Question Detail
Question
Three charges q1=1mC, q2=5mC and q3=9mC are arranged on a line with q2 at the middle of the other two. If the distance between q1 and q2 is 7cm while between q1 and q3 is 16cm, determine the total force acting on q2 as well as the direction.
Correct Answer
Option A is the correct answer.
Detailed Explanation
Given:q1 = 1 mC = 1 x 10-3 Cq2 = 5 mC = 5 x 10-3 Cq3 = 9 mC = 9 x 10-3 CDistance between q1 and q2, r12 = 7 cm = 0.07 mDistance between q1 and q3, r13 = 16 cm = 0.16 mSince q2 is in the middle, the distance between q2 and q3 is r23 = r13 - r12 = 0.16 m - 0.07 m = 0.09 m.Using Coulomb's Law, F = k |q1 q2| / r2, where k = 9 x 109 Nm2/C2.1. Force on q2 due to q1 (F12): Both q1 and q2 are positive, so F12 is repulsive. q2 is repelled by q1, so F12 acts towards q3.F12 = (9 x 109) * (1 x 10-3) * (5 x 10-3) / (0.07)2F12 = (45 x 103) / 0.0049 = 9.1837 x 106 N.2. Force on q2 due to q3 (F32): Both q2 and q3 are positive, so F32 is repulsive. q2 is repelled by q3, so F32 acts towards q1.F32 = (9 x 109) * (9 x 10-3) * (5 x 10-3) / (0.09)2F32 = (405 x 103) / 0.0081 = 5.0 x 107 N.The forces F12 and F32 are in opposite directions (F12 towards q3, F32 towards q1). The net force on q2 is the difference between these two forces, directed towards the stronger force.F_net = F32 - F12 = (5.0 x 107 N) - (9.1837 x 106 N)F_net = (50 x 106 N) - (9.1837 x 106 N) = 40.8163 x 106 N ≈ 4.1 x 107 N.Since F32 is stronger and acts towards q1, the net force is towards q1.Thus, the total force is 4.1 x 107 N, towards q1.
Hint
First, calculate the distance between q2 and q3. Then, use Coulomb's Law to find the magnitude and direction of the force exerted by q1 on q2 and by q3 on q2. Remember that like charges repel, and unlike charges attract. Finally, sum the forces vectorially.