Question Detail

Given:q1 = 1mC = 1 x 10-3 Cq2 = 5mC = 5 x 10-3 Cq3 = 9mC = 9 x 10-3 CCoulomb's constant k = 9 x 109 N m2/C2The charges are on a line with q2 in the middle. Let's assume q1 is to the left, q2 in the middle, and q3 to the right.Distance r12 (between q1 and q2) = 7cm = 0.07mDistance r13 (between q1 and q3) = 16cm = 0.16mDistance r23 (between q2 and q3) = r13 - r12 = 0.16m - 0.07m = 0.09m1. Force on q2 due to q1 (F12): Since both are positive, it's repulsive. F12 acts on q2 away from q1 (towards q3).F12 = k * |q1 * q2| / r122 = (9 x 109) * (1 x 10-3) * (5 x 10-3) / (0.07)2F12 = (4.5 x 104) / 0.0049 ≈ 9.18 x 106 N (towards q3)2. Force on q2 due to q3 (F32): Since both are positive, it's repulsive. F32 acts on q2 away from q3 (towards q1).F32 = k * |q2 * q3| / r232 = (9 x 109) * (5 x 10-3) * (9 x 10-3) / (0.09)2F32 = (4.05 x 105) / 0.0081 ≈ 5.0 x 107 N (towards q1)The total force on q2 is the vector sum of F12 and F32. Since they are in opposite directions, the net force is their difference, in the direction of the larger force.Fnet = F32 - F12 = (5.0 x 107 N) - (9.18 x 106 N) = (50 x 106 N) - (9.18 x 106 N) = 40.82 x 106 N ≈ 4.08 x 107 N.The direction is towards q1 (the direction of F32, which is larger).Rounding to one decimal place, the force is 4.1 x 107 N, towards q1.