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Homework Help: Dissipating charge

  1. Aug 3, 2006 #1
    A friend of mine has given me a physics problem to solve. Here it is:
    Two small equally charged spheres, each of mass [itex]m[/itex], are suspended from the same point by silk threads of length [itex]\ell[/itex]. The distance between the spheres [itex]x << \ell[/itex]. Find the rate [tex]\frac{dq}{dt}[/tex] with which the charge leaks off each sphere if their approach velocity varies as [tex]v = \frac{a}{\sqrt{x}}[/tex], where [itex]a[/itex] is a constant. ​
    I started by writing down the forces:
    [tex]mg\tan{\theta} = \frac{q^2}{4\pi \epsilon_0 x^2}[/tex]
    Since [itex]x << \ell[/itex], I concluded that [itex]\sin{\theta} \approx \tan{\theta}[/itex]. I rewrote the above force equation with sin instead of tan.

    After this, I started doing things randomly. At first, I tried using angular speed with [itex]\ell[/itex] as the radius of rotation, but that seemed unnecessarily complicated.

    So I went back and differentiated both sides of the first equation. I don't know if this is correct, but I let [itex]\cos{\theta}[/itex] be 1 since [itex]\theta \approx 0[/itex]. So, using the quotient rule, I had:
    [tex]mg = \frac{1}{2\pi \epsilon_0}\frac{q^2(\frac{dx}{dt}) - q\frac{dq}{dt}}{x^3}[/tex]
    At this point, since [itex]\frac{dx}{dt} = \frac{a}{\sqrt{x}}[/itex], I started substituting things in. However, I ended up with charge as a function of time, distance as a function of time, and the time derivative of the charge function all in one equation - unsolvable.

    The other way that I tried it was to start from the premise that [itex]\ell \frac{d\theta}{dt} = \frac{a}{\sqrt{x}}[/itex] and go from there. But I don't know if this is the correct way either.

    Basically, I have no idea what I'm doing. My friend said this is an easy problem, but I am stumped. Am I approaching this incorrectly?

    Thanks for the assistance.
     
    Last edited: Aug 3, 2006
  2. jcsd
  3. Aug 4, 2006 #2

    Tide

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    Solve your original equation for q and approximate [itex]\theta[/itex] with x/l. Then differentiate q with respect to time. You should end up with dq/dt being constant within this approximation.

    [Note: you may need to make minor adjustments to the preceding depending on whether your angle is the half-angle or full angle.]
     
  4. Aug 4, 2006 #3

    andrevdh

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    I don't think one should assume that the spheres are in equilibrium. That is

    [tex]T\sin(\theta) - F_Q = m\ddot{x}[/tex]

    since

    [tex]v_x(x) \Rightarrow v_x(t)[/tex]
     
    Last edited: Aug 4, 2006
  5. Aug 4, 2006 #4
    Thanks - I have found my errors now. I had to approximate [itex]\theta[/itex] with [itex]\frac{x}{2\ell}[/itex].

    For those who are interested, the answer is:
    [tex]
    \frac{dq}{dt} = \frac{3}{2}a\sqrt{\frac{2\pi \epsilon_0 mg}{\ell}}
    [/tex]
     
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