Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2

Question

Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.246 m to the right of Q1. Q3 is located 0.180 m to the right of Q2 The force on Q2 due to its interaction with Q3 is directed to the.. True: Left if the two charges are negative. False: Left if the two charges have opposite signs. True: Right if the two charges have opposite signs True: Left if the two charges are positive You are correct. Your receipt no. is 155-5396Previous Tries In the above problem, Q1-1.90-10.6 C, Q2=-3.22-10-6 C, and Q3-3.33-10-6 C. Calculate the total force on Q2. Give with the plus sign for a force directed to the right. 2.38N Use Coulomb's Law and superposition of electric forces Sutomit Ansmer Incorrect. Tries 3/10 Previous Tries Now the charges Q1-1.90-10-6 C and Q2--3.22-10-6 C are fixed at their positions, distance 0.246 m apart, and the charge Q3- 3.33-10-6 C is moved along the straight line. For what position of Q3 relative to Q1 is the net force on Q3 due to Q1 and Q2 zero? Use the plus sign for 3 to the right of Q1 Suhmit Answwer Tries 0/10

Explanation / Answer

(b)

Force on Q2 due to Q1,

F21 = kQ1Q2 / r^2

F21 = 9*10^9*1.9*10^(-6)*3.22*10^(-6) / (0.246)^2 = 0.909 N

Force on Q2 due to Q3,

F23 = 9*10^9*3.33*10^(-6)*3.22*10^(-6) / (0.18)^2 = 2.97 N

Net force on Q2,

F = F23 - F21

F = 2.06 N

(c)

Let the position of Q3 = x

Net electric field on Q3 = 0

so,E1 = E2

k*1.90*10^(-6) / x^2 = k*3.22*10^(-6) / (x + 0.246)^2

1.378 / x = 1.794 / (x + 0.246)

x = 0.824 m

Position of Q3 = -x

position of Q3 = -0.824 m

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