1. Given data

• Mass of each sphere:
  $m = 10\,\text{mg} = 10\times 10^{-6}\,\text{kg} = 10^{-5}\,\text{kg}$
• Length of each thread:
  $l = 0.5\,\text{m}$
• Separation between the spheres:
  $r = 0.20\,\text{m}$
• Acceleration due to gravity:
  $g = 10\,\text{m s}^{-2}$

2. Geometry of the arrangement

Since the spheres are equally charged and suspended from the same point, the system is symmetric.

Each sphere is displaced horizontally by $x = \frac{r}{2} = \frac{0.20}{2} = 0.10\,\text{m}$

If the thread makes angle $\theta$ with the vertical, then $\sin\theta = \frac{x}{l} = \frac{0.10}{0.5} = 0.2$

So, $\sin\theta = 0.2$ $\cos\theta = \sqrt{1-0.2^2} = \sqrt{0.96}$

Hence, $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{0.2}{\sqrt{0.96}} \approx 0.204$

3. Forces on one sphere

For one sphere in equilibrium:

• Weight downward: $mg$
• Tension along thread: $T$
• Electrostatic repulsion horizontally: $F_e$

Resolving tension: $T\cos\theta = mg$ $T\sin\theta = F_e$

Therefore, $\tan\theta = \frac{F_e}{mg}$ $F_e = mg\tan\theta$

Now, $mg = 10^{-5}\times 10 = 10^{-4}\,\text{N}$

So, $F_e = 10^{-4}\times 0.204 \approx 2.04\times 10^{-5}\,\text{N}$

4. Using Coulomb's law

For equal charges $q$ separated by $r=0.20\,\text{m}$, $F_e = \frac{kq^2}{r^2}$

Taking $k = 9\times 10^9\,\text{N m}^2\text{/C}^2$

Thus, $q^2 = \frac{F_e r^2}{k}$

Substitute values: $q^2 = \frac{(2.04\times 10^{-5})(0.20)^2}{9\times 10^9}$ $q^2 = \frac{(2.04\times 10^{-5})(0.04)}{9\times 10^9}$ $q^2 = \frac{8.16\times 10^{-7}}{9\times 10^9}$ $q^2 \approx 9.07\times 10^{-17}$

Therefore, $q \approx \sqrt{9.07\times 10^{-17}} \approx 9.52\times 10^{-9}\,\text{C}$

5. Match with given form

Given, $q = \frac{a}{21}\times 10^{-8}\,\text{C}$

We found $q \approx 9.52\times 10^{-9} = 0.952\times 10^{-8}\,\text{C}$

So, $\frac{a}{21} = 0.952$ $a \approx 21\times 0.952 \approx 20.0$

6. Final answer

$\boxed{a=20}$

7. Comparison with stored answer

Stored correct answer = $20$.

My derived answer matches the stored answer.