# Find the components of the resultant force on it along&perpendicular to the string?

## Two similar balls, having charge +q and mass m gram are suspended from a common point by two insulating strings l meters long. The balls are held at a separation l/10 apart and released.

Feb 8, 2017

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.

It is not given clearly but assumed that the system is in equilibrium when distance between the charged balls is $= \frac{l}{10}$
Therefore, distance between vertical drawn from common hanging point and each ball is $\frac{l}{20}$

One force is tension $T$ in the insulating string. 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.

${F}_{\text{elec}} = {k}_{e} {q}^{2} / {\left(\frac{l}{10}\right)}^{2} = \frac{100 {k}_{e} {q}^{2}}{l} ^ 2$
where ${k}_{e}$ is Coulomb's Constant

It is clear from geometry that

1. $\sin \theta = \frac{\frac{l}{20}}{l} = \frac{1}{20} = 0.05$
and
2. $\cos \theta = \sqrt{1 - {\sin}^{2} \theta} = \sqrt{1 - \frac{1}{20} ^ 2} = 0.9987$

3. Component of weight along the string$= m g \cos \theta$
Component of weight perpendicular to the string$= m g \sin \theta$

4. Component of tension along the string$= T$
Component of tension perpendicular to the string$= 0$

5. Component of electrostatic force along the string$= \frac{100 {k}_{e} {q}^{2}}{l} ^ 2 \sin \theta$
Component of electrostatic force perpendicular to the string$= \frac{100 {k}_{e} {q}^{2}}{l} ^ 2 \cos \theta$

From (1) to (5) in the case of all forces being in equilibrium we get
$T = 0.9987 m g + 0.05 \frac{100 {k}_{e} {q}^{2}}{l} ^ 2$

$T = 0.9987 m g + \frac{5 {k}_{e} {q}^{2}}{l} ^ 2$ .....(1)

$0.05 m g = 0.9987 \frac{100 {k}_{e} {q}^{2}}{l} ^ 2$

$m g = \frac{1997.5 {k}_{e} {q}^{2}}{l} ^ 2$ ......(2)