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 ______.

Options

• 1 m

• 5 m

• 10 m

• 0.5 m

Answer 1

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 0.5 m.

Explanation: Given: 

u = Initial velocity of charged particle = 200 m/s

m = mass of charged particle = 100 mg

= 100 × 10^–6 kg = 10^–4 kg

q ⇒ charge on particle = 40 uC = 40 × 10^–6 C = 4 × 10^–5 C

|E| = Electric field intensity = 1 × 10⁵ N/C

Electrostatic force will act on the particle opposed to velocity as it is projected in the opposite direction of the electric field. Its mobility will therefore be delayed.

As electric field is constant, F = qE will be constants

Hence, a = `"F"/"m" = "qE"/"m"` = will also be constant.

Thus, equations of motion can be used

v² = u² + 2as    …(1)

`(0)^2 = (200)^2 + 2 xx ((-"qE")/"m" xx "S")`

`(2xx4xx10^-5xx1xx10^5)/10^-4 xx "S"`

= 4 × 10⁴ 

S = `(4xx10^4xx10-4)/8 = 1/2`

m = 0.5 m

Thus, S = 0.5 m