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 starts from rest and accelerates through a uniform electric field E over distance L; find its final velocity v.

Step 2: Key Formula or Approach:
Use Work-Energy Theorem: Work done by electric field = qEL = ½mv² – 0.

Step 3: Detailed Explanation:
qEL = ½mv² → 2qEL = mv² → v = √(2qEL/m).

Step 4: Final Answer:
v = √(2qEL/m), matching option (A).