Question

A particle of mass \(m\) and charge \(q\) traverses distance \(d\) from rest in a uniform electric field \(E\). Prove that the velocity \(v\) attained by the particle is \( v = \sqrt{\frac{2qEd}{m}} \).

Hint:

The velocity of a particle in an electric field depends on its charge, mass, the electric field strength, and the distance it travels.

Answer 1

When a charged particle of charge \(q\) and mass \(m\) moves in a uniform electric field \(E\), the work done by the electric field is equal to the change in kinetic energy of the particle.
The work done \(W\) by the electric field is given by: \[ W = F \times d = qE \times d \] where \(F = qE\) is the force acting on the particle.
Since the particle starts from rest, the work done is equal to the kinetic energy gained by the particle: \[ K.E. = \frac{1}{2} m v^2 \] Equating the work done to the kinetic energy: \[ qEd = \frac{1}{2} m v^2 \] Rearranging for \(v\), we get: \[ v = \sqrt{\frac{2qEd}{m}} \] Thus, the velocity \(v\) attained by the particle is: \[ v = \sqrt{\frac{2qEd}{m}} \]