Let's break down the problem step by step: --- ## **Given Data** - Side of the triangle: \( a = .48\,\text{m} \) - Charges: - \( S_1: +5.\,\\text{C} \) - \( S_: -5.\,\mu\text{C} \) - \( S_3: +5.\,\mu\text{C} \) - Coulomb's Constant: \( k = 8.99 \times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2 \) --- ## **Part (a): Net Force on \( S_3 \) (at its triangle vertex)** ### **Step 1: Find Positions** - \( S_1: (, a/2) \) - \( S_2: (, -a/2) \) - \( S_3: (a, ) \) ### **Step 2: Calculate Forces on \( S_3 \)** #### **Force from \( S_1 \) on \( S_3 \) (\( \vec{F}_{13} \))** - Distance: \[ r_{13} = \sqrt{(a-)^2 + ( - a/2)^2} = \sqrt{a^2 + (a^2/4)} = \sqrt{5a^2/4} = \frac{a\sqrt{5}}{2} \] - Magnitude: \[ F_{13} = k \frac{|q_1 q_3|}{r_{13}^2} = k \frac{(5. \times 10^{-6})^2}{(a\sqrt{5}/2)^2} \] \[ = k \frac{25 \times 10^{-12}}{(5a^2/4)} = \frac{25k \times 10^{-12} \times 4}{5a^2} = \frac{20k \times 10^{-12}}{a^2} \] - Direction: From \( S_1 \) to \( S_3 \) (repulsive, both positive), along vector \( (a, -a/2) \). #### **Force from \( S_2 \) on \( S_3 \) (\( \vec{F}_{23} \))** - Distance is the same as above. - Magnitude: \[ F_{23} = k \frac{|q_2 q_3|}{r_{23}^2} = F_{13} \] - Direction: From \( S_3 \) to \( S_2 \) (attractive, opposite charges), along vector \( (-a, -a/2-) = (-a, -a/2) \). ### **Step 3: Write the Force Vectors** Let’s normalize direction vectors: - From \( S_1 \) to \( S_3 \): \( \vec{r}_{13} = (a, -a/2) \) - Unit vector: \( \hat{r}_{13} = \frac{(2, -1)}{\sqrt{5}} \) - From \( S_3 \) to \( S_2 \): \( \vec{r}_{32} = (-a, -a/2) \) - Unit vector: \( \hat{r}_{32} = \frac{(-2, -1)}{\sqrt{5}} \) So: \[ \vec{F}_{13} = F_{13} \hat{r}_{13} = F_{13} \left( \frac{2}{\sqrt{5}}, -\frac{1}{\sqrt{5}} \right) \] \[ \vec{F}_{23} = F_{23} \hat{r}_{32} = F_{23} \left( -\frac{2}{\sqrt{5}}, -\frac{1}{\sqrt{5}} \right) \] ### **Step 4: Add Forces** \[ \vec{F}_{\text{net}} = \vec{F}_{13} + \vec{F}_{23} \] \[ = F_{13} \left( \frac{2}{\sqrt{5}}, -\frac{1}{\sqrt{5}} \right) + F_{23} \left( -\frac{2}{\sqrt{5}}, -\frac{1}{\sqrt{5}} \right) \] \[ = F_{13} \left( \frac{2}{\sqrt{5}} - \frac{2}{\sqrt{5}}, -\frac{1}{\sqrt{5}} - \frac{1}{\sqrt{5}} \right) \] \[ = F_{13} \left( , -\frac{2}{\sqrt{5}} \right) \] So total force is **straight down** (\(-y\) direction). ### **Step 5: Plug in Values** \[ F_{13} = \frac{20k \times 10^{-12}}{a^2} \] \[ a = .48\,\text{m} \] \[ F_{13} = \frac{20 \times 8.99 \times 10^9 \times 10^{-12}}{(.48)^2} \] \[ = \frac{179.8 \times 10^{-3}}{.2304} = \frac{.1798}{.2304} \approx .780\,\text{N} \] Now, the total force: \[ F_{\text{net}} = F_{13} \cdot \frac{2}{\sqrt{5}} = .780 \cdot \frac{2}{2.236} = .780 \cdot .894 \approx .698\,\text{N} \] But as calculated above, the net force is in the \(-y\) direction, magnitude is: \[ F_{\text{net}} = 2 \cdot F_{13} \cdot \frac{1}{\sqrt{5}} = 2 \cdot .780 \cdot .447 = .697\,\text{N} \] **Direction:** Directly down (\(270^\circ\) counterclockwise from \(+x\)-axis). --- ## **Final Answers** ### **(a) Net Force on \( S_3 \):** - **Magnitude:** \( .697\,\text{N} \) - **Direction:** \( 270^\circ \) counterclockwise from the \(+x\)-axis --- ### **(b) If \( S_3 \) Moves to the Origin** At the origin, the distances to \( S_1 \) and \( S_2 \) are the same but only along the \( y \)-axis (closer than previously). The net force will be **greater than** in part (a), because the force magnitude increases as distance decreases. **Answer:** - **Greater than in part (a)** --- ### **(c) Net Force on \( S_3 \) at the Origin** At the origin, the forces from \( S_1 \) and \( S_2 \) are along the \( y \)-axis, at a distance \( a/2 \): \[ r = a/2 = .24\,\text{m} \] \[ F_{13} = k \frac{(5. \times 10^{-6})^2}{(.24)^2} = 8.99 \times 10^9 \times \frac{25 \times 10^{-12}}{.0576} = 8.99 \times 10^9 \times 4.34 \times 10^{-10} \approx 3.90\,\text{N} \] - Both forces are along the \( y \)-axis, but in opposite directions (one up, one down). - The charges: \( S_1 \) is above (repulsive, so down), \( S_2 \) is below (attractive, so up). - Both have the same magnitude but opposite direction, so they add up. \[ F_{\text{net}} = F_{13} + F_{23} = 3.90\,\text{N} + 3.90\,\text{N} = 7.80\,\text{N} \] - **Direction:** Down (\(270^\circ\) counterclockwise from \(+x\)-axis). --- ## **Summary Table** | Part | Magnitude (N) | Direction (\(^\circ\)) | |------|---------------|------------------------| | a | .697 | 270 | | b | Greater than (a) | — | | c | 7.80 | 270 | --- **Let me know if you need the vector diagrams or further clarifications!**