1. **State the problem:** Calculate the electrostatic force between a +5.0 C charge and a +9.0 C charge separated by 4 m using Coulomb's Law. 2. **Formula:** Coulomb's Law is given by $$F_{elec} = k \cdot \frac{q_1 \cdot q_2}{d^2}$$ where - $k = 9 \times 10^9$ N·m²/C² (Coulomb's constant), - $q_1 = 5.0$ C, - $q_2 = 9.0$ C, - $d = 4$ m. 3. **Substitute the values:** $$F_{elec} = 9 \times 10^9 \cdot \frac{5.0 \times 9.0}{4^2}$$ 4. **Calculate the denominator:** $$4^2 = 16$$ 5. **Calculate the numerator:** $$5.0 \times 9.0 = 45$$ 6. **Rewrite the expression:** $$F_{elec} = 9 \times 10^9 \cdot \frac{45}{16}$$ 7. **Simplify the fraction:** $$\frac{45}{16} = 2.8125$$ 8. **Calculate the force:** $$F_{elec} = 9 \times 10^9 \times 2.8125 = 2.53125 \times 10^{10}$$ 9. **Final answer:** $$\boxed{F_{elec} = 2.53 \times 10^{10} \text{ N}}$$ This force is repulsive since both charges are positive.