Introduction:
In this problem, we are given a particle of mass 'm' carrying a charge '-q' moving around a charge 'q/2' along a circular path with radius 'r'. We need to find the time period of revolution of the charge '-q'.

Key Points:
1. A charged particle moving in a magnetic field experiences a force called the magnetic Lorentz force.
2. The magnitude of the magnetic Lorentz force is given by F = qvBsinθ, where q is the charge, v is the velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field.
3. This force provides the necessary centripetal force for the particle to move in a circular path.
4. The centripetal force is given by F = mv^2/r, where m is the mass, v is the velocity, and r is the radius of the circular path.
5. Equating the magnetic Lorentz force to the centripetal force, we can solve for the velocity of the particle.
6. The time period of revolution can be calculated as the time taken for the particle to complete one full revolution around the circular path.

Calculation:
1. The magnetic Lorentz force is given by F = (-q)vBsinθ.
2. The centripetal force is given by F = (mv^2)/r.
3. Equating the two forces, we have (-q)vBsinθ = (mv^2)/r.
4. Rearranging the equation, we get v = (-qBr)/(m*sinθ).
5. The time period of revolution can be calculated using the formula T = (2πr)/v.
6. Substituting the value of v from the previous equation, we have T = (2πr)/((-qBr)/(m*sinθ)).
7. Simplifying further, we get T = (2πmr*sinθ)/(-qB).
8. The negative sign indicates that the charge '-q' is moving in the opposite direction compared to the positive charge.
9. Therefore, the time period of revolution for the charge '-q' is given by T = (2πmr*sinθ)/(-qB).

Conclusion:
The time period of revolution for the charge '-q' moving around a charge 'q/2' along a circular path with radius 'r' is given by T = (2πmr*sinθ)/(-qB). The negative sign indicates the opposite direction of motion compared to the positive charge.