Question

An electron of mass '$m$' and charge '$q$' is accelerated from rest in a uniform electric field of strength '$E$'. The velocity acquired by the electron when it travels a distance '$L$' is

• A. $\sqrt{\frac{2qEL}{m}}$
• B. $\sqrt{\frac{2Em}{qL}}$
• C. $\frac{2qEL}{m}$
• D. $\sqrt{\frac{qEL}{m}}$

Hint:

You can also solve this using kinematics! The constant acceleration is $a = \frac{F}{m} = \frac{qE}{m}$. Plucking this value directly into Newton's third equation of motion ($v^2 = u^2 + 2aS$) with $u=0$ gives $v^2 = 2\left(\frac{qE}{m}\right)L$, leading to the same result in a single step!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A charged particle starting from rest ($u = 0$) accelerates through a uniform electric field $E$. We need to find its final linear velocity $v$ after it has traveled a total displacement distance $L$ along the direction of the field lines.

Step 2: Key Formula or Approach:
We can solve this problem using the Work-Energy Theorem, which states that the work done by the net force equals the change in kinetic energy: $$W = \Delta K.E. = \frac{1}{2}mv^2 - 0$$ The constant electrostatic force acting on a charge $q$ is $F = qE$, so the work done over a distance $L$ is $W = F \cdot L = qEL$.

Step 3: Detailed Explanation:
Let's equate the work done by the electric field to the kinetic energy gained by the particle: $$W = \frac{1}{2}mv^2$$ $$qEL = \frac{1}{2}mv^2$$ Multiply both sides by 2 to isolate the velocity terms: $$2qEL = mv^2$$ $$\text{Isolate } v^2: \quad v^2 = \frac{2qEL}{m}$$ Take the principal square root of both sides to get the final expression for velocity: $$v = \sqrt{\frac{2qEL}{m}}$$

Step 4: Final Answer:
The velocity acquired by the electron is $\sqrt{\frac{2qEL}{m}}$, which corresponds to option (A).