Question

A charge particle of mass 'm' and charge 'q' is released from rest in an uniform electric field 'E'. Neglecting the effect of gravity, the kinetic energy of the charged particle after 't' second is 

• A. Eqm/t
• B. 2E²t²/mq
• C. E²q²t²/2m
• D. Eq²m/2t²

Answer 1

The Correct Option is C

Using the equation of motion, calculate the kinetic energy of a particle moving with constant acceleration, where \( v = u + at \) and the energy involved is \( E \).

Step 1: Start with the equation of motion:

\( v = u + at \)

Rearrange the equation to solve for \( v \):

\( v = u + at \quad \Rightarrow \quad v = 0 + \frac{E}{m}t \quad \Rightarrow \quad v = \frac{E}{m} t \)

Step 2: Kinetic energy (K) of the particle:

The kinetic energy is given by:

\( K = \frac{1}{2} m v^2 \)

Substitute the value of \( v \) from earlier:

\( K = \frac{1}{2} m \left( \frac{E}{m} t \right)^2 \)

Step 3: Simplify the equation:

\( K = \frac{1}{2} m \left( \frac{E^2}{m^2} t^2 \right) = \frac{1}{2} \frac{E^2 t^2}{m} \)

Therefore, the correct expression for the kinetic energy is: \( K = \frac{E^2 t^2}{2m} \), which corresponds to option (C) \( \frac{E^2 t^2}{2m} \).