Question

Two Styrofoam balls with the same mass m = 7.90 x 10-0 kg and the same positive charge Q are suspended from the same point by insulating threads of lengthL0.980 m (the figure below). The separation of the balls is d = 0.0290 m, what is the charge Q ? no

If you could please do a step by step ? thank you!

Answer 1

Let the tension in the thread is T.
So, the vertically upward component of the tension is
     $T\cos\frac{\theta}{2}$
And the gravitational force on the ball acting vertically downward is
   $mg$
So, we get at the equilibrium
   $T\cos\frac{\theta}{2}=mg$
And horizontal component of the tension is
   $T\sin\frac{\theta}{2}$
And the electrical force on the charge is
   $\frac{Q^2}{4\pi\epsilon_0 d^2}$
And so, at the equilibrium, we have
   $T\sin\frac{\theta}{2}=\frac{Q^2}{4\pi\epsilon_0 d^2}$
So, taking the ratio of two equations, we get
   $\tan\frac{\theta}{2}=\frac{Q^2}{4\pi\epsilon_0 d^2mg}$
And as
   $d<<L$, so, we get
   $\tan\frac{\theta}{2}\approx \frac{d}{2L}$
So, we get
   $\frac{Q^2}{4\pi\epsilon_0 d^2mg}=\frac{d}{2L}$
  $\Rightarrow Q=\sqrt{\frac{4\pi\epsilon_0 d^3mg}{2L}}$

Now putting the given values,
   $m=7.90\times 10^{-8}~kg$, $L=0.980~m$, $d=0.029~m$, $g=9.80~m/s^2$, and
   $4\pi\epsilon_0=\frac{1}{9\times 10^9}~\text{SI unit}$
So, we get
  $Q=\sqrt{\frac{(0.029)^3\times 7.90\times 10^{-8}\times 9.8}{2\times 0.98\times 9\times 10^9}}$
  $\Rightarrow Q=3.27\times 10^{-11}~C$
  $\Rightarrow Q=0.0327\times 10^{-9}~C$
  $\Rightarrow Q=0.0327~nC$