there is a particle A with positive charge q1and mass m1. There is another particle B with negative charge q2 and mass m2 particles A and B are moving along circular paths under electrostatic attraction in such a manner that separation between them remains d and location of their center of mass is centre of circular path. what will be the angular velocity of particles,?
Answers
l = 1/2 √ q1q2(m1+m2) / π∈₀E(q2m1 + q1m2)
Explanation:
Given: There is a particle A with positive charge q1and mass m1. There is another particle B with negative charge q2 and mass m2. Particles A and B are moving along circular paths under electrostatic attraction in such a manner that separation between them remains d and location of their center of mass is centre of circular path.
Find: Angular velocity of particles.
Solution: Let initial separation between the particles be l and this remains constant during the motion. If relative separation remains constant then relative velocity should be zero; hence relative acceleration of the particles should also be zero.
ar = a1 - a2 = 0
or (q1E -Fe) / m1 = (Fe - q2E) / m2
(q1m2 + q2m1) / m1m2 = Fe [ 1/m1 + 1/m2]
or (q1m2 + q2m1) E = Fe (m1+m2) = 1/4π∈₀ . q1q2/l² (m1 + m2)
Solving, we get: l = 1/2 √ q1q2(m1+m2) / π∈₀E(q2m1 + q1m2)