1. **State the problem:** Calculate the net electrostatic force on charge $q_1 = 2\,\mu C$ located at a vertex of an equilateral triangle with side length $0.1$ m, due to charges $q_2 = -3\,\mu C$ and $q_3 = 4\,\mu C$ at the other vertices. 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 $k_e = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2$, $q_a$ and $q_b$ are the charges, and $r$ is the distance between them. 3. **Important rules:** - Forces are vectors; direction matters. - The force on $q_1$ due to $q_2$ and $q_3$ must be added vectorially. - Since the triangle is equilateral, all sides are $0.1$ m. 4. **Calculate magnitudes of forces:** - Force on $q_1$ due to $q_2$: $$F_{12} = k_e \frac{|q_1 q_2|}{r^2} = 8.99 \times 10^9 \times \frac{2 \times 10^{-6} \times 3 \times 10^{-6}}{(0.1)^2} = 5.394 \text{ N}$$ - Force on $q_1$ due to $q_3$: $$F_{13} = k_e \frac{|q_1 q_3|}{r^2} = 8.99 \times 10^9 \times \frac{2 \times 10^{-6} \times 4 \times 10^{-6}}{(0.1)^2} = 7.192 \text{ N}$$ 5. **Determine directions:** - $q_2$ is negative, $q_1$ positive, so force $F_{12}$ is attractive, directed from $q_1$ towards $q_2$. - $q_3$ is positive, same sign as $q_1$, so force $F_{13}$ is repulsive, directed from $q_1$ away from $q_3$. 6. **Set coordinate system:** Place $q_1$ at origin $(0,0)$. Let $q_2$ be at $(0.1,0)$ along positive x-axis. Then $q_3$ is at $(0.05, 0.0866)$ (height of equilateral triangle $= 0.1 \times \sqrt{3}/2 = 0.0866$ m). 7. **Express forces as vectors:** - $\vec{F}_{12}$ points from $q_1$ to $q_2$ along positive x-axis: $$\vec{F}_{12} = (5.394, 0) \text{ N}$$ - $\vec{F}_{13}$ points away from $q_3$ towards $q_1$, so from $q_1$ to $q_3$ is vector $(0.05, 0.0866)$, unit vector: $$\hat{r}_{13} = \frac{(0.05, 0.0866)}{0.1} = (0.5, 0.866)$$ Force is repulsive, so direction is opposite: $$\vec{F}_{13} = -7.192 \times (0.5, 0.866) = (-3.596, -6.227) \text{ N}$$ 8. **Calculate net force:** $$\vec{F}_{net} = \vec{F}_{12} + \vec{F}_{13} = (5.394 - 3.596, 0 - 6.227) = (1.798, -6.227) \text{ N}$$ 9. **Calculate magnitude of net force:** $$F_{net} = \sqrt{1.798^2 + (-6.227)^2} = \sqrt{3.233 + 38.78} = \sqrt{42.013} = 6.48 \text{ N}$$ 10. **Calculate direction angle relative to positive x-axis:** $$\theta = \tan^{-1} \left( \frac{-6.227}{1.798} \right) = -74.5^\circ$$ This means $74.5^\circ$ below the positive x-axis. **Final answer:** The net force on $q_1$ has magnitude $6.48$ N and points $74.5^\circ$ below the positive x-axis.