Let's first transcribe the problem for clarity: --- ### Problem Statement Three point charges lie along a straight line as shown in the figure. (Assume q₁ = +1.15 μC, q₂ = -3.15 μC, q₃ = +1.15 μC, and d = .195 m.) **Find the magnitude and direction of the net electrostatic force on the middle charge q₂.** --- ## Step 1: List Given Values - \( q_1 = +1.15 \ \mu C = +1.15 \times 10^{-6} \ C \) - \( q_2 = -3.15 \ \mu C = -3.15 \times 10^{-6} \ C \) - \( q_3 = +1.15 \ \mu C = +1.15 \times 10^{-6} \ C \) - \( d = .195 \ \text{m} \) ## Step 2: Visualize the Setup Charges are arranged in a straight line: ``` q1 q2 q3 |-------|-------| d d ``` ## Step 3: Calculate Force on \( q_2 \) Due to \( q_1 \) Coulomb's Law: \[ F = k \frac{|q_1 q_2|}{r^2} \] Where \( k = 8.99 \times 10^9 \ \text{N} \cdot \text{m}^2/\text{C}^2 \). \[ F_{21} = k \frac{|q_2 q_1|}{d^2} \] \[ F_{21} = 8.99 \times 10^9 \frac{|-3.15 \times 10^{-6} \times 1.15 \times 10^{-6}|}{(.195)^2} \] \[ F_{21} = 8.99 \times 10^9 \frac{3.6225 \times 10^{-12}}{.038025} \] \[ F_{21} = 8.99 \times 10^9 \times 9.535 \times 10^{-11} \] \[ F_{21} \approx .856 \ \text{N} \] **Direction:** \( q_1 \) is positive, \( q_2 \) is negative, so the force is attractive: **to the left** (toward \( q_1 \)). --- ## Step 4: Calculate Force on \( q_2 \) Due to \( q_3 \) Distance between \( q_2 \) and \( q_3 \) is also \( d \). \[ F_{23} = k \frac{|q_2 q_3|}{d^2} \] \[ F_{23} = 8.99 \times 10^9 \frac{|-3.15 \times 10^{-6} \times 1.15 \times 10^{-6}|}{(.195)^2} \] \[ F_{23} = .856 \ \text{N} \] **Direction:** \( q_3 \) is positive, \( q_2 \) is negative, so the force is attractive: **to the right** (toward \( q_3 \)). --- ## Step 5: Net Force on \( q_2 \) Both forces have the same magnitude but opposite directions: - \( F_{21} \) is **to the left** - \( F_{23} \) is **to the right** \[ F_{\text{net}} = F_{23} - F_{21} = .856 \ \text{N} - .856 \ \text{N} = \] --- ## Step 6: Final Answer ### **Magnitude and direction of the net force on \( q_2 \):** \[ \boxed{ \ \text{N}, \ \text{no direction (forces cancel out)}} \] --- **Explanation:** The two positive charges are equidistant from the negative middle charge, and both exert equal and opposite attractive forces. The net force is therefore zero.