1.Three particles are fixed in place on the x -axis. Particle 1 of charge q 1 is

Question

1.Three particles are fixed in place on the x-axis. Particle 1 of charge q1 is at the coordinate x = ?a. Particle 2 of charge q2 is at x = +1.5a. (The signs of q1 and q2 are unknown.)

(a)A third particle of charge +Q is fixed in place at x = +0.6a. For what ratio q1/q2 will the net force on particle 3 be zero? Derive a solution of the form q1/q2(Q,a). Be sure to include the correct sign with your answer.

(b)Now the third particle of charge +Q is fixed in place at x = +2.5a. For what ratio q1/q2 will the net force on particle 3 be zero? Derive a solution of the form q1/q2(Q,a). Be sure to include the correct sign with your answer.

Explanation / Answer

(a) The third particle +Q, will experience a net force of repulsion from q1 and q2 in opposite directions, as they are positive.

F(net-+Q) = F1 - F2

Condition given F(net -+Q) = 0

F1 = F2

We know that force of attraction/repulsion between charged pariccle is given as,

F = K Q1Q2 /r2

dist. of +Q from q1 = 0.6 a and that of q2 is 1.5 - 0.6 = 0.9

K Q q1 / (0.6a)2 = K Q q2 / (0.9a)2 (K Q and a2 gets cancel here)

q1 / q2 = 0.36 / 0.81 = 0.443

Hence, q1/q2 = 0.44

(b)now +Q is at +2.5 a

so dist. between q1 and +Q = 2.5a and +Q and q2 = 1a

Similarly as (a)

K Q q1 / (2.5a)2 = K Q q2 / (1a)2 (K Q and a2 gets cancel here)

q1/q2 = 6.25 / 1

.Hence, the ratio = q1/q2 = 6.25.

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