Let's break down the problem and **verify the answer** step by step: ### Given: - \( q_1 = -14.5\, \text{nC} = -14.5 \times 10^{-9}\, \text{C} \) at \( x_1 = -1.650\, \text{m} \) - \( q_2 = -33.0\, \text{nC} = -33.0 \times 10^{-9}\, \text{C} \) at \( x_2 = 0 \) - \( q_3 = 55.0\, \text{nC} = 55.0 \times 10^{-9}\, \text{C} \) at \( x_3 = -1.070\, \text{m} \) - \( \epsilon_0 = 8.854 \times 10^{-12}\, \frac{\text{C}^2}{\text{N} \cdot \text{m}^2} \) - \( K = \frac{1}{4\pi \epsilon_0} = 8.99 \times 10^9\, \text{N} \cdot \text{m}^2/\text{C}^2 \) --- ### **Step 1: Find distances** - Distance between \( q_1 \) and \( q_3 \): \[ d_{13} = |x_3 - x_1| = |-1.070 - (-1.650)| = |0.580| = 0.580\, \text{m} \] - Distance between \( q_2 \) and \( q_3 \): \[ d_{23} = |x_3 - x_2| = |-1.070 - 0| = 1.070\, \text{m} \] --- ### **Step 2: Calculate forces** #### **Force on \( q_3 \) due to \( q_1 \) (\( F_{13} \))** \[ F_{13} = K \frac{|q_1 q_3|}{d_{13}^2} \] \[ F_{13} = 8.99 \times 10^9 \cdot \frac{|(-14.5 \times 10^{-9}) (55.0 \times 10^{-9})|}{(0.580)^2} \] \[ |q_1 q_3| = 14.5 \times 55.0 \times 10^{-18} = 797.5 \times 10^{-18} = 7.975 \times 10^{-16} \] \[ F_{13} = 8.99 \times 10^9 \cdot \frac{7.975 \times 10^{-16}}{0.3364} \] \[ = 8.99 \times 10^9 \cdot 2.371 \times 10^{-15} \] \[ = 21.31 \times 10^{-6} = 2.13 \times 10^{-5}\, \text{N} \] **Direction:** Both are opposite signs, so force is **attractive**; \( q_3 \) is to the right of \( q_1 \), so the force is to the **left** (negative x-direction). --- #### **Force on \( q_3 \) due to \( q_2 \) (\( F_{23} \))** \[ F_{23} = K \frac{|q_2 q_3|}{d_{23}^2} \] \[ F_{23} = 8.99 \times 10^9 \cdot \frac{33.0 \times 55.0 \times 10^{-18}}{(1.070)^2} \] \[ = 8.99 \times 10^9 \cdot \frac{1815 \times 10^{-18}}{1.1449} \] \[ = 8.99 \times 10^9 \cdot 1.587 \times 10^{-15} \] \[ = 14.27 \times 10^{-6} = 1.43 \times 10^{-5}\, \text{N} \] **Direction:** Both are opposite signs, so force is **attractive**; \( q_3 \) is to the left of \( q_2 \), so the force is to the **right** (positive x-direction). --- ### **Step 3: Net Force on \( q_3 \) (x-direction)** - \( F_{13} \): negative (left) - \( F_{23} \): positive (right) \[ F_{\text{net},x} = -F_{13} + F_{23} \] \[ F_{\text{net},x} = (-2.13 \times 10^{-5}) + (1.43 \times 10^{-5}) = -0.70 \times 10^{-5} = -7.0 \times 10^{-6}\, \text{N} \] --- ### **Check Against Your Answer** Your answer: \(-3.156 \times 10^{-5}\, \text{N}\) #### **Possible Error** - Double-check all steps and significant digits. - Recalculate with more precision if needed. --- ### **Final Answer (to three significant digits):** \[ \boxed{-7.00 \times 10^{-6}\, \text{N}} \] Your answer seems **too large in magnitude** (about 4.5 times my result). Please double-check the calculations for any misplacement of decimals or errors in the use of the formula. If you want, you can show your arithmetic and I'll help you find the discrepancy!