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Electrostatics of balls on a string

  1. Feb 5, 2008 #1
    1. The problem statement, all variables and given/known data
    In Fig. 21-42 (I attached an MSPaint rendition of it), two tiny conducting balls of identical mass m and Identical charge q hang from non-conducting threads of length L. Assume [tex]\theta[/tex] is so small that [tex]tan \theta[/tex] can be replaced by its approximate equal, [tex] sin \theta[/tex]. (a) Show that [tex]x=(\frac{q^2L}{2\pi\epsilon_{0}mg})^{1/3}[/tex] gives the equilibrium separation of the balls. (b) If L=120cm, m=10 g, and x=5.0 cm, what is |q|?

    2. Relevant equations
    Well, I think Coulomb's law is clearly involved because of the [tex]\frac{1}{2\pi\epsilon_{0}}[/tex] bit, which is equal to 2*k.
    So then, restating Coulomb's law: [tex]F=\frac{1}{4\pi\epsilon_{0}}*\frac{\left|q_1\right|\left|q_2\right|}{r^2}[/tex]

    3. The attempt at a solution
    The problem here is I don't know how to start the problem. It appears simple enough but I can't seem to get an answer out of it. Here is what I was thinking: since [tex]q_1[/tex] and [tex]q_2[/tex] are both the same, the top bit of coulomb's law becomes [tex]q^2[/tex] since any number squared is positive (no need for abs. value). splitting out coulomb's law from the full equasion within parenthesis gives [tex]x=(\frac{1}{4\pi\epsilon_{0}}*\frac{q^2}{r^2}*\frac{2Lr^2}{mg})^{1/3}[/tex]. The problem with this is that I had to work in [tex]\frac{r^2}{r^2}[/tex], when I'm pretty sure there's an equivalency in there somewhere.

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    Last edited: Feb 5, 2008
  2. jcsd
  3. Feb 5, 2008 #2

    Doc Al

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    Staff: Mentor

    Pick one of the balls (the left one, say) and analyze the forces acting on it. The Coulomb force is just one of the forces involved. Hint: Consider vertical and horizontal components separately.
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