B

Step 1: Understanding the Problem
We have two identical positive charges of magnitude 5 µC each fixed at a distance of 8 cm apart. A negatively charged particle of mass 9 x 10^-6 kg and charge -10 µC is placed at equal distances (5 cm) from each of these positive charges.

These two positive charges will exert a force on the negatively charged particle due to electrostatic interaction as described by Coulomb's Law.

Step 2: Using Coulomb's Law
The force exerted by a single charge on the particle is given by Coulomb's Law:
$$ F = k \frac{|q_1 q_2|}{r^2} $$
where:
- $k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2$ (Coulomb's constant)
- $q_1 = 5 \times 10^{-6} \text{C}$ (charge of the fixed charge)
- $q_2 = 10 \times 10^{-6} \text{C}$ (charge of the particle)
- $r = 0.05 \text{m}$ (distance from the charge to the particle)

Calculating the force from one charge:
$$ F = 8.99 \times 10^9 \frac{5 \times 10^{-6} \times 10 \times 10^{-6}}{(0.05)^2} $$
$$ F = 8.99 \times 10^9 \frac{50 \times 10^{-12}}{0.0025} \approx 179800 \text{ N} $$

Since the particle is equidistant from both charges, the net force is aligned along the line connecting the charges and is equal to the sum of forces from both charges, which can be found similarly.

Step 3: Calculating Potential Energy
Next, we find the potential energy (PE) of the system when the particle is at the midpoint (when it is closest) which is when it will be at its minimum potential energy:
$$ U = k \sum_{i=1}^{n} \frac{q_1 q_2}{r} $$
At 5 cm from each charge (0.05 m), we have:
$$ U = 2 * k * \frac{(5 \times 10^{-6})(10 \times 10^{-6})}{0.05} $$
Substituting the values:
$$ U = 2 * 8.99 \times 10^9 * \frac{50 \times 10^{-12}}{0.05} \approx 179800 ext{ J} $$

Step 4: Conservation of Energy
At the initial position, all energy is potential energy (U) and after it reaches the closest point, it transforms to kinetic energy (K):
$$ K = \frac{1}{2} mv^2 \Rightarrow K + U = 0 $$
$$ \frac{1}{2} mv^2 = U \Rightarrow v^2 = \frac{2U}{m} $$
Substituting the values:
$$ v^2 = \frac{2 * 179800}{9 \times 10^{-6}} \Rightarrow v \approx \sqrt{39960000000} \approx 19990. ext{m/s} $$

Step 5: Result
The final speed of the particle when it is nearest to the two charges is approximately 19990 m/s.
Thus, the correct answer is option B which states the speed is approximately 20000 m/s.