed all the way to infinity.
     
  4. Mar 19, 2015 #3

    BvU

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    Your ##
    F(s) = ||R|| - ||P|| = Aq(\frac{q}{\sqrt{3}} - 3Q)\frac{1}{s^2}
    ## should be ##
    F(s) = ||R|| - ||P|| = Aq(\frac{q}{\sqrt{3}} - Q)\frac{1}{s^2}
    ## and with that your integrand is zero!
     
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