1. **State the problem:** We have two spheres with charges $q$ and $2q$ separated by a distance of 6 m. They exert a force of 20 N on each other. We need to find the values of $q$ and $2q$. 2. **Formula used:** The force between two point charges is given by Coulomb's law: $$F = k \frac{|q_1 q_2|}{r^2}$$ where $F$ is the force, $k = 8.99 \times 10^9 \ \text{Nm}^2/\text{C}^2$ is Coulomb's constant, $q_1$ and $q_2$ are the charges, and $r$ is the distance between them. 3. **Apply the values:** Here, $q_1 = q$, $q_2 = 2q$, and $r = 6$ m. Substitute into the formula: $$20 = 8.99 \times 10^9 \times \frac{|q \times 2q|}{6^2}$$ 4. **Simplify the equation:** $$20 = 8.99 \times 10^9 \times \frac{2q^2}{36}$$ 5. **Isolate $q^2$:** $$20 = \frac{8.99 \times 10^9 \times 2q^2}{36}$$ Multiply both sides by 36: $$20 \times 36 = 8.99 \times 10^9 \times 2q^2$$ $$720 = 8.99 \times 10^9 \times 2q^2$$ 6. **Divide both sides by $8.99 \times 10^9 \times 2$:** $$q^2 = \frac{720}{8.99 \times 10^9 \times 2}$$ Show cancellation: $$q^2 = \frac{\cancel{720}}{\cancel{8.99 \times 10^9} \times 2}$$ 7. **Calculate $q^2$:** $$q^2 = \frac{720}{1.798 \times 10^{10}} = 4.004 \times 10^{-8}$$ 8. **Find $q$ by taking the square root:** $$q = \sqrt{4.004 \times 10^{-8}} = 2.0 \times 10^{-4}$$ 9. **Final charges:** $$q = 2.0 \times 10^{-4}$$ $$2q = 4.0 \times 10^{-4}$$ Thus, the charges of the spheres are $2.0 \times 10^{-4}$ C and $4.0 \times 10^{-4}$ C respectively.