Question

A positive charge particle of 100 µg 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:

पर्याय

• 1 m

• 5 m

• 10 m

• 0.5 m

Answer 1

0.5 m

Explanation:

Here a = `"qE"/"m"`

= `(40xx10^-6xx10^5)/(100xx10^-6)`

= 40000 m/s²

As the charge particle moves in the opposite direction to that of `vec"E"`

As a result, `vec"E"` will declerate the charge, which will then come to rest.

Now, v² = u² + 2as

0 = 200² + 2 × (-40 × 10³) × s

80 × 10³s = 4 × 10⁴

∴ s = `(4xx10^4)/(80xx10^3)`

= `40/80`

= 0.5 m