Question

The time period of revolution of a charge \(q_1\) and of mass \(m\) moving in a circular path of radius \(r\) due to Coulomb force of attraction with another charge \(q_2\) at its centre is

• A. \(\sqrt{\frac{16\pi^3 \epsilon_0 mr^3}{q_1 q_2}}\)
• B. \(\sqrt{\frac{8\pi^2 \epsilon_0 mr^3}{q_1 q_2}}\)
• C. \(\sqrt{\frac{\epsilon_0 mr^3}{16\pi q_1 q_2}}\)
• D. \(\sqrt{\frac{16\pi^3 \epsilon_0 mr^3}{q_1 q_2}}\)
• E. \(\sqrt{\frac{\pi^2 \epsilon_0 mr^3}{8q_1 q_2}}\)

Hint:

This derivation is identical to Kepler's Third Law for planets, where \(T^2 \propto r^3\). The constant of proportionality here simply uses electrical units instead of gravitational ones!

Answer 1

The Correct Option is D

Concept: For a charge to move in a stable circular orbit, the electrostatic (Coulomb) force must provide the necessary centripetal force.
• Coulomb Force: \(F_c = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}\).
• Centripetal Force: \(F_p = \frac{mv^2}{r} = m\omega^2r\).
• Time Period: \(T = \frac{2\pi}{\omega}\).

Step 1: Equate the forces to find angular velocity \(\omega\).
\[ m\omega^2r = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \] Solving for \(\omega^2\): \[ \omega^2 = \frac{q_1 q_2}{4\pi\epsilon_0 mr^3} \]

Step 2: Relate \(\omega\) to the Time Period \(T\).
We know \(T = 2\pi/\omega \implies T^2 = 4\pi^2/\omega^2\). Substituting the expression for \(1/\omega^2\): \[ T^2 = 4\pi^2 \left( \frac{4\pi\epsilon_0 mr^3}{q_1 q_2} \right) \] \[ T^2 = \frac{16\pi^3\epsilon_0 mr^3}{q_1 q_2} \]

Step 3: Take the square root for final answer.
\[ T = \sqrt{\frac{16\pi^3 \epsilon_0 mr^3}{q_1 q_2}} \] This matches Option (D).