Question

A particle of mass \( m \) and charge \( q \) is fastened to one end \( A \) of a massless string having equilibrium length \( l \), whose other end is fixed at point \( O \). The whole system is placed on a frictionless horizontal plane and is initially at rest. If a uniform electric field is switched on along the direction as shown in the figure, then the speed of the particle when it crosses the x-axis is:

• A. \( \sqrt{\frac{qEI}{m}} \)
• B. \( \sqrt{\frac{2qEI}{m}} \)
• C. \( \sqrt{\frac{qEI}{4m}} \)
• D. \( \frac{qEI}{2m} \)

Hint:

In problems involving electric fields, apply the work-energy principle to calculate the kinetic energy gained by the charged particle.

Answer 1

The Correct Option is B

The particle is subjected to an electric field. The work-energy theorem will be applied to determine its velocity upon crossing the x-axis.

Step 1: The electric force acting on the particle is calculated as:

\[ F_{\text{electric}} = qE \]

Here, \( E \) represents the electric field.

Step 2: The work performed by this force to move the particle a distance \( l \) along the x-axis is:

\[ W = F_{\text{electric}} \times l = qEl \]

Step 3: The particle's kinetic energy gain equals the work done:

\[ K = \frac{1}{2} m v^2 \]

Therefore, by equating work and kinetic energy:

\[ qEl = \frac{1}{2} m v^2 \]

Step 4: Solve for \( v \):

\[ v = \sqrt{\frac{2qEl}{m}} \]

Final Conclusion: The speed of the particle at the x-axis crossing is \( \sqrt{\frac{2qEl}{m}} \), corresponding to Option (2).