Orbital Period of a Charged Particle - Interactive Physics Problem

Orbital Period of a Charged Particle

A charge \( q_{2} \) of mass \( m \) revolves around a stationary charge \( q_{1} \) in a circular orbit of radius \( r \). Determine the orbital periodic time (period) of \( q_{2} \).

Select the correct option:

• A) \( \left[\frac{4\pi^2 m r^3}{k\,q_{1}\,q_{2}}\right]^{\frac{1}{2}} \)
• B) \( \left[\frac{k\,q_{1}\,q_{2}}{4\pi^2 m r^3}\right]^{\frac{1}{2}} \)
• C) \( \left[\frac{4\pi^2 m r^4}{k\,q_{1}\,q_{2}}\right]^{\frac{1}{2}} \)
• D) \( \left[\frac{4\pi^2 m r^2}{k\,q_{1}\,q_{2}}\right]^{\frac{1}{2}} \)

Detailed Explanation

For a charge \( q_{2} \) of mass \( m \) revolving in a circular orbit of radius \( r \) under the influence of the Coulomb force due to a stationary charge \( q_{1} \), the centripetal force required for circular motion is provided by the electrostatic attraction/repulsion given by: \[ F = \frac{1}{4\pi\epsilon_0}\frac{q_{1}q_{2}}{r^2}. \]

The centripetal force needed to keep the particle in circular motion is: \[ F_{\text{centripetal}} = \frac{m v^2}{r}. \]

Equating the electrostatic force to the centripetal force: \[ \frac{m v^2}{r} = \frac{1}{4\pi\epsilon_0}\frac{q_{1}q_{2}}{r^2}. \]

Solve for the orbital speed \( v \): \[ v^2 = \frac{1}{4\pi\epsilon_0}\frac{q_{1}q_{2}}{m\,r} \quad \Longrightarrow \quad v = \sqrt{\frac{q_{1}q_{2}}{4\pi\epsilon_0\,m\,r}}. \]

The orbital period \( T \) is the time taken to complete one full orbit, and is given by: \[ T = \frac{2\pi r}{v}. \]

Substituting the expression for \( v \): \[ T = \frac{2\pi r}{\sqrt{\frac{q_{1}q_{2}}{4\pi\epsilon_0\,m\,r}}} = 2\pi r \sqrt{\frac{4\pi\epsilon_0\,m\,r}{q_{1}q_{2}}}. \]

Simplify the expression: \[ T = 2\pi \sqrt{\frac{4\pi\epsilon_0\,m\,r^3}{q_{1}q_{2}}}. \]

Recognizing that the Coulomb constant \( k \) is defined as \( k=\frac{1}{4\pi\epsilon_0} \), we have: \[ T = \sqrt{\frac{4\pi^2\,m\,r^3}{k\,q_{1}q_{2}}}. \]

This matches Option A.