# Question #54e33

Dec 6, 2016

We draw a free-body diagram for one of the charged balls, say left. This is shown in the figure below.

All the forces acting on the charged ball are in equilibrium.

One force is tension $T$ in the silk thread. The other force is weight $m g$ of ball acting downward and third is electrostatic force of repulsion ${F}_{\text{elec}}$ between the two charged balls.

Equating the magnitude of vertical component of tension with weight of the ball we get
$T \cos \theta = m g$ .....(1)
and equating the magnitude of horizontal component of tension with the magnitude of electrostatic force we get
$T \sin \theta = {F}_{\text{elec}} = {k}_{e} {q}^{2} / {x}^{2}$ .....(2)
where $x$ is the distance between the two charged balls.

Dividing (2) by (1) we get
$\tan \theta = {k}_{e} {q}^{2} / {x}^{2} / m g$ ....(3)

It is given that the angle between the silk threads is small ($2 \theta$ in our figure). We know that for small angles $\theta$

$\tan \theta \approx \sin \theta$ ....(4)

From the geometry of the figure we see that
$\sin \theta = \frac{\frac{x}{2}}{l} = \frac{x}{2 l}$ .....(5)
Using (4) and (5) equation (3) reduces to
$\frac{x}{2 l} \approx {k}_{e} {q}^{2} / \left({x}^{2} m g\right)$

Substituting the value of Coulomb's constant ${k}_{e}$ in terms of electric permittivity of free space ${\epsilon}_{\circ}$and solving for $x$ we get
${x}^{3} = \frac{1}{4 \pi {\epsilon}_{\circ}} \times 2 l {q}^{2} / \left(m g\right)$
$\implies x = {\left(\frac{{q}^{2} l}{2 \pi {\epsilon}_{\circ} m g}\right)}^{\frac{1}{3}}$