In Fig. 21-39, two tiny conducting balls of identical mass mand identical charge hang from non-conducting threads of length L. Assume that u is so small that tan u can be replaced by its approximate equal, sin u.

(a) Show that$\mathit{x}\mathbf{=}{\left(\frac{q^{2}L}{2\pi {\epsilon }_{0}mg}\right)}^{\mathbf{1}\mathbf{/}\mathbf{3}}$gives the equilibrium separation xof the balls.

(b) If $\mathit{L}\mathbf{}\mathbf{=}\mathbf{120}\mathbf{}\mathit{c}\mathit{m}\mathbf{,}\mathbf{}$and, $\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.0}\mathbf{}\mathit{c}\mathit{m}$what is$\mathbf{\left|}\mathbf{q}\mathbf{\right|}$?

1. Two tiny conducting balls of identical mass and identical charge hang from non-conducting threads of length, L .
2. Given values to get the charge,$L=120cm,m=10g,andx=5.0cm$

Using the concept of Newton's laws of motion and the force due to Coulomb's law, we can get the formula of the separation equilibrium of the system. Using this formula and the given data, we can get the magnitude of the unknown charge.

Formula:

The magnitude of the electrostatic force between any two particles, $F=k\frac{\left|q_1\right|\left|q_2\right|}{r^2}$ (i)

A force diagram for one of the balls is shown below. The force of gravity,$mg$acts downward, the electrical force $F_e$of the other ball acts to the left, and the tension in the thread acts along the thread, at the angle 6 to the vertical. The ball is in equilibrium, so its acceleration is zero.

The y component of forces acting in the system using Newton’s law of motion yields,

$Tcos\theta –mg=\mathrm{0............................}\left(a\right)$

The x component of the forces acting using Newton’s law of motion yields,

$Tsin\theta -F_e=\mathrm{0.............................}\left(b\right)$

Now, solving both equations (a) and (b), we get:

$T=mg/cos\theta$

$\begin{array}{c}mgtan\theta –F_e=0.\text{(aftersubstitutingtheabovevalueoftension)}\\ F_e=mgtan\theta \mathrm{.................................}\left(c\right)\end{array}$

Examination of the geometry of the figure shown leads to the value of the tangent angle as:

$tan\theta =\frac{x/2}{\sqrt{L^2-{(x/2)}^2}}$

For large values of length,$L(>>>>>x)$we can get that

$tan\theta \approx x/2L\mathrm{............................}\left(d\right)$

Thus, the electrostatic charge of one ball one other ball is given using equation (i) as follows:

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

Substituting the value of equation (d) in equation (c) and equating that to the above equation, we get that

$\begin{array}{c}\frac{mgx}{2L}\approx \frac{q^2}{4\pi {\epsilon }_0{x}^2}\\ x={\left(\frac{q^{2}L}{2\pi {\epsilon }_{0}mg}\right)}^{1/3}\end{array}$

Hence, the value of the equilibrium separation of the balls is${\left(\frac{q^{2}L}{2\pi {\epsilon }_{0}mg}\right)}^{1/3}$

Using the above formula of separation equilibrium in the part (a) calculations, we can get the magnitude of the charge by substituting the given values as:

$\begin{array}{c}q=\sqrt{\frac{mg{x}^3}{2kL}}\\ =\sqrt{\frac{(0.010kg)(9.8m/s^2){(0.050m)}^3}{2(9.0\times {10}^{9}N{m}^2/C^2)(1.20m)}}\\ =\pm 2.4\times {10}^{-8}C.\\ \left|q\right|=2.4\times {10}^{-8}C.\end{array}$

Hence, the value of the magnitude of the charge is$2.4\times {10}^{-8}C.$