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A particle of...

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A particle of charge q and mass m starts moving from the origin under the action of an electric field $\to E = E_{0}^i$ and $\to B = B_{0}^i$ with a velocity $\to v = v_{0}^i$ . The speed of the particle will become $2 v_{0}$ after a time.

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Solution

Final velocity of the particle, $v = \sqrt{v_{0}^{2} + {(\frac{q E t}{m})}^{2}} = 2 v_{0}$
$v_{0}^{2} + {(\frac{q E t}{m})}^{2} = 4 v_{0}^{2}$
${(\frac{q E t}{m})}^{2} = 3 v_{0}^{2}$
$t = \frac{\sqrt{3} m v_{0}}{q E}$

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