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A particle of mass $m$ and charge $q$ is thrown in the vertical direction with a velocity $v_{0}$ ill uniform horizontal electric field $E$ . Assuming the gravity force to be negligible, the equation the path followed by the particle will be

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$x = (\frac{q E}{2 m v_{0}^{2}}) y^{2}$

$x = \frac{q E}{2 m u_{0}^{2} y}$

$x = \frac{q E y}{2 m u_{0}^{2}}$

None of these

answer is A.

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Detailed Solution

$t = \frac{y}{v_{0}} \Rightarrow x = \frac{1}{2} α_{x} t^{2} = \frac{1}{2} (\frac{q E}{m}) {(\frac{y}{v_{0}})}^{2}$

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