Question

A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 × 105 NC–1. If the charge on the particle is 40 μC and the initial velocity is 200 ms–1, how much distance it will travel before coming to the rest momentarily?

• A. 1 m
• B. 5 m
• C. 10 m
• D. 0.5 m

Answer 1

Answer: (D)

0.5 m

Answer 2

The correct option is(D): 0.5 m

$v^2-u^2=2aS$

$⇒s=\frac{4}{2x4}m=0.5m$