Calculate the electrostatic force between two point charges q1 = 2 imes 10⁻⁷ C and q2 = 3 imes 10⁻⁷ C separated by 30 cm.

The first charge is $q_1 = 2 \times 10^{-7} \text{ C}$.
The second charge is $q_2 = 3 \times 10^{-7} \text{ C}$.
The distance between the charges is $r = 30 \text{ cm}$. Convert this to meters:

$$r = 30 \times 10^{-2} \text{ m} = 0.3 \text{ m}$$

The electrostatic constant (Coulomb's constant) is $k = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$.

Coulomb's Law states that the force ($F$) between two point charges is given by:

$$F = k \frac{q_1 q_2}{r^2}$$

$$F = (9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}) \frac{(2 \times 10^{-7} \text{ C})(3 \times 10^{-7} \text{ C})}{(0.3 \text{ m})^2}$$ $$F = (9 \times 10^9) \frac{(6 \times 10^{-14})}{0.09} \text{ N}$$ $$F = \frac{54 \times 10^{9-14}}{0.09} \text{ N}$$ $$F = \frac{54 \times 10^{-5}}{0.09} \text{ N}$$ $$F = 600 \times 10^{-5} \text{ N}$$ $$F = 6 \times 10^{-3} \text{ N}$$

The force between the two charged spheres is $6 \times 10^{-3} \text{ N}$.