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 follow these steps:

### Step 1: Understand the system
We have two small spheres, each with a mass of 0.1 g (which is 0.1 x 10^-3 kg) and each carrying a charge of \(10^{-9} C\). They are suspended by threads of equal length from the same point, and the distance between the centers of the spheres is 3 cm (which is 0.03 m).

### Step 2: Convert mass to kg
Convert the mass of the spheres from grams to kilograms:
\[ m = 0.1 \, \text{g} = 0.1 \times 10^{-3} \, \text{kg} = 1 \times 10^{-4} \, \text{kg} \]

### Step 3: Calculate the electrostatic force between the spheres
Using Coulomb's law, the electrostatic force \(F_e\) between the two charges is given by:
\[ F_e = k \frac{q_1 q_2}{r^2} \]
Where:
- \(k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\) (Coulomb's constant)
- \(q_1 = q_2 = 10^{-9} \, \text{C}\)
- \(r = 0.03 \, \text{m}\)

Substituting the values:
\[ F_e = 9 \times 10^9 \frac{(10^{-9})^2}{(0.03)^2} \]
\[ F_e = 9 \times 10^9 \frac{10^{-18}}{0.0009} = 9 \times 10^9 \times 1.1111 \times 10^{-15} = 10^{-5} \, \text{N} \]

### Step 4: Analyze forces acting on the spheres
Each sphere experiences three forces:
1. The gravitational force \(F_g = mg\) acting downwards.
2. The tension \(T\) in the thread acting along the thread.
3. The electrostatic force \(F_e\) acting horizontally (repulsive force).

### Step 5: Calculate the gravitational force
\[ F_g = mg = (1 \times 10^{-4} \, \text{kg})(10 \, \text{m/s}^2) = 1 \times 10^{-3} \, \text{N} \]

### Step 6: Set up the equations of motion
Using the equilibrium of forces:
- Vertically: \(T \cos \theta = F_g\)
- Horizontally: \(T \sin \theta = F_e\)

Dividing these two equations:
\[ \frac{T \sin \theta}{T \cos \theta} = \frac{F_e}{F_g} \]
\[ \tan \theta = \frac{F_e}{F_g} \]

### Step 7: Substitute the values
Substituting the values we calculated:
\[ \tan \theta = \frac{10^{-5}}{1 \times 10^{-3}} = 0.01 \]

### Step 8: Calculate the angle
Now, using the inverse tangent function:
\[ \theta = \tan^{-1}(0.01) \]

Using the given hint:
\[ \tan^{-1}(1/100) \approx 0.6^\circ \]
Thus, the angle made by the thread with the vertical is approximately \(0.6^\circ\).

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

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