1. **State the problem:** We have two point charges $q_1 = 3.0$ C at $x=+2.0$ m and $q_2 = -5.0$ C at $x=+4.0$ m. We want to find the total electric force on a charge $q_3 = 5.0$ C located at $x=0$ m. 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 distances:** - Distance between $q_3$ and $q_1$ is $r_1 = 2.0$ m. - Distance between $q_3$ and $q_2$ is $r_2 = 4.0$ m. 4. **Calculate forces magnitude:** $$F_1 = k \frac{|q_1 q_3|}{r_1^2} = 8.99 \times 10^9 \times \frac{3.0 \times 5.0}{2.0^2} = 8.99 \times 10^9 \times \frac{15}{4} = 8.99 \times 10^9 \times 3.75 = 3.37125 \times 10^{10} \text{ N}$$ $$F_2 = k \frac{|q_2 q_3|}{r_2^2} = 8.99 \times 10^9 \times \frac{5.0 \times 5.0}{4.0^2} = 8.99 \times 10^9 \times \frac{25}{16} = 8.99 \times 10^9 \times 1.5625 = 1.40406 \times 10^{10} \text{ N}$$ 5. **Determine force directions:** - $q_1$ and $q_3$ are both positive, so they repel. Since $q_1$ is at $+2.0$ m and $q_3$ at $0$, $F_1$ on $q_3$ points to the left (negative x-direction). - $q_2$ is negative and $q_3$ positive, so they attract. $q_2$ is at $+4.0$ m, so $F_2$ on $q_3$ points to the right (positive x-direction). 6. **Calculate net force:** $$F_{net} = F_2 - F_1 = 1.40406 \times 10^{10} - 3.37125 \times 10^{10} = -1.96719 \times 10^{10} \text{ N}$$ The negative sign means the net force is directed to the left (negative x-direction). **Final answer:** The total electric force on $q_3$ is approximately $$1.97 \times 10^{10}$$ N directed to the left (towards negative x-axis).