1. **State the problem:** We have three point charges: $q_1 = -1.50 \text{ nC}$ at $(0,6.00)$ m, $q_2 = +3.20 \text{ nC}$ at the origin $(0,0)$ m, and $q_3 = -5.00 \text{ nC}$ at $(2.00,-4.00)$ m. We want to find the total force (magnitude and direction) exerted on $q_3$ by $q_1$ and $q_2$. 2. **Formula used:** The force between two point charges is given by Coulomb's law: $$F = k_e \frac{|q_a q_b|}{r^2}$$ where $k_e = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2$, $q_a$ and $q_b$ are the charges, and $r$ is the distance between them. 3. **Calculate distances:** - Distance $r_{13}$ between $q_1$ at $(0,6)$ and $q_3$ at $(2,-4)$: $$r_{13} = \sqrt{(2-0)^2 + (-4-6)^2} = \sqrt{4 + 100} = \sqrt{104} = 10.198 \text{ m}$$ - Distance $r_{23}$ between $q_2$ at $(0,0)$ and $q_3$ at $(2,-4)$: $$r_{23} = \sqrt{(2-0)^2 + (-4-0)^2} = \sqrt{4 + 16} = \sqrt{20} = 4.472 \text{ m}$$ 4. **Calculate forces magnitudes:** - Force magnitude $F_{13}$: $$F_{13} = k_e \frac{|q_1 q_3|}{r_{13}^2} = 8.99 \times 10^9 \times \frac{1.50 \times 10^{-9} \times 5.00 \times 10^{-9}}{(10.198)^2} = 8.99 \times 10^9 \times \frac{7.5 \times 10^{-18}}{104} = 6.48 \times 10^{-10} \text{ N}$$ - Force magnitude $F_{23}$: $$F_{23} = k_e \frac{|q_2 q_3|}{r_{23}^2} = 8.99 \times 10^9 \times \frac{3.20 \times 10^{-9} \times 5.00 \times 10^{-9}}{(4.472)^2} = 8.99 \times 10^9 \times \frac{16 \times 10^{-18}}{20} = 7.19 \times 10^{-9} \text{ N}$$ 5. **Determine force directions:** - Vector from $q_1$ to $q_3$ is $(2-0, -4-6) = (2, -10)$. - Unit vector $\hat{r}_{13} = \frac{1}{10.198}(2, -10) = (0.196, -0.981)$. - Since $q_1$ and $q_3$ are both negative, force is repulsive, so $\vec{F}_{13}$ points away from $q_1$ along $\hat{r}_{13}$. - Vector from $q_2$ to $q_3$ is $(2-0, -4-0) = (2, -4)$. - Unit vector $\hat{r}_{23} = \frac{1}{4.472}(2, -4) = (0.447, -0.894)$. - $q_2$ is positive and $q_3$ negative, force is attractive, so $\vec{F}_{23}$ points toward $q_2$, opposite to $\hat{r}_{23}$, so direction is $(-0.447, 0.894)$. 6. **Calculate force vectors:** $$\vec{F}_{13} = F_{13} \times \hat{r}_{13} = 6.48 \times 10^{-10} (0.196, -0.981) = (1.27 \times 10^{-10}, -6.36 \times 10^{-10}) \text{ N}$$ $$\vec{F}_{23} = F_{23} \times (-\hat{r}_{23}) = 7.19 \times 10^{-9} (-0.447, 0.894) = (-3.21 \times 10^{-9}, 6.43 \times 10^{-9}) \text{ N}$$ 7. **Sum forces:** $$\vec{F}_{total} = \vec{F}_{13} + \vec{F}_{23} = (1.27 \times 10^{-10} - 3.21 \times 10^{-9}, -6.36 \times 10^{-10} + 6.43 \times 10^{-9}) = (-3.08 \times 10^{-9}, 5.79 \times 10^{-9}) \text{ N}$$ 8. **Calculate magnitude of total force:** $$F_{total} = \sqrt{(-3.08 \times 10^{-9})^2 + (5.79 \times 10^{-9})^2} = \sqrt{9.49 \times 10^{-18} + 3.35 \times 10^{-17}} = \sqrt{4.30 \times 10^{-17}} = 6.56 \times 10^{-9} \text{ N}$$ 9. **Calculate direction (angle) relative to positive x-axis:** $$\theta = \tan^{-1} \left( \frac{5.79 \times 10^{-9}}{-3.08 \times 10^{-9}} \right) = \tan^{-1}(-1.88) = -62.7^\circ$$ Since $x$ component is negative and $y$ positive, vector is in second quadrant, so add $180^\circ$: $$\theta = 180^\circ - 62.7^\circ = 117.3^\circ$$ **Final answer:** The total force on $q_3$ has magnitude approximately $6.56 \times 10^{-9}$ N and direction $117.3^\circ$ counterclockwise from the positive x-axis.