Question

A particle 'X' carrying a charge \( +Q \) is moving in a circular path of radius \( R \) around another particle 'Y' having a charge \( -Q \) with a frequency \( \nu \). Then the mass \( m \) of the charged particle is:

• A. \( M = \dfrac{Q^2}{16\pi^3 \varepsilon_0 R^3 \nu^2} \)
• B. \( M = \dfrac{Q^2}{16\pi^2 \varepsilon_0 R \nu^2} \)
• C. \( M = \dfrac{Q^2}{16\pi^2 \varepsilon_0 R^3 \nu^2} \)
• D. \( M = \dfrac{Q^2}{16\pi^3 \varepsilon_0 R^2 \nu^2} \)

Hint:

In circular motion under electrostatic force, always equate Coulomb force with centripetal force and use \( \omega = 2\pi \nu \) to simplify expressions.

Answer 1

The Correct Option is A

Step 1: Identify the forces acting.
The charged particle moves in a circular path due to electrostatic attraction between charges \( +Q \) and \( -Q \). This electrostatic force provides the required centripetal force.

Step 2: Write Coulomb force.
Electrostatic force between charges is given by: \[ F = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q^2}{R^2} \]

Step 3: Write centripetal force.
Centripetal force required for circular motion is: \[ F = m \omega^2 R \]

Step 4: Relate angular velocity with frequency.
\[ \omega = 2\pi \nu \quad \Rightarrow \quad \omega^2 = 4\pi^2 \nu^2 \]

Step 5: Equate both forces.
\[ \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q^2}{R^2} = m \cdot (4\pi^2 \nu^2) \cdot R \]

Step 6: Solve for mass \( m \).
\[ m = \frac{Q^2}{4\pi \varepsilon_0 R^2 \cdot 4\pi^2 \nu^2 \cdot R} \]
\[ m = \frac{Q^2}{16\pi^3 \varepsilon_0 R^3 \nu^2} \]

Step 7: Final Answer.
Thus, \[ m = \frac{Q^2}{16\pi^3 \varepsilon_0 R^3 \nu^2} \]
Hence, option (A) is correct.