Question

Two identical balls of mass 44.1 g are suspended from threads of length 1.11 m and carry equal charges as shown in the figure. Each ball is 0.70 cm from the centerline. Assume that the angle 9 is so small that its tangent can be replaced by its sine. This is called the small angle approximation and means that for small angles sin() tan(). Find the magnitude of charge on one of the balls 7.70x 10.9 C 4pts You are correct. Your receipt no. is 154-5142 Preous Tries Now, assume the two bails are losing charge to the air very slowly. That means they'll be slowly approaching each other. If a ball is moving at an instantaneous speed of 6.80E-5 m/s at what rate is the ball losing charge? Start by writing the speed of the ball and the rate of change of the charge as symbolic derivatives, and then relate those derivatives. Give your answer in Coulombs per second (C/s). Note that because the balls are losing charge so slowly, we can still use our results from the previous part, as the system is almost in equilibrium, Give your answer as a magnitude. 4pts Semit Ansae Incorrect. Tries 3/10 Previous Tries

Answer 1

I am assuming you have derived the following equilibrium condition

$\tan\theta=\frac{KQ^2}{4x^2mg}$

Because of small angle approximation

$\frac{x}{l}=\frac{KQ^2}{4x^2mg}$

$Q=\sqrt{\frac{4mgx^3}{Kl}}$

Take logarithm

$\log Q=\frac{3}{2}\log x+\textup{ constants}$

take time derivatives

$\frac{\dot{Q}}{Q}=\frac{3}{2}\frac{\dot{x}}{x}$

We also need to relate the instantaneous speed with $\dot{x}$, because the velocity is tangential to the string, so the relation should be

$v\cos\theta=\dot{x}$

but again because of small angle approximation

$v=\dot{x}$

Thus,

$\dot{Q}=\frac{3Qv}{2x}$

Now we have everything we need

$\dot{Q}=\frac{3\times7.7\times 10^{-7}\times 6.8\times 10^{-5}}{2\times7\times 10^{-3}}\textup{ C/s}$

$\dot{Q}\approx1.12\times 10^{-8}\textup{ C/s}$