Let's solve the problem step-by-step: --- ## **Given:** - Each side of the equilateral triangle \( l = 2.12\, \text{cm} = 0.0212\, \text{m} \) - Top charge: \( q = +4.00\, \mu\text{C} = 4.00 \times 10^{-6}\, \text{C} \) - Net force on \( 4.00\, \mu\text{C} \) charge: \( F_{\text{net}} = 666\, \text{N} \) (vertically downward) - We need to find \( q_A \) in \( \mu\text{C} \). --- ## **Step 1: Write the Forces** Let the charges at the bottom corners be \( q_A \) (left) and \( q_B \) (right). - **Force by \( q_A \) on \( 4.00\, \mu\text{C} \):** \[ F_A = k \frac{|q_A \cdot 4.00 \times 10^{-6}|}{l^2} \] - **Force by \( q_B \) on \( 4.00\, \mu\text{C} \):** \[ F_B = k \frac{|q_B \cdot 4.00 \times 10^{-6}|}{l^2} \] - Both forces act along the lines joining the top charge to \( q_A \) and \( q_B \), making \( 60^\circ \) angles with the horizontal. --- ## **Step 2: Net Force Direction** Since the net force is **vertically downward**, the horizontal components cancel each other, and the vertical components add up. - The vertical component of each force: \[ F_{A_y} = F_A \sin(60^\circ) \] \[ F_{B_y} = F_B \sin(60^\circ) \] - For the net force to be downward, **both \( q_A \) and \( q_B \) must be negative** (opposite to the positive charge at the top). - The horizontal components cancel: \[ F_{A_x} = -F_A \cos(60^\circ) \] \[ F_{B_x} = F_B \cos(60^\circ) \] \[ F_{A_x} + F_{B_x} = 0 \implies F_A = F_B \] So, \( q_A = q_B \). --- ## **Step 3: Net Vertical Force** \[ F_{\text{net}} = F_{A_y} + F_{B_y} = 2 F_A \sin(60^\circ) \] \[ 666 = 2 F_A \left(\frac{\sqrt{3}}{2}\right) \] \[ 666 = F_A \sqrt{3} \] \[ F_A = \frac{666}{\sqrt{3}} = 384.5\, \text{N} \] --- ## **Step 4: Plug into Coulomb's Law** \[ F_A = k \frac{|q_A \cdot 4.00 \times 10^{-6}|}{(0.0212)^2} \] \[ 384.5 = \frac{8.99 \times 10^9 \cdot |q_A| \cdot 4.00 \times 10^{-6}}{(0.0212)^2} \] Solve for \( |q_A| \): \[ |q_A| = \frac{384.5 \cdot (0.0212)^2}{8.99 \times 10^9 \cdot 4.00 \times 10^{-6}} \] Calculate numerator: \[ 384.5 \times (0.0212)^2 = 384.5 \times 0.00044944 = 0.1727 \] Calculate denominator: \[ 8.99 \times 10^9 \times 4.00 \times 10^{-6} = 8.99 \times 4.00 \times 10^{3} = 35.96 \times 10^3 = 35960 \] So, \[ |q_A| = \frac{0.1727}{35960} = 4.8 \times 10^{-6}\, \text{C} = 4.8\, \mu\text{C} \] --- ## **Step 5: Final Answer** Since \( q_A \) must be **negative** (to provide downward force): \[ \boxed{-4.8\, \mu\text{C}} \] --- ### **Summary** - **Charge \( q_A = q_B = -4.8\, \mu\text{C} \)** - **Direction: Negative (downward force on the positive charge at the top).**