Question

In Fig. two tiny conducting balls of identical mass m and identical charge q hang from nonconducting threads of length L. Assume that $\theta$ is so small that than $\theta$ can be replaced by its approximate equal, $\sin \theta$.
If $L=120cm,m=10g,andx=5.0cm,$ what is $|q|$?

Answer 1

A force diagram for one of the balls is shown below. The force of gravity $\overrightarrow{mg}$ acts downward, the electrical force ${\overrightarrow{F}}_e$ of the other ball acts to the left, and the tension in the thread acts along the the thread, at the angle $\theta$ to the vertical. The ball is in equilibrium so its acceleration is zero. The $y$ component of Newton's second law yield

$T\cos \theta -mg=0$ and the x component yield $T\sin \theta -F_e=0$. We solve the first equation for $T$ and obtain

$T=mg/\cos \theta$. We substitute the result into the second to obtain $mg\tan \theta -F_e=0$.

Examination of the geometry of Fig. leads to $\tan \theta =\frac{x/2}{\sqrt{L^2-(x/2{)}^2}}$

If L is much larger than x (which is the case if $\theta$ is very small), we may neglect $x/2$ in the denominator and write $\tan \theta \approx x/2L$. This is equivalent to approximating $\tan \theta$ by $\sin \theta$. The magnitude of the electrical force of one ball on the other is

$F_e=\frac{q^2}{4\pi {\epsilon }_0{x}^2}$

by Eq. When these two expressions are used in the equation $mg\tan \theta =F_e$, we obtain

$\frac{mgx}{2L}\approx \frac{1}{4\pi {\epsilon }_0{x}^2}\frac{q^2}{x^2}\Rightarrow x\approx {\left(\frac{q^{2}L}{2\pi {\epsilon }_{0}mg}\right)}^{1/3}$.

We slove $x^3=2k{q}^{2}L/mg$ for the charge $q=\sqrt{\frac{mg{x}^3}{2kL}}=\sqrt{\frac{\left(0.010kg\right)\left(9.8m/s^2\right)(0.050m{)}^3}{\left(2\right(8.99\times {10}^{9}N.m^2/C^2\left)\right(1.20m)}}=\pm 2.4\times {10}^{-8}C$
Thus, the magnitude is $|q|=2.4\times {10}^{-8}C$