1. **Problem statement:** Three point charges are arranged on a line: $q_3 = +5.00$ nC at the origin ($x=0$), $q_2 = -3.00$ nC at $x=+4.00$ cm, and $q_1$ at $x=+2.00$ cm. Find the magnitude and sign of $q_1$ 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: $$F = k_e \frac{|q_a q_b|}{r^2}$$ where $k_e = 8.99 \times 10^9$ N m$^2$/C$^2$, $q_a$ and $q_b$ are the charges, and $r$ is the distance between them. 3. **Important rules:** Forces are vectors; direction matters. Forces on $q_3$ from $q_1$ and $q_2$ must sum to zero. 4. **Calculate distances:** - Distance between $q_3$ and $q_1$ is $r_{31} = 2.00$ cm = 0.02 m. - Distance between $q_3$ and $q_2$ is $r_{32} = 4.00$ cm = 0.04 m. 5. **Calculate magnitudes of forces on $q_3$:** $$F_{31} = k_e \frac{|q_3 q_1|}{r_{31}^2} = 8.99 \times 10^9 \times \frac{5.00 \times 10^{-9} \times |q_1|}{(0.02)^2}$$ $$F_{32} = k_e \frac{|q_3 q_2|}{r_{32}^2} = 8.99 \times 10^9 \times \frac{5.00 \times 10^{-9} \times 3.00 \times 10^{-9}}{(0.04)^2}$$ 6. **Calculate $F_{32}$ numerically:** $$F_{32} = 8.99 \times 10^9 \times \frac{15 \times 10^{-18}}{0.0016} = 8.99 \times 10^9 \times 9.375 \times 10^{-15} = 8.43 \times 10^{-5} \text{ N}$$ 7. **Direction of forces:** - $q_2$ is negative, $q_3$ positive, so force on $q_3$ due to $q_2$ is attractive, towards $q_2$ (positive x direction). - $q_1$ is unknown; force direction depends on sign of $q_1$. 8. **Set net force on $q_3$ to zero:** $$F_{31} + F_{32} = 0$$ Since $F_{32}$ is towards positive x, $F_{31}$ must be towards negative x, so $q_1$ must be negative. 9. **Write equation with signs:** $$-F_{31} + F_{32} = 0 \implies F_{31} = F_{32}$$ 10. **Express $F_{31}$ numerically:** $$F_{31} = 8.99 \times 10^9 \times \frac{5.00 \times 10^{-9} \times |q_1|}{(0.02)^2} = 8.99 \times 10^9 \times \frac{5.00 \times 10^{-9} \times |q_1|}{0.0004}$$ 11. **Simplify:** $$F_{31} = 8.99 \times 10^9 \times 1.25 \times 10^{-5} \times |q_1| = 1.12375 \times 10^5 \times |q_1|$$ 12. **Set equal to $F_{32}$:** $$1.12375 \times 10^5 \times |q_1| = 8.43 \times 10^{-5}$$ 13. **Solve for $|q_1|$:** $$|q_1| = \frac{8.43 \times 10^{-5}}{1.12375 \times 10^5} = 7.5 \times 10^{-10} \text{ C} = 0.75 \text{ nC}$$ 14. **Determine sign:** $q_1$ must be negative to produce force opposite to $F_{32}$. **Final answer:** $$q_1 = -0.75 \text{ nC}$$