1. **Problem statement:** Three charges lie along the x-axis: $q_1 = 15.0$ nC at $x=2.00$ m, $q_2 = 6.0$ nC at $x=0$, and $q_3$ at an unknown position $x$. The resultant force on $q_3$ is zero. Find the $x$ coordinate of $q_3$. 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 rule:** Since $q_3$ experiences zero net force, the forces from $q_1$ and $q_2$ on $q_3$ must be equal in magnitude and opposite in direction. 4. **Set up the equation:** Let $x$ be the position of $q_3$. The distances are: - Distance between $q_3$ and $q_2$ is $|x - 0| = |x|$ - Distance between $q_3$ and $q_1$ is $|x - 2|$ The forces on $q_3$ due to $q_1$ and $q_2$ are: $$F_1 = k \frac{15.0 \times 10^{-9} \times q_3}{(x - 2)^2}$$ $$F_2 = k \frac{6.0 \times 10^{-9} \times q_3}{x^2}$$ Since $q_3$ is positive, forces repel, so for net force zero: $$F_1 = F_2$$ 5. **Equate and simplify:** $$k \frac{15.0 \times 10^{-9} q_3}{(x - 2)^2} = k \frac{6.0 \times 10^{-9} q_3}{x^2}$$ Cancel $k$ and $q_3$ (nonzero): $$\frac{15.0 \times 10^{-9}}{(x - 2)^2} = \frac{6.0 \times 10^{-9}}{x^2}$$ 6. **Simplify constants:** $$\frac{15.0}{(x - 2)^2} = \frac{6.0}{x^2}$$ 7. **Cross multiply:** $$15.0 x^2 = 6.0 (x - 2)^2$$ 8. **Expand right side:** $$(x - 2)^2 = x^2 - 4x + 4$$ So: $$15.0 x^2 = 6.0 (x^2 - 4x + 4)$$ 9. **Distribute 6.0:** $$15.0 x^2 = 6.0 x^2 - 24 x + 24$$ 10. **Bring all terms to one side:** $$15.0 x^2 - 6.0 x^2 + 24 x - 24 = 0$$ $$9.0 x^2 + 24 x - 24 = 0$$ 11. **Divide entire equation by 3 to simplify:** $$\cancel{3} \times (3.0 x^2 + 8 x - 8) = 0$$ $$3.0 x^2 + 8 x - 8 = 0$$ 12. **Use quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=3.0$, $b=8$, $c=-8$. Calculate discriminant: $$b^2 - 4ac = 8^2 - 4 \times 3.0 \times (-8) = 64 + 96 = 160$$ 13. **Calculate roots:** $$x = \frac{-8 \pm \sqrt{160}}{2 \times 3.0} = \frac{-8 \pm 4\sqrt{10}}{6}$$ 14. **Approximate values:** $$\sqrt{10} \approx 3.162$$ So: $$x_1 = \frac{-8 + 4 \times 3.162}{6} = \frac{-8 + 12.648}{6} = \frac{4.648}{6} \approx 0.775$$ $$x_2 = \frac{-8 - 12.648}{6} = \frac{-20.648}{6} \approx -3.441$$ 15. **Check physical location:** $q_3$ must be on the x-axis where forces balance. Since $q_1$ is at 2 m and $q_2$ at 0, $q_3$ cannot be between 0 and 2 because forces would not cancel. The solution $x \approx 0.775$ m lies between 0 and 2, so discard. The valid solution is: $$\boxed{x \approx -3.44 \text{ m}}$$ This means $q_3$ is located at approximately $-3.44$ meters on the x-axis for the net force to be zero.