Question

Two parallel plates separated by a distance \( d \) are kept at potential difference \( V \) volt. A charge \( q \) of mass \( m \) enters in parallel plates with some velocity. The acceleration of the charged particle will be

• A. \( \frac{md}{Vq} \)
• B. \( \frac{qm}{Vd} \)
• C. \( \frac{qV}{dm} \)
• D. \( \frac{qd}{Vm} \)

Hint:

The acceleration of a charged particle in an electric field is directly proportional to the charge and the potential difference, and inversely proportional to the mass and the distance between the plates.

Answer 1

The Correct Option is C

Step 1: Understand the force on the charged particle.
When the charge \( q \) enters the electric field between the parallel plates, the force \( F \) acting on it is given by: \[ F = qE \] where \( E \) is the electric field between the plates. The electric field is related to the potential difference \( V \) and the distance between the plates \( d \) by: \[ E = \frac{V}{d} \] Step 2: Apply Newton's second law.
The acceleration \( a \) of the charged particle is given by Newton's second law: \[ a = \frac{F}{m} = \frac{qE}{m} = \frac{qV}{dm} \] Step 3: Conclusion.
Thus, the acceleration of the charged particle is \( \frac{qV}{dm} \), which corresponds to option (C).