This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Welcome back Mukamwiza — missed you this week. To find the net force on the 2.0 × 10^-6 C charge, we will use Coulomb's Law and the principle of superposition. Let the charges be: q_1 = 8.0 × 10^-6 C (left charge) q_2 = 2.0 × 10^-6 C (middle charge) q_3 = 5.0 × 10^-6 C (right charge) The distances are: r_12 = 2.0 m (distance between q_1 and q_2) r_23 = 2.0 m (distance between q_2 and q_3) Coulomb's constant k = 9 × 10^9 \, N m^2/C^2. Since all charges are positive, the forces are repulsive. • The force F_12 (on q_2 due to q_1) pushes q_2 to the right. • The force F_32 (on q_2 due to q_3) pushes q_2 to the left. Step 1: Calculate the magnitude of the force F_12 exerted by q_1 on q_2. F_12 = (k |q_1 q_2|)/(r_12)^2 F_12 = (9 × 10^9 \, N m^2/C^2) (8.0 × 10^-6 \, C) (2.0 × 10^-6 \, C)(2.0 \, m)^2 F_12 = 9 × 8.0 × 2.0 × 10^9-6-64.0 \, N F_12 = 144 × 10^-34.0 \, N F_12 = 36 × 10^-3 \, N Step 2: Calculate the magnitude of the force F_32 exerted by q_3 on q_2. F_32 = (k |q_3 q_2|)/(r_23)^2 F_32 = (9 × 10^9 \, N m^2/C^2) (5.0 × 10^-6 \, C) (2.0 × 10^-6 \, C)(2.0 \, m)^2 F_32 = 9 × 5.0 × 2.0 × 10^9-6-64.0 \, N F_32 = 90 × 10^-34.0 \, N F_32 = 22.5 ×