1. **Problem statement:** We have three charges along the x-axis: $q_1 = 15.0$ nC at $x=2.00$ m, $q_2 = 6.0$ nC at $x=0$, and $q_3$ with unknown position $x$ such that the net force on $q_3$ is zero. 2. **Formula used:** The electrostatic force between two point charges is given by Coulomb's law: $$F = k \frac{|q_a q_b|}{r^2}$$ where $k = 8.99 \times 10^9$ N m$^2$/C$^2$, $q_a$ and $q_b$ are charges, and $r$ is the distance between them. 3. **Important rules:** Forces are vectors; since $q_3$ is negative and $q_1, q_2$ are positive, forces on $q_3$ due to $q_1$ and $q_2$ are attractive, pointing towards the positive charges. 4. **Set up the force balance:** Let $x$ be the position of $q_3$ along the x-axis, with $0 < x < 2.00$ m (between $q_2$ and $q_1$). - Distance between $q_3$ and $q_2$ is $x$. - Distance between $q_3$ and $q_1$ is $2.00 - x$. 5. **Write forces on $q_3$:** $$F_{23} = k \frac{|q_2 q_3|}{x^2}$$ $$F_{13} = k \frac{|q_1 q_3|}{(2.00 - x)^2}$$ 6. **Since net force on $q_3$ is zero, magnitudes of forces must be equal:** $$F_{23} = F_{13}$$ 7. **Substitute values and simplify:** $$k \frac{6.0 \times 10^{-9} |q_3|}{x^2} = k \frac{15.0 \times 10^{-9} |q_3|}{(2.00 - x)^2}$$ 8. **Cancel common factors $k$ and $|q_3|$:** $$\frac{6.0 \times 10^{-9}}{x^2} = \frac{15.0 \times 10^{-9}}{(2.00 - x)^2}$$ 9. **Cancel $10^{-9}$:** $$\frac{6.0}{x^2} = \frac{15.0}{(2.00 - x)^2}$$ 10. **Cross multiply:** $$6.0 (2.00 - x)^2 = 15.0 x^2$$ 11. **Divide both sides by 3 to simplify:** $$\cancel{6.0} (2.00 - x)^2 = \cancel{15.0} x^2 \Rightarrow 2 (2.00 - x)^2 = 5 x^2$$ 12. **Expand $(2.00 - x)^2$:** $$2 (4.00 - 4.00 x + x^2) = 5 x^2$$ 13. **Distribute 2:** $$8.00 - 8.00 x + 2 x^2 = 5 x^2$$ 14. **Bring all terms to one side:** $$8.00 - 8.00 x + 2 x^2 - 5 x^2 = 0$$ 15. **Simplify:** $$8.00 - 8.00 x - 3 x^2 = 0$$ 16. **Rewrite as standard quadratic:** $$-3 x^2 - 8.00 x + 8.00 = 0$$ 17. **Multiply both sides by $-1$ to simplify:** $$3 x^2 + 8.00 x - 8.00 = 0$$ 18. **Use quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a}$$ where $a=3$, $b=8.00$, $c=-8.00$. 19. **Calculate discriminant:** $$\Delta = 8.00^2 - 4 \times 3 \times (-8.00) = 64 + 96 = 160$$ 20. **Calculate roots:** $$x = \frac{-8.00 \pm \sqrt{160}}{6} = \frac{-8.00 \pm 12.65}{6}$$ 21. **Two solutions:** - $$x_1 = \frac{-8.00 + 12.65}{6} = \frac{4.65}{6} = 0.775$$ m - $$x_2 = \frac{-8.00 - 12.65}{6} = \frac{-20.65}{6} = -3.44$$ m (not between charges, discard) 22. **Final answer:** The $x$ coordinate of $q_3$ is approximately $$\boxed{0.78 \text{ m}}$$ This position balances the forces so the net force on $q_3$ is zero.