An electron of mass m and charge q is accelerated from rest in a uniform electric field of strength E. The velocity acquired by it as it travels a distance l is

Options

A. `sqrt(2Eql//m)`
B. `sqrt(2Eq//ml)`
C. `sqrt(2Em//ql)`
D. `sqrt(Eq//ml)`

Explanation

To find the velocity acquired by an electron of mass \( m \) and charge \( q \) when it is accelerated from rest in a uniform electric field of strength \( E \) over a distance \( l \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
The electron starts from rest, so its initial velocity \( u = 0 \).

2. **Understand the Forces Acting on the Electron**:
The force \( F \) acting on the electron due to the electric field is given by:
\[ F = qE \]
where \( q \) is the charge of the electron and \( E \) is the strength of the electric field.

3. **Calculate the Acceleration**:
According to Newton's second law, the acceleration \( a \) of the electron can be calculated using:
\[ a = \frac{F}{m} = \frac{qE}{m} \]
where \( m \) is the mass of the electron.

4. **Use the Kinematic Equation**:
We can use the kinematic equation that relates acceleration, distance, and velocity:
\[ v^2 = u^2 + 2as \]
Since the initial velocity \( u = 0 \), this simplifies to:
\[ v^2 = 2as \]

5. **Substitute the Values**:
Substitute \( a = \frac{qE}{m} \) and \( s = l \) (the distance traveled):
\[ v^2 = 2 \left(\frac{qE}{m}\right) l \]

6. **Solve for Final Velocity \( v \)**:
Taking the square root of both sides gives:
\[ v = \sqrt{\frac{2qEl}{m}} \]

### Final Result:
The velocity acquired by the electron after traveling a distance \( l \) in the electric field is:
\[ v = \sqrt{\frac{2qEl}{m}} \]