The sum of two point charges is 9 μC. They repel each otherforce of 2 N when kept 30 cm apart in free spaceCalculate thevalue of each charge?

Let q μC and (9 - q) μC be the two values of the two point charges.

Then,
F = 2 N
Q₁ = q × 10⁻⁶ C
Q₂ = (9 - q) × 10⁻⁶ C
r = 30 cm = 0.3 m

By Coulomb's law :
F = kₑ Q₁ Q₂ / r² .... where kₑ = ke = 8.99 × 10⁹ N m² C⁻²

2 = (8.99 × 10⁹) × (q × 10⁻⁶) × [(9 - q) × 10⁻⁶] / 0.3²
2 = (8.99 × 10⁻³) × q × (9 - q) / 0.09
2 × 0.09 / (8.99 × 10⁻³) = q (9 - q)
q (9 - q) = 20
q² - 9q + 20 = 0
(q - 4)(q - 5) = 0
q = 4 or q = 5

Hence, the value of the two point charges are 4 μC and 5 μC respectively.

Are you not overcomplicating things?

The sum of two is Nine, the value of each is 4.5 unless you show why they are not equal./