Question

Two small spheres, each of mass $0.1\text{gm}$ and carrying same charge of ${10}^{-9}\text{C}$ are suspended by threads of equal length from the same point. If the distance between the centers of the sphere is $3\text{cm}$, then find out the angle made by the thread with the vertical. $\left(g=10{\text{m/s}}^2\right)$ and $\left({\tan }^{-1}\frac{1}{100}=0.6^{\circ }\right)$

Answer 1

Step 1: The diagram can be drawn as follows:

Step 2: Equilibrium of force in the vertical direction

$T\cos \theta =mg$

Here, $T$ is the tension force, $m$ is the mass of the sphere, $g$ is the acceleration due to gravity, and $\theta$ is the angle.

The tension force can be calculated as follows:

$$\begin{gathered} T\cos \theta =0.1\text{gm}\times {10}^{-3}\text{kg}\times 10{\text{m/s}}^2 \\ T\cos \theta ={10}^{-3}\text{N} \end{gathered}$$

Step 3: Equilibrium of force in the horizontal direction

Due to the equilibrium of the forces, the electrical force will be equal to the tension force.

$T\sin \theta =\frac{k{q}^2}{r^2}$

Here, $k$ is the proportionality constant, $q$ is the charge in the spheres, and $r$ is the distance between the spheres.

Step 4: Calculation of angle.

$$\begin{gathered} \frac{T\sin \theta }{T\cos \theta }=\frac{{\displaystyle \frac{9\times {10}^9{\text{Nm}}^2/{\text{C}}^2\times {\left({10}^{-9}\text{C}\right)}^2}{{\left(3\text{cm}\times {\displaystyle \frac{1\text{m}}{100\text{cm}}}\right)}^2}}}{{10}^{-3}\text{N}} \\ \tan \theta =\frac{1}{100} \\ \theta ={\tan }^{-1}\left(\frac{1}{100}\right) \\ \theta =0.6^{\circ } \end{gathered}$$

Therefore, the angle made by the thread to the vertical is $0.6^{\circ }$.