Step 1: Understanding the Concept:
When a charged particle is placed in a uniform electric field, it experiences a constant force.
This constant force produces a constant acceleration.
We can use Newton's laws of motion and kinematic equations to find its final velocity and subsequently its kinetic energy.
Step 2: Key Formula or Approach:
Electric force $F = qE$.
Acceleration $a = \frac{F}{m} = \frac{qE}{m}$.
Kinematic equation for velocity starting from rest: $v = u + at \implies v = at$.
Kinetic energy $K = \frac{1}{2}mv^2$.
Step 3: Detailed Explanation:
The particle starts from rest, so its initial velocity $u = 0$.
The force acting on the particle is $F = qE$.
According to Newton's second law, the acceleration is:
\[ a = \frac{F}{m} = \frac{qE}{m} \]
Using the first equation of motion, find the velocity $v$ after time $t$:
\[ v = u + at = 0 + \left(\frac{qE}{m}\right)t \]
\[ v = \frac{qEt}{m} \]
The kinetic energy $K$ of the particle after time $t$ is:
\[ K = \frac{1}{2}mv^2 \]
Substitute the expression for velocity $v$ into the kinetic energy equation:
\[ K = \frac{1}{2}m \left(\frac{qEt}{m}\right)^2 \]
Expand the squared term:
\[ K = \frac{1}{2}m \left(\frac{q^2 E^2 t^2}{m^2}\right) \]
Simplify by canceling one mass term '$m$':
\[ K = \frac{q^2 E^2 t^2}{2m} \]
Step 4: Final Answer:
The kinetic energy is $\frac{E^2 q^2 t^2}{2m}$.