Charged Particles on a pendulum

patm95
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Homework Statement



What is the distance(d) separating the charged particles dangling on the pendulum?

Length of Nonconducting string=120cm
mass of particles= 10g
Charge q = 1.2*10^-8


Homework Equations



Fe=(k*q^2)/d^2



The Attempt at a Solution



I think I could get this problem easily if I had either the angle between verticle and the nonconducting string or the electric force opposing the masses. I have been using my knowledge of trigonometry to figure this out and trying to manipulate the above equation, but with no luck. Just needed to be pointed in the right direction I think. Thanks!
 
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Focus on one of the particles. You have three forces on the particle, gravity, electrical, and the tension force of the the string. They all have to balance (i.e. the sum is zero). Split each one of them into xy components and express them in terms of the unknown angle theta. Start writing down the equations you get, ok?
 
ok... I will show you what I have. I don't know Fe or Theta...That is where I am hitting a roadblock. mg= .098N

Fe=Tsin(theta)
mg=Tcos(theta)
Fe/mg=Tan(theta)

Am I still overlooking something?
 
You can also express Fe in terms of sin(theta), since the distance is a function of sin(theta).
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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