Let's break down the problem step by step. ### **Given:** - Three charges \( q = 2.50 \,\mu C = 2.50 \times 10^{-6} \) C. - Each charge is at the end of a 1.00 m string. - Strings make 120° angles at the center. - Find the magnitude of the force on the knot at the center. --- ### **Step 1: Diagram and Forces** Draw the three charges at the end of the strings, each 1.00 m from the center, separated by 120°. The knot at the center will feel forces from all three charges. Since all charges are the same and placed symmetrically, let's analyze the force on the **center knot** due to the three charges. --- ### **Step 2: Force Due to One Charge** Using Coulomb's Law: \[ F = k_e \frac{|q_1 q_2|}{r^2} \] - \( k_e = 8.99 \times 10^9 \) N·m²/C² - \( q_1 = q_2 = 2.50 \times 10^{-6} \) C - \( r = 1.00 \) m \[ F = (8.99 \times 10^9) \frac{(2.50 \times 10^{-6})^2}{(1.00)^2} \] \[ F = (8.99 \times 10^9) \times (6.25 \times 10^{-12}) \] \[ F = 56.1875 \times 10^{-3} \] \[ F = 5.62 \times 10^{-2} \text{ N} \] --- ### **Step 3: Net Force Calculation** At the knot, the three force vectors are **symmetrically** arranged at 120° angles. If you sum up three vectors of equal magnitude separated by 120°, the resultant vector is **zero** due to symmetry. --- ### **Step 4: Final Answer** \[ \boxed{0} \] #### **Explanation:** The forces from the three charges, symmetrically spaced, will **cancel each other out** at the center. The net force on the knot at the center is **zero**. --- **Final Answer:** **a. 0**