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A particle with charge $e$ and mass $m$, moving along the $X$ -axis with a uniform speed $u$, enters a region where a uniform electric field $E$ is acting along the $Y$ -axis. The particle starts to move in a parabola. Its focal length (neglecting any effect of gravity) is
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The correct answer is:
$\frac{m u^{2}}{2 e E}$
From parabola
$y=\frac{1}{2} \times\left[\frac{E e}{m}\right] \times \frac{x^{2}}{u^{2}}$
$=\frac{E \cdot e}{2 m u^{2}} \cdot X^{2}$
$\begin{array}{ll}\text { As } x=4 a y & \text { (for parabola) }\end{array}$
$\therefore \quad X^{2}=\frac{2 m u^{2}}{E \cdot e} \cdot y$
$a=\frac{2 m u^{2}}{4 E \cdot e}=\frac{m u^{2}}{2 E \cdot e}$
$y=\frac{1}{2} \times\left[\frac{E e}{m}\right] \times \frac{x^{2}}{u^{2}}$
$=\frac{E \cdot e}{2 m u^{2}} \cdot X^{2}$
$\begin{array}{ll}\text { As } x=4 a y & \text { (for parabola) }\end{array}$
$\therefore \quad X^{2}=\frac{2 m u^{2}}{E \cdot e} \cdot y$
$a=\frac{2 m u^{2}}{4 E \cdot e}=\frac{m u^{2}}{2 E \cdot e}$
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