1. **State the problem:** We have three charges on the x-axis: $q_1 = 3.0$ C at $x=+2.0$ m, $q_2 = -5.0$ C at $x=+4.0$ m, and $q_3 = 5.0$ C at $x=0$. We want to find the total electric force on $q_3$ due to $q_1$ and $q_2$, and the direction of this net force. 2. **Formula used:** The electric 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 the charges, and $r$ is the distance between them. 3. **Calculate force on $q_3$ due to $q_1$:** Distance $r_{13} = 2.0$ m. $$F_{13} = 8.99 \times 10^9 \times \frac{|3.0 \times 5.0|}{2.0^2} = 8.99 \times 10^9 \times \frac{15}{4} = 8.99 \times 10^9 \times 3.75 = 3.37125 \times 10^{10} \text{ N}$$ Since $q_1$ and $q_3$ are both positive, the force is repulsive, so $F_{13}$ points away from $q_1$ towards the left (negative x-direction). 4. **Calculate force on $q_3$ due to $q_2$:** Distance $r_{23} = 4.0$ m. $$F_{23} = 8.99 \times 10^9 \times \frac{|(-5.0) \times 5.0|}{4.0^2} = 8.99 \times 10^9 \times \frac{25}{16} = 8.99 \times 10^9 \times 1.5625 = 1.40406 \times 10^{10} \text{ N}$$ Since $q_2$ is negative and $q_3$ is positive, the force is attractive, so $F_{23}$ points towards $q_2$, which is to the right (positive x-direction). 5. **Determine net force direction and magnitude:** Net force $F_{net} = F_{23} - F_{13}$ because $F_{13}$ points left and $F_{23}$ points right. $$F_{net} = 1.40406 \times 10^{10} - 3.37125 \times 10^{10} = -1.96719 \times 10^{10} \text{ N}$$ The negative sign means the net force points to the left (negative x-direction). 6. **Final answer:** The total electric force on $q_3$ is approximately $$1.97 \times 10^{10} \text{ N}$$ directed to the left (negative x-axis).