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

The Correct Option is D

To determine the distance traveled by the charged particle before it momentarily comes to a rest, we begin by understanding the forces acting on it.

The problem provides:

• Mass of the particle, m = 100 \, \text{mg} = 0.1 \, \text{g} = 0.0001 \, \text{kg}.
• Electric field strength, E = 1 \times 10^5 \, \text{NC}^{-1}.
• Charge on the particle, q = 40 \, \mu\text{C} = 40 \times 10^{-6} \, \text{C}.
• Initial velocity, u = 200 \, \text{ms}^{-1}.

The electric force acting on the charged particle is given by:

F = qE

Substituting the known values, we get:

F = 40 \times 10^{-6} \times 1 \times 10^5 = 4 \, \text{N}

The particle moves against the electric force, which will do negative work to bring it to rest. Hence, the acceleration a can be found using Newton's second law:

F = ma \Rightarrow a = \frac{F}{m} = \frac{4}{0.0001} = 40000 \, \text{ms}^{-2}

Since the particle is moving opposite to the direction of the electric field, the acceleration will be negative relative to its velocity. Thus, a = -40000 \, \text{ms}^{-2}.

Using the kinematic equation, we can find the distance s traveled before the particle stops:

v^2 = u^2 + 2as

Since the particle comes to rest momentarily, we have v = 0. Therefore,

0 = 200^2 + 2 \times (-40000) \times s

Simplifying,

0 = 40000 - 80000s

80000s = 40000

s = \frac{40000}{80000} = 0.5 \, \text{m}

Thus, the distance traveled by the particle before it comes to rest is 0.5 m. This matches the given correct answer.