Two small spheres, each of mass 0.1 gm and carrying same charge `10^(-9) C` are suspended by threads of equal length from the same point. If the distance between the centres of the sphere is 3 cm, then find out the angle made by the thread with the vertical. `(g=10 m//s^(2))` & `tan^(-1) (1/100) =0.6^(º)`

Answer

To solve the problem, we will analyze the forces acting on the two charged spheres and use trigonometric relationships to find the angle made by the threads with the vertical.

### Step 1: Understand the setup
We have two small spheres, each with a mass of \(0.1 \, \text{g}\) (which is \(0.1 \times 10^{-3} \, \text{kg} = 10^{-4} \, \text{kg}\)) and each carrying a charge of \(10^{-9} \, \text{C}\). The distance between the centers of the spheres is \(3 \, \text{cm}\) (which is \(0.03 \, \text{m}\)).

### Step 2: Identify the forces acting on the spheres
Each sphere experiences three forces:
1. The gravitational force (\(mg\)) acting downwards.
2. The tension (\(T\)) in the thread acting upwards at an angle \(\theta\) with the vertical.
3. The electrostatic repulsive force (\(F\)) acting horizontally due to the charges.

### Step 3: Write the equations for the forces
From the free body diagram:
- The vertical component of the tension balances the weight:
\[ T \cos \theta = mg \]
- The horizontal component of the tension balances the electrostatic force:
\[ T \sin \theta = F \]

### Step 4: Calculate the electrostatic force
Using Coulomb's law, the electrostatic force \(F\) between the two charges is given by:
\[ F = k \frac{q^2}{r^2} \]
where \(k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\), \(q = 10^{-9} \, \text{C}\), and \(r = 0.03 \, \text{m}\).

Calculating \(F\):
\[ F = 9 \times 10^9 \cdot \frac{(10^{-9})^2}{(0.03)^2} = 9 \times 10^9 \cdot \frac{10^{-18}}{0.0009} = 9 \times 10^9 \cdot \frac{10^{-18}}{9 \times 10^{-4}} = 10^{-9} \, \text{N} \]

### Step 5: Substitute \(F\) into the tension equations
From the equations of tension:
1. \(T \cos \theta = mg\)
2. \(T \sin \theta = F\)

Dividing the second equation by the first:
\[ \frac{T \sin \theta}{T \cos \theta} = \frac{F}{mg} \]
This simplifies to:
\[ \tan \theta = \frac{F}{mg} \]

### Step 6: Substitute the known values
Substituting \(F = 10^{-9} \, \text{N}\), \(m = 10^{-4} \, \text{kg}\), and \(g = 10 \, \text{m/s}^2\):
\[ \tan \theta = \frac{10^{-9}}{10^{-4} \cdot 10} = \frac{10^{-9}}{10^{-3}} = 10^{-6} \]

### Step 7: Calculate \(\theta\)
Using the small angle approximation:
\[ \theta \approx \tan^{-1}(10^{-6}) \]
Given that \(\tan^{-1}(1/100) = 0.6^\circ\), we can conclude that:
\[ \theta \approx 0.6^\circ \]

### Final Answer
The angle made by the thread with the vertical is approximately \(0.6^\circ\).

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