Let's break down the problem step by step: --- ### **1. Understanding the Problem** - **Triangle:** Equilateral, side = 2.12 cm = 0.0212 m - **Charges:** At each vertex: - Top: \( +4.00\, \mu\text{C} \) - Bottom left: \( q_A \) - Bottom right: \( q_B \) - The **4.00 μC** charge experiences a **net force of 666 N, vertically downward** due to \( q_A \) and \( q_B \). - Find \( q_A \) (in μC). --- ### **2. Drawing Forces** The forces on the top charge due to \( q_A \) and \( q_B \) are along the lines joining the top charge with each of the bottom charges (i.e., along the sides of the triangle). Let’s call the top charge \( Q = +4.00\, \mu\text{C} \). By symmetry, if \( q_A = q_B \), the horizontal components would cancel, and the vertical components would add up. But the question says the total net force is vertically downward, so **\( q_A \) and \( q_B \) must be equal and both negative** (repulsive force would push the top charge up, so attractive force needed, so both \( q_A \) and \( q_B \) are negative). --- ### **3. Force Calculation** The force between two point charges: \[ F = k \frac{|q_1 q_2|}{r^2} \] where \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \). Distance between charges, \( r = 2.12\, \text{cm} = 0.0212\, \text{m} \). #### **Force due to each charge** Let \( F_A \) be the force on \( Q \) due to \( q_A \): \[ F_A = k \frac{|Q q_A|}{r^2} \] Similarly, \( F_B = k \frac{|Q q_B|}{r^2} \). --- ### **4. Components** The angle at the top of the equilateral triangle is \( 60^\circ \). The vertical (downward) component of each force is: \[ F_{\text{vertical}} = F_A \sin(30^\circ) + F_B \sin(30^\circ) \] But since both forces are symmetrically placed (bottom corners), their vertical components **add**, and their horizontal components **cancel**. \[ F_{\text{net, vertical}} = 2 \left( k \frac{|Q q_A|}{r^2} \right) \sin(30^\circ) \] But since \( q_A = q_B \), let's denote \( q = q_A = q_B \). \[ F_{\text{net, vertical}} = 2 \left( k \frac{|Q q|}{r^2} \right) \sin(30^\circ) \] \[ \sin(30^\circ) = 0.5 \] \[ F_{\text{net, vertical}} = 2 \left( k \frac{|Q q|}{r^2} \right) \cdot 0.5 = k \frac{|Q q|}{r^2} \] --- ### **5. Plug in the Values** Given: \( F_{\text{net, vertical}} = 666\, \text{N} \), \( Q = 4.00 \times 10^{-6}\, \text{C} \), \( r = 0.0212\, \text{m} \), \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \) So, \[ 666 = k \frac{|Q q|}{r^2} \] \[ |q| = \frac{666 \cdot r^2}{k \cdot Q} \] --- ### **6. Calculate \( q_A \)** Plug in the numbers: \[ |q| = \frac{666 \times (0.0212)^2}{8.99 \times 10^9 \times 4.00 \times 10^{-6}} \] Calculate numerator: \[ 666 \times (0.0212)^2 = 666 \times 0.00044944 = 0.299 \text{ (rounded)} \] Denominator: \[ 8.99 \times 10^9 \times 4.00 \times 10^{-6} = 8.99 \times 4.00 \times 10^{9-6} = 35.96 \times 10^{3} = 35960 \] Now, \[ |q| = \frac{0.299}{35960} = 8.32 \times 10^{-6}\, \text{C} = 8.32\, \mu\text{C} \] --- ### **7. Sign of \( q_A \) and \( q_B \)** Since the net force is downward, \( q_A \) and \( q_B \) must be **negative** (to attract the positive \( Q \) upward, but since the question says the force is downward, the charges must repel, which means \( q_A \) and \( q_B \) are **positive**). But, for the force to be **downward**, the force from \( q_A \) and \( q_B \) must both be **repulsive** (since the 4 μC charge is positive), so **\( q_A \) and \( q_B \) are positive**. --- ### **8. Final Answer** \[ \boxed{q_A = 8.32\, \mu\text{C}} \] --- **Summary of Steps:** 1. Use symmetry and geometry to determine the force components. 2. Set up the equation for the net vertical force. 3. Solve for \( q_A \) using the given force magnitude. 4. Plug in the values and solve. --- **Final Answer:** \[ \boxed{q_A = 8.32\, \mu\text{C}} \]