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Potential due to a curved rod

  1. Sep 12, 2009 #1
    1. The problem statement, all variables and given/known data

    problem.jpg

    2. Relevant equations

    V=kq/r

    3. The attempt at a solution

    V=kQ/r
    dV=kdQ/r

    [itex]dQ=\lambda d\theta[/itex] <---This is where I'll make a mistake if any.

    [itex]dV=k(\lambda d\Theta) /r[/itex]
    [tex]V=\frac{k\lambda}{r}\int_0^{\phi}d\Theta[/tex]
    [itex]\phi=2\pi/3[/itex]
    [itex]\lambda=Q/L= Q/(2\pi/3)[/itex]

    [tex]V=k\frac{3Q}{2\pi r}\int_0^{\phi}d\Theta[/tex]


    ...so far?
     
  2. jcsd
  3. Sep 12, 2009 #2
    Yerr, that is a mistake. You forgot that the length of the rod is [tex]R\cdot \phi[/tex] and not just [tex]\phi[/tex].

    [tex]dq=\lambda dl=\lambda R \cdot d\theta[/tex]
    [tex]\lambda \equiv \frac{Q}{R\cdot\phi}[/tex]

    [tex]dV=\frac{K}{R}\cdot dq[/tex]

    [tex]dV=\frac{K}{R}\cdot\lambda R \cdot d\theta[/tex]

    [tex]dV=K\lambda\cdot d\theta[/tex]

    [tex]dV=\frac{KQ}{R\cdot\phi}\cdot d\theta[/tex]

    And now it's just a question of taking the integral along the arc and you're done.

    The surprising result I got is that the potential is just [tex]V=\frac{KQ}{R}[/tex]
     
    Last edited: Sep 12, 2009
  4. Sep 12, 2009 #3
    See, that is where I knew I'd make a mistake. I can't understand why it's [tex]
    R\cdot \phi[/tex]. No problems I've found online explain it either. Where does the R come from?
     
  5. Sep 12, 2009 #4
    Just by definition. The length of a circular arc of radius [tex]R[/tex] resting on [tex]x[/tex] radians is [tex]R\cdotx[/tex]

    Just like how a circle of radius [tex]R[/tex] has a circumference of [tex]2\pi R[/tex], that's the same as saying you've got a circular arc of radius [tex]R[/tex] resting on [tex]2\pi[/tex] radians.
     
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