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) \( \frac{E^2 t^2}{2m} \).

Answer 2

A charged particle with mass \( m \) and charge \( q \) is released from rest (\( u=0 \)) in a uniform electric field \( \vec{E} \). We need to find its kinetic energy (KE) after a time \( t \). We neglect gravity.

Step 1: Force on the charged particle
The electric field exerts a force \( \vec{F} \) on the charged particle, given by: \[ \vec{F} = q\vec{E} \] The magnitude of the force is \( F = qE \). Since the electric field is uniform, this force is constant.

Step 2: Acceleration of the particle
According to Newton's Second Law, \( \vec{F} = m\vec{a} \), where \( \vec{a} \) is the acceleration. \[ m\vec{a} = q\vec{E} \] \[ \vec{a} = \frac{q\vec{E}}{m} \] The acceleration is constant because \( q, E, m \) are constant. The magnitude of the acceleration is \( a = \frac{qE}{m} \).

Step 3: Velocity of the particle after time \( t \)
Since the particle starts from rest (\( u=0 \)) and moves with constant acceleration \( a \), its velocity \( v \) after time \( t \) can be found using the kinematic equation \( v = u + at \): \[ v = 0 + at = at \] Substituting the expression for \( a \): \[ v = \left(\frac{qE}{m}\right)t = \frac{qEt}{m} \]

Step 4: Kinetic energy of the particle after time \( t \) 
The kinetic energy (KE) is given by the formula \( KE = \frac{1}{2}mv^2 \). \[ KE = \frac{1}{2}m \left(\frac{qEt}{m}\right)^2 \] \[ KE = \frac{1}{2}m \left(\frac{q^2E^2t^2}{m^2}\right) \] Simplifying the expression by canceling one factor of \( m \): \[ \mathbf{KE = \frac{q^2E^2t^2}{2m}} \] This can also be written as: \[ \mathbf{KE = \frac{E^2q^2t^2}{2m}} \]

Comparing this result with the given options:

• Eqm/t
• 2E²t²/mq
• E²q²t²/2m
• Eq²m/2t²

The calculated kinetic energy matches the third option.