Two small spheres each of mass 10 mg are suspended from a point by threads 0.5 m long. They are equally charged and repel each other to a distance of 0.20 m. The charge on each of the sphere is `(a)/(21) xx 10^(-8)C`. The value of 'a' will be _______.
[Given `g=10ms^(-2)`]

Answer

To solve this problem, we need to find the value of the charge \( q \) on each sphere and then determine the value of \( a \) in the given expression \( q = \frac{a}{21} \times 10^{-8} \, \text{C} \).

### Step-by-Step Solution:

1. **Given Data:**
- Mass of each sphere, \( m = 10 \, \text{mg} = 10 \times 10^{-6} \, \text{kg} \)
- Length of each thread, \( L = 0.5 \, \text{m} \)
- Distance between the spheres, \( d = 0.20 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Diagram and Geometry:**
- The spheres are in equilibrium, repelling each other due to their charges.
- The threads form an isosceles triangle with the vertical.
- The horizontal separation between the spheres is \( d = 0.20 \, \text{m} \).

3. **Calculate the Angle \( \theta \):**
- The horizontal separation is \( d \), so each half is \( \frac{d}{2} = 0.10 \, \text{m} \).
- Using trigonometry, \( \tan \theta = \frac{\frac{d}{2}}{L} = \frac{0.10}{0.5} = 0.2 \).

4. **Forces in Equilibrium:**
- The tension \( T \) in the thread has components:
- Vertical component: \( T \cos \theta \)
- Horizontal component: \( T \sin \theta \)
- The weight \( mg \) acts downward.
- The electrostatic force \( F \) acts horizontally.

5. **Balance of Forces:**
- Vertically: \( T \cos \theta = mg \)
- Horizontally: \( T \sin \theta = F \)

6. **Find \( T \):**
- From vertical balance: \( T = \frac{mg}{\cos \theta} \)

7. **Electrostatic Force \( F \):**
- Using Coulomb's law: \( F = \frac{k q^2}{d^2} \)
- Here, \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)

8. **Relate \( F \) and \( T \sin \theta \):**
- \( F = T \sin \theta \)
- Substitute \( T \) from the vertical balance: \( F = \left( \frac{mg}{\cos \theta} \right) \sin \theta \)
- Simplify using \( \sin \theta / \cos \theta = \tan \theta \): \( F = mg \tan \theta \)

9. **Calculate \( F \):**
- \( \tan \theta = 0.2 \)
- \( F = mg \tan \theta = (10 \times 10^{-6} \, \text{kg}) \times (10 \, \text{m/s}^2) \times 0.2 = 2 \times 10^{-5} \, \text{N} \)

10. **Relate \( F \) to \( q \):**
- \( F = \frac{k q^2}{d^2} \)
- \( 2 \times 10^{-5} = \frac{9 \times 10^9 \times q^2}{(0.20)^2} \)
- Solve for \( q^2 \): \( q^2 = \frac{2 \times 10^{-5} \times (0.20)^2}{9 \times 10^9} \)
- \( q^2 = \frac{2 \times 10^{-5} \times 0.04}{9 \times 10^9} = \frac{8 \times 10^{-7}}{9 \times 10^9} = \frac{8}{9} \times 10^{-16} \)

11. **Calculate \( q \):**
- \( q = \sqrt{\frac{8}{9} \times 10^{-16}} = \frac{2 \sqrt{2}}{3} \times 10^{-8} \, \text{C} \)
- \( q = \frac{2 \sqrt{2}}{3} \times 10^{-8} \, \text{C} \approx 0.94 \times 10^{-8} \, \text{C} \)

12. **Determine \( a \):**
- Given \( q = \frac{a}{21} \times 10^{-8} \, \text{C} \)
- \( 0.94 \times 10^{-8} = \frac{a}{21} \times 10^{-8} \)
- \( a = 0.94 \times 21 \approx 20 \)

### Final Answer:
The value of \( a \) is \( \boxed{20} \).