1. **Problem statement:** Calculate the net force (magnitude and direction) on charge $q_3$ due to charges $q_1$ and $q_2$. 2. **Given data:** - $q_1 = 5$ nC, $q_2 = -5$ nC, $q_3 = 5$ nC - Distance between $q_1$ and $q_2$ is 2 cm = 0.02 m - Distance between $q_2$ and $q_3$ is 2 cm = 0.02 m - Distance between $q_1$ and $q_3$ is 4 cm = 0.04 m 3. **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 charges, and $r$ is the distance between them. 4. **Calculate forces on $q_3$:** - Force by $q_1$ on $q_3$ ($F_{13}$): $$F_{13} = k_e \frac{|q_1 q_3|}{(0.04)^2} = 8.99 \times 10^9 \times \frac{(5 \times 10^{-9})(5 \times 10^{-9})}{0.0016}$$ $$= 8.99 \times 10^9 \times \frac{25 \times 10^{-18}}{0.0016} = 8.99 \times 10^9 \times 1.5625 \times 10^{-14} = 1.404 \times 10^{-4} \text{ N}$$ Direction: Both $q_1$ and $q_3$ are positive, so force is repulsive. Since $q_1$ is to the left of $q_3$, $F_{13}$ points to the right (+x direction). - Force by $q_2$ on $q_3$ ($F_{23}$): $$F_{23} = k_e \frac{|q_2 q_3|}{(0.02)^2} = 8.99 \times 10^9 \times \frac{(5 \times 10^{-9})(5 \times 10^{-9})}{0.0004}$$ $$= 8.99 \times 10^9 \times \frac{25 \times 10^{-18}}{0.0004} = 8.99 \times 10^9 \times 6.25 \times 10^{-14} = 5.619 \times 10^{-4} \text{ N}$$ Direction: $q_2$ is negative and $q_3$ positive, so force is attractive. $q_2$ is to the left of $q_3$, so $F_{23}$ points to the left (-x direction). 5. **Calculate net force on $q_3$:** $$F_{net} = F_{13} - F_{23} = 1.404 \times 10^{-4} - 5.619 \times 10^{-4} = -4.215 \times 10^{-4} \text{ N}$$ The negative sign means the net force points to the left (-x direction). 6. **Final answer:** The net force on $q_3$ has magnitude $$4.215 \times 10^{-4}$$ N and points to the left along the x-axis.