Three identical small balls, each of mass 0.1 g, are suspended at one point on silk thread having a length of `l = 20cm` . What charges should be imparted to the balls for each thread to form an angle of `alpha = 30^@` with the vertical?

Explanation

To solve the problem, we need to find the charge that should be imparted to each of the three identical small balls so that they form an angle of \( \alpha = 30^\circ \) with the vertical when suspended by silk threads of length \( l = 20 \, \text{cm} \).

### Step-by-Step Solution:

1. **Convert Units**:
- Convert the mass of each ball from grams to kilograms:
\[ m = 0.1 \, \text{g} = 0.1 \times 10^{-3} \, \text{kg} = 10^{-4} \, \text{kg} \]
- Convert the length of the thread from centimeters to meters:
\[ l = 20 \, \text{cm} = 0.2 \, \text{m} \]

2. **Determine the Geometry**:
- The angle \( \alpha = 30^\circ \) implies that each ball will be at a distance \( R \) from the vertical. Using trigonometry:
\[ R = l \sin(30^\circ) = 0.2 \times \frac{1}{2} = 0.1 \, \text{m} \]
- The distance between the two charges (balls) can be found using the geometry of the equilateral triangle formed by the three balls:
\[ d = \sqrt{3} R = \sqrt{3} \times 0.1 = 0.1732 \, \text{m} \]

3. **Calculate the Forces**:
- The gravitational force acting on each ball is:
\[ F_g = mg = 10^{-4} \times 10 = 10^{-3} \, \text{N} \]
- The electrostatic force \( F \) between any two charges \( q \) at a distance \( d \) is given by Coulomb's law:
\[ F = k \frac{q^2}{d^2} \]
- Here, \( k \) (Coulomb's constant) is approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

4. **Force Balance**:
- In the vertical direction, the tension \( T \) in the thread balances the weight of the ball:
\[ T \cos(30^\circ) = mg \]
- In the horizontal direction, the tension provides the centripetal force due to the electrostatic repulsion:
\[ T \sin(30^\circ) = F \]

5. **Express Tension**:
- From the vertical force balance:
\[ T = \frac{mg}{\cos(30^\circ)} = \frac{10^{-3}}{\frac{\sqrt{3}}{2}} = \frac{2 \times 10^{-3}}{\sqrt{3}} \, \text{N} \]

6. **Substituting into Horizontal Force Equation**:
- Substitute \( T \) into the horizontal force equation:
\[ \frac{2 \times 10^{-3}}{\sqrt{3}} \cdot \frac{1}{2} = k \frac{q^2}{d^2} \]
- Simplifying gives:
\[ \frac{10^{-3}}{\sqrt{3}} = k \frac{q^2}{(0.1732)^2} \]

7. **Solving for Charge \( q \)**:
- Rearranging gives:
\[ q^2 = \frac{10^{-3} \cdot (0.1732)^2}{k \cdot \sqrt{3}} \]
- Plugging in the values:
\[ q^2 = \frac{10^{-3} \cdot (0.1732)^2}{9 \times 10^9 \cdot \sqrt{3}} \]
- Calculating \( q \):
\[ q = \sqrt{\frac{10^{-3} \cdot 0.0300}{9 \times 10^9 \cdot 1.732}} \approx 3.33 \times 10^{-8} \, \text{C} \]

### Final Answer:
The charge that should be imparted to each ball is approximately \( q \approx 3.33 \times 10^{-8} \, \text{C} \).