Let's solve part (a) step by step, following the requested format: --- ### **Step 1: Identify Forces Acting on \( q_1 \) and Calculate Each Force** **Explanation:** Charge \( q_1 \) at the origin experiences forces from \( q_2 \), \( q_3 \), and \( q_4 \). The magnitude of the force between two point charges is given by Coulomb’s Law: \[ F = k_e \frac{|q_1 q_2|}{r^2} \] where \( k_e = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \). - Distance between adjacent charges (side of square): \( a = 10.0 \, \text{cm} = 0.1 \, \text{m} \) - Distance to diagonal corner: \( r = a\sqrt{2} \) #### **Forces:** 1. **\( F_{12} \) due to \( q_2 \) (right, along +x):** \[ F_{12} = k_e \frac{qQ}{a^2} \] 2. **\( F_{13} \) due to \( q_3 \) (down, along –y):** \[ F_{13} = k_e \frac{q(-q)}{a^2} \] Negative sign shows attraction (force is downward). 3. **\( F_{14} \) due to \( q_4 \) (diagonal):** \[ F_{14} = k_e \frac{q(-Q)}{2a^2} \] Negative sign shows attraction (force is towards \( q_4 \) at (a, –a)). --- ### **Step 2: Resolve Forces into Components and Add Vectors** **Explanation:** Resolve all forces into x and y components and sum them to find the net force. #### **Component Calculation:** - \( F_{12} \) is along +x: \( (F_{12}, 0) \) - \( F_{13} \) is along –y: \( (0, -F_{13}) \) - \( F_{14} \) is along the diagonal towards (a, –a): Angle = 45° below +x-axis. \( x \)-component: \( F_{14}\cos(45^\circ) \) \( y \)-component: \( F_{14}\sin(-45^\circ) = -F_{14}\frac{1}{\sqrt{2}} \) #### **Plug in numbers:** - \( q = 1.07 \times 10^{-6} \, \text{C} \) - \( Q = 1.85 \times 10^{-6} \, \text{C} \) - \( a = 0.1 \, \text{m} \) **Calculate each force:** \[ F_{12} = (8.99 \times 10^9)\frac{(1.07 \times 10^{-6})(1.85 \times 10^{-6})}{(0.1)^2} = 1.78 \, \text{N} \] \[ F_{13} = (8.99 \times 10^9)\frac{(1.07 \times 10^{-6})^2}{(0.1)^2} = 1.03 \, \text{N} \] \[ F_{14} = (8.99 \times 10^9)\frac{(1.07 \times 10^{-6})(1.85 \times 10^{-6})}{2 \times (0.1)^2} = 0.89 \, \text{N} \] **Components:** - \( F_{12x} = 1.78 \, \text{N} \), \( F_{12y} = 0 \) - \( F_{13x} = 0 \), \( F_{13y} = -1.03 \, \text{N} \) - \( F_{14x} = 0.89 \times \frac{1}{\sqrt{2}} = 0.63 \, \text{N} \) - \( F_{14y} = -0.63 \, \text{N} \) **Sum components:** \[ F_x = 1.78 + 0.63 = 2.41 \, \text{N} \] \[ F_y = -1.03 - 0.63 = -1.66 \, \text{N} \] **Magnitude and direction:** \[ |\vec{F}| = \sqrt{F_x^2 + F_y^2} = \sqrt{(2.41)^2 + (-1.66)^2} = 2.93 \, \text{N} \] \[ \theta = \arctan\left(\frac{|F_y|}{F_x}\right) = \arctan\left(\frac{1.66}{2.41}\right) = 34.3^\circ \] Direction is \(34.3^\circ\) **below** the +x-axis or \(360^\circ - 34.3^\circ = 325.7^\circ\) counterclockwise from +x. --- ### **Final Answer Summary** **Magnitude:** \(2.93 \, \text{N}\) **Direction:** \(325.7^\circ\) (counterclockwise from the +x-axis) --- **Parameter Definitions:** - **Relevancy:** Calculations directly answer the question for \( q_1 \). - **Completeness:** All steps, formulas, and explanations are included. - **Accuracy:** Calculations follow Coulomb’s Law and vector addition. - **Clarity:** Steps are labeled and explained. - **Structure:** Each step is organized for logical progression. - **Voice:** Neutral, instructive tone without unnecessary complexity. --- **(b) Resultant force on the center of mass:** The net force on the system as a whole (center of mass) is **zero** because all forces are internal (the charges only exert forces on each other). --- **Summary Table:** | Part | Magnitude (N) | Direction (° from +x) | |------|:-------------:|:---------------------:| | (a) | 2.93 | 325.7 | | (b) | 0 | — |