Question

A charge Q is fixed at a point and another equal and opposite charge is revolving around the fixed charge

With fixed angular velocity $\omega$, find the radius of circular path.

Answer 1

r = \left(\frac{k Q^2}{m\omega^2}\right)^{1/3}

Answer 2

For a charge of mass m (revolving charge) to move in a circular orbit of radius r under the influence of the Coulomb force from a fixed charge Q, equate the centripetal force to the Coulomb force:

$$m\omega^2 r = \frac{k Q^2}{r^2}$$

Solving for r:

$$m\omega^2 r^3 = k Q^2 \quad \Longrightarrow \quad r^3 = \frac{k Q^2}{m\omega^2}$$

Thus, the radius of the circular path is:

$$r = \left(\frac{k Q^2}{m\omega^2}\right)^{1/3}$$