Question

Particle of charge q and mass m starts moving from the origin under the action of an electric field e=ei and b=bj with a velocity v=vj.The speed of the particle will become 2v after a time?

Answer 1

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Answer 2

The speed of the particle will become 2v after a time t is (√3v₀m/qE).

We know under the application of external electric field, the final velocity (speed of particle after time t) of the particle is given as:

v = v₀ + qEt/m

So, the magnitude of the velocity is :

v = √([v₀]² + [qEt/m]²)

It is also mentioned that final veocity is 2v₀

So, 2v₀ = √([v₀]² + [qEt/m]²)

Squaring both sides, we get:

4v₀² = [v₀]² + [qEt/m]²

⇒q²E²t²/m² = 3v₀²

⇒ t² = (3v₀²m²)/(q²E²) =   (√3v₀m/qE)²

Taking the square root,

t = (√3v₀m/qE)